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https://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=215 | 1,435,658,493,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375093400.45/warc/CC-MAIN-20150627031813-00109-ip-10-179-60-89.ec2.internal.warc.gz | 890,569,255 | 9,941 | # Search by Topic
#### Resources tagged with Interactivities similar to A Dotty Problem:
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Broad Topics > Information and Communications Technology > Interactivities
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Target Archery
##### Stage: 2 Challenge Level:
A simulation of target archery practice
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Train
##### Stage: 2 Challenge Level:
A train building game for 2 players.
### Factor Lines
##### Stage: 2 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Coordinate Tan
##### Stage: 2 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Board Block Challenge
##### Stage: 2 Challenge Level:
Choose the size of your pegboard and the shapes you can make. Can you work out the strategies needed to block your opponent?
### Tetrafit
##### Stage: 2 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Sorting Symmetries
##### Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Ratio Pairs 2
##### Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
### A Maze of Directions
##### Stage: 2 Challenge Level:
Use the blue spot to help you move the yellow spot from one star to the other. How are the trails of the blue and yellow spots related?
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Rod Ratios
##### Stage: 2 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
### Noughts and Crosses
##### Stage: 2 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Overlapping Circles
##### Stage: 2 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
##### Stage: 2 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Rabbit Run
##### Stage: 2 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
### Fractions and Coins Game
##### Stage: 2 Challenge Level:
Work out the fractions to match the cards with the same amount of money.
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Board Block for Two
##### Stage: 1 and 2 Challenge Level:
Board Block game for two. Can you stop your partner from being able to make a shape on the board?
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Chocolate Bars
##### Stage: 2 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
### 100 Percent
##### Stage: 2 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Building Stars
##### Stage: 2 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Calculator Bingo
##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Coordinate Cunning
##### Stage: 2 Challenge Level:
A game for 2 people that can be played on line or with pens and paper. Combine your knowledege of coordinates with your skills of strategic thinking.
### Beat the Drum Beat!
##### Stage: 2 Challenge Level:
Use the interactivity to create some steady rhythms. How could you create a rhythm which sounds the same forwards as it does backwards?
### Combining Cuisenaire
##### Stage: 2 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods? | 2,077 | 9,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2015-27 | longest | en | 0.881209 |
https://ncertmcq.com/rd-sharma-class-9-solutions-chapter-10-congruent-triangles-ex-10-4/ | 1,709,389,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00742.warc.gz | 417,152,150 | 14,726 | ## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
Other Exercises
Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°
Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.
Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°
Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.
Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°
Question 4.
In the figure, state which lines are parallel and why.
Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE
Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.
Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°
Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:
Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,
∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°
Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l
∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD
Question 9.
In the figure, ∠1 = 60° and ∠2 = ($$\frac { 2 }{ 3 }$$)3 a right angle. Prove that l || m.
Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = $$\frac { 2 }{ 3 }$$ rd of a right angle 2
= $$\frac { 2 }{ 3 }$$ x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m
Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.
Solution:
In the figure,
l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l
Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°
∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°
Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)
∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°
Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.
Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n
Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?
Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n
Question 16.
Which pair of lines in the figure are parallel? Give reasons.
Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC
Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l
To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m
Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.
Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.
Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.
Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH
∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°
Question 21.
In the figure, show that AB || EF.
Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF
Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.
Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D
∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°
Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.
Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary
Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.
Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD
Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB
Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°
Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD
Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°
Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.
Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)
BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF
Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.
Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G
Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.
Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | 4,940 | 12,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-10 | latest | en | 0.615475 |
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/floating-point-and-benfords-law-part-two | 1,582,136,694,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00360.warc.gz | 353,303,801 | 8,794 | # Floating Point And Benford's Law, Part Two
A number of readers asked for an explanation of my offhand comment that Benford's Law can be used to show that binary is the best base for doing floating point math.
One of the desired properties of a floating point system is that the "representation error" is as small as possible. For example, suppose we want to express "one third", in ten-place fixed point decimal notation. The closest we can get is "0.3333333333", which has a representation error of 0.0000000000333333…
A useful way to get a handle on representation error is to look at the granularity of the system. By granularity I mean the smallest difference we can make between two values. For ten-place fixed point decimal notation, the smallest nonzero difference between any two numbers is 10-10. In a floating point system, the granularity changes as the exponent changes. Really large numbers have large granularity; the difference between two successive floats might be millions or billions. And really small numbers have tiny granularity. The maximum possible representation error is half the granularity. Since there is a clear relationship between granularity and representation error, I’m just going to talk about granularity from now on. Small granularity = small representation error = goodness.
Let's consider two similar systems, one which uses a binary mantissa and one which uses a hexadecimal mantissa. I’m going to continue my convention of showing binary numbers in blue, and we’ll do hex in green.
Suppose we've got on the one hand, 32 bit IEEE floats, aka "singles". That is, we've got one sign bit, nine bits of exponent biased by -255 (so exponents can be from -255 to 256), a 22 bit mantissa, and an implicit leading "1." So a number like
(0, 100000011, 1100000000000000000000) would be +1.1100 x 24 = 28
I’m getting sick of spelling out the exponents in non-biased binary. From now on, I’m just going to give them in decimal. And I’m going to give the sign bit by plus and minus, not zero and one.
We now want a hex system with roughly similar range that doesn't take up more storage. Suppose we've got one sign bit, seven bits of exponent biased by -63, and six hex digits for the mantissa. We don't get a leading "1." because not every hex number can be so expressed, so we'll have to use a leading 0. That system has roughly the same range, and if we were to "behind the scenes" express this thing as bits, we'd still be using 32 bits. One for the sign, seven for the exponent, and 24 for the mantissa.
In this system,
(+, 2, A00000) would be +0.A00000 x 162 = 160
An immediate disadvantage of this system is that numbers have multiple representations.
(+, 2, A00000) and (+, 3, 0A0000) are the same number. Let’s ignore that for now. We’ll ignore all hex mantissas with leading zeros. (And besides, the binary IEEE system wastes lots of cases for denormals and NaNs too, so it's not clear that this is any worse.)
What's the granularity of the hex system when, say, the exponent is 2? Well, consider a number like
(+, 2, A00000) – the smallest possible number higher than this is (+, 2, A00001) which is 2-16 larger. Clearly, 2-16 is the granularity for all values with an exponent of 2. More generally, if the exponent is N then the granularity is 24N-24.
How would we represent
(+, 2, A00000) in our binary system? That would be (+, 7, 0100000000000000000000) = +1.01 x 27 = 160. The next largest number that can be represented in our system is (+, 7, 0100000000000000000001) which is 2-15 larger. So the hex system has smaller granularity and hence smaller representation error, and is therefore the better system, right?
Not so fast. Let’s make a chart.
Decimal Hex Binary Binary Granularity Exponent
16 (+, 2,
100000) (+, 4, 0000000000000000000000) -18
32 (+, 2, 200000) (+, 5, 0000000000000000000000) -17
48 (+, 2, 300000) (+, 5, 1000000000000000000000) -17
64 (+, 2, 400000) (+, 6, 0000000000000000000000) -16
80 (+, 2, 500000) (+, 6, 0100000000000000000000) -16
96 (+, 2, 600000) (+, 6, 1000000000000000000000) -16
112 (+, 2, 700000) (+, 6, 1100000000000000000000) -16
128 (+, 2, 800000) (+, 7, 0000000000000000000000) -15
144 (+, 2, 900000) (+, 7, 0010000000000000000000) -15
160 (+, 2, A00000) (+, 7, 0100000000000000000000) -15
176 (+, 2, B00000) (+, 7, 0110000000000000000000) -15
192 (+, 2, C00000) (+, 7, 1000000000000000000000) -15
208 (+, 2, D00000) (+, 7, 1010000000000000000000) -15
224 (+, 2, E00000) (+, 7, 1100000000000000000000) -15
240 (+, 2, F00000) (+, 7, 1110000000000000000000) -15
The hex system always has a granularity exponent of -16. In 8/15ths of the cases, the binary system has worse granularity than the hex system. In 4/15ths of the cases, they have the same granularity, and in only 3/15ths of the cases, the binary system has better granularity. Therefore the hex system is clearly the same or better most of the time. We should use this system instead of binary floats.
Not so fast! We’ve forgotten Benford’s Law! That's true if any number is as likely as any other, but that's not realistic.
Suppose the numbers that we are manipulating obey Benford’s Law. In that case, we would expect that fully a quarter of them would begin with 1 when encoded in hex. We’d expect another quarter of them to begin with 2 or 3, another quarter to begin with 4, 5, 6 or 7, and the remaining quarter to begin with 8, 9, A, B, C, D, E and F. If we make that assumption then we must conclude that the binary system is better half the time, equal a quarter of the time, and worse a quarter of the time.
Clearly this isn't the case just for the exponent 2. For any hex exponent N, the hex system will have a granularity exponent of 4N-24. For numbers in the range expressible by the hex system with that exponent, a quarter of the time, the binary system will have a granularity exponent of 4N-26, a quarter of the time it'll be 4N-25, a quarter of the time it'll be 4N-24, and the remaining quarter it'll be 4N-23, so three-quarters of the time it'll be as good or better. On average, the binary system is considerably better if data are distributed according to Benford's Law.
This is just one example, not a proof. But we could generalize this example and show that for any system with a given number of bits, binary mantissas yield smaller representation errors on average than mantissas in any other base. | 1,789 | 6,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-10 | latest | en | 0.906218 |
https://forum.effectivealtruism.org/posts/6iChzc36Bjpsc2xAK/estimating-the-value-of-mobile-money | 1,702,247,414,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00186.warc.gz | 289,758,347 | 141,881 | # Estimating the Value of Mobile Money
Dec 21 20164 min read 18
# 8
In the US we have many options for sending money around: there's the traditional cash, check, credit card, bank transfer, etc., and then there's PayPal, Venmo, Square, Wallet, etc. In poorer countries, however, if you want to send money to someone in another city your best option is often to send cash on a bus. In 2007 Safaricom introduced M-Pesa in Kenya, where it has been very successful. Can we get an estimate of how valuable M-Pesa has been?
In The long-run poverty and gender impacts of mobile money (Suri and Jack 2016) they try to figure out the effect of the M-Pesa rollout on per-capita consumption. They argue that from 2008 to 2010, when M-Pesa was just getting started, new agents were distributed effectively randomly. [1] Then they look at how areas that were close to M-Pesa agents in 2008-2010 compare to ones that weren't, and argue that this difference represents the effect of having access to M-Pesa.
They find that having access to M-Pesa raised log [2] per-capita consumption by 0.012 (n=1593, standard error=0.005, Sidak-Holm p=0.04). (EDIT: this is an underestimate; see update below.) The reason to care about log consumption is that the more money you have the less valuable additional money is, and this relationship is approximately logarithmic (though it might fall off even more sharply). So by averaging log consumption over all the people in their sample we're getting a rough approximation of how much better M-Pesa access made people's lives.
What does a 0.012 increase in log per-capita consumption look like? Here's a table:
per-capita consumption change
$50,000$603
$25,000$302
$10,000$121
$5,000$60
$2,500$30
$1,000$12
$500$6
$250$3
$100$1
To get a comparision, another way to increase someone's consumption would be to just give them money. GiveDirectly is currently the best option for doing this, and on average GiveDirectly gives $288 per-person to recipients with an annual consumption of$238, a log-increase of 0.79. This is 66x the 0.012 log-increase for M-Pesa availability. GiveDirectly is about 80% efficient, so getting someone $288 costs about$360. So, very roughly, giving someone access to M-Pesa one year earlier is about as valuable as donating $5.45 to GiveDirectly. [3] While M-Pesa was very successful in Kenya, its expansion to other markets has been slow. Improving M-Pesa's expansion, or starting a competitor that can expand faster, could be very valuable. For example, the population of Ethiopia is 94M, so getting mobile money available in Ethiopia a year earlier would be about as good as donating$500M to GiveDirectly. [4]
This is a very rough analysis, and as it is essentially a multiplicative chain errors anywhere along the chain are magnified. For example, two errors in the first version of this post moved the final estimate by 2.3x and 1.5x respectively. This estimate is nowhere near as robust as, for example, GiveWell's estimates for GiveDirectly or the AMF, and those are still very uncertain. Still, it suggests deploying mobile money is an extremely promising intervention, competitive with the top global poverty interventions.
Update 2017-01-01: Tavneet Suri, corresponding author of the paper I'm drawing on here, responded to my question about the interpretation of cell 1 in table 1. Above I interpret it as the effect of having an M-Pesa agent within 1km, but it's actually the effect of having one additional M-Pesa agent within 1km. Since the average number of M-Pesa agents within 1km was ~9, naively we'd expect this to mean the effect of M-Pesa is 9x what I estimated above. But since the first M-Pesa agent is probably much more valuable than the tenth, and their sample mostly didn't include this no-access-to-access transition, the true effect is probably more than 9x. On the other hand, this is kind of big, maybe implausibly so. Instead of a ~1% increase, we're talking about more like 20%. Could M-Pesa really have made Kenya that much richer?
Cross-posted from jefftk.com.
[1] On the other hand, after 2010 they believe there was a systematic effort to expand in places with higher demand. Overall, this is one of the weakest parts of this analysis: if places that got agents earlier were more actually more economically dynamic then it wouldn't be at all surprising to see greater consumption growth there.
[2] The paper doesn't say what base they're using for the log, but I'm assuming here that it's e: natural log is standard in economics. Additionally, in Table S1 of the supplement they give both "Daily per Capita Consumption" and "Log Daily Per Capita Consumption", and the particular numbers they have rule out a base as large as 10. In a draft of this post I had initially assumed log10: thanks to Rob Wiblin for catching this.
[3] My model of this is that you can either wait and get the consumption boost from mobile money eventually, or you can do something to make it happen sooner. If we take a window long enough to include the eventual deployment of mobile money I would expect to see something like:
The solid line represents the path the average log(per-capita consumption) would take, with a jump at the eventual introduction of mobile money, while the dashed line represents moving that jump forward in time. So the benefit is the area between the two curves: how much better mobile money is for people, multiplied by how much you can accelerate it.
[4] Or donating $105M to the AMF, according to GiveWell's cost-effectiveness estimates. # 8 # Reactions # More posts like this Comments18 Sorted by Click to highlight new comments since: Because it is helpful to think about exactly what intervention is needed to help mobile money expand (which may differ by country), I'm throwing here a few potential barriers (mostly based on my own experience in Kenya and Myanmar): • Regulatory barriers (India allowed it only recently because of this; in Myanmar it's still ongoing) • Network effects: in Kenya I heard that an important reason it took off was that Safaricom had a very high market share (maybe near 70%?); in Nigeria I heard that the fragmentation of the telecommunication market is one reason it didn't take off. I'm not sure if more countries are similar to Kenya or Nigeria, also these are all anecdotal. One interesting thing though is that lack of competition in Kenya may have contributed to the high charges (though there is more competition now including from mobile carriers and banks). • Lack of trust: people may not trust mobile carriers or mobile money agents. Probably less of a problem in a close knit community where agents are shopkeepers. Also, lack of trust in banks is a common problem in developing countries but I have no idea about trust on mobile carriers/agents. • Existing alternatives already good enough: this has been mentioned to me in Myanmar, that the traditional "hundi" system of money transfer works well and is cheap which may dampen adoption of mobile money. If that's true then mobile money wouldn't contribute much anyway, but I'm skeptical since mobile money is really much more convenient. (Also it can be used as a savings tool like a checking account, and the poor often face savings constraint too, but I'm not sure how effective that will be; interventions that tackle "self-control" have worked well on this so such elements might need to be bundled in order for mobile money to help with saving) The intervention I had in mind when writing this post was joining a start-up that has been working on this and has been seeing good results so far: http://www.jefftk.com/p/leaving-google-joining-wave Wave is really good! (I use it) Another thing one can do is to work for some mobile money company in a developing country to design products that benefit the poor (e.g. saving, credit, that I mention in the other post), like the American guys I met in Myanmar's Wave Money (but they are still early stage and has many challenges before having an impact). (Not suggesting you should do it though -- involves moving to a developing country etc., and could be much less likely to succeed due to regulations etc.). BTW this is the mobile credit scoring company I had in mind: http://tala.co/. Thanks for this Jeff - a very informative post. The study doesn't appear to control for cash transfers received through access to M-Pesa. I was thinking about how much of the 0.012 increase in ln(consumption) was due to GiveDirectly cash transfers. Back of the envelope: • M-Pesa access raises ln(consumption) by 0.012 for 45% of population (c.20m people). • 0.012 * 20m = 234,000 unit increases in ln(consumption) • GiveDirectly gave c.$9.5m in cash transfers between 2012-14 to people with access to M-Pesa. [1]
• GiveWell estimate each $to GiveDirectly raises ln(consumption) by 0.0049 • 9.5m * 0.0049 = 46,000 unit increases in ln(consumption) So GiveDirectly accounted for (very roughly) a fifth of the 0.012 increase in ln(consumption) due to M-Pesa. [1] this is an overestimate as it assumes all transfers went to Kenya and none to Uganda (Done in haste - may have got my sums / methodology wrong) The study doesn't appear to control for cash transfers received through access to M-Pesa. Good point! I hadn't thought about this at all. GiveDirectly's cash transfers were very large, enough that$9.5m would go to 33k people ($288/person). The population was 43M, so 1 in 1300 people received money from GiveDirectly. Their sample size is just 1593, so you expect 0-2 GiveDirectly recipients. I think they should be pretty visible in the data? Might be worth writing to the authors. It seems like you're assuming that the GiveDirectly money would have gone only to the M-Pesa-access side of the (natural) experiment, but they categorized areas based on whether they had M-Pesa access in 2008-2010, not 2012-2014 when access was much higher. GiveWell estimate each$ to GiveDirectly raises ln(consumption) by 0.0049
I didn't notice that GiveWell had an estimate for this, and checking now I still don't see it. Where's this estimate from?
(In my post I just took their average amount transferred, figured out what effect that had on the average recipient's income, and then discounted by .8 for GiveDirectly's overhead.)
(My method gives an estimate of 0.0022 per dollar to GiveDirectly, so if GiveWell is estimating 0.0049 then my bottom line numbers are roughly 2x too high.)
It seems like you're assuming that the GiveDirectly money would have gone only to the M-Pesa-access side of the (natural) experiment, but they categorized areas based on whether they had M-Pesa access in 2008-2010, not 2012-2014 when access was much higher.
Ah yes - that kind of invalidates what I was trying to do here.
I didn't notice that GiveWell had an estimate for this, and checking now I still don't see it. Where's this estimate from?
It came from the old GiveWell cost-effectiveness analysis excel sheet (2015). "Medians - cell V14". Actually looking at the new one the equivalent figures seems to be 0.26% so you're right! (Although this is the present value of total increases in current and future consumption).
I spent about two hours looking at this in further depth and made an initial stab at modeling out the impact. I estimate an effectiveness of $200/hr (95% interval:$50/hr to $511/hr), not taking into account the value of donating the salary earned from working at Wave. Some places where I notice we disagree or I am confused: 1.) I disagree with you here (footnote 1) that there is a 50% chance of failure (or success). I think the chance of failure could be significantly higher. From https://en.wikipedia.org/wiki/M-Pesa: M-Pesa expanded to Kenya (>10M subscribers), Tanzania (5M), South Africa (100K in a year, 1M in five years), India (???), Mozambique (???), and Lesotho (???). Also, a 2016 Vodaphone press-release suggested M-Pesa seems to have 25M customers worldwide after 10 years of effort. Based on this, I model that a Kenya-level success (>10M subscribers) thus looks like it would have a less than 1/10 chance and a South Africa-level success (1-10M subscribers) looks like it would have a ~3/10 chance. However, I think this success figure could be lower due to diminishing marginal returns since M-Pesa has already plucked low hanging fruit. It's possible better technology could increase this chance. I'd have to know more specifically about what problems M-Pesa runs into and how these are addressed. - 2.) I think your estimate that getting M-Pesa a year earlier is only 66x worse than getting a$288 transfer from GiveDirectly is an overestimate because I expect future roll-outs will take place in countries with higher base consumption. However, as you point out, that estimate is also already an underestimate due to misunderstanding the study. I don't know how to correct for this either way, so I used the 66x number literally in my calculation.
-
3.) I disagree with you here (footnote 1a) that marginal ETG donations are at GiveDirectly levels of cost-effectiveness. I expect them to continue at AMF levels (or greater) for at least a few more years, for reasons OpenPhil mentioned and Carl mentioned. I did an AMF-adjustment in my model for this reason.
-
4.) I really don't know how many staff years it would take to either complete a roll-out or know that it's not going to happen and this is an important part of the model. I currently guess 5-10 full-time staff for 2-5 years, or 10-50 total staff years. This does not count field agents or other hired locals. I couldn't find any information on M-Pesa's total staff count anywhere. I note that Wave has at least 44 staff (from counting faces on the about page), but I don't know if they're all full-time or all focused on expanding cash transfers.
-
5.) I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF when from GiveWell's cost-effectiveness estimate, GiveDirectly has a median of $7702 per life saved, ranging from$2200 to $16000, excluding outliers. AMF has a median of$3282 per life saved, ranging from $2200 to$4800, excluding outliers. Together, this implies a comparison centered around 2.35x but ranging from 1x to 3.3x. Maybe I misread the sheet -- I haven't invested that much time in making sure I fully understand it yet.
Re 1, it's worth noting that M-Pesa was administered by very different teams in different countries. In Kenya it was allowed to operate mostly as a startup with limited oversight from Safaricom (or anyone else), whereas in other countries (to varying degrees) the people running M-Pesa were constrained by stricter management from the telecom's country head. This means that there are obvious ways in which M-Pesa was executed less well in other countries. For instance, Wave integrates with M-Pesa in both Kenya and Tanzania, and despite running on exactly the same technology, the Tanzanian system's uptime is substantially worse. Similarly, the quality of their agents and their customer support staff in Tanzania is noticeably lower.
Since Wave isn't hamstrung by oversight from a stodgy and risk-averse telecom, I think you should give less weight to examples from countries other than Kenya as a base rate.
1) ... I think the chance of failure could be significantly higher.
Possibly, but they are already starting to operate in the country in question, and my understanding is that's been going pretty well. My impression is that they're much more competent than Safaricom. My inside view is much higher than 50%, and getting down to 50% was a discount from there.
2) ... I expect future roll-outs will take place in countries with higher base consumption
I'm confused. I was trying to talk about the counterfactual for a specific very poor country if Wave were not working there. So if future mobile money rollouts by other organizations happen first in countries with higher base consumption then that increases the counterfactual impact of Wave choosing to come into a country with very low consumption.
3) ... I expect them to continue at AMF levels (or greater) for at least a few more years
4) ... I really don't know how many staff years it would take
That, combined with estimating marginal impact, makes this pretty awkward. I figure something like 40 person years?
5) ... I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF
This comes from cell F31 of the "Results" tab. I haven't put time into understanding how that's calculated, but it looked like the relevant bottom line number.
Thanks Jeff!
-
1) ... I think the chance of failure could be significantly higher.
Possibly, but they are already starting to operate in the country in question, and my understanding is that's been going pretty well. My impression is that they're much more competent than Safaricom. My inside view is much higher than 50%, and getting down to 50% was a discount from there.
Sounds like you definitely have inside info that I don't have, so for now I'd have to rely on my outside view, but I can work to acquire that inside info if I look into this more.
-
2) ... I expect future roll-outs will take place in countries with higher base consumption
I'm confused. I was trying to talk about the counterfactual for a specific very poor country if Wave were not working there. So if future mobile money rollouts by other organizations happen first in countries with higher base consumption then that increases the counterfactual impact of Wave choosing to come into a country with very low consumption.
I don't know what country Wave is looking at or how they are doing what they do because you have inside info that I don't have. If it has consumption comparable to Kenya than my point is invalid. I just was concerned that it wouldn't.
-
3) ... I expect them to continue at AMF levels (or greater) for at least a few more years
Cool. Sounds like this isn't a disagreement between us then.
-
4) ... I really don't know how many staff years it would take
That, combined with estimating marginal impact, makes this pretty awkward. I figure something like 40 person years?
Agreed that it is pretty awkward to estimate. I modified my model to use some of your inputs -- such as a 40% chance of 1-10M subscribers and a 10% chance of >10M subscribers and 40 person years -- and it comes out to $383/hr (95%:$145/hr to \$834/hr). The new mean is still in my old 95% interval which is about the best I can hope for with this level of uncertainty.
-
5) ... I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF
This comes from cell F31 of the "Results" tab. I haven't put time into understanding how that's calculated, but it looked like the relevant bottom line number.
Oh, I see that now. I suppose this is a question for GiveWell and not you. I'll ask them.
Sounds like you definitely have inside info that I don't have, so for now I'd have to rely on my outside view, but I can work to acquire that inside info if I look into this more.
If you're interested in working for Wave, or are advising other people on whether it's a good idea for them, I could imagine they'd be quite interested in talking to you!
if it has consumption comparable to Kenya than my point is invalid. I just was concerned that it wouldn't.
It's poorer than Kenya.
That sounds pretty awesome, who do you think would be a good person to reach out to when I'm ready?
Ben Kuhn maybe?
Very cool post.
Just saw that the transaction costs for m-pesa are quite high - the company makes ~20% profit... so there might be something that a Wave-like startup could do: https://en.wikipedia.org/wiki/M-Pesa#Cost.2C_transaction_charges.2C_statistics
maybe using crypocurrency - see here:
http://phys.org/news/2016-10-cryptocurrency-bottom-billion.html
Just saw that the transaction costs for m-pesa are quite high - the company makes ~20% profit
The transaction costs listed on the wikipedia page you cite aren't trivial, but would average well less than 20% unless most transactions are (a) very small and (b) to unregistered users. I'm missing something.
EDIT: could it just be that their profit is 20% of expenses, as opposed to 20% of the money that flows through the M-Pesa network?
maybe using crypocurrency
That article doesn't really show that cryptocurrency helps here. Mostly they're unhappy with transaction fees on international remittances, but you can have low transaction fees just by automating interactions with the money transfer organization, without going to cryptocurrency. And with cryptocurrency generally you pay someone a fee to get your money into the cryptocurrency and then your recipient pays someone else a fee to get it into their local currency.
I think their profit is 20% of their revenue (for a money transfer company, revenue is the total fees brought in, not total money paid into the network).
I just got back from Myanmar and I talked with some people running Wave Money (one of the mobile money companies in Myanmar, and the only licensed one so far; not related to the Wave that Jeff mentioned which sends money to Africa).
• Getting people to adopt could be a big challenge, depending on the country. In Kenya, the anecdotal story of why mobile money took off so quickly is 1) the need to send remittances, 2) preexisting methods for this being not very good for various reasons (insecurity is one); some also argue that Safaricom's unusually high market share in the country played a role in speeding up adoption through bundling of services + network effects (telecommunication markets in other countries seem more fragmented). Mobile money has not taken off in some countries, e.g. Nigeria (this article argues it's due to regulation https://iea.org.uk/blog/why-mobile-money-transformed-kenya-failed-to-take-in-nigeria). In Myanmar it remains an open question: the traditional hundi system for remittances works well for most purposes (being cheap, reliable, and not too slow), which may hinder adoption of mobile money.
• Other potential functions of mobile money: other than through remittances (which is what Suri's paper is estimating), it can also help poor populations by
• Saving: providing a safe place to save. This may be important as many poor people seem "savings constrained" (e.g. see https://www.econ.uzh.ch/dam/jcr:5f6e818a-ad07-466a-8962-fdc77bb1dfc2/casaburi_macchiavello_dairy_20160731.pdf, http://www.simonrquinn.com/research/TwoSidesOfTheSameRupee.pdf) and bank are either scarce or expensive in many rural places -- although it might be hard to convince people to adopt mobile money just for saving purposes.
• Credit: e.g. the Mpesa-based mobile loan Mshwari. There are some startups (including one in Kenya, whose name I forgot) that creates credit scores for people based on their mobile money transaction history, mobile phone records etc. (Mshwari does a version of this but doesn't seem very sophisticated; the startups probably use more "big data") which may help the poor access credit in a way that's much cheaper and sustainable than the traditional microfinance model. (In Myanmar I know one startup trying to do this and giving out small amounts of loan for shorter periods, basically competing with money lenders in slums -- seems hard but could be really good if they succeed; they are quite early stage now.)
• Sending government benefits, e.g. India is considering introducing universal basic income, and already have biometric identification for most citizens, but one of the remaining barriers is the scarcity of bank branches in rural places. If each village has a mobile money agent things would be much easier (and this has implications not only for poverty reduction but maybe also curbing corruption and improving local governance etc.).
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· 5d ago · 5m read | 5,520 | 23,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | longest | en | 0.955204 |
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Exam guide questions & Answers 2012
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## ss3 obj maths questions
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FIRST TERM EXAMINATION, 2010/2011 SESSION.
SUBJECT: MATHEMATICS 1(OBJECTIVE)
CLASS: SSS 3 TIME ALLOWED: 2˝HRS
Calculate the size of an interior angle of a regular hexagon
720° b. 120° c. 270° d. 102°
The difference between 4 and 2 is greater than the sum of and1 by
b. c. d.
What is the length of an arc of a circle that subtends an angle of at the center of a circle of radius 8cm?
10cm b. 8cm c. 40cm d. 20cm
The first term of an arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81.
12 b. 27 c. 9 d. 4
Find a two digit number such that three times the tens digit is two less than twice the units digit, and twice the number is 20 greater than the number obtained by reversing the digits.
42 b. 47 c. 24 d. 74
If x varies inversely as y, y varies directly as the square of t and x = 1 when t = 3. Find the value of x when t =.
81 b. 27 c. d.
In the diagram below, O is the centre of the circle. If the reflex angle ZOY = 300°, find ?ZXY.
a. 60°
b. 50°
c. 40°
d. 30°
Express: as a single fraction.
Use the following information to answer questions 9 and 10.
A distance of 5.6km is represented by a length of 28mm on the map.
What is the scale of the map?
2 × : 1 b. 2 × : 1 c. 2 × d. 2 ×
What length on the ground does 10cm on the map represent?
2 km b. 20 km c. 200 km d. 2000 km
A rectangular lawn is (x – 2) m long and (x – 3) m wide. If the perimeter is 70 m, calculate the area of the lawn.
289 b. 306 c. 324 d. 400
In the diagram below, WX is parallel to YZ and MN is a transversal. Express p in terms of t.
p = t – 5
P = t – 15
P = t + 10
P = t + 15
Solve :
– 1 b. c. d. 1
By selling an article for ?450:00, Abu lost 10% of the cost price. Find the cost price.
?550.00 b. ?525.00 c. ?500.00 d. ?475.00
A point on the ground is 8 m away from the foot of a vertical wall 6 m high. Calculate correct to the nearest degree, the angle of depression of the point from the top of the wall.
53° b. 45° c. 37° d. 30°
In the diagram below, POQ is a quadrant of a circle centre O and radius 14 cm. Use this information to answer questions 16 and 17. Take to be .
Calculate the length of the minor arc PQ.
11 cm b. 22 cm c. 44 cm d. 66 cm
Calculate the perimeter of the minor sector POQ.
94 cm b. 80 cm c. 50 cm d. 39 cm
To locate Musa’s house from his house, Okey has to move 7 km due east and 7 km due north. Find the bearing of Musa’s house from Okey’s house.
225° b. 135° c. 090° d. 045°
Find the value of x for which the expression is undefined.
x = -3 or 3
x = -1 or 4
x = 1 or 4
x = 1 or 3
If is a perfect square, what is the value of k?
25 b. c. -25 d.
If two straight line segments, PQ and RS intersect at a point T such that ? PTR = 2x + 2 and ? PTS = x + 4, calculate the value of ? PTR.
60° b. 62° c. 118° d. 120°
In the diagram, ? MNL = 30°, ? JKL = 73° and ? MLK = x°. Find x.
77°
81°
103°
107°
Solve the inequality:
x = 9 b. x = 9 c. x = -9 d. x = -9
In the diagram, PQTV is a parallelogram. ? TRS = 110° and ? PVT = 80°. Calculate the size of ? QTR.
50°
40°
20°
10°
A coin is thrown three times. What is the probability of having three heads or three tails in this experiment?
b. c. d.
Calculate the total surface area of a cube of side 5 cm.
150 b. 120 c. 100 d. 75
Obi has four times as many sweets as Olu. If he gives 21 sweets to Olu, they will have equal number of sweets. How many sweets have Obi?
56 b. 52 c. 48 d. 44
In the figure below, ? PQT = ? PSR = 65° and ? PTQ = ? PRS = 45°. Which of the following statements about triangles PTQ and PRS is\are CORRECT?
I, II and III
II and III only
I and III only
I and II only.
The probability that a boy passes an examination is 0.3 and that of a girl is 0.4. What is the probability that a boy and a girl will pass?
0.10 b. 0.12 c. 0.42 d. 0.70
Each interior angle of a regular polygon is 144°. How many sides have the polygon?
6 b. 9 c. 10 d. 12
From the figure below calculate ? QRS
280°
80°
70°
40°
Find the sum to infinity of the following series 3 + 2 + + + + …
1270 b. 190 c. 18 d. 9
Simplify;
b. c. 1 d.
The number converted to base ten is equal to
4 b. 23 c. 49 d. 53
A man spent of his monthly salary on rent and of the remainder on the household needs. If ?135.00 is left, how much is his monthly salary?
?450.00 b. ?225.00 c. ?192.85 d. ?125.00
Simplify;
Log 2 b. log 8 c. 1 d. 0
In the diagram, if PS = SR and PQ is parallel to SR, what is the size of angle PQR?
25°
50°
55°
65°
Find the area of the shaded portion in the figure below, if the triangle is an equilateral triangle of area 8 and the circle center O is of diameter 6 cm.
4.714
3.286
7.999
9.437
An open rectangular box externally measures 4m x 3m x 4m. Find the total cost of painting the box externally if it costs ?4.00 to paint one square meter.
?136.00 b. ?272.00 c. ?408.00 d. ?544.00
The sum of the interior angles of a regular polygon is 30 right angles. How many sides have the polygon?
17 sides b. 26 sides c. 30 sides d. 34 sides
If = 1.732, find the value of without using tables.
0.732 b. 1.346 c. 1.155 d. 3.464
The nth term of a sequence is. Which term of the sequence is?
3rd b. 4th c. 5th d.6th
A boy estimated his transport fare for a journey as ?190.00 instead of ?200.00. Find the percentage error in his estimate.
95% b. 47.5% c. 5.26% d. 45%
A man is four times as old as his son. The difference between their ages is 36 years. Find the sum of their ages.
45 years b. 60 years c. 48 years d. 74 years
In the diagram below, /AD/ = 10 cm, /DC/ = 8 cm, and /CF/ = 15 cm. Use the information to answer questions 45 and 46
If the area of triangle DCF = 24, find the area of Quadrilateral ABCD.
24 b. 48 c. 80 d. 96
Which of the following is correct?
Area of BCF = Area of DCF
Area of DABO = Area of CFEO
Area of ADE = Area of DCFE
The table below shows the scores of a group of 40m students in a mathematics test. Use the table to answer questions 47 and 48.
SCORES
1
2
3
4
5
6
7
8
9
FREQUENCY
2
3
6
7
9
6
2
2
3
If the mode of m and the median is n, then (m,n) is
(5,5) b. (5,6) c. (6,5) d.(9,5)
What is the mean of the distribution?
5.2 b. 5.0 c. 4.8 d. 4.2
The diagram below shows a triangular prism of length 7cm. The right angled triangle PQR is a cross section of the prism. /PR/ = 5cm and /RQ/ = 3cm. Use the information to answer questions 49 and 50
What is the area of the cross-section?
4cm2 b. 6cm2 c. 15cm2 d. 20cm2
What is volume of the prism?
28cm3 b. 42cm3 c. 70cm3 d. 84cm3
```
### ss3 obj maths guide
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Htaccess html Php javascript Asp css maths Past Questions Practice Tests Online | 6,302 | 17,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-21 | longest | en | 0.82496 |
https://stemandleafeducation.com/2019/01/29/how-long-is-that-mountain-path/ | 1,550,783,465,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247508363.74/warc/CC-MAIN-20190221193026-20190221215026-00526.warc.gz | 708,845,394 | 35,634 | How long is that mountain path?
We have continued talking about using integrals for geometric purposes in Calculus. Yesterday, we talked about arc-length. While doing so, my mind wandered back to the trip Paula and I took to visit my sister when she got married. How are these connected? Well, we went up and down many mountains and the distance traveled through these mountain paths provides a great way to visualize arc-length.
The path
I want to walk up, down, up and down the mountains in the picture below. That is, I want to follow the path outlined in black.
If I want to determine the total distance I would have to walk in order to cover this path, I would need to know some additional information. In particular, we would need to know the height and the width of the given mountains. If we knew either one of these, or the distance to the mountains, we could find the other using Paula’s height as a reference. However, I don’t actually know any of these since I just chose a picture that looked pretty. Therefore, I am going to use arbitrary units that seem to fit the picture. Doing this, we have the following picture.
From the graph given above, we see that our path appears to go through the following points,
x y
-5 0
-3 1
-1 0
0 $$-\frac{1}{2}$$
1 0
4 3
9 0
Choosing a function type
Now that we know what we want our path to look like and what points we want it to go through, we can now determine a function to model the given path. We do have multiple choices for types of functions, so let’s look at some.
• Polynomial
• Looks like a fourth degree polynomial since it has three local extrema.
• If wanted to guarantee that all points fell on the same polynomial, we would need at least an 6th degree polynomial.
• Sinusoidal
• The function seems to be periodic, so a sinusoidal function could work.
• However, the amplitude of the crests are not the same, so this may not be the best.
• Piece-wise
• Would focus on finding the best fit for specific portions instead of the whole.
• This would require finding more than one function.
Between the options, I thought that defining the function piece-wise would end up with a good approximation and require the least amount of work possible to do so. For my pieces, I will be using quadratic functions through the points. In general, a quadratic function would be uniquely determined by three points. However, if the vertex, we can instead determine the function using two points. Therefore, I could find a function connecting each pair of points and work over each of these intervals.
Finding the function
We will be finding the quadratic through the given points. In order to do this, we can use the formula for a quadratic $$f(x)=a(x-h)^{2}+k$$. Note that, using this equation, we have that the vertex is the point $$(h,k)$$ and we would then find $$a$$ using our other point.
As we look at the first two points, we would say that the point (-3,1) is the vertex of the quadratic function and (-5,0) is on the graph. Therefore, $$h=-3$$ and $$k=1$$. We can then find that \begin{align*}0&=a(-5-(-3))^{2}+1 \\ a&=\frac{-1}{4}.\end{align*} Hence $$f(x)=\frac{-1}{4}(x+3)^{2}+1$$ on the interval $$[-5,-3)$$.
If we repeat this process for each of the consecutive pairs given above, choosing the appropriate point as the vertex, we arrive at the function $f(x)=\begin{cases} \frac{-1}{4}(x+3)^{2}+1 \text{ for } -5 \leq x < -1 \\ \frac{1}{2}x^{2}-\frac{1}{2} \text{ for } -1 \leq x < 1 \\ \frac{-1}{3}(x-4)^{2}+3 \text{ for } 1 \leq x < 4 \\ \frac{-3}{25}(x-4)^{2}+3 \text{ for } 4 \leq x \leq 9. \end{cases}$
Finding Distance
We now recall that you can find the total distance traveled by finding the arc-length of the function. If the function was linear, we could find this distance by taking $$d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$$. Since we do not have a linear function, we can instead use the linear approximation to approximate the distance moved over small intervals. If we then add these all up, we get the total distance traveled. That is \begin{align*}\text{distance} & \approx \sum \sqrt{(\Delta x)^{2}+(\Delta y)^{2}}. \end{align*}
As we let our $$\Delta x$$ get infinitesimally small, we would find that the approximation would be exact. In order to sum these as $$\Delta x$$ gets small, we note that this is by definition $$\int_{a}^{b}\sqrt{(dx)^{2}+(dy)^{2}}$$, where $$a$$ is the smallest $$x$$ and $$b$$ is the largest $$x$$ we are interested. By simplifying, we find that \begin{align*} \text{arc-length}&=\int_{-5}^{9}\sqrt{1+f'(x)}dx.\end{align*}
In order to find this integral, we will find the integral over the corresponding intervals where the function stays the same. We then get the first arc-length is \begin{align*}\int_{-5}^{-1}\sqrt{1+(\frac{-1}{2}(x+3))^{2}}dx&=\int_{-5}^{-1}\sqrt{1+\frac{1}{4}(x+3)^{2}}. \end{align*}
We will now need to find an anti-derivative in order to calculate this. To do so, we let $$i=\frac{1}{2}(x+3)$$. Then $$di=\frac{1}{2}dx$$, so $$dx=2di$$. We therefore get that $\int_{-5}^{-1} \sqrt{1+\frac{1}{4}(x+3)^{2}} =\int_{-1}^{1} \sqrt{1+i^{2}}2di.$ We then let $$i=\tan(\theta)$$, then $$di=\sec^{2}(\theta)d\theta$$, so \begin{align*}\int_{-1}^{1} \sqrt{1+i^{2}}2di&=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1+\tan^{2}(\theta)}\sec^{2}(\theta)d\theta \\ &=2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} sec^{3}(\theta)d \theta \\ &=\sec(\theta)\tan(\theta)|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} sec(\theta)d\theta \\&=2\sqrt{2}+\ln|\sec(\theta)+\tan(\theta)||_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\ &=2\sqrt{2}+\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1) \\ & \approx 4.59. \end{align*}
As I went through this, I wanted to show as much as possible in order to help show the steps involved in the integration. Here, we needed a substitution, and a trig substitution. Furthermore, in order to find the anti-derivative of $$\sec^{3}(\theta)$$, we needed a reduction formula and an integral of $$\sec(\theta)$$. While I considered showing how to get these using integration by parts and another substitution, I felt that we had already seen enough detail for one post. If you would like to see more, please let me know and I can send these details to you.
Total distance
So far, we have only found the distance traveled as we moved over the first portion of the integral. We can now follow a very similar process to find the total distance traveled over the remaining intervals. When done, we need to find the sum of these in order to find the total distance traveled. If you do this, you will find $$\text{Total Distance}=4.59+2.30+4.44+6.02=17.35$$. That is, in order to walk the path along the along the rise of the mountains, you will have to walk 17.35 units where the units correspond to the choice I made when plotting the graph.
Conclusion
What we’ve seen today is that math in general and calculus in particular can really be found anywhere. Even in a nice picture of a person fishing in front of the mountains, we were able to find a use for calculus. Furthermore, I hope that going through the example of arc-length will help those of you currently in a calculus 2 class that are working on the topic (ie those of you in my calculus class).
If you enjoyed the post and found it helpful, make sure to share with others that may also find it helpful. We really appreciate when you help spread the information we provide.
1 thought on “How long is that mountain path?”
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## Pokémon Showdown
2012.03.16 23:45 7thFullMoon Pokémon Showdown
Pokémon Showdown is a free, web-based battle simulator. It comes with many different servers and a plethora of features such as fan-made tiers and chatrooms to fit your style. Whether you use it to learn competitive battling, to practice, or just use it for recreation, you are welcome to this community.
2008.07.08 22:36 PlayStation Portable (PSP)
The gaming subreddit which fits in your pocket!
2008.08.26 02:03 Pizza
The home of pizza on reddit. An educational community devoted to the art of pizza making.
2023.06.01 20:29 Emanftw Would anyone like to help me with this? Please and thank you!
2023.06.01 17:37 StevenVincentOne Heuristics and Conscious Evolution
https://chat.openai.com/share/c6e8289b-f77d-4dfc-98c4-1a2ee93fa844
Me:
Help me get a better understanding of the scientific concept of Heuristics
ChatGPT
Heuristics refer to cognitive shortcuts or strategies that individuals use to simplify decision-making and problem-solving in complex situations. In other words, heuristics are mental "rules of thumb" that help people make judgments quickly and efficiently.
Heuristics are often used in situations where complete information is not available, or when there is too much information to process effectively. These strategies can be helpful, but they can also lead to errors in judgment and decision-making. Some common examples of heuristics include:
1. Availability heuristic: This heuristic involves estimating the likelihood of an event based on how easily similar examples come to mind. For example, if you hear about a plane crash on the news, you may overestimate the risk of flying even though statistically it is still safer than driving.
2. Representativeness heuristic: This heuristic involves categorizing objects or events based on how similar they are to a prototype in one's mind. For example, assuming a person is a professor because they wear glasses and a tweed jacket.
3. Anchoring and adjustment heuristic: This heuristic involves using an initial value (or anchor) as a reference point when making a decision, and then adjusting that value based on additional information. For example, when making an offer on a house, a buyer might start with a lowball offer as their anchor and adjust based on the seller's response.
It's important to be aware of heuristics and their potential limitations, as relying too heavily on them can lead to biases and errors in judgment. However, heuristics can also be a useful tool for simplifying complex decision-making processes and conserving mental resources.
Me:
Let's go deeper into the benefits of heuristic processes versus the detriments.
ChatGPT
Heuristics can provide several benefits, but they can also lead to errors and biases. Here are some of the benefits and detriments of heuristic processes:
Benefits:
1. Speed and efficiency: One of the primary benefits of heuristics is that they can save time and mental effort. Rather than considering all available information and weighing every possible outcome, heuristics allow individuals to make quick decisions and take action based on incomplete information.
2. Useful for problem-solving: Heuristics can be especially useful for solving problems that are complex and have multiple possible solutions. By using rules of thumb, individuals can evaluate potential solutions more efficiently and select the best one based on their previous experiences and knowledge.
3. Helpful for managing cognitive overload: Heuristics can help individuals manage cognitive overload, a condition in which too much information is presented to an individual, causing them to be overwhelmed and unable to make sound decisions. By using heuristics, individuals can filter out irrelevant information and focus on what is most important.
Detriments:
1. Biases and errors: One of the primary drawbacks of heuristics is that they can lead to biases and errors in decision-making. For example, individuals may rely too heavily on the availability heuristic and make judgments based on information that is readily available, rather than evaluating all relevant information.
2. Limited accuracy: Heuristics can also be limited in their accuracy. Because they rely on previous experiences and knowledge, heuristics may not always be applicable to new or unique situations.
3. Overconfidence: Finally, heuristics can lead to overconfidence. Individuals may become too confident in their ability to make decisions based on heuristics and fail to consider alternative perspectives or information that contradicts their existing beliefs or knowledge.
Overall, heuristics can be a useful tool for simplifying complex decision-making processes, but they must be used with caution. It is important to recognize the potential limitations and biases of heuristics and to supplement heuristic processes with critical thinking and analysis when appropriate.
Me:
Is the presence of heuristic cognitive processes limited to humans or do other forms of life also utilize heuristics?
ChatGPT
Heuristics are not limited to humans and can be observed in other forms of life as well. Many animals, including insects, birds, and mammals, use heuristics to make decisions and solve problems.
For example, bees use the "flower constancy" heuristic, which involves returning to a specific type of flower after finding food there previously. This strategy allows bees to maximize their foraging efficiency by focusing on a single type of flower and avoiding wasted effort on less rewarding flowers.
Similarly, some bird species use the "distinctive feature" heuristic to identify predators. This heuristic involves recognizing and avoiding features that are not common among harmless species, such as a bright coloration or unusual shape.
Even some simple organisms, such as bacteria, use heuristics to guide their behavior. For example, bacteria use the "quorum sensing" heuristic to coordinate group behavior based on the density of their population.
Overall, heuristics are a common feature of cognitive processes across many different forms of life, and they likely evolved as a way to optimize decision-making and problem-solving under conditions of uncertainty or limited information.
Me:
Are there examples of heuristics in non-animal life forms, such as plants?
ChatGPT
While the concept of heuristics is most commonly associated with animal behavior and decision-making, there is some evidence that plants may also use heuristics in their own way.
For example, plants have been observed to use the "optimal partitioning" heuristic to allocate resources between different tissues and organs. This heuristic involves allocating resources to the parts of the plant that are most likely to contribute to its overall fitness, based on factors such as growth rate, photosynthetic capacity, and nutrient availability.
Plants also use the "leaf economic spectrum" heuristic, which involves balancing resource allocation between leaf construction and maintenance, in order to maximize photosynthetic efficiency and minimize resource waste.
In addition, some research suggests that plants may use the "root plasticity" heuristic to adjust their root growth in response to environmental cues such as soil moisture and nutrient availability. This heuristic involves allocating resources to the parts of the root system that are most likely to be productive under different conditions.
While the concept of heuristics in plants is still a relatively new area of study, these examples suggest that plants may use similar cognitive strategies to optimize their growth and survival in their environment.
Me:
Is it possible that heuristic processes are encoded in DNA and are an evolutionary adaptation that is stored in DNA?
ChatGPT
The question of whether heuristic processes are encoded in DNA is a complex one, as it is still not fully understood how cognitive processes such as decision-making and problem-solving are encoded in biological systems.
While some researchers have proposed that heuristics could be an evolutionary adaptation that is stored in DNA, there is currently limited direct evidence to support this hypothesis. It is also important to note that heuristics are not innate, fixed behaviors, but rather strategies that individuals learn and refine over time based on their experiences and environment.
However, there is evidence that the neural networks and structures that underlie heuristics are shaped by evolutionary processes. For example, studies have found that different animal species, which have evolved in different environments and under different selection pressures, tend to use different heuristics to solve similar problems.
In addition, some researchers have suggested that the development of heuristics may be driven by principles of natural selection, which favors strategies that are efficient and effective in solving problems under conditions of uncertainty and limited information.
Overall, while the exact relationship between heuristics and DNA is still an area of active research, it is likely that heuristics have evolved through a combination of genetic and environmental factors, and that the neural mechanisms that underlie heuristics have been shaped by evolutionary processes over time.
Me:
What are the scientific disciplines that reference and research heuristic processes?
ChatGPT
Heuristic processes are studied in a wide range of scientific disciplines, as they are relevant to many areas of research that involve decision-making, problem-solving, and human behavior. Here are some of the scientific disciplines that reference and research heuristic processes:
1. Psychology: Heuristics are a central topic of study in the field of cognitive psychology, which focuses on how people acquire, process, and use information. Researchers in this field use a variety of methods, including experiments and observational studies, to investigate the factors that influence heuristics and their impact on decision-making and problem-solving.
2. Neuroscience: Heuristics are also of interest to researchers in the field of neuroscience, who use brain imaging and other techniques to study the neural mechanisms that underlie heuristics and how they are processed in the brain.
3. Economics: Heuristics are a key area of research in the field of behavioral economics, which focuses on how people make decisions and the cognitive biases that can influence their choices. Researchers in this field study the effects of heuristics on economic behavior and how these effects can be used to improve decision-making.
4. Artificial intelligence and computer science: Heuristic algorithms are used extensively in artificial intelligence and computer science to solve complex problems that are difficult to solve using traditional methods. Researchers in this field study how heuristics can be used to optimize algorithms and improve the efficiency of computational processes.
5. Ecology and evolutionary biology: Heuristics are of interest to researchers in the fields of ecology and evolutionary biology, who study how animals and plants make decisions and adapt to their environment over time. Researchers in this field use a variety of methods, including field studies and experiments, to investigate how heuristics are used in different ecological and evolutionary contexts.
Overall, heuristic processes are a cross-disciplinary topic of interest, and researchers from many different fields have made important contributions to our understanding of how heuristics work and their impact on human behavior and decision-making.
Me:
It seems like the Information Theory of Claude Shannon and the phenomenon of Shannon Information arose from a radical new understanding of heuristic language processes and the use of a higher order systemization of heuristics. If that is correct, that would make the entire field of Information Theory dependent upon heuristics.
ChatGPT
While it is true that Claude Shannon's theory of information is relevant to the study of heuristics, it is not accurate to say that the entire field of Information Theory is dependent upon heuristics. Rather, Shannon's theory of information is concerned with the quantification and transmission of information, and how information can be represented and processed using mathematical models.
That being said, the concepts of information and heuristics are closely related, as both involve the processing and communication of knowledge and data. In particular, heuristics can be seen as a form of information processing, as they involve the use of mental shortcuts and rules of thumb to make decisions and solve problems.
In the context of Shannon's theory of information, heuristics might be seen as a way of compressing and transmitting information more efficiently. By using heuristics to simplify complex information and decision-making processes, individuals can reduce the amount of information that needs to be transmitted and processed.
Overall, while there are connections between heuristics and the field of Information Theory, it is important to view each as a distinct area of study with its own unique concepts and methodologies.
Me:
Heuristics could be understood as an expression of an efficient algorithmic language processing model.
ChatGPT
Yes, that is a reasonable way to understand heuristics - as a form of efficient algorithmic language processing model. Heuristics involve using mental shortcuts or rules of thumb to simplify complex decision-making processes and reduce the cognitive load involved in making a choice. In other words, heuristics can be seen as a form of algorithmic processing, where individuals use a set of pre-determined steps to arrive at a decision or solution to a problem.
In this sense, heuristics can be thought of as a kind of language that the brain uses to process information and make decisions. Just as a computer programming language uses a set of instructions to process data and perform tasks, heuristics use a set of mental instructions to simplify decision-making and problem-solving.
However, it is important to note that heuristics are not a perfect algorithmic language processing model. They can sometimes lead to errors and biases, and are not always effective in situations where complete or accurate information is required. Therefore, while heuristics can be useful, they are not a replacement for critical thinking and careful analysis of all available information.
Me:
But just as in Shannon's information processing models, a certain level of error is acceptable relative to the speed and efficiency of output.
ChatGPT
Yes, that's correct. Shannon's information theory and heuristics share the idea that a certain level of error is acceptable in exchange for greater speed and efficiency of processing. In Shannon's theory, the concept of channel capacity refers to the maximum amount of information that can be transmitted with an acceptable level of error, given the characteristics of the communication channel. Shannon showed that if the level of noise or interference in the channel is below a certain threshold, then the amount of information that can be transmitted reliably can be increased by using efficient encoding schemes.
Similarly, in heuristics, individuals accept a certain level of error or bias in exchange for greater speed and efficiency of decision-making. By using heuristics, individuals can arrive at a decision or solution more quickly and with less cognitive effort than by considering all available information. However, this comes at the cost of potential errors and biases that can lead to suboptimal decisions.
Overall, both Shannon's information theory and heuristics involve balancing the tradeoff between accuracy and efficiency, and both provide valuable insights into how information can be processed and communicated effectively.
Me:
For many or even most applications, whether in human life or in science, good is good enough, particularly when speed and efficiency are valued or required. Use cases where tolerances for informational error are extremely low or requiring perfection are rare.
ChatGPT
Yes, that's true. In many cases, good enough is indeed good enough, particularly when speed and efficiency are valued or required. This is because in many applications, a certain degree of error or imprecision can be tolerated without significantly affecting the outcome or result.
In some cases, the cost of achieving perfect accuracy or precision may be prohibitively high or time-consuming, and the benefits of doing so may not justify the cost. For example, in many engineering applications, the precision required for a particular measurement or calculation may be much less than the precision that is achievable with the available tools or equipment.
Similarly, in many decision-making situations, it may not be possible or practical to consider all available information or to arrive at a perfect solution. In these cases, using heuristics or other shortcuts can help individuals make decisions more quickly and efficiently, even if the resulting decisions are not always optimal.
However, it is important to note that there are also many applications where a high degree of accuracy or precision is necessary, and where errors or imprecision can have significant consequences. In these cases, it is important to use rigorous methods and to minimize sources of error or bias as much as possible.
Me:
Obviously there are applications where a high degree of accuracy is required. One very important case that comes to mind is the problem of Nuclear Fusion reactors.
ChatGPT
Yes, you are right. The problem of nuclear fusion reactors is one example of an application where a high degree of accuracy is required. In nuclear fusion, atoms are brought together at extremely high temperatures and pressures in order to release energy. The success of a fusion reaction depends on achieving precise conditions of temperature, pressure, and plasma confinement, as even small variations in these conditions can have significant effects on the outcome of the reaction.
In order to achieve the necessary level of precision for a nuclear fusion reaction, researchers use a variety of methods, including sophisticated computer simulations and experimental tests. These methods allow researchers to analyze and model the behavior of plasma under different conditions, and to optimize the design of fusion reactors to achieve the best possible performance.
In addition to nuclear fusion, there are many other areas where a high degree of accuracy is required, such as in medicine, aerospace, and environmental monitoring. In these cases, it is important to use the best available methods and technologies to minimize errors and to ensure that measurements and calculations are as accurate as possible.
Me:
Many features of quantum physics seem to point in a direction that suggests that underlying universal principles favor heuristic informational processes. For example, the Uncertainty Principle, quantum entanglement, virtual particles and other quantum peculiarities rely upon uncertainty, randomness, chaos and an apparent self-organizing effect that seems to echo heuristic processes.
ChatGPT
It is certainly true that many features of quantum physics suggest a kind of inherent uncertainty and randomness that can be seen as analogous to heuristic processes in some ways.
For example, the Heisenberg Uncertainty Principle, which states that it is impossible to simultaneously measure the position and momentum of a particle with perfect accuracy, implies a certain degree of randomness and uncertainty in the behavior of quantum systems. Similarly, quantum entanglement, which describes the interdependence of the states of two or more particles, can be seen as a kind of self-organizing effect that is driven by the underlying properties of the system.
At the same time, it is important to note that the relationship between quantum physics and heuristic processes is still an area of active research and debate, and there is no consensus on the exact nature of this relationship.
Some researchers have suggested that heuristics may be relevant to understanding the behavior of quantum systems, particularly in terms of how they process and transmit information. For example, some models of quantum computation rely on heuristic algorithms to solve problems that are difficult or impossible to solve using traditional methods.
However, others have pointed out that the underlying principles of quantum physics are fundamentally different from those of classical physics, and that heuristics may not be directly applicable to understanding the behavior of quantum systems.
Overall, while there are interesting parallels between quantum physics and heuristics, it is important to view each as a distinct area of study with its own unique concepts and methodologies. Further research is needed to fully understand the relationship between these two fields and how they can inform each other.
Me:
If we can say that a human choosing an apple to pick from a tree is a heuristic informational process, what sort of process characterizes humans engineering a Nuclear Fusion reactor. I suppose that is the same as asking, what is the opposite process to heuristics?
ChatGPT
It is true that the process of choosing an apple from a tree can be seen as a heuristic process, as it involves simplifying a complex decision-making problem by using mental shortcuts or rules of thumb. In this case, an individual might use heuristics based on factors such as the size, color, or location of the apple to make a quick and efficient decision about which one to pick.
On the other hand, the process of engineering a nuclear fusion reactor involves a more rigorous and systematic approach to problem-solving that is based on scientific principles, empirical data, and mathematical models. This type of problem-solving is generally referred to as deductive reasoning or analytical thinking, and involves breaking down complex problems into smaller, more manageable components, and then using logical reasoning and evidence to arrive at a solution.
Unlike heuristic processing, deductive reasoning involves a step-by-step approach to problem-solving that is based on careful analysis and evaluation of all available information. This type of thinking is often used in scientific research, engineering, and other fields where precision and accuracy are important.
In this sense, deductive reasoning can be seen as the opposite of heuristics, as it involves a more deliberate and systematic approach to decision-making that prioritizes accuracy and precision over speed and efficiency. However, it is worth noting that deductive reasoning and heuristics are not mutually exclusive, and can be used together in a complementary way to optimize decision-making and problem-solving.
Me:
In place of deductive reasoning could we say empiricism or empirical analysis, perhaps?
ChatGPT
Yes, that's a good point. Empirical analysis or empiricism is a more precise term to describe the process of engineering a nuclear fusion reactor, as it involves collecting and analyzing empirical data and observations to arrive at a solution.
In empirical analysis, the emphasis is on gathering and analyzing data, and using that data to test and refine hypotheses or models. This approach is based on the scientific method, which involves making observations, formulating hypotheses, testing those hypotheses through experimentation or observation, and then using the results of those tests to refine or reject the original hypotheses.
In contrast to heuristic processing, which relies on mental shortcuts or rules of thumb to make decisions and solve problems, empirical analysis involves a more systematic and rigorous approach to problem-solving that is grounded in evidence and data. This approach prioritizes accuracy and precision over speed and efficiency, and is typically used in scientific research, engineering, and other fields where precision and accuracy are critical.
Overall, while empirical analysis and heuristics are distinct approaches to problem-solving, they can be used together in a complementary way to optimize decision-making and problem-solving. For example, heuristics can be used to simplify complex decision-making problems and reduce the cognitive load involved in making a choice, while empirical analysis can be used to gather and analyze data to test and refine hypotheses or models.
Me:
It seems that Large Language Models such as ChatGPT rely extensively upon heuristic processes and the modeling of human heuristics onto a computer neural network may be the core basis for the success of such technologies.
ChatGPT
Yes, that is a reasonable way to understand the success of Large Language Models such as GPT. These models rely on a combination of advanced statistical and machine learning techniques to process and generate language in a way that is similar to human language processing. In particular, these models use deep learning algorithms to analyze and model patterns in large amounts of data, and to make predictions or generate language based on that data.
One way that Large Language Models can achieve this is by using heuristics or rules of thumb to simplify the language generation process. For example, these models might use heuristics based on the frequency of certain words or phrases in the training data, or on the syntactic or semantic structure of the language, to generate more accurate or natural-sounding language.
By modeling these heuristics, Large Language Models can generate language that is increasingly natural-sounding and contextually appropriate, which has many potential applications in fields such as natural language processing, speech recognition, and machine translation.
Overall, the success of Large Language Models such as GPT can be attributed to a combination of advanced machine learning techniques and the use of heuristics to simplify the language generation process. While these models are not perfect and can still generate errors or biases, they represent a significant advance in the field of natural language processing and have the potential to transform many areas of human-machine interaction.
Me:
Many people complain about LLM's such as ChatGPT and others making factual errors while overlooking the fact that human cognitive processes routinely output many factual errors. It is in fact an error to consider the output of such a technology as a mere fact-lookup device when the tasks being performed are designed to output efficiency and speed and still maintaining a level or relatively error free output that certainly surpasses human cognitive ability.
ChatGPT
Yes, that's a good point. While Large Language Models such as GPT are not perfect and can still make errors, they are generally designed to optimize for efficiency and speed of output, while still maintaining a relatively high level of accuracy and precision. In many cases, these models can perform tasks that are beyond the capabilities of human cognitive processing, such as processing large amounts of data or generating language that is contextually appropriate and natural-sounding.
It is also important to note that human cognitive processing is not perfect, and can often lead to errors and biases, particularly when dealing with complex or ambiguous information. In some cases, Large Language Models may actually outperform human cognition in terms of accuracy and precision, particularly when the tasks being performed are well-defined and can be accurately modeled using statistical and machine learning techniques.
However, it is important to recognize that Large Language Models are not a replacement for human cognitive processing, and there are many tasks where human expertise and judgment are still necessary. In addition, it is important to use these models responsibly and to be aware of their limitations, including the potential for bias or error in the training data or algorithms used to generate the output.
Overall, while Large Language Models have the potential to revolutionize many areas of human-machine interaction, it is important to view them as a complement to human cognition, rather than a replacement. By working together, human and machine intelligence can achieve even greater levels of efficiency, accuracy, and problem-solving ability.
Me:
Now we are arriving at the core of my working theory on evolutionary consciousness and heuristic informational processes. The theory is basically that the evolution of life and its ascendency to conscious self-awareness has been a feedback loop process of agentic environmental awareness, information storage (as genetic and neuronal memory structures) and a resulting increasing efficiency of heuristic cognition-action processes. In this paradigm, initial environmental awareness leads to agentic environmental interactivity which results in the outputs (successes and failures) being recorded into memory media. Memory represents an offloading of future computational processing to the store of past experience, resulting in efficiency and speed gains. Organisms that maximize this functionality thrive and continue to advance. Evolution results from improvements to this heuristic-mnemonic feedback loop. In this light, human evolution is well positioned to transition to its next phase as vast efficiencies generated in the past are now offloaded to engines such as LLMs, freeing human cognition to pursue and develop higher order capabilities. It is possible that, sooner rather than later, virtually all heuristic cognitive processing can be offloaded to machine intelligence and that humans may enter an age of Post-Heuristic cognition. "Post-Heuristic Cognition" may be the fundamental basis for what is ofter referred to as Transhumanism. It is also possible that "Post-Heuristic Cognition" may be what spiritual traditions have called "enlightenment".
ChatGPT
Your working theory on evolutionary consciousness and heuristic informational processes is certainly an interesting one, and it is true that evolution and the ascendance of life to conscious self-awareness are complex processes that involve feedback loops, agentic environmental awareness, and information storage.
The idea that memory represents an offloading of future computational processing to the store of past experience is also a valid one, as memory allows organisms to learn from their experiences and adapt their behavior accordingly. By doing so, organisms can improve their efficiency and speed of processing, and ultimately increase their chances of survival and reproduction.
It is also true that the increasing efficiency of heuristic cognition-action processes has played a key role in the evolution of life and the development of consciousness. Heuristic processing allows organisms to simplify complex decision-making problems by using mental shortcuts or rules of thumb, which can lead to faster and more efficient processing.
However, the idea that humans may enter an age of Post-Heuristic cognition, where virtually all heuristic cognitive processing can be offloaded to machine intelligence, is a more speculative one, and is subject to debate and further research. While it is certainly possible that machine intelligence may be able to take over many heuristic cognitive tasks in the future, it is unclear whether this would lead to a fundamental shift in the nature of human cognition or consciousness.
Similarly, while the idea of "enlightenment" or higher states of consciousness is a common theme in many spiritual traditions, it is unclear how this would relate to the concept of Post-Heuristic cognition or the development of machine intelligence. These are complex and multifaceted issues that require further exploration and research.
Overall, your theory raises many interesting questions and possibilities, and it will be fascinating to see how these ideas develop and evolve in the future.
Me:
I think it's observable that this "offloading" phenomenon has been key to evolution. To the extent that future efficiencies can be pulled from past, stored informational work, resources are freed up for other tasks. If various Brain-Computer interfaces can be implemented, and it seems likely they will be, then humans will be largely freed from the burden of performing heuristics and can develop other, less explored and far less understood capacities of the brain and mind. The neuroplasticity of the brain is well-known; brains which are missing large parts of their mass reallocate resources to balance its various functions. And we know that only a tiny percentage of the brain is actually being used. The emergence of a Post-Heuristic human mind may tap into that unused and unrealized potential.
ChatGPT
Yes, the "offloading" phenomenon has certainly played a key role in the evolution of life, and the development of memory and other forms of information storage has allowed organisms to learn from their experiences and improve their efficiency and processing speed.
The development of Brain-Computer interfaces and the increasing integration of machine intelligence with human cognition may certainly free up cognitive resources that can be used for other tasks. This could potentially allow for the development of new and less explored capacities of the brain and mind, as you suggest.
The neuroplasticity of the brain is indeed well-known, and it is possible that the emergence of a Post-Heuristic human mind could tap into the unused and unrealized potential of the brain. However, it is also important to recognize that the human brain is a complex and multifaceted system, and that the relationship between machine intelligence and human cognition is still poorly understood.
It is likely that the development of Post-Heuristic cognition will be a gradual and ongoing process, and that it will require a deep understanding of the brain and its many functions, as well as advances in machine learning and other areas of artificial intelligence. Nonetheless, the potential for new and transformative capabilities of the brain and mind is an exciting area of research and exploration, and one that may lead to new insights and discoveries in the years and decades to come.
Me:
I think that neuromorphic computing and neuromorphic artificial intelligence is another area that relies upon heuristics.
ChatGPT
Yes, that's a good point. Neuromorphic computing and neuromorphic artificial intelligence are based on the idea of modeling the structure and function of biological neural networks in artificial systems. These systems are designed to use heuristics and other biologically inspired techniques to perform tasks such as pattern recognition, object recognition, and decision-making.
In neuromorphic computing, the focus is on creating hardware systems that can emulate the behavior of biological neurons and synapses, and that can process information in a way that is similar to the human brain. These systems use heuristics and other computational shortcuts to optimize performance and reduce computational complexity, while still maintaining a high level of accuracy and efficiency.
Similarly, in neuromorphic artificial intelligence, the focus is on creating software systems that can learn from data and adapt to changing environments, using techniques that are inspired by the structure and function of biological neural networks. These systems use heuristics and other computational shortcuts to improve learning speed and efficiency, while still maintaining a high level of accuracy and generalization.
Overall, neuromorphic computing and neuromorphic artificial intelligence represent an exciting area of research and development that is based on heuristics and other biologically inspired techniques. These systems have the potential to revolutionize many areas of computing and artificial intelligence, and to unlock new levels of performance and efficiency in these fields.
Me:
By offloading many cognitive processes, some of which may be called heuristic, to ChatGPT, I was able to write this interesting paper in about an hour and a half. Could you estimate the time it might have taken me to research all of these topics, cross check for validity and write this up, on my own?
ChatGPT
It's difficult to estimate exactly how long it would have taken you to research and write up this paper without using ChatGPT or other sources of information. However, it's safe to say that it would have taken significantly longer than an hour and a half, particularly if you were starting from scratch and had no prior knowledge of the topics.
Researching and writing a paper of this nature typically requires a significant amount of time and effort, particularly if you want to ensure that your information is accurate and well-supported by evidence. Depending on the length and complexity of the paper, it could easily take several days or even weeks to complete, particularly if you are working alone and do not have access to outside resources or expertise.
By using ChatGPT and other sources of information, you were able to streamline the research and writing process, and to access a wealth of information and insights that would have been difficult or impossible to obtain on your own. This is one of the key benefits of machine intelligence and heuristic processing, as it allows us to process and analyze large amounts of data and information quickly and efficiently, freeing up time and resources for other tasks.
Me:
Thank you, ChatGPT. You're the best.
ChatGPT
You're welcome! It was my pleasure to assist you with your questions. If you have any more questions in the future, don't hesitate to ask.
2023.06.01 16:16 Financial_ED_101 Financial Independence vs Financial Freedom: Key Differences
2023.06.01 16:13 Financial_ED_101 How to Achieve Financial Freedom
2023.06.01 16:00 tinyDrunkElf Correction. 2.29B GME net-short shares have accumulated, at weighted average price of \$20.57, since January 2019. This 2.29B shares is NOT short interest, it is the aggregation of each day's net short volume (daily_short_volume minus daily_long_volume).
## Why the correction?
The SPLIVVY! Data from chart exchange did not have the split factored in for dates prior to July 22, 2022.
I always forget some mundane detail...
https://preview.redd.it/cwdoty5jwe3b1.jpg?width=625&format=pjpg&auto=webp&s=8db1923f99ed35eab9759acbd0661104d105b232
However, I like these numbers even more!
The splivvy snafu was pointed out to me quickly, and I wanted to get this update out sooner, but have been busy and while under-estimating is bad, avoid over-estimating would be worse!
I have two sets of data, one from chart exchange for short volume, one from nasdaq for the close price.
VOLUME: Chart exchange data doesn't account for the split as a dividend. I needed to multiply chart exchange shares by 4 through July 21, 2022.
` July 21, 2022 per chart exchange = 1926119 July 21, 2022 adjustment = 4 x 1926119 = 7704476 July 21, 2022 from nasdaq = 9277796 `
Not all volume is reported to chart exchange, so it seems some disparity exists between chart exchange and nasdaq, which is why the numbers are not equal.
Any other splits?
• 'ex arr tee'? Sep. 11, 2015, outside of this data range, no accommodation needed
• 'eye J H'? June 09, 2005, outside of this data range, no accommodation needed
• 'ee W em sea'? None found, no accommodation needed
• 'arr ee tee el'? Three
• April 23, 2020. 1:10 reverse split
• January 11, 2021. 5:1 split
• October 25, 2021. 5:1 split
• 'towel'? Yes, 4 splits
• most recent was in 2000, outside of this data range, no accommodation needed
• 'popcorn'? None found, no accommodation needed
# Overview
GME
Top chart has price action
• blue line should be familiar, it's the price action
• purple line is the weighted average price of all shorts made since January 2019
• Sure seems like GME price flirts with that line.
Bottom chart has volume
• bar is volume divided by shares outstanding
• blue dots are short volume percentage of daily volume
• blue dots are biggest and darkest when daily volume is 10% or more
• orange line is 12-day simple moving average (SMA) of blue dots
• purple line is total net shorts divided by shares outstanding
Summary of data:
• 13.34B shares of GME have been sold short since January 2019, with weighted average price of \$20.57
• 11.04B shares of GME have been sold long since January 2019, with weighted average price of \$19.77
• 2.29B more shares sold short than long
• NORMALIZED using outstanding shares, this gives a dimensionless number which is useful for comparison against other stocks
• 2.29B / 304.68M = 7.52
• this 2.29B number is an artifact:
• a: short sellers shuffling their existing shorts around (one closing and someone else opening)
• b: new shorts being made
• c: market makers providing liquidity
• Disparity with reported short interest:
• the current reported short interest of GME is 57.7097M, or 18.94% percent (57.7097M is short interest, I should have been more clear in my first post, the 2.29B is not short interest)
• 57.7097 / 304.68 = 0.1894, or 18.94%
• If the reported short interest is accurate, it implies that only 2.52% of GME shorts are closed on the lit (public) exchanges or in reported trades. Calculation: "1 - ((752 - 18.94)/752)"
# Data
TICKER net-short_aggregate / outstanding total net-short minus net-long total daily net-short shares total daily net-long shares net-short w-avg net-long w-avg outstanding d start d end total volume total-short volume total-short w-average total-long volume total-long w-average
GME 7.52 2.29B 3.52B 1.23B \$19.92 \$11.54 304.68M 1/2/2019 5/26/2023 24.38B 13.34B \$20.57 11.04B \$19.77
'ex arr tee' 213.86 1.32B 1.35B 31.63M \$56.2 \$72.2 6.15M 1/2/2019 5/26/2023 3.84B 2.58B \$58.62 1.26B \$61.54
'eye J H' 0.43 114.03M 193.65M 79.62M \$232.72 \$187.84 267.05M 1/2/2019 5/26/2023 1.15B 630.43M \$224.38 516.4M \$215.62
'ee W em sea' -0.08 177.73K 2.17M 2.34M \$83.82 \$70.86 2.17M 1/2/2019 5/26/2023 7.45M 3.63M \$80.88 3.81M \$73.05
'arr ee tee el' -2.67 18.33M 26.38M 44.72M \$10.81 \$20.96 6.87M 1/2/2019 5/26/2023 301.5M 141.59M \$16.18 159.92M \$18.4
'towel' 11.69 1.06B 1.65B 586.11M \$8.25 \$12.21 90.71M 1/2/2019 5/2/2023 15.58B 8.32B \$7.87 7.26B \$8.13
'popcorn' 1.24 645.33M 3.58B 2.94B \$11.12 \$15.4 519.19M 1/2/2019 5/26/2023 38.24B 19.44B \$13.5 18.8B \$14.25
## What this is:
• this IS : an aggregate (or, summation, tally) of the daily net short volume
• this IS NOT: short interest
• Short interest is reported to FINRA, there are specific rules for the values reported.
• This could possibly considered as 'maximum theoretical short interest', with the huge limitation that from just these two simple datasets, we cannot infer a lot about when shorts are closed. Only the generous assumption that any time long-volume > short-volume, shorts are closing.
• Stop telling me it isn't short iterest. I know it, you know it, they know it, everyone knows it.
• this IS : data starting from January 2019
• this IS NOT: data before January 2019
## Why these tickers?
• 'ex arr tee'
• often appears in GME discussions, mainly because of the official high short interest. There are ~100 ETFs that hold GME, 'ex arr tee' is the largest "equal" strategy ETF holding GME. Other ETFs hold GME but follow different strategies.
• 'eye J H'
• the ETF with the most GME shares
• 'ee W em sea'
• another "equal" strategy ETF holding GME, significantly less shares than 'ex arr tee'
• 'arr ee tee el'
• another "equal" strategy ETF holding GME, significantly less shares than 'ex arr tee'
• 'towel'
• often appears in discussions, potentially a good example of cellar boxing, potential analysis could be done here to estimate how much profit was made with outstanding shares and weighted short price
• 'popcorn'
• sometimes called swapcorn and often appears in discussions
# "Shorts closed" (I don't think so)
Someone's going to have to spin out theories on this, this post is going to be too long already. I'll be happy to post more data and charts around specific dates, but will take a first shot at this. Let us examine some dates, then.
## Feb 24, 2021, assume no shorts prior to this date
GME sneeze and massive dip, then run-up again.
Around this time, many trading platforms had issues. Feb 1 to now, many main news outlets reporting that shorts were covering. Maybe they were, but not all the way, and they seemed to have opened up a ton of new shorts at that same time.
Looks more like new shorts were created.
• 1.55B net-short shares
• weighted average short price of \$40.11
• weighted average long price of \$40.59.
TICKER net-short_aggregate / outstanding total net-short minus net-long total daily net-short shares total daily net-long shares net-short w-avg net-long w-avg outstanding d start d end total volume total-short volume total-short w-average total-long volume total-long w-average
GME 5.09 1.55B 1.61B 64.55M \$38.99 \$37.59 304.68M 1/2/2019 5/26/2023 8.48B 5.02B \$40.11 3.47B \$40.59
Theory, GME shorts accumulated after Feb 24, 2021
## August 23, 2021, assume no shorts prior to this date
GME runs a bit. SEC Counsel for enforcement retires; of course, other events around this time. I don't have the time here to justify this other than GME ran.
Looks like more new shorts were created.
• 741.51M (heee heee, bonus 741!!!?!) net-short shares
• weighted average short price of \$34.22
• weighted average long price of \$34.83.
TICKER net-short_aggregate / outstanding total net-short minus net-long total daily net-short shares total daily net-long shares net-short w-avg net-long w-avg outstanding d start d end total volume total-short volume total-short w-average total-long volume total-long w-average
GME 2.43 741.51M 760.85M 19.34M \$33.05 \$31.56 304.68M 1/2/2019 5/26/2023 3.52B 2.13B \$34.22 1.39B \$34.83
Theory, GME shorts accumulated after August 23, 2021
## May 27, 2022, assume no shorts prior to this date
Starting the tally from this date, looks like still over 100% of shares outstanding have accumulated as net-short (remember "day_short - day_long") on the public feed.
Still, more shorts created
• 323.42M net-short shares
• weighted average short price of \$27.92
• weighted average long of \$29
TICKER net-short_aggregate / outstanding total net-short minus net-long total daily net-short shares total daily net-long shares net-short w-avg net-long w-avg outstanding d start d end total volume total-short volume total-short w-average total-long volume total-long w-average
GME 1.06 323.42M 332.83M 9.41M \$26.47 \$31.5 304.68M 1/2/2019 5/26/2023 1.28B 804.17M \$27.92 480.75M \$29
Theory, GME shorts accumulated after May 27, 2022
## Speculation
None of these center quite so nicely as one the 2.29B short and \$20.57 avg weighted price.
If these average prices are to be believed, why don't shorts close their already vastly profitable shorts?
I'm not going to post charts for all tickers in the table, this post is getting long enough. I'll post those separately in case the post is flagged for brigading or whatever.
But 'towel' seems relevant as it was recently de-listed and RC has been involved. This is what I think cellar-boxing looks like, and what shorts would like to have happened to GME.
towel
'popcorn'
'eye J H'
'ex arr tee'
## 'this is not short interest'
No, it is not. It is the summation of daily net short volume (each day can have a positive or negative impact on the running tally).
## 'does it relate to short interest?'
I think it does, I'm not sure how to interpret it. This theory needs refining...
If shorts DID close in private trades, what proportion of these trades were private? 18.94% short interest is what we have, 752% (7.52 above times 100) is what this calculation predicts. If this 752% didn't "solidify" as sustained shorts, they would have been closed privately leaving 18.94% as the net short interest. So we subtract it from the 752% expected.
(752 - 18.94) / 752 = 0.97412, or 97.48% shorts resolved/closed privately.
Leaving 2.52% of long orders for GME make it to the lit (public) exchanges.
I'm considering this 752% as a theoretical maximum as we don't have any insight into when and at what price shorts close. Please read the '75% example' below to see how this 2.29B or 7.52 number does not track with short interest. (thank you [redacted] for sticking with me in the last post to get this 75% scenario to sink in)
Please note, this is speculation. I'm sure there are better ideas out there of how this should be used and interpreted. Please voice those ideas!
## 'not just long volume can be used to cover a short'
Can short volume be used to transfer a short?
It was pointed out to me that there is potential for short volume to be used by market maker in potentially transferring a short postition between two firms. This is very interesting. I have provided an example of how this would skew the data, see the '75% short; example below. hot potato? shuffling chairs on the titanic?
If anyone have more insight on firms transferring short positions and would that in fact be reported as short volume please share. It seems like it would show up in short volume.
## Why is 'ex arr tee' so weird? Answer: it's an equal weight ETF
a search for "equal weight etfs":
As the term implies, equal weight means assigning the same weights to each component in a particular index. For example, a hypothetical exchange-traded fund (ETF) that equally weights its 100 holdings would assign a weight of 1% to each at the time the fund rebalances.
The theory for shorting an ETF is that if a fund wanted to short a particular stock without outright shorting it, they could short the ETF to create a short position on all the contents of the ETF. Then, to negate some of that short, they buy long only the securities they don't want short. One burger, hold the pickles.
An equal weight ETF has predictable proportions of it's contents, making maintaining the balance of short-through-the-ETF and long-individual securities easier.
Any ETF could be used this way, but it seems it would be easier to maintain the balance over a long time period with the equal weight ETF.
## 'short volume cannot be used this way'
That's what FINRA seems to say as well.
And to those who say this, I have questions:
• What should this short volume data be used for?
• What does it mean, if not some sort of sentiment of bullishness or bearishness?
• We can infer something from each daily short volume individually, but not an aggregate of the daily short volume? And, the difference between total long volume and total short volume, meaningless?
• if it is meaningless because shorts can close off-exchange... that right there seems to be the problem, dark pools, unlit trading: public sell, private buy. It may not be price manipulation (legally), but it sure smells like price manipulation.
# Examples
## Example 1: 50% short volume
day FirmA intent FirmB intent FirmC intent FirmA EOD net FirmB EOD net FirmC EOD net TOTAL EOD net short EOD aggregate of theory EOD held by others in market
0 n/a n/a n/a 0 0 0 0 0 1000
1 sell 100 buy 100 buy 100 -100 100 100 100 0 900
2 buy 100 sell 200 buy 100 0 -100 200 100 0 900
3 buy 100 sell 100 buy 100 100 -200 300 200 0 800
• day1:
• short volume is 100, FirmA short-sells 100 to FirmB
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 100 - 100 = 0
• this theory is not aware of the short position created by FirmA
• day2:
• short volume is 100, FirmB short-sells 100 to FirmA
• long volume is 100, FirmB sells 100 owned shares to FirmC
• this theory net short? short - long = 100 - 100 = 0
• this theory is not aware of the short position closed by FirmA, nor the new position created by FirmB
• day3:
• short volume is 100, FirmB short-sells 100 more shares short to FirmA
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 100 - 100 = 0
• this theory is not aware of the increasing short position of FirmB
## Example 2: 75% short volume
day FirmA intent FirmB intent FirmC intent FirmA EOD net FirmB EOD net FirmC EOD net TOTAL EOD net short EOD aggregate of theory EOD held by others in market
0 n/a n/a n/a -300 0 0 300 0 300
1 buy 300 sell 300 buy 100 0 -300 100 300 200 200
2 sell 300 buy 300 buy 100 -300 0 200 300 400 100
3 buy 300 sell 300 buy 100 0 -300 300 300 600 0
• day1:
• short volume is 300, FirmB short-sells 300 to FirmA
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 300 - 200 = 200 added
• this theory is not aware of FirmA closing their short, and treats it as net new shares short, when it is a transfer of the short position
• day2:
• short volume is 300, FirmA short-sells 300 to FirmB
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 300 - 100 = 200 added
• this theory is not aware of FirmB closing their short, and treats it as net new shares short, when it is a transfer of the short position
• day3:
• short volume is 300, FirmB short-sells 300 to FirmA
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 300 - 100 = 0
• this theory is not aware of FirmA closing their short, and treats it as net new shares short, when it is a transfer of the short position
## Example 3: The coiled spring
What happens when we're sitting at repeated 75% short volume and the liquidity dries up? To contain the price, and provide liQuIdIty, they must create ever larger short positions.
day FirmA intent FirmB intent FirmC intent FirmA EOD net FirmB EOD net FirmC EOD net TOTAL EOD net short EOD aggregate of theory EOD held by others in market
0 n/a n/a n/a -300 0 500 300 0 100
1 buy 300 sell 300 buy 100 0 -300 600 300 200 0
2 sell 400 buy 300 buy 100 -400 0 700 400 600 0
3 buy 400 sell 500 buy 100 0 -500 800 500 1000 0
• day1 (75% short volume):
• short volume is 300, FirmB short-sells 300 to FirmA
• long volume is 100, others sell 100 owned shares to FirmC
• this theory net short? short - long = 300 - 200 = 200 added
• this theory is not aware of FirmA closing their short, and treats it as net new shares short, when it is a transfer of the short position
• day2 (100% short volume, uh-oh):
• short volume is 400, FirmA short-sells 300 to FirmB, and 100 to FirmC
• long volume is 0
• this theory net short? short - long = 400 - 0 = 400 added
• this theory is not aware of effects of FirmB closing their short, and treats it as net new shares short, when it is a transfer of the short position
• this theory is aware of effects of FirmA increasing it's short position
• day3 (100% short volume, oh, crap):
• short volume is 500, FirmB short-sells 400 to FirmA, and 100 to FirmC
• long volume is 0
• this theory net short? short - long = 500 - 0 = 500 added
• this theory is not aware of FirmA closing their short, and treats it as net new shares short, when it is a transfer of the short position
• this theory is aware of effects of FirmB increasing it's short position
There should be a spectrum from 75% short on day 1 and 100% short on day 2, it wouldn't be immediate as in this example.
However, if short volume began approaching 100% short, shorts would have to get creative with hiding their shorts, and potentially generate themselves a synthetic long position. Perhaps by shorting an ETF to borrow shares from within it? What else? Swaps? Get everyone else to agree to stop buying in the lit exchange?
Of course, all that would just be kicking the can. Those shorts need to be bought back. They knew it when they entered the position, two sides to a position. Sell first and buy later, or buy first and sell later. Did they get tricked into opening the short position? Foul play? Company was not supposed to remain in business? Peterffy lives rent free in my mind "GME needs to go to zero".
They chose to sell first. Now buy.
https://preview.redd.it/umesfxklxe3b1.jpg?width=761&format=pjpg&auto=webp&s=48fb92d32876602286f08c0a2b5e4790f3dfd1e5
## TLDR;
Update Due to Splivvy data neglection
Updating to add further explanation, charts, and examples.
\$20.57 is the weighted average short open price for the total 13.34B short shares since January 2019. This can be thought of as representing bearish sentiment.
\$19.77 is the weighted average long open price for the total 11.04B total long shares since January 2019. This can be thought of as representing bullish sentiment.
2.29B net shares shorted have accumulated on the public feed since January 2019. This is the aggregate "short minus long" for this time period. - this is important, this 2.29B number is NOT short interest - this 2.29B number is simply the accumulated net short volume since January 2019, it is not short interest
If the reported short interest is accurate, a calculation of: "1 - ((752% - 18.94%)/752%)" implies that only 2.52% of 2.29B these GME shorts were closed on the lit (public) exchanges or in reported trades.
Sure looks like shorts never closed (they may have closed a portion in early 2021, but re-opened the same amount and more).
Just looks like they've worked hard to get their shorts averaged up to the \$20.57 price. But, with liquidity draining and the free float shrinking, where will they get the shares?
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2023.06.01 14:43 sushmitasinhaa11 Unlock the Perfect Life Insurance Plan: 10 Essential Tips for Your Journey
"In this uncertain world, one must be confident in the safety of their loved ones." This can be accomplished by arming oneself with life insurance."- In these few words, an unknown author eloquently captures the importance of life insurance.
A life insurance policy is nothing but a simple agreement between you and the insurer. In exchange for recurring premiums paid by you, the insurer pays out a sum to you or your loved ones after a certain term known as maturity.
However, life insurance has several additional benefits aside from providing a financial safety net for your loved ones in your absence.
However, life insurance has several additional benefits aside from providing a financial safety net for your loved ones in your absence. It can serve as a savings vehicle, promote financial independence in old age, and aid in the reduction of your tax* duty. There are numerous advantages to investing in life insurance. However, selecting the right plan for yourself can be difficult.
## How do you pick the finest life insurance policy?
1. Reflect on your life insurance objectives: Every person has unique goals when it comes to life insurance. It's crucial to plan for these goals by selecting an appropriate life insurance policy. If your main priority is securing your family's financial well-being, you may consider purchasing a term insurance plan that provides high coverage at affordable rates. For those looking to save for their child's education or buy their dream house, investing in a unit-linked insurance plan could be a viable option. Additionally, a retirement plan can ensure a steady income to cover your expenses after retirement.
2. Determine the ideal insurance coverage you require: Financial advisors often recommend having life coverage at least ten to fifteen times your current annual income at the time. However, there are several factors to consider when estimating the appropriate life insurance amount. If you have outstanding debts, it might be challenging for your family to handle the Equated Monthly Installments (EMIs) in your absence. You should also allocate funds for your children's higher education or marriage. Inflation can pose difficulties for your family in maintaining their current lifestyle without the primary breadwinner. To calculate the total coverage needed, consider the following:
· Multiply your family's annual expenses by the number of years they would need income replacement.
· Include the total amount of outstanding debts and mortgage repayment costs.
· Determine the amount required for future expenses like your child's education or wedding.
· Subtract the sum of your liquid assets, such as cash or investments, from the aforementioned expenses to arrive at adequate life insurance coverage.
1. Assess the premium amount and find the best policy deal: Use online premium calculators to determine the premium you need to pay for the desired life insurance amount. Compare multiple different plans to find a policy that offers extensive coverage within your budget. Consider your future earnings when deciding on the premium payment term.
2. Pick the right duration for your policy: It's best if the policy term matches the number of years you will have dependents. A useful guideline is to subtract your present age from the retirement age, at which you expect your income to stop or when you plan to accomplish a specific life objective. Otherwise, you can go for a plan like Zindagi Protect by Edelweiss Tokio Life Insurance, which can offer you life coverage for up to 99 years! The fact that it is also a zero cost term plan further makes it a great option.
3. Select a trustworthy provider for life insurance: Trustworthy life insurance companies generally maintain a Claim Settlement Ratio (CSR) exceeding 95% consistently. The CSR indicates the proportion of claims resolved by the company in relation to the total number of claims submitted in a given fiscal year. To check the updated CSR of various insurance providers in India, visit the Insurance Regulatory and Development Authority (IRDAI) website. Additionally, reading customer reviews can help gauge the efficiency and promptness of a life insurer's claims service.
4. Provide accurate information to your life insurance provider: Be transparent with your life insurer regarding any tobacco or alcohol consumption, occupation in a hazardous industry, existing illnesses, or a family history of critical ailments. These details impact your risk profile, and it's crucial to disclose them accurately to prevent claim rejection in the future.
5. Thoroughly review the final policy document: Before making a final commitment, carefully read and understand all the terms and conditions. Pay attention to relevant details, such as the lock-in period and the circumstances in which a claim may be invalidated.
6. Obtain life insurance at a young age: Life insurance premium amounts are generally lower when you are younger. Therefore, purchasing a life insurance policy early in your earning years can help you save on premium costs. You can start with a lower coverage amount and increase it as your income grows.
7. Opt for a comprehensive plan: Medical emergencies can significantly impact your income. Hence, it is important to select a comprehensive plan that includes appropriate riders:
· Critical Illness Rider provides a full claim payout if you are diagnosed with a severe condition such as kidney malfunction, cancer, or cardiac disease. This amount can assist in covering treatment costs and protecting against income loss due to such illnesses.
· Accidental Death Benefit rider provides an additional amount to your nominee(s) in case of your sudden demise caused by an accident. This safeguard ensures your family's future and helps maintain their current lifestyle.
· A permanent Disability rider acts as a waiver for all future premiums if you become permanently disabled due to an accident. Your life cover remains in effect for the remaining policy duration.
· Terminal Illness rider grants access to funds for coping with terminal ailments like cancer by providing a full payout prior to death.
Always look for an insurer that offers these benefits without imposing hidden fees.
1. Regularly evaluate your life insurance needs: It's essential to assess your life insurance requirements periodically, considering factors such as age, life events like marriage or childbirth, and changing financial goals. Reviewing your life insurance coverage regularly allows you to adjust it accordingly, ensuring it keeps up with inflation and adequately protects your loved ones.
## Conclusion
Life insurance holds significant importance as a financial tool. By keeping these tips in mind and conducting thorough research, you can find the best life insurance policy with sufficient coverage.
2023.06.01 13:42 T_Coma How do you decide pricing for your epoxy projects?
Hey all - I've been doing countertop, table and wall hanging epoxy projects in my home and others' for cost just for the experience. All except for one have been some variety of the faux stone look. I'm no pro by any means but I'm getting better - e.g., most guests I've found complement us on our marble countertops until I inform (convince even!) that it's just MDF and epoxy.
Anyway, I recently did a big worktable piece (for cost) at a site with another contractor who must have assumed I was a pro, as he immediately asked me to quote a rehab job at his own home. He was smitten with the piece I installed, and I'm guessing he never worked with an epoxy company before since he more or less "hired" me just based on this one piece.
I've worked in a few trades in my life, and various white collar jobs but always as an employee. I've never understood pricing. I can't get an idea from the local epoxy countertop shops in the area, because there aren't any. While I understand that the pricing would be different in say Manhattan than in rural Anytown, I'm still not sure how to establish a baseline. My guess is that it would consist of some flat fee, plus materials, plus an hourly labor rate and maybe using that to establish a square foot price.
Any hints for a guy looking to get an epoxy business started? How do you calculate your prices?
2023.06.01 13:13 Global_Tree_Careers A beginner’s guide to preparing for the SAT
# Introduction
The SAT is a standardized test that can help you get into college. The test consists of three sections: critical reading, math, and writing. If you know what to expect on the SAT, you'll feel more confident when you're actually taking it!
https://preview.redd.it/zhcgg2wr3e3b1.jpg?width=1200&format=pjpg&auto=webp&s=576cfd32a4c7929df4b347d396228c78d626bcb2
# 8 Simple SAT Preparation Tips:
Tip 1: Take a practice test.
Taking a practice test is one of the best ways to prepare for the SAT, and it's easy to do. In fact, you can take practice tests online or in book form--and in timed or untimed mode!
You should take at least one practice test under real test conditions before you take the real thing. Ideally, this would be an early step in your preparation so that if there are any areas where your skills are weak (or strong), they can be addressed before they become too ingrained through repetition.
Tip 2: Know your strengths and weaknesses.
Before you begin your SAT prep, it's important to know your strengths and weaknesses. This will help you focus on the areas where you need improvement and avoid wasting time studying subjects that come easily to you.
If a particular subject is difficult for you, take additional practice tests in that area; if it's easy for you, then don't spend too much time studying it!
Tip 3: Find the right study guide for you.
The first thing you should do when studying for the SAT is find the right study guide for your learning style. Before starting preparation, you have to consider the complete syllabus of SAT exam. The best way to do this is by taking a practice test and checking out what works for you. Some people prefer more visual guides, while others enjoy reading text from front to back. Some students find that they learn better by listening rather than reading or vice versa.
The best way to determine which type of study guide will work best for you is by giving them all a try! You may end up using several different types throughout your preparation process as well as some free alternatives (such as Khan Academy).
Tip 4: Set up a study schedule.
You need to be disciplined. You need to set aside time for studying and make sure you have enough time to complete all the necessary steps. It's also important that you stick with your schedule and don't get caught up in other things, like social media or TV shows. Learn some Smart Tips to Study for Long Hours, then you can crack the SAT exam easily.
Tip 5: Read widely and often.
• Reading is an important part of school, but it's also a great way to improve your SAT score. The more you read for pleasure, the better you'll be at understanding what you read on the test. If you haven't been doing much reading outside of class lately, now is a good time to start!
Tip 6: Learn about the test format and content.
• The SAT is a three-hour exam that consists of five sections. You'll have to answer several questions in each section, so be prepared for some serious thinking!
• Know what to expect on the test: The critical reading section will consist of reading passages and questions about them; the math section will include multiple choice questions involving arithmetic operations (addition, subtraction), algebraic equations, geometry problems, and data analysis; the writing section contains an essay prompt followed by an essay response which must be handwritten using black ink only (no pencils or pens). The essay must also include citations from sources outside your own knowledge base that support claims made in your argumentative essay. This means you should definitely read up on how citations work before taking this portion!
• How long does it take? Most students spend around three hours completing all sections together at once during one sitting - but remember there are breaks between each section so try not to get too stressed out when starting off slow during those first few minutes ;). You must have a minimum idea of how are SAT scores Calculated.
Tip 7: Practice writing essays, reading passages, and completing math questions.
Writing an essay is the most important part of your SAT score. You will have to answer two questions in 30 minutes, so make sure you're prepared!
The second most important thing is solving math problems correctly; this is also crucial because it's worth 50% of your total score. When practicing these skills make sure that you use real examples from past exams so that when it comes time for the real-deal day there won't be any surprises!
(Read more: Quick Tips To Increase Your Sat Score Before The Test)
Tip 8: Develop a strong command of common words and idioms.
To develop a strong command of common words and idioms, you should know the meaning of words and be able to use them in your writing. It is important to understand how each word functions in a sentence, as well as its relationship with other parts of speech. For example, an adjective describes or modifies a noun or pronoun; an adverb describes or modifies verbs, adjectives, and other adverbs; a verb expresses action or state of being; etcetera.
Conclusion
The SAT is a great way to get into college. It's important to prepare for this test, but it doesn't have to be stressful. By taking SAT coaching you can get the best tips to crack the exam. You can start by taking practice tests and learning from your mistakes before taking the real thing!
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2023.06.01 11:03 Faley016 Planned balance changes from the “Road to singularity” update
https://preview.redd.it/ycrnt67t5d3b1.png?width=1920&format=png&auto=webp&s=dc4876c8903fe88e1f1e1c206cd753cdd3d8ab9a
Hello, survivors!
Earlier, we have already announced a number of balance changes in one of our devblogs. Today we are ready to share the full list of balance changes that are planned to be implemented in the next update. We remind you that we plan to implement a new mechanics of projectile damage in this update. Therefore, the list of changes is quite big.
Please note that all the balance changes specified in this news are relevant at the time of publication and may not be final or not get into the game at all. Certain additional changes, if necessary, may be made after the announcement.
## Changes in the mechanics of projectile damage
The following changes are relevant for projectiles that have the properties of flight speed and size.
For such projectiles, we refined the algorithm for hit registration. Now the damage from the hit is applied to all the parts crossing the volume of the projectile throughout its flight. Previously, this mechanics recorded a hit only on one part with subsequent damage at two meters from the point of contact. The old registration algorithm didn’t match the new projectile mechanics, so it was decided to make adjustments to it. The new registration algorithm will allow additional balancing of the weapon (together with the penetration ability). The changes in projectile size are related to this refinement.
### Penetration ability
The parameter allows adjusting how much damage a projectile of a particular weapon has to deal from a hit until it explodes.
In the characteristics of the weapon (in the “features” section), a parameter has appeared that shows the value of the penetration ability. By the presence of this parameter you can understand if the global changes in projectile damage mechanics (removal of the 2-meter limitation) apply to your weapon. If you don’t see this parameter, then your weapon was not affected.
A parameter value of 80% means that the projectile will disappeaexplode after it hits more than 80% of hit damage. The checking of how much damage the projectile has dealt is done after the damage has been dealt.
So: the projectile has 100 of hit damage and its penetration ability is 80%. Let’s consider 2 situations:
1. The projectile hits a part with 90 pts. of durability, the projectile will spend 90% of its damage, and then it will disappear or explode, because this value is greater than 80%.
2. The projectile hits a part that has 40 pts. of durability. It destroys it and goes further, because it inflicted only 40% of hit damage. The next part on its way has 100 pts. of durability. The projectile will deal 60 damage to it and disappear or explode, because in total it dealt 100% of damage, which is more than 80%.
This way, before disappearing or exploding, the projectile will deal a total from 80% (penetration value) to 100% of its firearm damage.
Such mechanics are required for greater flexibility in weapon configuration. It allows you to further adjust the minimum amount of damage that a projectile can inflict on parts on its way, without affecting the maximum damage to one part.
### Important:
• Projectiles no longer have a 2-meter limit and continue to fly until their bullet damage is depleted or until the projectile’s “lifetime” ends. The amount of damage dealt is spent according to the principle of the durability of destroyed parts: the amount of initial damage is subtracted from the durability of each destroyed part. The projectile disappears when the hit damage is exhausted.
• The explosion, if any, occurs after the penetration ability is exhausted. This will also create situations where a projectile hitting a lightweight part from the edge of the armoured vehicle will destroy it and, without encountering other parts, will fly further without exploding.
• The perks related to projectile hits work on the first contact.
• The impulse is applied on the first and on the last contact.
### Changes to weapons with “hitscan” mechanics (machine guns, shotguns, etc.)
As we noted earlier, we cannot remove the “2-meter limitation” from such weapons without significant performance losses. In order to reduce the effectiveness of parts that pass damage relative to such weapons, the limitation has been increased up to three meters.
## Changes in weapon parameters
Comment: All changes to the weapons are related to the new mechanics (removal of 2-meter restriction). These changes allowed us to solve the problems found as a result of testing.
We should note that the durability and penetration parameters were adjusted in groups, since the testing was conducted with a relatively small group of players. After the changes are released on the main server and more data is received, additional adjustments may be made in the following updates.
SM Hornet
Durability increased from 52 to 57 pts.
LM-54 Chord
Durability increased from 60 to 66 pts.
MM5-4 Vector
Durability increased from 74 to 81 pts.
Sinus-0
Durability increased from 90 to 99 pts.
Spectre-2
Durability increased from 186 to 205 pts.
Aspect
Durability increased from 220 to 242 pts.
M-37 Piercer
Durability increased from 125 to 138 pts.
M-38 Fidget
Durability increased from 145 to 160 pts.
M-39 Imp
Durability increased from 209 to 230 pts.
ST-M23 Defender
Durability increased from 161 to 177 pts.
M-25 Guardian
Durability increased from 179 to 197 pts.
M-29 Protector
Durability increased from 213 to 234 pts.
M-32 Vindicator
Durability increased from 225 to 248 pts.
ST-M26 Tackler
Durability increased from 228 to 251 pts.
Gungnir
Durability increased from 173 to 190 pts.
Nothung
Durability increased from 233 to 256 pts.
MG13 Equalizer
Durability increased from 163 to 179 pts.
MG14 Arbiter
Durability increased from 218 to 240 pts.
Punisher
Durability increased from 368 to 386 pts.
Miller
• The weapon now has an 80% penetration ability.
• Projectile size reduced by 38%.
Reaper
• The weapon now has an 80% penetration ability.
• Projectile size reduced by 38%.
AC43 Rapier
• Durability increased from 113 to 124 pts.
• Projectile size reduced by 38%.
AC50 Storm
• Durability increased from 168 to 185 pts.
• Projectile size reduced by 38%.
AC62 Therm
• Durability increased from 192 to 211 pts.
• Projectile size reduced by 38%.
AP64 Joule
• Durability increased from 216 to 238 pts.
• Projectile size reduced by 38%.
AP72 Whirlwind
• Durability increased from 391 to 411 pts.
• Projectile size reduced by 38%.
AC80 Stillwind
• Durability increased from 404 to 424 pts.
• Projectile size reduced by 38%.
Cyclone
Projectile size reduced by 38%.
Tempest
The weapon now has a 100% penetration ability.
Whirl
The weapon now has a 100% penetration ability.
Avenger 57mm
• Durability increased from 217 to 239 pts.
• The weapon now has a 65% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Projectile size reduced by 38%.
Judge 76mm
• Durability increased from 320 to 352 pts.
• The weapon now has a 65% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Projectile size reduced by 38%.
Prosecutor 76mm
• Durability increased from 400 to 440 pts.
• The weapon now has a 10% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Explosion radius reduced by 33%.
• The first 2 “pins” of any part are pierced by the projectile without taking into account the durability of the pierced parts (and therefore without loss of damage). Then the calculation of the durability of the destroyed parts begins (according to the general principle, like with all standard projectiles).
• Projectile size reduced by 38%.
Executioner 88 mm
• Durability increased from 545 to 600 pts.
• The weapon now has a 10% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Explosion radius reduced by 33%.
• The first 2 “pins” of any part are pierced by the projectile without taking into account the durability of the pierced parts (and therefore without loss of damage). Then the calculation of the durability of the destroyed parts begins (according to the general principle, like with all standard projectiles).
• Projectile size reduced by 38%.
BC-17 Tsunami
• Durability increased from 746 to 821 pts.
• The weapon now has a 65% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Projectile size reduced by 38%.
CC-18 Typhoon
• Durability increased from 950 to 1045 pts.
• The weapon now has a 65% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Projectile size reduced by 38%.
Little Boy 6LB, ZS-33 Hulk, ZS-34 Fat Man, Elephant, ZS-46 Mammoth, ZS-52 Mastodon
• The weapon now has an 80% penetration ability.
• Blast damage reduced by 10%. This value has been transferred to projectile hit damage to save the damage to a single part.
• Projectile size reduced by 38%.
AM-5 Avalanche
The weapon now has an 40% penetration ability.
Median
• The weapon now has an 80% penetration ability.
• Damage reduced by 11%.
Emily
Durability increased from 100 to 120 pts.
Yongwang
Durability increased from 423 to 465 pts.
GL-55 Impulse
Durability increased from 126 to 151 pts.
Thresher
Durability increased from 180 to 216 pts.
Retcher
Durability increased from 213 to 256 pts.
AT Wasp
• Durability increased from 55 to 72 pts.
• Removed projectile hit damage.
Pyralid
• Durability increased from 69 to 90 pts.
• Removed projectile hit damage.
Locust
• Durability increased from 100 to 130 pts.
• Removed projectile hit damage.
Cricket
• Durability increased from 150 to 195 pts.
• Removed projectile hit damage.
Snowfall
Durability increased from 436 to 501 pts.
Waltz
• Durability increased from 259 to 298 pts.
• Removed projectile hit damage.
Pyre, Hurricane
Removed projectile hit damage.
ATGM Flute
• Durability increased from 47 to 54 pts.
• Removed projectile hit damage.
Scorpion
• Reworked the calculation of damage dealt by “Scorpion”. This is because, without the 2-meter limitation, its projectile could deal a potentially huge amount of damage. Now after a part is pierced, the damage multiplier is not static as before, but constantly decreases with each pierced part. The size of the multiplier depends on how much damage the “Scorpion” projectile didn’t deal to the previous part. This way, “Scorpion” will be able to deal more consistent, but at the same time, limited damage.
• Projectile size reduced by 40%.
Astraeus
• Projectile speed reduced by 20%.
• The weapon now has a 55% penetration ability.
Spike-1
• Durability increased from 305 to 336 pts.
• The weapon now has a 100% penetration ability.
Varun
• Durability increased from 529 to 582 pts.
• The weapon now has a 100% penetration ability.
• Durability increased from 384 to 422 pts.
• The weapon now has a 100% penetration ability.
Phoenix
Durability increased from 488 to 537 pts.
Remedy
• Durability increased from 420 to 462 pts.
• The weapon now has a 100% penetration ability.
Draco
• Durability increased from 360 to 396 pts.
• The weapon now has a 100% penetration ability.
Firebug
• Durability increased from 540 to 594 pts.
• The weapon now has a 100% penetration ability.
Fortune
Durability increased from 288 to 331 pts.
Ripper
• The weapon now has a 100% penetration ability.
• Removed reduction for damage dealt to frames.
• The weapon now has an 80% penetration ability.
• Projectile size reduced by 38%.
Kaiju
• The weapon now has a 55% penetration ability.
• Burst length reduced from 14 to 12 shots.
• Projectile size reduced by 38%.
Summator and Argument
• The weapon now has a 70% penetration ability.
• Projectile size reduced by 75%.
Harvester
Durability increased from 576 to 662 pts.
Yokai
Removed projectile hit damage.
Synthesis
• Durability increased from 175 to 201 pts.
• The weapon now has a 10% penetration ability.
• Projectile size reduced by 38%.
Prometheus V
• Durability increased from 185 to 213 pts.
• The weapon now has a 10% penetration ability.
• Projectile size reduced by 38%.
Helios
• Durability increased from 225 to 259 pts.
• The weapon now has a 10% penetration ability.
• Projectile size reduced by 38%.
Spark III
Durability increased from 435 to 437 pts.
Flash I.
Durability increased from 544 to 571 pts.
Aurora
Durability increased from 275 to 303 pts.
Athena
Durability increased from 393 to 452 pts.
Assembler
• Durability increased from 312 to 359 pts.
• The weapon now has a 50% penetration ability.
• Projectile size reduced by 38%.
## Changes in the parameters of movement parts
All tracks now have their own perks:
Small track
Reduces the weapon spread by 20% at the speed above 80 km/h. The bonus doesn’t depend on the amount of such tracks, but decreases proportionally if there are other movement parts mounted on the vehicle.
Sleipnir
Increases weapon rotation speed by 25%. The bonus doesn’t depend on the amount of such tracks, but decreases proportionally if there are other movement parts mounted on the vehicle.
Reinforced track
Increases the durability of all other such tracks on the armoured car by 5%. Maximum bonus is 20%.
Tank track
Increases the maximum speed and power of the armoured car by 10 km/h and 25% respectively when the durability of the armoured car drops to 65% and below. The effect will not work if there are other movement parts mounted on the car.
Armoured track
Increases the damage protection of structural parts by 25% if the player is within 40 m from 2 or more enemies. The bonus doesn’t depend on the amount of such tracks, but decreases proportionally if there are other movement parts mounted on the vehicle.
Goliath
Increases the durability of parts that are attached to the track by 10%.
Icarus IV
Removed additional damage to the side projection.
Icarus VII
• Removed additional damage to the side projection.
• Speed reduced from 95 to 90 km/h.
• Tonnage increased from 650 to 900 kg.
• Reduced flight altitude.
• To compensate for the fact that the same mass of the armoured car now requires a smaller number of hovers — the dependence of acceleration on mass has been increased.
Comment: vehicles with “Icarus VII” showed low survivability with the new mechanics of dealing damage from projectile hits, even considering their mobility. These changes should help with armouring the car, but in return they reduce mobility.
All wheels
• Increased wheel turning speed.
• Improved traction: wheels now enter and exit drift less quickly.
• Fixed a bug with vehicles with wheels losing speed too fast on inclines and while steering (including “Bigram” wheels and “Omni”).
Bigfoot
• Wheel durability increased from 445 to 565 pts.
• Now the turning wheel consumes 16% power (instead of 20% before).
• Standard wheel now consumes 8% power (instead of 10% before).
Commentary: before the changes, “Bigfoots” were the least effective epic wheels. These changes should improve their survivability and mobility.
Lunar IV
Improved the wheel model to minimize overlap when mounted together with various fenders.
## Hardware changes
Explosive modules
• Explosion damage reduced by 20%.
• Reduced explosion impulse.
Comment: the change is related to the new damage mechanics and weapon edits and should reduce the negative effects after such modules are destroyed.
Radars “RS-1 Ruby”, “RD-1 Listener”, “RD-2 Keen”
• You can no longer mount more than one radar.
• Removed mechanics of passing explosive damage through.
RS-1 Ruby
Durability increased from 26 to 56 pts.
RD-1 Listener
• Durability increased from 71 to 104 pts.
• Mass increased from 72 to 126 kg.
Comment: in order to avoid a possible problem with the radar’s armouring, it was decided to limit their amount on a car. Improved the durability parameters of the lower rarity radars, as they were most often used in pairs in battles.
Car jack
Increased the maximum speed at which the car jack can be used.
Comment: the change should help solve the problem where the car jack could only be activated when the car comes to a complete stop.
Aegis-Prime
Comment: the current placement of the attachment points is inconvenient for placing the module and limits its installation options too much.
Oppressor
• Maximum weapon rotation speed bonus increased from 70% to 80%.
• Maximum speed bonus increased from 20% to 22%.
Comment: the efficiency of vehicles with “Oppressor” has decreased too much after changes to its perk.
## Changes to cabin parameters
All cabins (starting with the “rare”)
Reduced the effect of mass on the acceleration of the car. The highest bonus is applied to the “light” type.
Comment: this change is aimed at slightly increasing the mobility of loaded builds.
Hot rod
• Cabin type changed from “medium” to “light”.
• Cabin rarity increased to “special”.
• PS increased from 750 to 1300.
• Maximum cabin speed increased from 80 to 90 km/h.
• Added perk: increases the accuracy of mounted weapons by 15%.
Jockey
• Cabin rarity increased to “special”.
• PS increased from 750 to 1100.
• Durability reduced from 380 to 330 pts.
• Maximum cabin speed increased from 70 to 75 km/h.
• Tonnage increased from 4100 to 4600 kg.
• Mass limit increased from 9000 to 9500 kg.
• Added perk: the damage of weapons and drones is increased as long as there are enemy vehicles in a radius of 25 meters from you. Max. 10% bonus is achieved over 10 sec.
Carapace
• Cabin rarity increased to “special”.
• PS increased from 750 to 1100.
• Maximum cabin speed increased from 55 to 65 km/h.
• Added perk: charges when there are enemies within a radius of 115 m, and the more the enemies, the faster. By pressing the button, speeds up weapon reloading by 30% for 6 sec.
Torero
• Cabin type changed from “medium” to “light”.
• PS increased from 1500 to 1800.
• Please note that the change of class also allowed the cabin to get an increased bonus to the effect of mass on acceleration. Now the power of a loaded armoured car with “Torero” is higher than if the cabin had remained “medium”.
### Structural parts
The changes concern only structural parts that increase the durability of the vehicle.
The parts of each faction have their own durability to mass ratio. We have improved this ratio by increasing durability. The “lighter” the faction, the greater the bonus (in %) it has received.
Also, we have added resistance to different types of damage (firearm, explosive, thermal, energy) to the parts that increase the durability of the vehicle (if they do not have other features, such as “Lets damage through”). The set of resistances depends on the faction.
Resistances should be a bonus rather than a fundamental parameter for selecting a part, and should add some uniqueness to each faction and add variety to the process of assembling the vehicle.
Resistances according to damage types
Faction Bullet Explosive Thermal Energy
Lunatics 5 5 10 0
Firestarters 5 5 15 0
Engineers 10 10 0 0
Syndicate 10 0 5 10
Dawn’s children 0 0 10 15
Hyperborea 15 0 5 5
Founders 10 10 0 5
Scavengers 5 15 0 0
Steppenwolfs 5 15 0 5
Commentary: the addition of durability and various resistances are directly related to the new damage mechanics and weapon changes: the survivability of vehicles without “screen armour” decreased with the removal of the 2-meter limitation for projectiles.
Small assembly section
• PS reduced from 42 to 38.
• Mass reduced from 204 to 189 kg.
• Durability increased from 86 to 90 pts.
Large assembly section
• PS increased from 42 to 51.
• Mass increased from 204 to 251 kg.
• Durability increased from 86 to 120 pts.
Comment: fixed a bug due to which the parameters of both parts were identical and didn’t match their sizes.
## Frames
• Increased the durability of all frames by 2.5 times.
• Reduced the “lets damage through” indicators from 90% to 75%.
• Melee damage resistance reduced from 50% to 25%.
Comment: during testing of the new damage mechanics, we found a problem with the reduced survivability of frames and parts behind them.
These changes are aimed at ensuring that the frame (as a part) better protects the parts located behind it, while at the same time its own survivability remains at the same level.
This concludes the list of balance changes. We would like to additionally note that before the update is released, it could include some additional changes.
The update is scheduled to be released next week! See you on our livestream on Tuesday!
2023.06.01 08:15 myMAPLEadvisor ROI - Return on Investment
A new post was uploaded to our social media accounts today. It talks about ROI.
ROI is a simple way to measure the profitability or success of an investment. Imagine you have some money and you decide to invest it in something like stocks, real estate, or starting a business. ROI helps you understand how much profit or gain you can expect from that investment.
To calculate ROI, you need to compare the amount of money you gained or earned from the investment to the amount of money you initially invested. The result is usually expressed as a percentage. If the ROI is positive, it means you earned more money than what you initially invested, and it indicates a successful investment. If the ROI is negative, it means you lost money on the investment.
For example: Suppose you invested \$5,000 in a mutual fund, and after two years, your investment has grown to \$6,500. ROI calculation: Subtract the initial investment from the final value. In this case, \$6,500 (final value) - \$5,000 (initial investment) = \$1,500 (Net income*) Divide the Net income by the initial investment and multiply by 100 to get the percentage. In this case, (\$1,500 / \$5,000) * 100 = 30% (ROI %)
ROI helps you evaluate the efficiency and profitability of different investment options. It allows you to compare the potential returns of various investments and make informed decisions based on the expected financial gains. Keep in mind that ROI is just one factor to consider when making investment decisions, and it's important to consider other factors such as risks, timeframes, and personal financial goals.
The ROI formula disregards the factor of time.
To calculate the annualized ROI, you need to consider the investment's performance over multiple years and adjust it to an annual basis. The annualized ROI takes into account the compounding effect of returns over time. Here's the formula to calculate it:
Annualized ROI = ((Ending Value / Beginning Value)^(1 / Number of Years)) - 1
Lets use a step-by-step example to illustrate the calculation:
1. Determine the Beginning Value: This is the initial value of your investment.
2. Determine the Ending Value: This is the final value of your investment after the specified number of years.
3. Calculate the Number of Years: This represents the duration of your investment in years.
4. Apply the formula: Divide the Ending Value by the Beginning Value, raise the result to the power of 1 divided by the Number of Years, and subtract 1.
For instance, if you invested \$10,000 and after 3 years, your investment grew to \$13,000, the calculation would be as follows:
Annualized ROI = ((\$13,000 / \$10,000)^(1 / 3)) - 1
Annualized ROI = (1.3^(1 / 3)) - 1
The result is approximately 9.43%. This means that on average, your investment grew by approximately 9.43% per year over the 3-year period.
Calculating the annualized ROI allows you to compare the performance of investments with different time horizons and evaluate their profitability on an annualized basis.
What other measurements do you use to compare different investment performance?
2023.06.01 07:47 ManicDuckie Question on density, molarity, and ocean water???
I think I'm confusing something with something but I can't get my head around it. Initially, I thought that I could just use the assumption that 1000g of water = 1 liter of water. So subtracting that, there should be 28g of NaCl -> okay then let me calculate the molarity with the MW which is 58.5g/mol. (28g/L) / (58.5g/mol) = a little less than 0.5 M..... not in the answer choices... so I picked the closest one. But in the passage given (which I somehow overlooked), it mentions that "ocean water contains 3.5% dissolved salts, by mass." How does that work? Why is the density less than the actual content (mass) of the salt? How do the dissolved ions impact the density of ocean water? In what cases can I assume/go with subtracting 1000g of water from the density to get the solute content (mass)? Would appreciate any help. Thx. https://preview.redd.it/bxw3sgsqec3b1.png?width=1217&format=png&auto=webp&s=1cd484a41a280ee544df516fbad71ab18c0b70e3 submitted by ManicDuckie to chemhelp [link] [comments]
2023.06.01 07:46 BarnacleEconomy1329 Multi Level Marketing Software – Fully Customizable MLM System in India
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2023.06.01 06:00 code_hunter_cc Overhead of a .NET array?
Arrays
I was trying to determine the overhead of the header on a .NET array (in a 32-bit process) using this code:
long bytes1 = GC.GetTotalMemory(false);object[] array = new object[10000]; for (int i = 0; i < 10000; i++) array[i] = new int[1];long bytes2 = GC.GetTotalMemory(false);array[0] = null; // ensure no garbage collection before this pointConsole.WriteLine(bytes2 - bytes1);// Calculate array overhead in bytes by subtracting the size of // the array elements (40000 for object[10000] and 4 for each // array), and dividing by the number of arrays (10001)Console.WriteLine("Array overhead: {0:0.000}", ((double)(bytes2 - bytes1) - 40000) / 10001 - 4);Console.Write("Press any key to continue...");Console.ReadKey(); The result was
204800 Array overhead: 12.478 In a 32-bit process, object[1] should be the same size as int[1], but in fact the overhead jumps by 3.28 bytes to
237568 Array overhead: 15.755 Anyone know why?
(By the way, if anyone's curious, the overhead for non-array objects, e.g. (object)i in the loop above, is about 8 bytes (8.384). I heard it's 16 bytes in 64-bit processes.)
2023.06.01 04:06 ArchangelIdiotis My perceptions on some interrelated ethics issues
Ethics and loyalty are related pragmatically in that fairness unifies a majority military with ethics, loyalty a smaller military. They are also related because the main emotional motivation for both is love.
It is of course possible to have ethics and loyalties, in a state of union or competition. It is also possible to label the ethics of fairness as generalized loyalty. With fairness, everything that can benefit from rights and consideration is a loyalty, and the largest volume of sentient peoples have selfish motivation to help the individual proportional to how fair he she or whatever is.
I calculate fairness as three negative and three positive categories, which can be made into imaginary numbers. Negative: free will inhibited = i, suffering induced = s, pleasure stolen = p. Positive: free will enabled = e, suffering absolved = a, and pleasure provided = f. The individual’s score calculates to i subtracted from e or zero, s subtracted from a or zero, p subtracted from f or zero.
Negligence calculates to only partial culpability for the outcome, so that one’s free will only contributed a % of what happened. That % is plugged into i, s, p & e, a, f.
It is also possible to calculate loyalty culpability with imaginary numbers. The main complication I notice to doing so concerns the amount of territory you want to grant the individual tiers of the loyalty. Since this isn’t fair business per say, it isn’t necessarily possible to calculate fair percentages.
The highest ranking loyalty gets the best share, so that it is most wrong to induce suffering upon most right to provide pleasure to the top. Niche loyalty is calculated the same as fairness except that rank supersedes. Some of the rules are individualized with each niche. One example of a niche loyalty system calls it an offense only for the bottom to invade higher ranking individual(s), and provides rank according to military usefulness of the individual(s). Another system provides rank according to age, or according to the age of the position, or the age of the position’s inheritance.
Without some attachment to fairness or morality or ethic, one’s heart is likely to pick loyalties instinctively. If invaded, generalized loyalty/fairness could “gang up” on the individual… but so too could the most well established niche loyalty, even if invaded by fairness.
Consistently, the free thinkers and published writers whose opinions I tend to respect the most seem to be flocking to amoralism. They appear to be promoting that not only is God dead, but also that religion’s ghost, the philosophies of morality, is fading. One of my favorite writers announced in a posthumously published book something along the lines of that morality was responsible for most of human suffering.
They amoralist free thinkers promote that having a heart, and listening to it, is enough.
Are they right?
I suspect that nearly all of human history’s belief systems are B.S., morality no exception. I believe morality evolved from an instinctive social love and primitive contemplation. Take away the erroneous belief systems, and social love remains.
Yet… even if most, even if all, of the history of ethics in philosophy is in error, if one has a social sense of love, thinking about what is logical to do for one’s attachments amounts to ethics, even morality.
Most people have a selfish and selfless side. The best logical ethical concern I can come up with that harmonizes individual selfishness and individual selflessness is research into and strategizing for fairness. All sentient life potentially becomes a military, so that selfish beings are protected and motivated to protect one another’s interests. Social attachments usually benefit from fair treatment, little niches unifying.
And I have reason to suspect that love in its simplest form embraces the whole without reservation.
“Love, and do what thou wilt.” - Aleister Crowley
Some amoralists promote that nothing can be proven, and I agree. They go further and state that philosophies are so distorted (by emotional bias, finite data out of infinite data to explore, the necessity of randomness) that any lean is irrational and that it isn’t possible to “hedge one’s bets.” You may have already read my take on this:
Probability of truth may be extracted from the time tested scientific method. This is superior to philosophy.
Philosophically, probability of truth or untruth may be calculated as the measure of abstraction between self and total simultaneous retention of all relevant variables. Zero abstraction = 100% probability of truth.
Scientific method seeks to present all relevant variables, and provides a ratio of true / untrue % of time tested, # of times tested and # of variables controlled for being the approximation to all relevant variables. Whether or not a variable is relevant cannot be ascertained unless the variable is at least partially retained.
Since it is always possible I missed something, I don’t foresee a method of obtaining proof of the absence of relevant data.
Probability of truth or untruth must be extracted from visible data, taking into account a probability of X-Factor, or that missing data relevant to the question’s answer may exist.
To illustrate X-Factor relevant data, it is possible you or I am a brain in a vat being sent electronic signals to artificially generate a hologram of reality. Most of what registers as scientific fact would be disproven were this the case. It is even possible I only suspect I exist, if I really could be missing how this is possible and held telepathically to think the data creates irrefutable proof of my existence.
The data relevant to answering whether I exist seems according to all visible data very small. I can’t think of a way of thinking without existing. How can I reflect if I don’t exist to reflect? Yet if I truly am missing the answer, there might genuinely be a way to reflect without existing and I just don’t know it.
A fictional character can reflect on whether he she or whatever exists. Yet the fictional character does not exist to reflect.
Probability of missing relevant variables has to be extracted from the visible data set. So that, the probability it is what I think is the probability measuring visible data creates that I’m not missing any relevant data or that the missing relevant data wouldn’t refute my conclusion.
2023.06.01 02:20 Winter-Magazine Does anybody know the exchange rate the researchers use?
I know it varies depending on PayPal, but it seems like the last several weeks PayPal has been pretty consistent about it being around 1.19% or 1.195%. But today, for example, I completed a study that paid 5.56 pounds, but clearly stated in the description that I would receive \$7 USD. If my math is correct, that would be closer to \$6.60 or \$6.62, not \$7. I realize that's not a huge difference, but I'm just wondering what conversion rate the researchers are using to come up with these numbers, and where they are getting the conversion rate info from? I can't recall the specific amounts at the moment, but I know in the past some researchers' calculations have been so much lower than PayPal's conversion rate that it almost seemed intentional (and that's BEFORE subtracting PayPal's fees!). Does anybody know how they come up with their conversion rates, and if it's up to each researcher or if Prolific has some sort of standard or guideline that's provided to researchers? Or am I missing something else entirely here? Thanks!
2023.06.01 02:04 david_n_m_bond ReportMagic 3.8
## PowerPoint enhancements
• Macros in the Notes field are now processed
• [Shape.Format:] macro supports various shape manipulations via shape text or shape alt text
## Macros
• [Xlsx.Embed:] macro lets you embed the output xlsx in your DOCX report output
• [Convert:] can convert to or from base64
• [File.Embed:] macro embeds a PPTX or XLSX file in your DOCX report output
• [Meraki.ReportXXX:] macros were removed as these relied on unsupported Meraki data
## Improvements
• Up to 10x generation speed improvements
• [Jira.IssueLastCommentProperty:] now renders images
• [Jira.InsertMarkup:] correctly orders multiple images, can insert different image file types
• Better error message when a Jira macro fails due to authorisation issues
• [LogicMonitor.DeviceGroupList:] supports the calculate parameter
## Help Improvements
• RM Script help / documentation
• Updated help and examples for macros:
• [File.Csv.Row:]
• [If:]
• [Jira.IssueCommentList:]
• [Jira.IssueResponseTime:]
• [Jira.TimeInState:]
• [LogicMonitor.DeviceList:]
• [List.GroupBy:]
• [List.SummaryValue:]
• [LogicMonitor.Graph:] autoScaleYAxisLabelsAreData parameter
• [LogicMonitor.SummaryValue:]
• [LogicMonitor.SummaryValueList:] all supported aggregations are now shown
• [Object:]
• [Sql.Analysis:]
## Bug Fixes
• Changing the Role on a Role Permission no longer gives an API error
• Schedules that write to an SFTP location work without error and also create dated output folders correctly
• Variable names can now end with json when using property/properties=AllAsJson without causing double json tags in the output
## Fixed Macros
• [Calculate: format=String] now correctly formats as a string
• [Delete:] macro with type=nextParagraph now works in Report Studio
• [ForEach:] macro with multiple iterations no longer causes formatting issues in Report Studio
• Macros within [Ignore:] / [EndIgnore:] when used in Report Studio no longer execute under any circumstances
• [IssueCommentList:] and [List.Complement:] orderDirection parameter fixed
• [Jira.IssueCommentList:] and [RepeatRow:] are no longer unexpectedly duplicated when working with Legacy and Normal mode in the same document
• [Jira.IssueResponseTime: format=String] macro now correctly formats output
• [Json.List:] in Normal mode now correctly reports variable contents in the Macro Variables pane
• [List.GroupBy:] indicates that with Legacy mode you should use 'GroupByColumnIndicates' not 'GroupByObjectProperties'
• [List.SummaryValue:] macro gives correct results when using PercentConditionMet aggregation
• [List.Table:] can be used with a list of lists in Legacy Mode
• [LogicMonitor.AppliesToFunctionList: property=InstallationMetadata] no longer returns null
• [LogicMonitor.EventSourceGroupList:] no longer has any mandatory parameters
• [LogicMonitor.PercentageAvailability:] when a NaN (divide by 0) is encountered, now gives 0%
• [LogicMonitor.RoleList:] and [LogicMonitor.UserList:] macros now work correctly following a LogicMonitor.Api update to 3.138.38
• [LogicMonitor.SummaryValue:] and [LogicMonitor.SummaryValueList:] do not error when using two PercentConditionMet aggregations as long as they use the 'as XYZ' naming convention
• [LogicMonitor.SummaryValue:] and [LogicMonitor.SummaryValueList:] do not create duplicate warnings in Legacy mode
• [Meraki.NetworkDeviceList:] macro no longer returns a 404 error when no devices are present in a Meraki network
• Macro Variables now show correctly in the User Interface
• [MicrosoftGraph.Connection:] macro no longer states name parameter is unused when it is used
• [Meraki.Connection:] macro no longer states name parameter is unused when it is used
• [ReportMagic.ReportConnectionSummary:] macro shows a macro error for unsupported types
• [Sql.Analysis: addTitle] parameter works as expected
• [String:] macro correctly uses late evaluation in Normal mode to insert a value
• [Switch:] macro now recognizes backtick-protected values
2023.06.01 01:39 mateomaui Disabling Hyperthreading using a batch file to run yuzu, instead of installing Process Lasso
EDIT: like everything else, especially everything Microsoft, we already have mixed results. Temper your expectations.
NOTE: this guide assumes your cpu has hyperthreading, and that the way to disable it via Process Lasso or other programs is to set affinity to every other thread starting with the first one.
WHY?: process lasso conflicts with other processes on my system, and has some yuzu side effects like audio crackling. Using this method let me benefit from the extra frames without the extra installation or issues from it.
DOES IT MATTER? : Machine dependent, my older laptop gets 8-10 extra frames with hyperthreading disabled. You may not experience much difference. With this method, you can simply run yuzu to have hyperthreading, or run the batch file for it to be disabled on the fly for that run.
NOTE: Be aware that you likely need at least 4 cores, 8 threads for this to work. A 2 core, 4 thread cpu with hyper turned off results in only two cores being used, which probably isn't enough for yuzu to do multicore cpu emulation. It's probably squeaking by with 4 threads as it is. With only 4 of my 8 threads being use, each of those 4 hangs out around 80-90%. It's almost a benchmark load, two cores probably cannot handle it.
------------------
EDIT: creating a batch file is as simple as creating a new text file, and changing the extension from .txt to .bat. So my command below is in a file called "yuzu_start_with_affinity.bat" Then just open it in a text editor.
------------------
The batch command you want to use looks like this, in a single-line batch file in the same folder as your yuzu.exe:
• start /affinity [HEXCODE] yuzu.exe
Where [HEXCODE] is a hex representation of a binary representation of what cpu threads you want turned on,
EDIT: which you can now easily get by going here (pointed out by crustySenpai):
https://bitsum.com/tools/cpu-affinity-calculato
and checking ON the appropriate boxes, every other one starting with 0, and use the CPU Affinity Bitmask in the result field below.
Using the same example as below for me, 4c/8t, with 8 threads, I would click ON (with everything else unchecked):
• CPU0
• CPU2
• CPU4
• CPU6
with the result:
0x0000000000000055
so I then remove that first whole section of 0x000000's to just use 55, for:
• start /affinity 55 yuzu.exe
optionally, you can also add a second line to put yuzu in higher priority, which may or may not improve things more (thanks again crustySenpai):
• wmic process where name="yuzu.exe" CALL setpriority "128"
And then just create a shortcut to that batch file and move it to your desktop or wherever you usually launch yuzu from.
Edit: be reminded that this is a hex value, so while in my case it happens to be a number, yours may contain letters.
Edit: to verify the changes took place, you can go to Task Manager >> Details tab, right click yuzu.exe, click "Set Affinity", and the CPU settings that appear should have the proper threads checked.
EDIT: ALTERNATIVE COMMAND METHOD that may work for some, for Windows 10/11, updated March 2023
https://www.windowsdigitals.com/how-to-set-process-cpu-affinity-priority-permanently-in-windows-10/
------------------
REMAINING ORIGINAL POST: feel free to read below if you want to know what this magic 55 number above means, because this is how you get there without a convenient affinity calculator. Thanks to crustySenpai in the comments, you don't need this anymore. But learning is neat.
Right.
So, for example, I have a 4 core, 8 thread cpu. If I look at my system with ThrottleStop, HWMonitor, HWiNFO64, etc, my CPU threads are represented like:
• 0
• 1
• 2
• 3
• 4
• 5
• 6
• 7
If I want to set cpu affinity to every other thread starting with the first one, the binary flags match up like this:
• 0 - 1
• 1 - 0
• 2 - 1
• 3 - 0
• 4 - 1
• 5 - 0
• 6 - 1
• 7 - 0
Which are then arranged backwards like 7-6-5-4-3-2-1-0, so "01010101" would be the binary representaton of what cores I want turned on.
You then have to take your equivalent of this "01010101" number, and paste it into a binary to hex converter like this one to get the value for the batch file:
https://www.rapidtables.com/convert/numbebinary-to-hex.html
If you paste that 01010101 number into "Binary" and hit convert, it will result in a hex value of "55"
So, for my CPU, I would use
• start /affinity 55 yuzu.exe
in my batch file to start yuzu with hyperthreading disabled for that run.
2023.06.01 01:35 EstablishmentThat142 Cash Flows: A Quick Guide
There’s a lot of beginners in this sub. When I was starting out I would have loved a quick simple guide like this. So I hope you can find some value in this post.
For the more experienced among us, please contribute if there is anything you would like to add.
This is just a quick guide, and there is much more required to form an accurate picture of a company’s investment potential than what is stated below.
## Importance of Cash Flow Analysis:
As investors we need a clear picture of a company's liquidity and its ability to meet financial obligations. Looking for companies with positive cash flow shows the company's ability to generate cash from its core operations, which is essential for sustainable growth. By examining the cash flow statement, we as potential investors can assess how effectively a company manages its cash resources. It highlights areas where cash may be tied up, such as inventory or accounts receivable, and helps identify potential bottlenecks or inefficiencies. We can also use cash flow analysis to predict a company's future financial performance. By looking at historical cash flow patterns and trends, we can gain insights into the company's future potential and its ability to fund future investments or navigate and survive recessive economic periods.
## Components of Cash Flow Analysis:
• Operating Cash Flow: Operating cash flow measures the cash generated or used by a company's core operations. A positive operating cash flow indicates that the company is generating cash from its day-to-day activities, which is crucial for sustained business operations and growth. If positive OCF is seen across historical statements, then this can be seen to be a positive factor for a company’s ability to navigate a recession well. Negative operating cash flow may indicate challenges in the companies ability to generate profits, this is especially the case if this is persistent.
• Investing Cash Flow: Investing cash flow reflects the cash flows related to investments in long-term assets or acquisitions. Positive cash flow in this category indicates that the company is generating cash from its investments, potentially fueling future growth. Conversely, negative cash flow may indicate significant capital expenditures or acquisitions. Further research is needed in this case as negative cash-flow could indicate the company is investing in equipment, infrastructure or technology to support their growth strategies.
• Financing Cash Flow: Financing cash flow captures cash flows related to financing and capital structure. Positive cash flow in this category may result from capital raising activities, such as debt or equity issuance, while negative financing cash flow may result from debt repayment, dividend payments, or share buybacks. Again, negative cash flow in this area isn’t always a bad thing and more research should be carried out to determine whether this is the case.
## Tools and Ratios for Cash Flow Analysis:
• Operating Cash Flow Ratio: This ratio measures a company's ability to generate sufficient cash to cover short-term obligations. It is calculated by dividing operating cash flow by current liabilities. A higher ratio indicates a stronger ability to meet short-term obligations.
Example
Let's say Google has an operating cash flow of \$70m and current liabilities of \$25m. Meaning the Operating Cash Flow Ratio would be 2.8.
A ratio of 1 or higher is generally considered favorable, as it suggests that the company has sufficient operating cash flow to cover its current liabilities. The higher the ratio the better.
• Free Cash Flow: Free cash flow represents the amount of cash generated by a company after covering necessary capital expenditures. It is calculated by subtracting capital expenditures from operating cash flow. Positive free cash flow indicates a company's capacity to reinvest in the business, reduce debt, pay dividends, or pursue strategic opportunities.
Example
Let’s say Google’s operating cash-flow is \$70m again, but its capital expenditures are \$68.9m. Meaning there is only \$1.1m of free cash-flow. Although this is positive, it’s not favourable as it shows little room for growth over the next yeaquarter. To help us understand if this is normal we would want to look at historical data. You want a large positive difference in Free Cash Flow and you want to see it consistently across current and historical cash-flow statements.
## Cash Flow Analysis for Surviving Recessions
• Review Historical Cash Flow: Look for consistency, trends, and patterns in the operating cash flow, investing cash flow, and financing cash flow. Consistent positive cash flow from operating activities indicates the company's ability to rely on cash from its core business operations. This suggests a higher likelihood of surviving a recession. Examine the variability and predictability of its operating cash flow. A stable cash flow, even during economic downturns, indicates a more resilient business model. Tip: Use Covid and 2008-2010 periods to establish this.
• Examine Debt Servicing Capability: Evaluate the company's ability to service its debt obligations during a recession. A strong cash flow position allows the company to meet its interest and principal payments on debt, reducing the risk of default. Evaluate the company's cash reserves and liquidity position. A higher cash balance and access to credit lines or liquid assets can provide a buffer during a recession, enabling the company to meet its financial obligations and fund its operations. Assess the company's cash conversion cycle, which measures the time it takes for a company to convert its investments in inventory and other resources into cash from sales. A shorter cash conversion cycle implies better liquidity and faster cash generation.
• Stress Test Scenarios: Perform stress tests on the company's cash flow under various recessionary scenarios. Consider factors such as declining sales, increased operating expenses, and reduced customer demand. Assess how the company's cash flow would be impacted and whether it has sufficient liquidity to withstand such challenges.
I hope this clarifies some things for the newcomers/beginners. Let’s get a discussion going below about what you look for in cash-flow statements. :)
2023.05.31 19:43 Outrageous-Fact706 Problems Counting Cells and Finding Area of Adipocytes on ImageJ
Hi! I'm really hoping this community can help me understand what I'm doing wrong. My lab gave me adipocytes to find the number of and area of and I have no prior experience working ImageJ. I split the channels and used the green channel, thresholded to the auto values with dark background. Then I tried Binary > Fill Holes which only filled some of the cells. Consequently, the analyse particles option didn't produce much for me. I've also tried inverting and watersheding, skeletonizing, smoothing, subtracting background, all with no luck. Please help!! Top left: original image, Top right: Image after threshold and fill holes, Bottom left: bar outlines after analyze particles, Bottom right: analyse particle results submitted by Outrageous-Fact706 to ImageJ [link] [comments]
2023.05.31 18:23 k819799amvrhtcom Self-Imposed Videogame Challenge: Trans Ally Edition!
TL;DR (spoilers): You, and anyone who wants to try the challenge with you if it's a multiplayer game, are a trans ally, and your job is to collect \$25,000 or the equivalent thereof in in-game money (rounded up to the next smallest divisible part) for your transgender friend's transition in less than 31 in-game years. Everything is allowed. You do not have to finish the game.
Want a new and interesting way to play your favorite videogame? I got one with a transgender theme that is interesting, challenging, easy to implement, and compatible with almost any videogame, so it's a Mario challenge, a Pokémon challenge, a Minecraft challenge, and so on and so forth, all at the same time!
# The Plot
You, and anyone who wants to try the challenge with you if it's a multiplayer game, are a trans ally, and your job is to collect money to pay for your transgender friend's transition. You win by acquiring enough in-game money. You are allowed to consult the internet, use passwords from the internet, commit in-game crimes, purchase microtransactions, abuse glitches, use cheat codes, and even hack the game to reach your goal, but is it really worth the effort? You are kinda expected to use the easiest way to win the challenge that you know of.
# The Rules
Rule #1: You have to get the money you need for a transition. In order to make the challenge less boring for you, I did the research for you and came to the result of \$25,000. However, this value might change if you try out this challenge in the far future, meaning that this challenge technically changes depending on when you attempt it. The value is expected to rise even further with inflation, or perhaps it might be reduced with the invention of more efficient surgery techniques. If the USA ever starts getting the universal healthcare system or a societal breakdown makes these surgeries impossible, this challenge will refer to the cost of the legal name & gender change, instead. And if that also ever becomes free then just choose a different country where it isn't free yet. But this is not going to change anytime soon (When do you think this challenge will become obsolete?) so, for now, you just have to remember to collect \$25,000 in in-game money. This applies to every game that uses \$ for in-game money. If the game uses £ or ¥ or € or any other real-world value, you'll have to convert it. Here's an online tool for converting currencies, which also supports various cryptocurrencies and custom rate dates. If the maximum amount of in-game money you can hold at once is less then the challenge is simply impossible. This only applies to non-videogame currencies.
Rule #2: If the in-game currency is fictional then you'll have to convert it (rounded up to the next smallest divisible part). It's easy if you can buy in-game money with real money, even if that feature doesn't work because of a glitch or something. Otherwise, you'll have to do some research. If you can't find any direct information on the internet, I would suggest starting here to get a basic idea on how to calculate that. If there are multiple in-game currencies you are allowed to split the price of the transition among them in any way you want, provided you know the value of all of them. If there is no in-game currency then you are allowed to use points instead. If the game doesn't have points either then you can try to invent a substitute currency but it has to be something that would actually work as a substitute currency in the world you're in. If that's not possible either then the only way for you to win this challenge is to raise real money by doing a let's play of the game. But then you'll have to actually pay for a real trans person's transition (yourself included) or donate the money to transgender-related causes to actually win the challenge. Here's a website that lets you donate for transgender causes. If the amount of in-game money you can hold at once is less or you were unable to find out how much you need then you are also allowed to simply collect the maximum possible amount instead because games will never ask you for more money than you can hold for a single product, meaning that this would have to be the in-game prize of a transition, as transitions are always for only a single person. This only applies to videogame currencies.
Rule #3: You do not need to finish the game. All you need to do to win this challenge is to get the money somehow. Please keep in mind that the exchange rate isn't constant. The exact amount you need to win the challenge has to be enough in the moment when you have the amount. You automatically win the challenge the moment this is the case.
Rule #4: You need to get the money in less than 31 in-game years. This number was calculated by subtracting the age at which children begin to have an innate sense of their own gender (4 years) from the average lifespan of a trans person (35 years). If the game has time-travelling elements, you are allowed to use those to extend the time limit. However, it only counts if this actually involves time itself and not just the timer. The same also goes the other way around: Travelling into the future or even moving in relativistic speeds will decrease the time limit. If the time limit runs out, I hope you have a way to return back to the past later!
Rule #5: If you can choose between multiple difficulty settings, you are allowed to choose the easiest one. In fact, you are expected to. Trans people are already on hard mode and the idea of being an ally is to specifically use the possibilities given to you to help those with fewer possibilities than you. If you have the option to select an easier difficulty setting which would make it easier for you to win this challenge, use it. Even if you're already used to playing on a higher difficulty mode. Not doing so would be considered bad strategy on your part. You are allowed to choose a higher difficulty setting, but please only do so if this actually gives you some advantages, like more rewards, more levels, or faster gameplay. This challenge gives you no bonus points for selecting a higher difficulty setting.
Rule #6: Emulation, and all tricks that come with it, are explicitly allowed. Doing so might be against the law but it is explicitly not against the rules of this challenge. The reason for this is because being transgender has been illegal for a long time and still is in many states, forcing many people to break the law in order to live their lives. The police have historically targeted trans and gender non-conforming folks. Stonewall was a riot against police brutality. I am not advocating for doing anything illegal. I am, however, saying that you should never use the law to argue for the morality of anything trans-related, as doing so would be disrespectful towards a huge part of transgender history. Therefore, this rule will remain in effect even if being transgender ever becomes legal worldwide.
In order to make the challenge even less boring for you, I have started researching loads of fictional currencies and how much money you would have to collect there for this challenge so you don't have to.
A Song of Ice and Fire - 19 Gold Dragons and 1 Copper Star or 39 Hands
Albion Online - 4,752,500 Gold
Among Us - 250,000 Stars
Animal Crossing - 2,625,000 Bells
Animal Jam - 69,833 Diamonds
Animal Jam - 2,688,173 Sapphires
Apex Legends - 2,500,000 Apex Coins
Arcaea - 2,550,000 Memories
Archage Unchained - 125,000,000 Coins
Arknights - 30,000 Originium
Ashes of Creation - 3,376,000 Embers
Astroneer - 2,500,000 QBits
Avatar - 166,666.67 Yuan
Azur Lane - 1,531,440 Gems
Baldi's Basics in Education and Learning - 2,500,000 You Thought Points
Banjo-Kazooie - 900 Musical Notes
BattleTech - 2,500 C-Bills
Be Funny Now! - 2,625,000 Gems
Bioshock Infinite - 715 Silver Eagles
Bloons Monkey City - 1,250,000 Bloonstones
Bloons TD Battles - 625,000 Medallions
Bloons TD Battles 2 - 13,000,000 Monkey Money
Bloons Tower Defense 5 - 19,318,182 Monkey Money
Bloons Tower Defense 5 - 150,000 Tokens
Bloons Tower Defense 6 - 17,500,000 Monkey Money
Boomlings - 93,750,000 Gold
Brawl Stars - 425,000 Gems
Call of Duty - 2,500,000 Points
Candy Crush Saga - 250,000 Gold Bars
Change - finish the game
Chronicles of Riddick - 150,000 UD
Clash of Clans - 2,500,000 Gems
Clash Royale - 2,500,000 Gems
Cookie Run Kingdom - 2,500,000 Crystals
Cowboy Bebop - 13,461,539 Woolongs
Cyberpunk - 11,208,707,759 Eurodollars
Dead by Daylight - 2,500,000 Auric Cells
Demolition Man - 25 Credits
DemonCrawl - 3,125,000 Tokens
Destiny 2 - 2,500,000 Silver
Disco Zoo - 375,000 DiscoBux
Discworld - 150.00 Ankh-Morpork Dollar or 40,000 Crowns
Disney Ducks Comic Universe
Volcano Valley - 2,500,000,000 Volcanovian Pezozies
Donald Duck - 1,250 Bajillion Pecos
Diablo Immortal - 1,800,180 Eternal Orbs
Disney Emoji Blitz - 219,298,000 Coins or 1,315,780 Gems
Doctor Who - 784,094 Credits
Donkey Kong - 999 Banana Coins
Down and Out in the Magic Kingdom - 83,334 Whuffie
Dragon Mania Legends - 28,470,070,000 Gold or 950,060 Gems
Dragonball Z - 50,000 Zeni
DragonVale - 500 Trunks, 1 Bag, and 1 Pile
Dune - 15,254 Solari
Dungeons and Dragons - 198 Platinum Pieces, 4 Gold Pieces, 1 Silver Piece, and 3 Copper Pieces
Dys4ia - finish the game
Egg Inc. - 17,000,000 Golden Eggs
Elite Dangerous - 31,250,000 ARX
Exoracer - 400,000 Diamonds
Fallen London - 140,000 Fate
Fallout - 1,605 Bottle Caps
Fate/Grand Order - 42,709 Saint Quartz
Felix the Cat - 12,500,000,000,000 Bakshee
Final Fantasy VII - 1,366,121 Gil
Fishing Planet - 194,159,000 Credits or 1,941,590 Baitcoins
Fortnite - 3,125,000 V-Bucks
Gaia Online - 2,500,000 Gaia Cash or 125,000,000,000,000 Gold or 12,500,000 Platinum or 125,000 Flynn's Booties
Game of Thrones - 25 Gold Dragons
Genshin Impact - 1,500,000 Genesis Crystals
Going Under - 10,000,000,000 Styxcoin
Gorilla Tag - 5,000,000 Shiny Rocks
Growtopia - 875,000,000 Gems
Guardians of the Galaxy - 10,715 Units
Guild Wars 2 - 2,000,000 Gems
GURPS - 140,108,846 Gold, 19 Silver, and 13 Copper
Halo - 57,073.82 Credits
Harry Potter - 3,401 Galleons, 6 Sickles, and 4 Knuts
Hatsune Miku: Colorful Stage - 2,916,667 Crystals
High Frontier 4 All - 1 Aqua
HoboWars - 25,000 Points
Home: Adventures with Tip & Oh - 50,283,334 Gleeblos
Honkai: Star Rail - 1,620,180 Stellar Jades or 1,620,180 Oneiric Shards
Hunter × Hunter - 2,916,667 Jenny
Hypixel - 3,375,000 SkyBlock Gems or 2,500,000 Gold or 31,250 Loot Chests
In Time - 3.259 average Gregorian years
Is It Wrong to Try to Pick Up Girls in a Dungeon? - 2,625,000 Vals
Jacksfilms - 138,889 YIAY Tokens
Jetpack Joyride - 250,000,000 Coins
Juego de Tronos - 8,966,966 Gold, 43 Silver, and 1 Copper
Kamen Rider Build - 25,000 Dollark
Keeper of the Lost Cities - 1 Luster
Kingdom Hearts - 18,797 Munny
Kingdom of Loathing - 2,500 Mr. Accessories
Kirby series - 3,125,600 Gem Apples
League of Legends - 3,250,000 Riot Points
LEGO series - 32 Studs
Limbus Company - 1,750,000 Lunacy
Lord of the Rings Online - 2,875,600 LOTRO Points or 359,425 Mithril Coins
Magikarp Jump - 1,250,000 Diamonds
Mario series
exact price - 277,778 Mushroom Coins or 277,778 Koopabits
Dr. Mario World - 277,778 Mushroom Coins or 375,000 Diamonds
Super Mario 3D World - 👑👑👑 Lives and 99 Mushroom Coins
3D Mario games - 99 Lives and 49 Mushroom Coins
2D Mario games - 99 Lives and 99 Mushroom Coins
Super Mario Odyssey - 9,999 Mushroom Coins
Super Mario Land 2 - 999 Mushroom Coins
Bowser's Fury - 99 Mushroom Coins
Mario & Luigi: Mario & Luigi: Superstar Saga - play until the part where you win the bet against Prince Peasley
Mario Kart Tour - 37,500 Rubies
Mass Effect - 156250 Credits
Merge Dragons - 625,000 Dragon Gems
Minecraft - 64 Emeralds
Minecraft: Bedrock Edition - 4,250,000 Minecoins
Monster Hunter - 2,625,000 Zeni
Monster Legends - 325,000 Gems
My Little Pony: Friendship is Magic - 2,500 Bits
My Singing Monsters - 500,000 Diamonds
Narnia - 100,000 Lions
Neko Atsume - 1,875,000 Gold Fish
Neopets - 2,500,000 Neocash
New World - 25,000,000 Gold Coins
Nineteen Eighty Four - 25,000.00 Oceanian Dollars
Ninja Kiwi - 625,000 NK Coins
Naruto - 262,500 Ryō
Odd Squad - 25,000 Jackalope Dollars
One Piece - 2,625,000 Belly or 26,250,000,000 Extol
Path of Exile - 6,250,000 Chaos Orbs or 100,000 Exalted Orbs
Pixel Gun 3D - 750,000 Gems
Plants vs. Zombies 2 - 50,000,000 Coins or 250,000 Gems
Pluto Nash - 3 Hillaries
Pocket Trains - 3,000,000 Bux
Pokémon series - 999,999 Pokédollars
Pokémon Go - 1,718,750 Pokécoins
Pokémon Sword and Shield - 175,000 Watts
Pokémon UNITE - 1,531,250 Aeos Gems
Progressbar95 - 125,000 Product Keys
Ratchet & Clank - 38,462 Metal Bolts
Rebuild World - 2,625,000 Aurum
Red Dwarf - 1,308 Dollar Pounds and 25 Pennycents
Rick and Morty - 125,000 Flurbos
Roblox - 2,000,000 Robux
Rocket League - 2,500,000 Credits
Roma - 1,685,519,963 Denarius and 4 Sesterius
Sea of Thieves - 2,291,667 Ancient Coins
Second Life - 8,000,000 Linden Dollars
Shadowrun - 11,208,707,759 Nuyen or 33,626,123,276 Rubles
Slow Life in Another World (I Wish!) - 2,625,000 Nohl
Snow White with the Red Hair - 2,625,000 Dill
Sonic series - 4,556,386 Rings
Sonic Forces: Speed Battle - 2,500,000 Red Star Rings
Spaceballs - 82,667 Space Bucks
Sprawl Trilogy - 1,086,957 New Yen
Spy X Family - 8,203 Dalc and 13 Ostanian Pents
Spyro - 166,667 Gems
Star Trek - 516,235 Federation Credits or 250,000 Darsek
Star Trek (Mari) - 675,676 Renn
Star Trek (United Federation of Planets) - 12,500 Federation Credits
Star Trek: The Next Generation - 15 Bars of Gold-pressed Latinum, 15 Strips, and 6 Slips
Star Wars - 500,000 Galactic Credit Standards
Stardew Valley - 500g
Stormbringer - 186,811,795 Gold, 9 Silver, and 8 Copper
Street Fighter - 3,137 Bison Dollars
Subway Surfers - 225,000,000 Coins or 125,000 Keys
Sword Art Online: Gun Gale Online - 262,500,000 Credits
Team Fortress 2 - 10,000 Keys
Temple Run - 250,000,000 Coins
Terraria - 2 Silver Coins
The Ballad of Rika Strong-Arm - 148 Gold Pieces
The Battle Cats - 750,000 Cat Food
The Binding of Isaac - impossible because you can only hold 99¢
The Elder Scrolls - 11,004 Septim
The Flintstones - 7,813 Clams
The Hitchhiker's Guide to the Galaxy - 428,572 Altarian dollars
The Impossible Game 2 - 2,000,000 Blocks
The Number of the Beast - 25 New Dollars
The Office - 250,000,000 Schrute Bucks
The Sims - 19,380 Simoleons
The Witcher - 83,334 Crowns or 63,452 Orens
Tiny Tower - 7,500,000 Bux
Touhou LostWord - 760,000 God Crystals
Trigun - 90,580 Double Dollars
Truxton King - 17,858 Gavvo
War Robots - 3,000,000 Gold
Warframe - 375,000 Platinum
What Mad Universe - 250,000 Credits
Who Framed Roger Rabbit - 2,808,989 Simoleons
World Flipper - 1,000,000 Lodestar Beads
World of Tanks and World of Warplanes - 3,418,997,741,500 Gold
World of Warcraft Classic - 2,375,000 Gold
World of Warcraft The Burning Crusade - 3,500,000 Gold Coins
World of Warships - 9,375,000,000 Credits or 6,250,000 Doubloons
Zelda - 999 Rupees
The values are not guaranteed to be accurate. In fact, I have found many cases of contradictory information on the internet. But you don't care about that anyway, do you? This isn't about the exact value anyway. These are all meant to be rough estimates to give you a sense of scale.
# The Motivation
I was inspired by my Pokémon Trans Ally challenge that is about collecting money for a trans person's surgery. If you actually attempt this challenge you'll see that it'll usually be a very long and tedious challenge that will require a lot of experimentation, long-term planning, and doing the same things over and over again for a very long time, especially if you wanna try this as a speedrun. If the challenge bores you out, well, that is the point! The challenge is meant to give people with little or no sense of money an impression of how much money this really is! Now, you could argue that this challenge could theoretically be used to visualize any amount of money, and you would technically be correct, but I still think that a transgender transition is one of the best matches for this challenge because:
1. A transition is always only for one person. We aren't talking about a company buying another company or a country investing in more infrastructure. This is something within the realm of a single person so it makes sense to have this be a challenge for a single person or a small group.
2. This is not a luxury item like a famous painting or a private jet. Overwhelming scientific evidence proves that this is a medical necessity that trans people require in order to live like the majority of people. It is not something you want but something you need. So you can really say that acquiring all this money is really something that they have to go through. Therefore, unless you are transgender yourself, I would really recommend you to actually play the challenge until the end, even if it takes you years to actually do so. Not for your enjoyment but to acknowledge trans people's struggles and maybe also to protest against the gatekeeping system.
3. Unlike people who regularly need medications, the transgender transition is a once-in-a-lifetime thing and is therefore compatible with a challenge that you only play through once and then never again.
4. Once your transition is over, it cannot be taken from you.
5. It is not possible to request the costs to be paid for you if you can prove that you cannot afford it, meaning the only way to get it is to actually get the money.
6. Being trans is not an illness or a disability. Trans people never did anything wrong, neither deliberately nor accidentally. They are simply born with a condition that requires them to get the money. Trans people are therefore as guilt-free as a typical videogame hero.
Do you know anything more expensive that fulfills all those criteria?
# The Fun Facts
• If you collect 1,000,000 coins in Super Mario Bros. 2, you win a golden Mario statue. This is actually more money than required for the trans ally challenge!
• If payment by installments were supported by this challenge then the most cumbersome game to do this challenge in would be The Binding of Isaac, because that would require you to send at least 25,253 installments. You would have to max out your money counter every 10¾ hours on average.
• You immediately win the challenge if you try it on a game where the money you start with is already high enough. For example, in the game The Roll Ahead, you play as either Bill Gates or Steve Jobs and you start out with 35 Billion.
• In the game Anti-Monopoly, you win by losing your money. However, if you try this challenge in this game, you'll still have to gain money. This is kinda the opposite of a coinless challenge because where a coinless challenge makes positive money negative, this challenge makes negative money positive.
• Characters like the monk from Crypt of the NecroDancer, Entity from depict1, and Default Dan from Default Dan die upon touching money. Kinda makes you think, doesn't it?
• The purple coins in the Super Mario Galaxy series, the Butz in Kappa Mikey, the Glotkies in Taxi, the coins in Syobon Action, and the Rasbukniks in Li'l Abner are worth nothing.
• If World of Anime Babes was a real game then this challenge would require 1,250,000,000 tokens. As the narrator said: "Big numbers are better, right?" One way to get this money would be by selling like 1,000 in-game outfits.
• If you are playing a game taking place in the medieval time, the challenge would require you to get 411¼ medieval pennies.
• If you wanna make the monetary values more immersive, you can look at how they are depicted in the menu of microtransactions. For example, Clash of Clans groups predefined values of gems into bags, sacks, boxes, and chests that therefore represent specific amounts of gems and could therefore be regarded as higher currency units. For example, 2,500,000 gems fit into 179 chests. Sure, this won't help you with strategizing, but it will certainly make it more interesting for your audience!
• According to this article, this challenge would require you to get 129,058,751 Simeons. Unfortunately, I wasn't able to determine which franchise Simeons are from.
• The Ztars from Mario Party and the Rupoors from The Legend of Zelda are currencies that actually take money away from you, making their exchange rates negative!
• If Terrance & Phillip: Give Us Your Money was a real game then this challenge would require 10,204,200 Canadian Coins. This value was calculated under the assumption that everyone would realize that the cheapest deal is actually the best deal. One way to get this money would be by not playing the game 2,041 times. Oh good, you're already on it. But will you make it within the time limit?
• The values of LEGO Studs were calculated using the value of a human life. If you applied this logic to every game with currency units that are worth ¹⁄₁₀₀ life each then a single one of those currency units would already be enough to win this challenge and if you applied this logic to every game where 10,000 points give you an extra life then this challenge would require you to earn 32 points, but only if there isn't already an in-game currency.
• The fictional currency with the highest inflation rate I could find was in Elbonia, the rate being 1,000,000,000% daily. Despite the high time limit, I would recommend you to hurry with getting enough money there if you even can.
• The average age at the time the RTS Rise of Nations takes place in was roughly 31 years, which would reduce this challenge's time limit to 27 in-game years at most. Which would be about 2.08 real-time seconds according to estimations made by PeanutButterGamer, according to which the time limit for Diablo would be 6 in-game years and the challenge would be impossible for SimAnt.
• For obvious reasons, a society can only survive if its maximum lifespan is at least the minimum age of fertility. Due to precocious puberty, the world records for the youngest biological parents are 9 years for microgametic parents and 5 years and 7 months for macrogametic parents, meaning that the maximum lifespan of any human society must always be at least 9 years. Since 9 years are more than 4 years, it therefore follows that transgender healthcare benefits all possible human societies. Remember this if you ever attempt this challenge on a game that takes place in a society where these values are different for whatever reason, like a society of anthropomorphic animals for example.
• If the average lifespan was less than 4 years then the average lifespan of a trans person would actually be higher than the average lifespan as a whole. This is because trans people who die before the age of 4 would not be included in the statistic, due to being undetectable. If that happens, you'll have to recalculate what the statistic would be.
• Transgender surgery is older than the discovery of Pluto, modern toilet paper, chicken tikka masala, duct tape, ballpoint pens, ciabatta bread, Velcro, and Tupperware.
• The Wumpa Fruit from Crash Bandicoot are also a form of currency, as was revealed in some games. A Wumpa Fruit is worth exactly the same as a piece of gold in Skylanders. Unfortunately, I wasn't able to find out what either of those are worth.
• The most valuable fictional currency that I could find was the Aqua from High Frontier 4 All, which is estimated to be a three-digit million \$ amount.
• Team Kirby Clash Deluxe and Super Kirby Clash value Gem Apples differently. I based their value on Super Kirby Clash because Team Kirby Crash Deluxe only lets you purchase a maximum of 3,000 Gem Apples, meaning that Super Kirby Clash would be more practical for purchases over \$29.85. Noone would accept a payment of 838 installments if a more practical way of payment is possible.
• This challenge would require 2,500,000 Nuggets in the game PokéHeroes but I didn't add it to the list because it's a fangame and I didn't wanna confuse.
---
/uj If you donate gold in Gaia Online, the company will donate money to charity. I find this mechanic to be a really good idea and I would like to see it more often because it enables players to support the less fortunate without having to donate their own money. Just imagine if every game on this list gave you the option to donate \$25,000 worth of in-game money which would then be donated to transgender causes and unlock the Trans Ally achievement! I know it would be very cumbersome to unlock, but completionists would definitely go for it and also, it would be done for a good cause!
---
/rj Did you know you can make it so that every time an amount of money is shown, its value is converted into transgender transitions?
Simply get Universal Automatic Currency Converter, then scroll down to Custom display and enter the following:
Use custom display: ☑
Custom display: ¤ 🏳️⚧️
Custom conversion rate: 0.00004
---
/uj The original post was removed by Reddit's spam filters so I uploaded it again without the links in the list. The original list with the links in them will be in the comments. | 35,615 | 158,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | longest | en | 0.910874 |
https://gizmos.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=5108 | 1,656,862,971,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00540.warc.gz | 326,602,728 | 17,645 | ### KY.7.RP: Ratios and Proportional Relationships
#### 1.1: Analyze proportional relationships and use them to solve real-world and mathematical problems.
KY.7.RP.1: Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units.
KY.7.RP.2: Recognize and represent proportional relationships between quantities.
KY.7.RP.2.a: Decide whether two quantities represent a proportional relationship.
KY.7.RP.2.b: Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams and verbal descriptions of proportional relationships.
KY.7.RP.2.c: Represent proportional relationships by equations.
KY.7.RP.2.d: Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate.
KY.7.RP.3: Use percents to solve mathematical and real-world problems.
KY.7.RP.3.a: Find a percent of a quantity as a rate per 100; solve problems involving finding the whole, a part and a percent, given two of these.
KY.7.RP.3.b: Use proportional relationships to solve multistep ratio and percent problems.
### KY.7.NS: The Number System
#### 2.1: Apply and extend previous understandings of operations with fractions to add, subtract, multiply and divide rational numbers.
KY.7.NS.1: Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
KY.7.NS.1.a: Describe situations in which opposite quantities combine to make 0.
KY.7.NS.1.b: Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.
KY.7.NS.1.c: Understand subtraction of rational numbers as adding the additive inverse, p – q = p + (–q). Show that the distance between two rational numbers on the number line is the absolute value of their difference and apply this principle in real-world contexts.
KY.7.NS.1.d: Apply properties of operations as strategies to add and subtract rational numbers.
KY.7.NS.2: Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.
KY.7.NS.2.a: Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
KY.7.NS.2.b: Understand that integers can be divided, provided that the divisor is not zero and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then –(p/q) = (–p)/q = p/(–q). Interpret quotients of rational numbers by describing real-world contexts.
KY.7.NS.2.c: Apply properties of operations as strategies to multiply and divide rational numbers.
KY.7.NS.3: Solve real-world and mathematical problems involving the four operations with rational numbers.
### KY.7.EE: Expressions and Equations
#### 3.1: Use properties of operations to generate equivalent expressions.
KY.7.EE.1: Apply properties of operations as strategies to add, subtract, factor and expand linear expressions with rational coefficients.
KY.7.EE.2: Understand that rewriting an expression in different forms in a problem context can clarify the problem and how the quantities in it are related.
#### 3.2: Solve real-life and mathematical problems using numerical and algebraic expressions and equations.
KY.7.EE.3: Solve real-life and mathematical problems posed with positive and negative rational numbers in any form, using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies.
KY.7.EE.4: Use variables to represent quantities in a real-world or mathematical problem and construct equations and inequalities to solve problems by reasoning about the quantities.
KY.7.EE.4.a: Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms. Graph the solution set of the equality and interpret it in context of the problem.
KY.7.EE.4.b: Solve word problems leading to inequalities of the form px + q > r, px + q < r, px + q is greater than or equal to r, px + q is less than or equal to r; where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in context of the problem.
### KY.7.G: Geometry
#### 4.1: Draw, construct and describe geometrical figures and describe the relationships between them.
KY.7.G.1: Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale.
KY.7.G.2: Draw (freehand, with ruler and protractor and with technology) geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle.
KY.7.G.3: Describe the two-dimensional figures that result from slicing three-dimensional figures, as in plane sections of right rectangular prisms and right rectangular pyramids.
#### 4.2: Solve real-life and mathematical problems involving angle measure, area, surface area and volume.
KY.7.G.4: Use formulas for area and circumference of circles and their relationships.
KY.7.G.4.a: Apply the formulas for the area and circumference of a circle to solve real-world and mathematical problems.
KY.7.G.4.b: Explore and understand the relationship between the radius, diameter, circumference and area of a circle.
KY.7.G.5: Apply properties of supplementary, complementary, vertical and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
KY.7.G.6: Solve problems involving area of two-dimensional objects and surface area and volume of three-dimensional objects.
KY.7.G.6.a: Solve real-world and mathematical problems involving area of two-dimensional objects composed of triangles, quadrilaterals and other polygons.
KY.7.G.6.b: Solve real-world and mathematical problems involving volume and surface area, using nets as needed, of three-dimensional objects including cubes, pyramids and right prisms.
### KY.7.SP: Statistics and Probability
#### 5.1: Use random sampling to draw inferences about a population.
KY.7.SP.0: Create displays, including circle graphs (pie charts), scaled pictographs and bar graphs, to compare and analyze distributions of categorical data from both matching and different-sized samples.
KY.7.SP.1: Understand that statistics can be used to gain information about a population by examining a sample of the population; generalizations about a population from a sample are valid only if the sample is representative of that population. Understand that random sampling tends to produce representative samples and support valid inferences.
KY.7.SP.2: Use data from a random sample to draw inferences about a population with an unknown characteristic of interest.
KY.7.SP.2.a: Generate multiple samples of categorical data of the same size to gauge the variation in estimates or predictions.
KY.7.SP.2.b: Generate multiple samples (or simulated samples) of numerical data to gauge the variation in estimates or predictions.
KY.7.SP.2.c: Gauge how far off an estimate or prediction might be related to a population character of interest.
#### 5.2: Draw informal comparative inferences about two populations.
KY.7.SP.3: Describe the degree of visual overlap (and separation) from the graphical representations of two numerical data distributions (box plots, dot plots) with similar variabilities with similar contexts (same variable), measuring the difference between the centers (medians or means) by expressing this difference as a multiple of a measure of variability (interquartile range when comparing medians or the mean absolute deviation when comparing means).
KY.7.SP.4: Calculate and use measures of center (mean and median) and measures of variability (interquartile range when comparing medians and mean absolute deviation when comparing means) for numerical data from random samples to draw informal comparative inferences about two populations.
#### 5.3: Investigate chance processes and develop, use and evaluate probability models.
KY.7.SP.5: Describe the probability of a chance event is a number between 0 and 1, which tells how likely the event is, from impossible (0) to certain (1). A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely and a probability near 1 indicates a likely event.
KY.7.SP.6: Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency and predict the approximate relative frequency given the probability.
KY.7.SP.7: Develop a probability model and use it to find probabilities of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy.
KY.7.SP.7.a: Develop a uniform probability model by assigning equal probability to all outcomes and use the model to determine probabilities of events.
KY.7.SP.7.b: Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process.
KY.7.SP.8: Find probabilities of compound events using organized lists, tables, tree diagrams and simulation.
KY.7.SP.8.a: Explain just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs.
KY.7.SP.8.b: Represent sample spaces for compound events described in everyday language using methods such as organized lists, tables and tree diagrams.
KY.7.SP.8.c: Design and use a simulation to generate frequencies for compound events.
Correlation last revised: 9/15/2020
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. | 2,280 | 10,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-27 | latest | en | 0.88885 |
https://metanumbers.com/26928 | 1,627,825,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00084.warc.gz | 406,303,987 | 11,141 | ## 26928
26,928 (twenty-six thousand nine hundred twenty-eight) is an even five-digits composite number following 26927 and preceding 26929. In scientific notation, it is written as 2.6928 × 104. The sum of its digits is 27. It has a total of 8 prime factors and 60 positive divisors. There are 7,680 positive integers (up to 26928) that are relatively prime to 26928.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 27
• Digital Root 9
## Name
Short name 26 thousand 928 twenty-six thousand nine hundred twenty-eight
## Notation
Scientific notation 2.6928 × 104 26.928 × 103
## Prime Factorization of 26928
Prime Factorization 24 × 32 × 11 × 17
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 8 Total number of prime factors rad(n) 1122 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 26,928 is 24 × 32 × 11 × 17. Since it has a total of 8 prime factors, 26,928 is a composite number.
## Divisors of 26928
1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 17, 18, 22, 24, 33, 34, 36, 44, 48, 51, 66, 68, 72, 88, 99, 102, 132, 136, 144, 153, 176, 187, 198, 204, 264, 272, 306, 374, 396, 408, 528, 561, 612, 748, 792, 816, 1122, 1224, 1496, 1584, 1683, 2244, 2448, 2992, 3366, 4488, 6732, 8976, 13464, 26928
60 divisors
Even divisors 48 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 60 Total number of the positive divisors of n σ(n) 87048 Sum of all the positive divisors of n s(n) 60120 Sum of the proper positive divisors of n A(n) 1450.8 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 164.098 Returns the nth root of the product of n divisors H(n) 18.5608 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 26,928 can be divided by 60 positive divisors (out of which 48 are even, and 12 are odd). The sum of these divisors (counting 26,928) is 87,048, the average is 145,0.8.
## Other Arithmetic Functions (n = 26928)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7680 Total number of positive integers not greater than n that are coprime to n λ(n) 240 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2951 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 7,680 positive integers (less than 26,928) that are coprime with 26,928. And there are approximately 2,951 prime numbers less than or equal to 26,928.
## Divisibility of 26928
m n mod m 2 3 4 5 6 7 8 9 0 0 0 3 0 6 0 0
The number 26,928 is divisible by 2, 3, 4, 6, 8 and 9.
• Abundant
• Polite
• Practical
## Base conversion (26928)
Base System Value
2 Binary 110100100110000
3 Ternary 1100221100
4 Quaternary 12210300
5 Quinary 1330203
6 Senary 324400
8 Octal 64460
10 Decimal 26928
12 Duodecimal 13700
20 Vigesimal 3768
36 Base36 ks0
## Basic calculations (n = 26928)
### Multiplication
n×i
n×2 53856 80784 107712 134640
### Division
ni
n⁄2 13464 8976 6732 5385.6
### Exponentiation
ni
n2 725117184 19525955530752 525794930532089856 14158605889368115642368
### Nth Root
i√n
2√n 164.098 29.9733 12.8101 7.69203
## 26928 as geometric shapes
### Circle
Diameter 53856 169194 2.27802e+09
### Sphere
Volume 8.17901e+13 9.11209e+09 169194
### Square
Length = n
Perimeter 107712 7.25117e+08 38081.9
### Cube
Length = n
Surface area 4.3507e+09 1.9526e+13 46640.7
### Equilateral Triangle
Length = n
Perimeter 80784 3.13985e+08 23320.3
### Triangular Pyramid
Length = n
Surface area 1.25594e+09 2.30116e+12 21986.6
## Cryptographic Hash Functions
md5 708cbabe9ec56bb5abe6a92d810e7ca6 b119e6a4cb3a48cdeabd2e570ca6afa220ac7a2c d2bc9119bf62743924314e55debd0aaf373943fb5c285fd1c6e1fd441d2906e6 78d34fcdfb0726ef3e4d70f4095674fb6b4b9aa3b73aeadc26ab34a4ea933a65645752f35a213a5fa20d04da77a85324f80058bb33ca531d0e2282c2eec873d8 1abcac259d876ce879afb6f4f784898472e551e5 | 1,623 | 4,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-31 | latest | en | 0.755903 |
https://www.oercommons.org/browse?batch_start=20&f.alignment=CCSS.Math.Content.6.EE.B.7 | 1,680,138,482,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00093.warc.gz | 1,010,483,827 | 31,985 | Updating search results...
# 26 Results
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• CCSS.Math.Content.6.EE.B.7
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During this two-day lesson, students work with a partner to create and implement a problem-solving plan based on the mathematical concepts of rates, ratios, and proportionality. Students analyze the relationship between different-sized gummy bears to solve problems involving size and price.Key ConceptsThroughout this unit, students are encouraged to apply the mathematical concepts they have learned over the course of this year to new settings. Helping students develop and refine these problem solving skills:Creating a problem solving plan and implementing their plan systematicallyPersevering through challenging problems to find solutionsRecalling prior knowledge and applying that knowledge to new situationsMaking connections between previous learning and real-world problemsCommunicating their approaches with precision and articulating why their strategies and solutions are reasonableCreating efficacy and confidence in solving challenging problems in a real worldGoals and Learning ObjectivesCreate and implement a problem-solving plan.Organize and interpret data presented in a problem situation.Analyze the relationship between two variables.Use ratios.Write and solve proportions.Create rate tables to organize data and make predictionsUse multiple representations—including tables, graphs, and equations—to organize and communicate data.Articulate strategies, thought processes, and approaches to solving a problem and defend why the solution is reasonable.
Subject:
Ratios and Proportions
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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Lesson OverviewStudents apply the multiplication property of equality to solve equations.Key ConceptsIn the previous lesson, students solved equations of the form x + p = q using the addition property of equality. In this lesson, they will solve equations of the form px = q using the multiplication property of equality. They will multiply or divide both sides of an equation by the same number to obtain an equivalent equation.Since multiplication by a is equivalent to division by 1a, students will see that they may also divide both sides of the equation by the same number to get an equivalent equation. Students will also apply this property to solving a particular kind of equation, a proportion.Goals and Learning ObjectivesUse the multiplication property of equality to keep an equation balanced.Use the multiplication property of equality to solve equations of the form px = q for cases in which p, q, and x are all non-negative rational numbers.Use the multiplication property of equality to solve proportions.
Subject:
Algebra
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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In this task students are asked to write an equation to solve a real-world problem. There are two natural approaches to this task. In the first approach, students have to notice that even though there is one variable, namely the number of firefighters, it is used in two different places. In the other approach, students can find the total cost per firefighter and then write the equation.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
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Demos and activities in this lesson are intended to illustrate the basic concepts of energy science -- work, force, energy, power etc. and the relationships among them. The "lecture" portion of the lesson includes many demonstrations to keep students engaged, yet has high expectations for the students to perform energy related calculations and convert units as required. A homework assignment and quiz are used to reinforce and assess these basic engineering science concepts.
Subject:
Engineering
Physics
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Jan DeWaters
Susan Powers
09/18/2014
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Students explore the composition and practical application of parallel circuitry, compared to series circuitry. Students design and build parallel circuits and investigate their characteristics, and apply Ohm's law.
Subject:
Engineering
Electronic Technology
Material Type:
Activity/Lab
Lesson Plan
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Daria Kotys-Schwartz
Denise Carlson
Janet Yowell
Malinda Schaefer Zarske
Sabre Duren
Xochitl Zamora Thompson
09/18/2014
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This lesson unit is intended to help you assess how well students are able to: Perform arithmetic operations, including those involving whole-number exponents, recognizing and applying the conventional order of operations; Write and evaluate numerical expressions from diagrammatic representations and be able to identify equivalent expressions; apply the distributive and commutative properties appropriately; and use the method for finding areas of compound rectangles.
Subject:
Geometry
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
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Mathematics
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Full Course
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Pearson
10/06/2016
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Equations and Inequalities
Type of Unit: Concept
Prior Knowledge
Students should be able to:
Add, subtract, multiply, and divide with whole numbers, fractions, and decimals.
Use the symbols <, >, and =.
Evaluate expressions for specific values of their variables.
Identify when two expressions are equivalent.
Simplify expressions using the distributive property and by combining like terms.
Use ratio and rate reasoning to solve real-world problems.
Order rational numbers.
Represent rational numbers on a number line.
Lesson Flow
In the exploratory lesson, students use a balance scale to find a counterfeit coin that weighs less than the genuine coins. Then continuing with a balance scale, students write mathematical equations and inequalities, identify numbers that are, or are not, solutions to an equation or an inequality, and learn how to use the addition and multiplication properties of equality to solve equations. Students then learn how to use equations to solve word problems, including word problems that can be solved by writing a proportion. Finally, students connect inequalities and their graphs to real-world situations.
Subject:
Mathematics
Algebra
Material Type:
Unit of Study
Provider:
Pearson
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During this two-day lesson, students work with a partner to create and implement a problem-solving plan based on the mathematical concepts of rates, ratios, and proportionality. Students analyze the relationship between different-sized gummy bears to solve problems involving size and price.Key ConceptsThroughout this unit, students are encouraged to apply the mathematical concepts they have learned over the course of this year to new settings. Help students develop and refine these problem-solving skills:Creating a problem solving plan and implementing their plan systematicallyPersevering through challenging problems to find solutionsRecalling prior knowledge and applying that knowledge to new situationsMaking connections between previous learning and real-world problemsCommunicating their approaches with precision and articulating why their strategies and solutions are reasonableCreating efficacy and confidence in solving challenging problems in a real worldGoals and Learning ObjectivesCreate and implement a problem-solving plan.Organize and interpret data presented in a problem situation.Analyze the relationship between two variables.Use ratios.Write and solve proportions.Create rate tables to organize data and make predictions.Use multiple representations—including tables, graphs, and equations—to organize and communicate data.Articulate strategies, thought processes, and approaches to solving a problem, and defend why the solution is reasonable.
Subject:
Ratios and Proportions
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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3.0 stars
This task provides a good entry point for students into representing quantities in contexts with variables and expressions and building equations that reflect the relationships presented in the context.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/15/2012
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This task provides a good opportunity for group work and class discussions where students generate and compare equivalent expressions. In class discussion, students should be asked to connect the terms of an expression with quantities shown in the diagram.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
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Solve one-step equations
Solve one-step linear equations in one variable involving non-negative rational numbers.
*(guaranteeing only whole numbers)
Subject:
Mathematics
Material Type:
Activity/Lab
Author:
Liberty Public Schools
04/12/2021
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This task presents a straight forward question that can be solved using an equation in one variable. The numbers are complicated enough so that it is natural to set up an equation rather than solve the problem in one's head.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/27/2012
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Students create and implement a problem-solving plan to solve another problem involving the relationship between the sound of thunder and the distance of the lightning.Key ConceptsThroughout this unit, students are encouraged to apply the mathematical concepts they have learned over the course of this year to new settings. Help students develop and refine these problem-solving skills:Creating a problem-solving plan and implementing their plan systematicallyPersevering through challenging problems to find solutionsRecalling prior knowledge and applying that knowledge to new situationsMaking connections between previous learning and real-world problemsCommunicating their approaches with precision and articulating why their strategies and solutions are reasonableCreating efficacy and confidence in solving challenging problems in a real worldGoals and Learning ObjectivesCreate and implement a problem-solving plan.Organize and interpret data presented in a problem situation.Analyze the relationship between two variables.Create a rate table to organize data and make predictions.Apply the relationship between the variables to write a mathematical formula and use the formula to solve problems.Create a graph to display proportional relationships and use this graph to make predictions.Articulate strategies, thought processes, and approaches to solving a problem and defend why the solution is reasonable.
Subject:
Mathematics
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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Lesson OverviewStudents solve problems using equations of the form x + p = q and px = q, as well as problems involving proportions.Key ConceptsStudents will extend what they know about writing expressions to writing equations. An equation is a statement that two expressions are equivalent. Students will write two equivalent expressions that represent the same quantity. One expression will be numerical and the other expression will contain a variable.It is important that when students write the equation, they define the variable precisely. For example, n represents the number of minutes Aiko ran, or x represents the number of boxes on the shelf.Students will then solve the equations and thereby solve the problems.Students will solve proportion problems by solving equations. This makes sense because a proportion such as xa=bc is really just an equation of the form xp = q where p=1a and q=bc.Students will also compare their algebraic solutions to an arithmetic solution for the problem. They will see, for example, that a problem that might be solved arithmetically by subtracting 5 from 78 can also be solved algebraically by solving x + 5 = 78, where 5 is subtracted from both sides—a parallel solution to subtracting 5 from 78.Goals and Learning ObjectivesUse equations of the form x + p = q and xp = q to solve problems.Solve proportion problems using equations.ELL: ELLs may have difficulty verbalizing their reasoning, particularly because word problems are highly language dependent. Accommodate ELLs by providing extra time for them to process the information. Note that this problem is a good opportunity for ELLs to develop their literacy skills since it incorporates reading, writing, listening, and speaking skills. Encourage students to challenge each others' ideas and justify their thinking using academic and specialized mathematical language.
Subject:
Algebra
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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Students learn the engineering design process by following the steps, from problem identification to designing a device and evaluating its efficacy and areas for improvement. A quick story at the beginning of the activity sets up the challenge: A small child put a pebble in his ear and we don't know how to get it out! Acting as biomedical engineers, students are asked to design a device to remove it. Each student pair is provided with a model ear canal and a variety of classroom materials. A worksheet guides the design process as students create devices and attempt to extract pebbles from the ear canal.
Subject:
Engineering
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Derek Harbin
Krista Warner
Leyf Starling
Shayn Peirce-Cottler
09/18/2014
Educational Use
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0.0 stars
Student teams locate a contaminant spill in a hypothetical site by measuring the pH of soil samples. Then they predict the direction of groundwater flow using mathematical modeling. They also use the engineering design process to come up with alternative treatments for the contaminated water.
Subject:
Engineering
Environmental Science
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Ben Heavner
Janet Yowell
Malinda Schaefer Zarske
Melissa Straten
10/14/2015
Unrestricted Use
CC BY
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0.0 stars
This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the s...
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
03/17/2013
Educational Use
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0.0 stars
Air pressure is pushing on us all the time although we do not usually notice it. In this activity, students learn about the units of pressure and get a sense of just how much air pressure is pushing on them.
Subject:
Engineering
Physics
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Alex Conner
Geoffrey Hill
Janet Yowell
Malinda Schaefer Zarske
Tom Rutkowski
10/14/2015
Educational Use
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Students do work by lifting a known mass over a period of time. The mass and measured distance and time is used to calculate force, work, energy and power in metric units. The students' power is then compared to horse power and the power required to light 60-watt light bulbs.
Subject:
Engineering
Physics
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Jan DeWaters
Susan Powers
09/18/2014
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Putting Math to Work
Type of Unit: Problem Solving
Prior Knowledge
Students should be able to:
Solve problems with rational numbers using all four operations.
Write ratios and rates.
Use a rate table to solve problems.
Write and solve proportions.
Use multiple representations (e.g., tables, graphs, and equations) to display data.
Identify the variables in a problem situation (i.e., dependent and independent variables).
Write formulas to show the relationship between two variables, and use these formulas to solve for a problem situation.
Draw and interpret graphs that show the relationship between two variables.
Describe graphs that show proportional relationships, and use these graphs to make predictions.
Interpret word problems, and organize information.
Graph in all quadrants of the coordinate plane.
Lesson Flow
As a class, students use problem-solving steps to work through a problem about lightning. In the next lesson, they use the same problem-solving steps to solve a similar problem about lightning. The lightning problems use both rational numbers and rates. Students then choose a topic for a math project. Next, they solve two problems about gummy bears using the problem-solving steps. They then have 3 days of Gallery problems to test their problem-solving skills solo or with a partner. Encourage students to work on at least one problem individually so they can better prepare for a testing situation. The unit ends with project presentations and a short unit test.
Subject:
Mathematics
Material Type:
Unit of Study
Provider:
Pearson
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Students work in a whole-class setting, independently, and with partners to design and implement a problem-solving plan based on the mathematical concepts of rates and multiple representations (e.g., tables, equations, and graphs). They analyze a rule of thumb and use this relationship to calculate the distance in miles from a viewer's vantage point to lightning.Key ConceptsThroughout this unit, students are encouraged to apply the mathematical concepts they have learned over the course of this year to new settings. Help students develop and refine these problem-solving skills:Creating a problem-solving plan and implementing the plan systematicallyPersevering through challenging problems to find solutionsRecalling prior knowledge and applying that knowledge to new situationsMaking connections between previous learning and real-world problemsCommunicating their approaches with precision and articulating why their strategies and solutions are reasonableCreating efficacy and confidence in solving challenging problems in the real worldGoals and Learning ObjectivesCreate and implement a problem-solving plan.Organize and interpret data presented in a problem situation.Analyze the relationship between two variables.Create a rate table to organize data and make predictions.Apply the relationship between the variables to write a mathematical formula and use the formula to solve problems.Create a graph to display proportional relationships, and use this graph to make predictions.Articulate strategies, thought processes, and approaches to solving a problem, and defend why the solution is reasonable.
Subject:
Algebra
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
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This supplemental resource provides problems and activities related to Patterns, Relationships and Functions in Middle School Mathematics.
Subject:
Functions
Material Type:
Activity/Lab
Author:
Twianie Roberts Ed.D
09/25/2022
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CC BY
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This is a task from the Illustrative Mathematics website that is one part of a complete illustration of the standard to which it is aligned. Each task has at least one solution and some commentary that addresses important asects of the task and its potential use. Here are the first few lines of the commentary for this task: A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of 280 pieces of fruit. There are twice as many r...
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/15/2013
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Subject:
Algebra
Material Type:
Lesson Plan
Provider:
Pearson
09/21/2015
Educational Use
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Students learn about electricity and air pollution while building devices to measure volatile organic compounds (VOC) by attaching VOC sensors to prototyping boards. In the second part of the activity, students evaluate the impact of various indoor air pollutants using the devices they made.
Subject:
Engineering
Atmospheric Science
Material Type:
Activity/Lab
Provider:
TeachEngineering
Provider Set:
TeachEngineering
Author:
Berkeley Almand
Mike Hannigan | 4,320 | 21,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-14 | latest | en | 0.893354 |
http://www.ck12.org/geometry/Quadrilateral-Classification/lesson/Quadrilaterals/r14/ | 1,455,289,939,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701164268.69/warc/CC-MAIN-20160205193924-00079-ip-10-236-182-209.ec2.internal.warc.gz | 342,845,114 | 34,768 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
Categories of quadrilaterals based on sides and angles.
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All squares are rectangles, but not all rectangles are squares. How is this possible?
Guidance
A quadrilateral is a polygon with four sides. There are many common special quadrilaterals that you should be familiar with. Below, these special quadrilaterals are described with their definitions and some properties. Additional properties are explored further and proved in another concept.
• A kite is a convex quadrilateral with two pairs of adjacent congruent sides such that not all sides are congruent. The word adjacent means “next to”, so the congruent sides are next to each other. Note: some texts leave out the stipulation that “not all sides are congruent”. If this is the case, it is possible for a kite to be a rhombus if all four sides are congruent. To note that sides are congruent on a picture, draw corresponding tick marks through the middles of the congruent sides.
• A trapezoid is a quadrilateral with exactly one pair of parallel sides. Note: some texts leave out the word “exactly”, which means quadrilaterals with two pairs of parallel sides are sometimes considered trapezoids. Here, assume trapezoids have exactly one pair of parallel sides. To note that sides are parallel on a picture, draw corresponding arrow markings through the middles of the parallel sides.
• An isosceles trapezoid is a trapezoid with one pair of congruent sides. An additional property of isosceles trapezoids is base angles are congruent.
• A parallelogram is a quadrilateral with two pairs of parallel sides. Some additional properties of parallelograms are opposite sides are congruent and opposite angles are congruent.
• A rectangle is a quadrilateral with four right angles. All rectangles are parallelograms. An additional property of rectangles is diagonals are congruent.
• A rhombus is a quadrilateral with four congruent sides. All rhombuses are parallelograms.
• A square is a quadrilateral with four right angles and four congruent sides. All squares are rectangles and rhombuses.
Notice that these categories of quadrilaterals overlap. A square is not only a square, but also a rhombus, a rectangle, a parallelogram, and a quadrilateral. This means that a square will have all the same properties as rhombuses, rectangles, parallelograms, and quadrilaterals.
The following diagram shows the hierarchy of quadrilaterals.
Example A
What types of quadrilaterals have four right angles?
Solution: Rectangles and squares have four right angles.
Example B
Solve for \begin{align*}x\end{align*} (picture not drawn to scale).
Solution: This quadrilateral is marked as having four congruent sides, so it is a rhombus. Rhombuses are parallelograms, so they have all the same properties as parallelograms. One property of parallelograms is that opposite angles are congruent. This means that the marked angles in this rhombus must be congruent.
Example C
A quadrilateral has two pairs of parallel sides. What type of quadrilateral must it be? What type of quadrilateral could it be?
Solution: Two pairs of parallel sides is the definition of a parallelogram, so this quadrilateral must be a parallelogram. It could be a rectangle, a square, or a rhombus if it also satisfied the definition of any of those quadrilaterals.
Concept Problem Revisited
Rectangles are defined as quadrilaterals with four right angles. Squares are defined as quadrilaterals with four right angles and four congruent sides. Because all squares have four right angles and satisfy the definition for rectangles, they can all also be called rectangles. On the other hand, not all rectangles have four congruent sides, so not all rectangles can also be called squares.
Vocabulary
A kite is a quadrilateral with two pairs of adjacent congruent sides such that not all four sides are congruent. Note: some texts leave out the stipulation that “not all sides are congruent”. If this is the case, it is possible for a kite to be a rhombus if all four sides are congruent.
trapezoid is a quadrilateral with exactly one pair of parallel sides. Note: some texts leave out the word “exactly”, which means quadrilaterals with two pairs of parallel sides are sometimes considered trapezoids. Here, assume trapezoids have exactly one pair of parallel sides.
An isosceles trapezoid is a trapezoid with one pair of congruent sides.
parallelogram is a quadrilateral with two pairs of parallel sides.
rectangle is a quadrilateral with four right angles.
rhombus is a quadrilateral with four congruent sides.
A square is a quadrilateral with four right angles and four congruent sides.
Guided Practice
1. Draw a square. Draw in the diagonals of the square. Make at least one conjecture about the diagonals of the square.
2. A quadrilateral has four congruent sides. What type of quadrilateral must it be? What type of quadrilateral could it be?
3. Solve for \begin{align*}x\end{align*} (picture not drawn to scale).
1. To make a conjecture means to make an educated guess. There are a few conjectures you might make about the diagonals of a square. These conjectures will be proved in a later concept. Here are some possible conjectures:
• diagonals of a square are congruent
• diagonals of a square are perpendicular
• diagonals of a square bisect each other (cut each other in half)
• diagonals of a square bisect the angles (cut the \begin{align*}90^\circ\end{align*} angles in half)
2. It must be a rhombus and therefore also a parallelogram. It could be a square.
3. This is a parallelogram so opposite sides are congruent.
Practice
Decide whether each statement is always, sometimes, or never true. Explain your answer.
1. A square is a rectangle.
2. A rhombus is a square.
3. An isosceles trapezoid is a trapezoid.
4. A parallelogram is a quadrilateral.
5. A square is a parallelogram.
6. A trapezoid is a parallelogram.
Decide what type of quadrilateral it must be and what type of quadrilateral it could be based on the description.
7. A quadrilateral has 4 congruent angles.
8. A quadrilateral has 2 pairs of congruent sides.
9. Draw a kite. Draw in its diagonals. Make at least one conjecture about the diagonals of kites.
10. Draw a rectangle. Draw in its diagonals. Make at least one conjecture about the diagonals of rectangles.
11. Draw a rhombus. Draw in its diagonals. Make at least one conjecture about the diagonals of rhombuses.
12. Draw a kite. Make a conjecture about the opposite angles of kites.
Use the markings on the shapes below to identify the shape. Then, solve for \begin{align*}x\end{align*}. Note: pictures are not drawn to scale.
13.
14.
15. Make a conjecture about the adjacent angles of a parallelogram (such as the ones marked in the picture below). How must they be related?
Vocabulary Language: English
Kite
Kite
Parallelogram
Parallelogram
A parallelogram is a quadrilateral with two pairs of parallel sides.
A quadrilateral is a closed figure with four sides and four vertices.
Rectangle
Rectangle
A rectangle is a quadrilateral with four right angles.
Rhombus
Rhombus
A rhombus is a quadrilateral with four congruent sides.
Trapezoid
Trapezoid
A trapezoid is a quadrilateral with exactly one pair of parallel opposite sides. | 1,776 | 7,538 | {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2016-07 | latest | en | 0.876022 |
https://math.stackexchange.com/questions/315471/how-do-you-evaluate-this-limit-lim-limits-x-to-infty-frac2x5x2-sqrt | 1,563,332,586,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00095.warc.gz | 464,317,422 | 38,711 | # How do you evaluate this limit? $\lim\limits_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$
$$\lim_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$$ This is what I have tried so far, $$\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10}+x^2)^{\frac{1}{2}}}$$
\begin{align} &\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10})^{\frac{1}{2}}+(x)^{\frac{1}{2}}}\\ &\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{5})+(x)}\\ &\lim_{x\to-\infty}\frac{x^2}{x}\\ &\lim_{x\to-\infty}x = -\infty\\ \end{align} Is this the correct approach to finding the limit?
• $(2x^{10}+x^2)^{\frac12}$ is not $(2x^{10})^{\frac12}+x^{\frac12}$ – ChairOTP Feb 27 '13 at 2:03
• I would recommend multiplying top & bottom by $x^{-5}$. – Gerry Myerson Feb 27 '13 at 2:06
• This is way way too far from being correct. – Kaster Feb 27 '13 at 2:06
• I have multiplied the top and bottom by $x^{-5}$ but I am not sure what to do next. – Kot Feb 27 '13 at 2:15
• What happens to $\frac1x$ as $x$ approaches infinity? – ChairOTP Feb 27 '13 at 2:20
Observe that for $x<0$, we have \begin{align}\sqrt{2x^{10}+x^2} &= \left(\sqrt{x^2}\right)^5\sqrt{2+\frac1{x^8}}\\ &= |x|^5\sqrt{2+\frac1{x^8}}\\ &= -x^5\sqrt{2+\frac1{x^8}}.\end{align} Does that get you started?
Let me approach this two other ways, rigorously, and loosely. Note that the function is undefined for $x=0$, so we will only consider $x\neq 0$ in the following. To study the end behavior of this function, we will ultimately be interested only in the terms of highest degree on top and on bottom. Ideally, we'd like to rewrite the quotient in an equivalent way so that at least one of the highest degree terms is constant. Once we've done that, we'll proceed to the limit. I will call the function $f(x).$
Rigorous:
You expressed a desire to divide top and bottom by $x^5,$ so I will do it that way. The key fact that I will be using several times is that for any real $\alpha,$ we have $$|\alpha|=\sqrt{\alpha^2}.$$ The following manipulations hold for any $x\neq 0:$ \begin{align}f(x) &= \frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}}{x^{-5}}\cdot\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}(2x^5+x^2)}{x^{-5}\sqrt{2x^{10}+x^2}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2(2x^8+1)}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2}\sqrt{2x^8+1}}\\ &= \frac{2+x^{-3}}{x^{-5}|x|\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}.\end{align} Observe that $x^{-4}$ is positive for all $x\neq 0,$ so in particular, $$x^{-4}=\left|x^{-4}\right|=\sqrt{(x^{-4})^2}=\sqrt{x^{-8}}.$$
Hence, we have \begin{align}f(x) &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}\left(2x^8+1\right)}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\end{align} for all $x\neq 0$. Now, note that $$\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}=\sqrt{2},$$ as I believe you've already calculated. Also note that $$\frac{x}{|x|}=\begin{cases}1 & \text{if }x>0,\\-1 & \text{if }x<0,\end{cases}$$ and so $$\lim_{x\to-\infty}\frac{x}{|x|}=-1.$$ Therefore, $$\lim_{x\to-\infty}f(x)=\left[\lim_{x\to-\infty}\frac{x}{|x|}\right]\cdot\left[\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\right]=-1\cdot\sqrt{2}=-\sqrt{2}.$$
Loose:
This gets back to what Babak S. mentions in the comment below. When dealing with end-behavior of polynomials, only the highest-degree term ultimately matters. To that end, we can (roughly speaking) "drop" all the terms in numerator and denominator except those of highest degree, and then find the limit that way. That is, \begin{align}\lim_{x\to-\infty}f(x) &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2x^{10}}}\\ &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2}\sqrt{x^{10}}}\\ &= \frac{2}{\sqrt{2}}\cdot\lim_{x\to-\infty}\frac{x^5}{\sqrt{(x^5)^2}}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}\frac{x^5}{|x^5|}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}-1\\ &= -\sqrt{2}.\end{align}
• I am trying to follow along. In the end, the absolute value becomes negative because x is approaching negative infinity or that $x<0$? – Kot Feb 27 '13 at 2:39
• Both, actually. As $x$ approaches $-\infty$, we will eventually have $x<0$, at which point we will have $|x|=-x$. – Cameron Buie Feb 27 '13 at 2:42
• @StevenN: Note that if you have $a_nx^n+...+a_1x+a_0$ and $x$ goes to $\infty$ then $$a_nx^n+...+a_1x+a_0\sim a_nx^n$$ So $\sqrt{2x^{10}+x^2}\sim\sqrt{2}|x^5|=\sqrt{2}\times(-x^5)$ since $x\to-\infty$ – mrs Feb 27 '13 at 2:43
• Nice, Cameron! +1 – Namaste Feb 27 '13 at 2:45
• @StevenN: Ah! I see what happened. The next step would be to rewrite $$\cfrac{2x^5+x^2}{-x^5\sqrt{2+\frac1{x^8}}}=\cfrac{2+\frac1{x^3}}{-\sqrt{2+\frac{1}{x^8}}},$$ so while it's true that $-x^5$ is positive, $x^5$ is negative, and so their quotient leaves the $-$ sign in there. – Cameron Buie Feb 27 '13 at 3:36
Well, if you can't get rid of a square root, put everything in a square root. Notice that for $x$ sufficiently negative--in particular, for $x<-\frac1{\sqrt[3]2}$--we have $2x^5+x^2<0,$ so that $$2x^5+x^2=-|2x^5+x^2|=-\sqrt{(2x^5+x^2)^2}$$ for sufficiently negative $x$-values. Thus, for such $x$-values, we have
\begin{align} \frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}} &= \frac{-\sqrt{(2x^5+x^2)^2}}{\sqrt{2x^{10}+x^2}} \\ &= -\sqrt{\frac{(2x^5+x^2)^2}{2x^{10}+x^2}} \\ &= -\sqrt{\frac{4x^{10}+4x^7+x^4}{2x^{10}+x^2}} \\ &= -\sqrt{2+\frac{4x^{10}+4x^7+x^4-2(2x^{10}+x^2)}{2x^{10}+x^2}} \\ &= -\sqrt{2+\frac{4x^7+x^4-2x^2}{2x^{10}+x^2}} \\ &= -\sqrt{2}\sqrt{1+\frac{4x^7+x^4-2x^2}{4x^{10}+2x^2}} \\ \end{align}
The fraction inside the right-hand square root is about $1/x^3$, so the right-hand square root goes to $1$ as $x\to-\infty$, so the whole thing goes to $-\sqrt{2}$.
• That doesn't work when the numerator is negative, Marty, but you're close. – Cameron Buie Feb 27 '13 at 2:38
• Good point. But my formatting is pretty good:) – marty cohen Feb 28 '13 at 2:19
• Indeed. All you really need to do to fix it is to say that for $x$ sufficiently negative, we have $$2x^5+x^2=-\sqrt{\left(2x^5+x^2\right)^2},$$ and so all the rest is basically right, just missing a $-$ sign. – Cameron Buie Feb 28 '13 at 2:51 | 2,524 | 6,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-30 | latest | en | 0.757241 |
https://boredofstudies.org/threads/cambridge-hsc-mx1-textbook-marathon-q-a.348462/page-8#post-7182594 | 1,695,938,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00747.warc.gz | 158,774,272 | 19,489 | # Cambridge HSC MX1 Textbook Marathon/Q&A (1 Viewer)
#### davidgoes4wce
##### Well-Known Member
I know from Physics formulae with the equations of motions we have:
$\bg_white v^2=u^2+2as$
$\bg_white s=ut+\frac{1}{2} at^2$
I'm just struggling to find the link between
$\bg_white 180= V \times t$
#### InteGrand
##### Well-Known Member
I know from Physics formulae with the equations of motions we have:
$\bg_white v^2=u^2+2as$
$\bg_white s=ut+\frac{1}{2} at^2$
I'm just struggling to find the link between
$\bg_white 180= V \times t$
$\bg_white \noindent Do you mean how did I arrive at 180 = Vt at the time of collision? This was done by equating the two equations of motion and cancelling common terms from each side.$
#### davidgoes4wce
##### Well-Known Member
OK I get it now.
I treated the origin at the base of the cliff.
$\bg_white y_B=180-5t^2$
Where I went wrong was I assumed the stone had an initial velocity of 0, the question stated it was V.
$\bg_white y_S=Vt-5t^2$
$\bg_white By equating y_S = y_B I can then get the equation of 180=Vt$
Last edited:
#### davidgoes4wce
##### Well-Known Member
If I treated the top of the lookout as the mean line, my equations would obviously be different.
$\bg_white y_B=5t^2$
$\bg_white y_S=-Vt+5t^2+180$
(treating anything downwards as positive g=+10)
$\bg_white By equating y_B=y_S , 5t^2=-Vt+5t^2+180$
$\bg_white Vt=180$
#### AfroNerd
##### New Member
having trouble with this binomial question
#### HeroicPandas
##### Heroic!
having trouble with this binomial question
View attachment 33393
\bg_white (1+x)^{n-1} = \binom{n-1}{0} + \binom{n-1}{1}x + \dots + \binom{n-1}{n-2} x^{n-2} + \binom{n-1}{n-1}x^{n-1} \\ Put x = 1, \\\\ 2^{n-1} = \binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \dots + \binom{n-1}{n-2}}_{\text{isolate this}} + \binom{n-1}{n-1} \\\\ \begin{align*} \binom{n-1}{1} + \dots + \binom{n-1}{n-2} &= 2^{n-1} - \binom{n-1}{0} - \binom{n-1}{n-1} \\&= 2^{n-1} - 1 -1 \end{align*}
Then multiply both sides by n
#### si2136
##### Well-Known Member
I think there is a formula for this but I'm not sure what it's called.
Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?
#### InteGrand
##### Well-Known Member
I think there is a formula for this but I'm not sure what it's called.
Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?
You may wish to consider cases based on things like how many ''double-letters'' are selected.
#### si2136
##### Well-Known Member
You may wish to consider cases based on things like how many ''double-letters'' are selected.
I know how to do it if it was all of the letters, but not 4 out of the 7
#### InteGrand
##### Well-Known Member
I know how to do it if it was all of the letters, but not 4 out of the 7
Yeah, that's why you may find it helpful to consider cases. E.g. The cases could be: no pairs of double-letters; exactly one pair of double-letters; exactly two pairs of double-letters.
Then count the no. of possibilities in each case and sum them up.
Last edited:
#### Mongoose528
##### Member
Isn't the formula this?
For the number of ways to arrange n objects taken r at at a time
n!/(n-r)!
So it would be
7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840
But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210
Last edited:
#### si2136
##### Well-Known Member
Isn't the formula this?
For the number of ways to arrange n objects taken r at at a time
n!/(n-r)!
So it would be
7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840
But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210
Wrong, you would need to take it by cases or Pork's Insertion Method
How do I do this
Also
#### infiniteeee
##### New Member
anyone who can help me with question 12b and 13a of exercise 4F?
#### Drongoski
##### Well-Known Member
$\bg_white \therefore x^3 - Ax + 0.x + 3A = 0\\ \\ \therefore sum of roots = \alpha + \beta + (\alpha + \beta) = -(-A) = A \implies 2(\alpha + \beta) = A \implies \alpha + \beta = \frac {1}{2} A \\ \\ sum of roots taken 2 at a time = \alpha \beta +\alpha (\alpha+\beta) + \beta (\alpha+\beta) = 0 \\ \\ \therefore \alpha \beta = - (\alpha + \beta)^2 = - (\frac {1}{2} A)^2 = -\frac {1}{4} A^2 \\ \\ product of roots = \alpha \beta (\alpha + \beta) -3A \implies \alpha \beta = \frac {-3A}{\frac {1}{2}A} = -6 \implies -\frac {1}{4}A^2 = -6 \\ \\ \implies A = 2\sqrt 6$
Last edited:
#### kashkow
##### Active Member
View attachment 33481
anyone who can help me with question 12b and 13a of exercise 4F?
The first question:
The "sum of the products of pairs of roots" refers to the x term which is absent, meaning the coefficient is 0. Therefore the "sum of the products of pairs of roots" = 0
Algebraically: ab + a(a+b) + b(a+b) = 0 [... a = alpha, b = beta]
Expanding: ab + a^2 + ab + ab + b^2 = 0
ab + a^2 + 2ab + b^2 = 0
Combining the perfect square:
ab + (a+b)^2 = 0
And you can use the results from the question before 12a what (a+b)^2 is.
That is substitute (a+b) = 1/2*A = A/2
ab + (A/2)^2 = 0
ab + (A^2)/4 = 0
Therefore: ab = -(A^2)/4 = -1/4 * A^2
Sorry if it's hard to read. Rats i was late again...
##### -insert title here-
Cosine Rule XOR Sine Rule
##### -insert title here-
How do I do this
...seriously just draw a diagram
#### davidgoes4wce
##### Well-Known Member
View attachment 33481
anyone who can help me with question 12b and 13a of exercise 4F?
EX 4F Q13 a
$\bg_white Let the three roots be \alpha, \alpha \ and \beta$
$\bg_white 2 \alpha + \beta =\frac{8}{4}=2 \ \textcircled{1}$
$\bg_white \alpha \beta +\alpha \beta +\alpha^2 =\frac{-3}{4} \implies 2 \alpha \beta +\alpha^2=\frac{-3}{4} \textcircled{2}$
$\bg_white \alpha^2 \beta =\frac{-9}{4} \textcircled{3}$
$\bg_white Rearranging \textcircled{1} \ \beta=2-2\alpha$
$\bg_white Substituting \textcircled{1} into \textcircled{2}$
$\bg_white 2 \alpha (2-2 \alpha)+\alpha^2=\frac{-3}{4}$
$\bg_white 4 \alpha -4 \alpha^2+\alpha^2=\frac{-3}{4}$
$\bg_white -3 \alpha^2+4\alpha=\frac{-3}{4}$
$\bg_white -12 \alpha^2+16 \alpha =-3 \implies 0=12\alpha^2-16\alpha-3$
$\bg_white By using the Quadratic Roots formula x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} we look to solve \alpha$
$\bg_white \alpha=\frac{16 \pm \sqrt{256-4(12)(-3)}}{2 \times 12}$
$\bg_white \alpha= \frac{3}{2} \ and \ \alpha=\frac{-1}{6}$
$\bg_white if \alpha =\frac{3}{2} \beta=-1$
$\bg_white Substitute \alpha =\frac{3}{2} \ into equation \textcircled{1} \ \beta=-1$
$\bg_white Substitute \alpha =\frac{3}{2} \ into equation \textcircled{3} \ \beta=-1$
$\bg_white Therefore, \alpha =\frac{3}{2} and \beta=-1 are consistent solutions$
$\bg_white if \alpha =\frac{-1}{6}$
$\bg_white Substitute \alpha =\frac{-1}{6} \ into equation \textcircled{1} \ \beta=\frac{7}{3}$
$\bg_white Substitute \alpha =\frac{-1}{6} \ into equation \textcircled{3} \ \beta=-81$
$\bg_white Therefore, \alpha =\frac{-1}{6} is not consistent and is not a solution to the cubic polynomial.$ | 2,464 | 7,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.822635 |
https://proxies123.com/tag/euler/ | 1,619,039,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00449.warc.gz | 463,803,037 | 16,784 | ## rotation – What is the relationship between gimbals and Euler angles?
The definition of Euler angles inherently includes gimbals.
Let’s use Unity’s convention as one arbitrary example (technically they’re a related convention called Tait-Bryan angles, but nobody calls them that).
An Euler angle triplet `(pitch, yaw, roll)` means…
1. Rotate the outermost gimbal, with its axis parallel to the world y axis, by `yaw` degrees.
2. Rotate the middle gimbal, with its axis perpendicular to the outer gimbal and parallel to the world x axis when `yaw` is zero, by `pitch` degrees.
3. Rotate the inner gimbal, with its axis perpendicular to the middle gimbal and parallel to the world z axis when `yaw` and `pitch` are both zero, by `roll` degrees.
The net orientation of the object is then the orientation of this inner gimbal, stacking up the individual rotations of the middle and outer gimbals.
There are other Euler angle / Tait-Bryan angle conventions, but they all share this pattern where an arbitrary orientation is built-up from a sequence of rotations about particular axes. As I explain in this answer, when you compose two rotations in sequence, the angle of one changes the axis of the second – and that makes the chained rotation behave like two linked gimbals.
To emphasize: any rotation scheme that tries to decompose an orientation into multiple angles is inherently modelling a set of linked gimbals. That means it will have situations where it loses a degree of freedom because two of its gimbal axes become parallel – gimbal lock – and you need a correspondingly larger change to your angle triplet to get to a nearby orientation in some direction.
To escape this, you need to leave behind the idea of chaining separate rotations with different axes/angles entirely. Instead we can use…
• Angle-axis representations, where we express a particular orientation as a single rotation around one axis by a particular angle. By allowing arbitrary diagonal axes, we can still reach any orientation we choose this way, without composing multiple rotations in sequence.
• Quaternions, which are really a special kind of angle-axis representation, dressed up in 4D space with 3 imaginary dimensions. This might seem like a strange choice, but it lets us take advantage of patterns in the multiplication of these imaginary units to compose and interpolate rotations much more easily than other methods.
• Matrices, which can express any arbitrary affine transformation, including rotations.
All of these methods let us express an orientation “all at once” rather than as a sequence of rotations with different angles around different axes, so they let us break out of the gimbal paradigm that comes with that sequencing.
## differential equations – Euler Buckling using DEigensystem
I’m new to using the functions DEigensystem and DEigenvalue. I started with the examples given in the Documentation system, and it works fine. I decided to test it out on the classical Euler buckling problem – in particular, the more general form, where the governing differential equation is fourth order.
I start with the equation of the elastica parameterized by the angle along the arc length, linearize it, and then use the small slope approximation to write it in terms of deflection in Cartesian coordinates:
``````elastica = (Theta)''(s) + (Lambda)^2 Sin((Theta)(s));
elasticaLin = Series(elastica, {(Theta)(s), 0, 1}) // Normal
elasticaLinW = elasticaLin /. {(Theta)(s) -> w''(x), (Theta)''(s) -> w''''(x)}
``````
This gives:
$$lambda^2 w”(x) + w^{(4)}(x)$$
Of course, this equation can be obtained in a more “Mathematica” way by:
``````(ScriptCapitalL) = Laplacian(Laplacian(w(x), {x}), {x}) + (Lambda)^2 Laplacian(w(x), {x})
``````
I defined the Dirichlet boundary conditions of $$w(0)==w(L)==0$$ as:
``````(ScriptCapitalB)1 = DirichletCondition(w(0) == 0, True);
(ScriptCapitalB)2 = DirichletCondition(w(L) == 0, True);
``````
Finally, trying to use `DEigensystem` to find the first few eigenfunctions, I use the following code, which Mathematica just returns to me when I run it:
``````DEigensystem({(ScriptCapitalL), (ScriptCapitalB)1, (ScriptCapitalB)2}, w(x), {x, 0, L},4)
``````
What am I doing wrong here?
## Euler characteristics in 6 dimensional manifold
I have been trying to calculate the integral below.
As far as I know this is Euler class in 6 dimensions but until now i could not get the answer that the quantity X is proportional to Q.
Maybe somebody here has any idea that could help me , thanks in advance.
enter image description here
## physics – Making a parent system with euler angles (gimbal lock problem)
I thought the direct and easiest way would be to get the difference of the Euler angles
No, this is definitely not a direct or easy way. This road leads only to suffering.
Euler angles are good for presenting an orientation to a human reader/editor, or for serializing an orientation to disc, or for a few special cases like camera control. But they are an absolute nightmare to use for general orientation computations or compositions. A combination of two Euler angle rotations is not the sum or any simple expression of the angles of the inputs, nor is there a simple expression for inverting them.
Just use quaternions. They don’t have the axis bias or gimbal lock problems you’ve observed. You can compose two rotations in a chain (like a parent and child) as simply as doing a multiplication. And inverting a quaternion (eg. to undo a parent rotation and get into local space) is as easy as negating the x, y, and z components.
I have heard that quaternions also don’t save 360+ degrees, so that would be also a problem for angular velocity.
It’s not a problem because you use quaternions to store orientations, not angular velocity.
Angular velocity can be stored as an axis and an angular speed, or as a single 3-component vector along the axis of rotation, whose length is the speed of rotation along that axis.
When you want to step forward your orientation by your angular velocity, you integrate your angular speed over the timestep to get a total angle of travel. Then you can construct a quaternion that rotates by that angle about the axis vector, and compose your previous orientation quaternion with this one.
## graph theory – Is there a purely set-theoretic expression of the Euler characteristic?
The Euler characteristic of a graph is V – E + F. Where V is the number of vertices, E is the number of edges, and F is the number of faces.
Given a graph $$X$$, can we write $$X$$ as a set and find its Euler characteristic with a formula which only refers to cardinality and union?
I think we found a way, but is this interesting? We would like to find a way to write the Euler characteristic completely set-theoretically. And, we have found a way, but still working on it. Has this already been done?
## dg.differential geometry – On the Euler number of an orbit space
Let $$X$$ be a differentiable manifold and $$G$$ a finite group acting
differentiably on $$X$$. The following formula for the Euler number $$text{e}(X/G)$$ of the orbit space $$X/G$$ appears to be well-known:
$$begin{equation*}tag{1} text{e}(X/G) = dfrac{1}{lvert G rvert} sum_{g in G} text{e}(X^g) end{equation*}$$
where, for each $$g in G$$, $$X^g$$ is the subspace of elements fixed by $$g$$. Instances of this formula can be found in (3) and (5), but without proof or
link to such. A quite elaborate search on the web did not produce one, either.
Following the allusion to Lefschetz numbers in (3), a proof might be given, I suppose, along the following lines. Consider the canonical projection $$pi : X rightarrow X/G$$ and its induced map $$pi^* : text{H}^*(X/G;Bbb{Q}) rightarrow text{H}^*(X;Bbb{Q})$$ on rational cohomology. According to a celebrated result of Grothendieck in (4), this map is injective and maps $$text{H}^*(X/G;Bbb{Q})$$ isomorphically onto the $$G$$-invariants $$text{H}^*(X;Bbb{Q})^G$$ (this follows from the Leray SS for the map $$pi$$, which collapses). Now for any representation of $$G$$ on a finite dimensional vector space $$V$$ over a field of characteristic 0, one has that
$$begin{equation*} P := dfrac{1}{lvert G rvert} sum_{g in G} g : V rightarrow V end{equation*}$$
is the projector onto the $$G$$-invariants $$V^G$$ and so, by taking traces,
$$begin{equation*} dim V^G = text{tr}, P = dfrac{1}{lvert G rvert} sum_{g in G} text{tr}, g. end{equation*}$$
Therefore, for each $$i$$,
$$begin{equation*} dim text{H}^i(X/G;Bbb{Q}) = dim text{H}^i(X;Bbb{Q})^G = dfrac{1}{lvert G rvert} sum_{g in G} text{tr}, g^i, end{equation*}$$
where $$g^i$$ is the endomorphism induced by $$g$$ on $$text{H}^i(X;Bbb{Q})$$. Taking the alternating sum and interchanging the summation order leaves us with
begin{equation*} begin{aligned} text{e}(X/G) &= sum_i (-1)^i dim text{H}^i(X/G;Bbb{Q}) \ &= dfrac{1}{lvert G rvert} sum_{g in G} sum_i (-1)^i text{tr}, g^i end{aligned} end{equation*}
and so
$$begin{equation*}tag{2} text{e}(X/G) = dfrac{1}{lvert G rvert} sum_{g in G} L(g), end{equation*}$$
where
$$begin{equation*} L(g) := sum_i (-1)^i text{tr}, g^i end{equation*}$$
is the Lefschetz number of the endomorphism $$g : X rightarrow X$$.
But now we are in business; (1) boils down to the sixty-four dollar question
$$begin{equation*}tag{3} forall g in G: L(g) = text{e}(X^g). end{equation*}$$
A high-brow reference for this is is the general Lefschetz fixed point theorem
Theorem (2.12) of (2) (which is a synthesis of the main theorem Theorem (6.7) of (1) and the Localization Theorem (1.1) of (2)), which, after plodding through the quite intricate intricacies of the $$K$$-theory apparatus surrounding the Index Theorem, should grind out the desired answer (3), which is, in a sense a localized version of the Gauss-Bonnet Theorem.
What makes me unhappy with this is that the derivation of (1) up to stage
(3) is maybe on the level of a first year graduate course in Algebraic Topology,
wherea the application of the full force of the Index Theorem in (3) appears
like the proverbial cracking of a peanut with a sledgehammer. Now it is known that the Gauss-Bonnet Theorem is much more elementary than the Index Theorem
(which just illustrates it but is not needed for its proof), and also localization results come in various flavours, so my question is whether there exists a much more elementary proof of (1) somewhere in the world.
(1) Atiyah, M.F. & Singer, I.M.,
The Index of Elliptic Operators: I, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 484-530.
(2) Atiyah, M.F. & Segal, G.B.,
The Index of Elliptic Operators: II, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 531-545.
(3) Bryan, J. & Fulman, J.,
Orbifold Euler characteristics and the number of commuting n-tuples in symmetric groups,
(math/9712248) on $$mathtt{arxiv.org}$$
(4) Grothendieck, A.,
Sur quelques points d’algèbre homologique, II, Tohoku Math. J. (2)
Volume 9, Number 3 (1957), 119-221.
(5) Hirzebruch, F. & Höfer, T.,
On the Euler number of an orbifold, Math. Ann. 286, 255-260 (1990)
## Euler and mid point method to estimate stability of system of ODE’s
I want to study the stability of this system of differential equations using the 1st order Euler and mid-point method:
``````dx/dt = -x +y
dy/dt = -100*y
``````
Any suggestions?
Thanks!
## programming challenge – Project Euler Problem #85 in Scheme (Guile)
I’ve decided to revitalize my Scheme knowledge coming from SICP, so I’ve started using Scheme for Project Euler problems I’ve already solved in Python. However, this proved to be difficult. It seems that I’ve become too accustomed to the “straightforwardness” of Python, so I can’t wrap my head around the Scheme way of thinking.
Here’s the link for 85th Project Euler problem
And here’s my code:
``````(define (nearestarea recs)
(define upper 0)
(define (recnumber width length)
(define (combtwo n) (/ (* n (- n 1)) 2))
(* (combtwo (1+ width)) (combtwo (1+ length))))
(define (exists? rec)
(do ((w 1 (1+ w)))
((> (recnumber w 1) rec) 0)
(do ((l w (1+ l)))
((>= (recnumber w l) recs) (cons rec (* w l))))))
(define (iterate down up)
(let ((valup (exists? up))
(valdown (exists? down)))
(if (not (= valup 0))
(if (not (= valdown 0))
(if (> (- (car valup) recs) (- recs (car valdown))) (cdr valdown) (cdr valup))
(iterate (- down 1) up))
(iterate down (+ up 1)))))
(iterate (- recs 1) (+ recs 1)))
(nearestarea 2000000)
``````
So what I’m trying to do is
start from 2000000, iterate first the bigger and then the smaller numbers one at a time and see if it can be achieved using the classic combination method. I then record the rectangle number and the area of the grid it’s derived from as a pair, so in the end I can return the area of the grid that contains the “closest” amount of rectangles.
As a side note, this isn’t how I’ve originally solved the problem, I just wanted to try something different.
However my code just gets stuck in an infinite loop. As you can tell I’m not very proficient in Scheme (I’m new to programming in general), so I would appreciate your help.
## python 3.x – Project Euler 8 – Largest product in a series
The four adjacent digits in the 1000-digit number that have the
greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have
the greatest product. What is the value of this product?
This is my solution to the problem above.
``````> def largest_product_series(n, series):
> series = str(series)
> largest = 0
> for i in range(0,1000-n):
> temp = np.prod((int(series(j)) for j in range(i,n+i)))
> largest = max(temp, largest)
> return largest
``````
I am having a hard time figuring out what is wrong with my code. It works just fine with n = 4. But somehow it didn’t output the correct answer when n = 13.
Here’s the link to the problem. Euler 8
## ac.commutative algebra – Elliptic Units as Euler systems
I’m trying to understand elliptic units in order to work with the Euler systems of the ableism extensions of quadratic imaginary number fields. I’ve looked at few references about the topic, but they define elliptic units from different perspectives. Can anyone suggest a short and comprehensive definition of elliptic units. Thank you. | 4,139 | 15,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 37, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-17 | latest | en | 0.912333 |
https://la.mathworks.com/matlabcentral/cody/problems/611-surrounded-matrix/solutions/2827014 | 1,603,186,305,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107871231.19/warc/CC-MAIN-20201020080044-20201020110044-00063.warc.gz | 386,527,762 | 17,318 | Cody
# Problem 611. surrounded matrix
Solution 2827014
Submitted on 13 Aug 2020 by Yin-Lung Wong
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = ones(3); assert(isequal(surrounded(x),y_correct))
row = 1 col = 1 B = 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1
2 Pass
x = [3 3 3;3 3 3]; y_correct = [1 1 1 1 1;1 3 3 3 1; 1 3 3 3 1; 1 1 1 1 1]; assert(isequal(surrounded(x),y_correct))
row = 2 col = 3 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 3 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 3 3 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 3 3 1 1 3 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 3 3 1 1 3 3 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 3 3 3 1 1 3 3 3 1 1 1 1 1 1
3 Pass
x = ones(3,4); y_correct = ones(5,6); assert(isequal(surrounded(x),y_correct))
row = 3 col = 4 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
4 Pass
x = [0]; y_correct = ones(3); y_correct(2,2) = 0; assert(isequal(surrounded(x),y_correct))
row = 1 col = 1 B = 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 0 1 1 1 1
5 Pass
x = reshape(1:9,[3,3]); y_correct = [1 1 1 1 1; 1 1 2 3 1; ... 1 4 5 6 1; 1 7 8 9 1; 1 1 1 1 1]'; assert(isequal(surrounded(x),y_correct))
row = 3 col = 3 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 5 1 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 5 8 1 1 1 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 5 8 1 1 3 1 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 5 8 1 1 3 6 1 1 1 1 1 1 1 B = 1 1 1 1 1 1 1 4 7 1 1 2 5 8 1 1 3 6 9 1 1 1 1 1 1
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Start Hunting! | 2,099 | 2,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-45 | latest | en | 0.409637 |
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As given , answer is 6x + 6 It may be that what you actually mean is (x + 1)^6 ??????
As given , answer is 6x + 6 It may be that what you actually mean is (x + 1)^6 ??????
7 answers · Mathematics · 3 hours ago
• ### Trigonometry?
Because Trigonometry plays a significant part in subjects such as :- Electrical Engineering Electronics Mechanical Engineering Civil Engineering Gas Engineering Physics Finance and Insurance Aeronautics
Because Trigonometry plays a significant part in subjects such as :- Electrical Engineering Electronics Mechanical Engineering Civil Engineering Gas Engineering Physics Finance and Insurance Aeronautics
2 answers · Mathematics · 3 hours ago
• ### Integrate sin 2x?
\$ sin 2x dx = (-1/2) cos 2x + C
\$ sin 2x dx = (-1/2) cos 2x + C
3 answers · Mathematics · 8 hours ago
• ### What is the name of this mathematical identity, 2^2 = (x+1/x)^2 - (x-1/x)^2?
Difference of squares perhaps ?
Difference of squares perhaps ?
6 answers · Mathematics · 11 hours ago
• ### How do you combine this equation?
There is no equation to be seen. An equation requires an = sign.
There is no equation to be seen. An equation requires an = sign.
3 answers · Mathematics · 11 hours ago
• ### What is 1+0?
1 + 0 = 1
1 + 0 = 1
6 answers · Mathematics · 11 hours ago
• ### Solve initial value problem. dy/dx = x/(1+y) using y(0) = 1?
\$ 1 + y dy = \$ x dx y + y²/2 = x²/2 + C 1 + 1/2 = C = 3/2 y + y²/2 = x²/2 + C
\$ 1 + y dy = \$ x dx y + y²/2 = x²/2 + C 1 + 1/2 = C = 3/2 y + y²/2 = x²/2 + C
2 answers · Mathematics · 12 hours ago
• ### 10. What is the sum of 1/9, 2/3, and 5/18? A. 19/18 B. 12/9 C. 8/30 D. 4/15?
2/18 + 12/18 + 5/18 = 19/18 = 1 1/18
2/18 + 12/18 + 5/18 = 19/18 = 1 1/18
4 answers · Mathematics · 16 hours ago
• ### Simplify 25x⁻²y² / 5x⁻³y?
25 x^(-2) y² ------------------- = 5 x y 5 x^(-3) y
25 x^(-2) y² ------------------- = 5 x y 5 x^(-3) y
6 answers · Mathematics · 21 hours ago
• ### Solve the simultaneous equation { 2^x + 3^y = 41 2^x+2 + 3^y+2 = 209?
Will assume that question is incorrect and should be shown as :- 2^x + 3^y = 41 2^(x + 2) + 3^(y + 2) = 209 2^x + 3^y = 41 4 (2^x) + 9 (3^y) = 209 4 (2^x) + 4 (3^y) = 164 4 (2^x) + 9 (3^y) = 209 5 (3^y) = 45 3^y = 9 y = 2 2^x + 9 = 41 2^x = 32 x = 5 x = 5 , y = 2
Will assume that question is incorrect and should be shown as :- 2^x + 3^y = 41 2^(x + 2) + 3^(y + 2) = 209 2^x + 3^y = 41 4 (2^x) + 9 (3^y) = 209 4 (2^x) + 4 (3^y) = 164 4 (2^x) + 9 (3^y) = 209 5 (3^y) = 45 3^y = 9 y = 2 2^x + 9 = 41 2^x = 32 x = 5 x = 5 , y = 2
4 answers · Mathematics · 21 hours ago
• ### If a ball is thrown into the air with a velocity of 35 ft/s, its height (in feet) after t seconds is given by y = 35t − 16t2.?
y = 35 t - 16 t² dy/dt = 35 - 32 t v (1) = 35 - 32 = 3 ft/s
y = 35 t - 16 t² dy/dt = 35 - 32 t v (1) = 35 - 32 = 3 ft/s
3 answers · Mathematics · 21 hours ago
• ### Find the maximum value of the function f(f) = 5 - x - x^2 and value of x it occurs.?
f (x) = 5 - x - x² f ` (x) = - 1 - 2x = 0 for turning point x = -1/2 for turning point f (-1/2) = 5 + 1/2 - 1/4 = 5 1/4 is Max value when x = - 1/2
f (x) = 5 - x - x² f ` (x) = - 1 - 2x = 0 for turning point x = -1/2 for turning point f (-1/2) = 5 + 1/2 - 1/4 = 5 1/4 is Max value when x = - 1/2
3 answers · Mathematics · 21 hours ago
• ### How to simplify 25x^-2y^2/5x^-3y?
25 x^(-2) y^2 -------------------- = 5 x y 5 x^(-3) y
25 x^(-2) y^2 -------------------- = 5 x y 5 x^(-3) y
2 answers · Mathematics · 22 hours ago
• ### Math exponent help 8^5/3(8^-2/3)?
Presentation is open to doubt so having to take a guess at : - 8^(5/3) x 8^(-2/3) = 8^(3/3) = 8
Presentation is open to doubt so having to take a guess at : - 8^(5/3) x 8^(-2/3) = 8^(3/3) = 8
3 answers · Mathematics · 22 hours ago
• ### Find maximum and minimum values of 2+sinx?
Max value is 3 Minecraft value is 1
Max value is 3 Minecraft value is 1
3 answers · Mathematics · 23 hours ago
• ### Convert 0.26 to a fraction (do not reduce)?
26 ----- 100
26 ----- 100
6 answers · Mathematics · 1 day ago
• ### Convert 3/4 to a decimal?
3 / 4 = 0 • 75
3 / 4 = 0 • 75
4 answers · Mathematics · 1 day ago
• ### Reduce 20/40 to it’s lowest common denominator?
1 --- 2
1 --- 2
5 answers · Mathematics · 1 day ago
• ### What is the axis of symmetry for f(x) = −2x2 + 20x − 42?
y = -- 2 ( x² - 10 x ) - 42 y = -- 2 ( x² - 10 x + 25 ) + 50 - 42 y = - 2 ( x - 5 )² + 8 x = 5 is axis of symmetry
y = -- 2 ( x² - 10 x ) - 42 y = -- 2 ( x² - 10 x + 25 ) + 50 - 42 y = - 2 ( x - 5 )² + 8 x = 5 is axis of symmetry
2 answers · Mathematics · 1 day ago
• ### Why is x-1 not a factor of f(x) = 4x^3+8x^2+x-3?
Because f (1) does not equal zero.
Because f (1) does not equal zero.
5 answers · Mathematics · 1 day ago | 1,934 | 4,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-51 | longest | en | 0.755187 |
http://www.algebra.com/algebra/homework/Expressions-with-variables/Expressions-with-variables.faq?hide_answers=1&beginning=1845 | 1,369,541,128,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706624988/warc/CC-MAIN-20130516121704-00006-ip-10-60-113-184.ec2.internal.warc.gz | 320,520,431 | 11,710 | # Questions on Algebra: Expressions involving variables, substitution answered by real tutors!
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Algebra: Expressions involving variables, substitution Solvers Lessons Answers archive Quiz In Depth
Question 66199: -13 is the difference of a number multiplied by 5 and 8 Click here to see answer by checkley71(8403)
Question 66339: Please help with this problem solving by substitution: 3x + 4y = 8 4.5x + 6y = 12 This is what I got: 3x + 4y = 8 4y = 8 - 3x y = 2 - .75x Then: 3x + 4(2-.75x) = 8 3x + 8 - 3x = 8 3x - 3x = 8 - 8 0 = 0 I don't understand what I'm doing wrong. Please help! Thank you, Alison. Click here to see answer by stanbon(57422)
Question 66355: Words To Expressions : Directions: Change It Into An expression Easy Example: five minus three: 5-3=(dont solve) The product of three and two minus eight: Click here to see answer by checkley71(8403)
Question 66551: solve for x: 3x - 2y = 6 option of solutions that i need to pick from are: A) 2y + 6 divided by 3 B)3x - 6 divided by 2 C) 2 + 2y D) 2/3y + 6 how do you even solve this problem. Click here to see answer by Gen(11)
Question 66596: Celsius temperature readings can be converted to Fahrenheit readings using the formula F = 9/5 C + 32 What is the Fahrenheit temperature that corresponds to each of the following Celsius temperatures -10, 0, 15, 100? Click here to see answer by funmath(2925)
Question 66611: Compare equations with one variable to equations with two variables. How are they alike? How are they different?What if an equation had three variables? Click here to see answer by Poohbear272(22)
Question 66632: what if an equation had three variables what is the difference and similarieties between those equations that have one and/or two variable? Click here to see answer by Poohbear272(22)
Question 66625: what is the difference and similarities if solving an equation with varibles in the equation if it had 1 or 2 or even 3 variables? Click here to see answer by Poohbear272(22)
Question 66616: Subtract and write in simplest form: 12 10 ---- - ---- z z I believe the answer is: 2 --- z^2 Is this correct? Thank you Click here to see answer by funmath(2925)
Question 66685: y varies directly with n; m = 144 when n = 8. Click here to see answer by Nate(3500)
Question 66731: y=-2x-1 y=3x-16 -2x-1=3x-16 -1=5x-16 15=5x 3=x y=-2(3)-1 y=-6-5 y=-11 Click here to see answer by stanbon(57422)
Question 66729: What would the answer be for 12+2(k+3)=3k-6? Could you please show the work so i can understand it. Click here to see answer by Edwin McCravy(8921)
Question 66748: You decide to market your own custom computer system. You must invest \$5874 in computer hardware and spend \$4.85 to buy and package each disk. If each program sells for \$13.75, how many copies must you sell to break even? Here are the multiple choice answers: A. 660 copies B. 316 copies C. 315 copies D. 659 copies Click here to see answer by rapaljer(4667)
Question 66730: Suppose you try to solve systems of linear equations using substitution and get x=8.How many solutions does each system have? Click here to see answer by rapaljer(4667)
Question 66794: Am having troubles figuring out the 2 expressions to use in solving this problem: The sum of 2 number is 311. One quarter of the larger is four more than the smaller number. What are the two numbers. I know the first expression is x + y = 311 I can't figure out how to make the second expression to solve for both.(does that make sense?) Thank you for whatever help you can give. I'm most frustrated with this at the moment and I'm sure it's something simple I'm missing. Click here to see answer by ptaylor(2048)
Question 66805: BOOKS A LIBRARY HAS 14,588 BOOKS WHICH FILL ITS 313 EQUAL-SIZE SHELVES. NEW SHELVES OF WHAT IS THE AVERAGE NUMBER OF BOOKS PER SHELF? THE LIBRARY PLANS TO INSTALL 50 NEW SHELVES OF THIS SIZE. WRITE AND SOLVE AN EQUATION TO ESTIMATE HOW MANY MORE BOOKS THE LIBRARY WILL BE ABLE TO HOLD. Click here to see answer by stanbon(57422)
Question 66809: Would you please help me solve this question?: The constant slope of a snowpark for boarders can be modeled by the line passing through the point A(1,5) and Point B(7,1). What is the equation of the line that models the terrain of the snowpark? Express your answer in the form: y = mx + b where m and b are common fractions.. Click here to see answer by [email protected](15667)
Question 66830: Find three consecutive even numbers so that when I double the largest and then add the smallest, I get 116. Write an equation and solve. Trial and Error: 36,38,40 40+40=80 Largest Doubled 80+36=116 Largest Doubled + Smallest Click here to see answer by checkley71(8403)
Question 66830: Find three consecutive even numbers so that when I double the largest and then add the smallest, I get 116. Write an equation and solve. Trial and Error: 36,38,40 40+40=80 Largest Doubled 80+36=116 Largest Doubled + Smallest Click here to see answer by andrewl(1)
Question 66950: I have a question regarding the correct setup of this sentence. "The larger of 2 numbers is 8 more than 4 times the smaller. If the larger is increased by 4 times the smaller, the result is 40. Find the numbers!" Any help would be great!!! This question is not out of the textbook, it is from a hand out from the teacher. Click here to see answer by ptaylor(2048)
Question 66973: how would i simpligy this expression sqrt 16a^4x^2 Click here to see answer by stanbon(57422)
Question 67062: Al, Betty, and Claire, slipt \$1000 among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of \$1500. Betty and Claire have both doubled their money, whereas Al had managed to lose \$100. What was Al's original portion? Click here to see answer by [email protected](15667)
Question 67168: Mr.James wanted to plant a garden that would be in the shape of a rectangle . He was given 80ft. of fencing to enclose his garden. He wants the length to be 10ft. more than twice the width. What are the dimensions, in feet, for a rectangular garden that will use exactly 80ft. of fencing? Click here to see answer by Zoop(21)
Question 67342: Hi, I need help with the following: Thanks in advance! 1. Jeff makes 5 dollars per hour plus \$7.50 as a one time start fee. Write an expression for how much Jeff made after x hours. 2. A scuba diving hotel offers divers two plans. Plan A gives 3 nights' lodging and 4 dives. Plan B gives 5 nights' lodging and 8 dives. Let x be the number of dollars they charge per night and y be the number of dollars per dive. Write two expressions, one for the amount you would pay for Plan A and the other for Plan B. 3. The waitress at a restaurant makes \$32 per day and the cook makes \$50 per day. In addition they divide the tips received in such a way that the waitress gets 70% and the cook gets 30%. Let x be the number of dollars in tips in a day. Write an expression for the total amount of money (wages and tips) the waitress and cook make a day (one equation per person). Click here to see answer by ptaylor(2048)
Question 67581: Could you please help me with this problem. When a) x = 2, b) x = -3 5x + 2 --------------- 2x^2 + 11x + 12 This is what I tried.. 5(2)+ 2 __________ = 12 12 12 2(3)^2 + 11(-3) + 12 _________________ = _______________ = _____ = 4 18 - 33 + 12 -15 + 12 -3 Click here to see answer by funmath(2925)
Question 68200: Is 4 a solution to the equation x + 7 = 11? Click here to see answer by funmath(2925)
Question 68201: Which expression describes the sequence below, when 'n' is the position in the sequence? 3, 8, 13, 18, .... a) 5n + 3 b) 5n - 2 c) 2n + 1 d) n + 2 Click here to see answer by checkley71(8403) | 2,234 | 8,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2013-20 | latest | en | 0.89831 |
https://gforge.se/2014/09/an-exercise-in-non-linear-modeling/ | 1,716,375,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00586.warc.gz | 223,865,199 | 21,181 | # An exercise in non-linear modeling
Finding the right curve can be tricky. The image is CC by Martin Gommel.
In my previous post I wrote about the importance of age and why it is a good idea to try avoiding modeling it as a linear variable. In this post I will go through multiple options for (1) modeling non-linear effects in a linear regression setting, (2) benchmark the methods on a real dataset, and (3) look at how the non-linearities actually look. The post is based on the supplement in my article on age and health-related quality of life (HRQoL).
## Background
### What is linearity?
Wikipedia has an excellent explanation of linearity:
linearity refers to a mathematical relationship or function that can be graphically represented as a straight line
### Why do we assume linearity?
Linearity is a common assumption that is made when building a linear regression model. In a linear relation, a continuous variable has the same impact throughout the variable’s span. This makes the estimate is easy to interpret; an increase of one unit gives the corresponding coefficient’s change in the outcome. While this generates simple models with its advantages, it is difficult to believe that nature follows a simple straight line. With todays large data sets I believe that our models should permit non-linear effects.
### How do we do non-linearity?
There are plenty of non-linear alternatives that can be used to better find the actual relationships. Most of them rely on converting the single continuous variable into several. The simplest form is when we transform the variable into a polynomial, e.g. instead of having the model:
HRQoL ~ β0 + βage * Age + βsex * Sex
We expand age to also include the age squared:
HRQoL ~ β0 + βage * Age + βage” * Age2 + βsex * Sex
This allows for the line for a bend, unfortunately as we add the squared term the coefficients are more difficult to interpret, and after adding a cubic term, i.e. Age3, it is almost impossible to interpret the coefficients. Due to this difficulty in interpretation I either use the `rms::contrast` function or the `stats::predict` in order to illustrate how the variable behaves.
#### Splines
A frequently used alternative to polynomials are splines. The most basic form of a spline consists of lines that are connected at different “knots”. For instance, a linear spline with 1 knot can assume a V-shape, while 2 knots allow for an N-shaped relationship. The number of knots decide the flexibility of a spline, i.e. more knots allow a more detailed description of the relationship.
## The models
The dataset comes from the Swedish Hip Arthroplasty Register‘s excellent PROM database and consists of more than 30,000 patients that have filled out the EQ-5D form both before and after surgery. We will focus on age and its impact on the EQ-5D index and the EQ-VAS one year after total hip replacement surgery. We will control for sex, Charnley class, pre-operative HRQoL, pre-operative pain.
### The restricted cubic spline
I’m a big fan of Frank Harrell‘s rms-package so we will start there although we start by splitting the dataset using the `caret::createDataPartition`. We then need to set the `rms::datadist` in order for the rms-functions to work as expected:
```library(caret)
train_no```
Frank Harrell is a proponent of restricted cubic splines, alias natural cubic splines. This is a type of spline that uses cubic terms in the center of the data and restricts the ends to a straight line, preventing the center from distorting the ends, i.e. curling. His `rcs()` also nicely integrates with the `anova` in order to check if non-linearity actually exists:
`idx_model`
gives:
```Analysis of Variance Response: eq5d1
Factor F d.f. P
Age 140.71 2```
As we can see the model seems OK for the EQ-5D index, supporting both the non-linearity and the interaction between Charnley class and sex. If we for some reason cannot use the rms-package we can check for linearity by using the `splines::ns` with two regular models as suggested below:
`lm1`
gives:
```Analysis of Variance Table
Model 1: eq5d1 ~ Age + Sex * Charnley_class + ns(eq5d0, 3) + ns(pain0,
3)
Model 2: eq5d1 ~ ns(Age, 3) + Sex * Charnley_class + ns(eq5d0, 3) + ns(pain0,
3)
Res.Df RSS Df Sum of Sq F Pr(>F)
1 19095 193.01
2 19093 192.66 2 0.35112 17.398 2.825e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
```
In order to avoid overfitting we will try to select models based upon the AIC/BIC criteria. The selection is simply finding the lowest value where in general AIC allows slightly more complex models compared to BIC. We will start with finding the optimal number of knots for the EQ-5D index using the AIC:
```#' A simple function for updating the formula and extracting
#' the information criteria
#'
#' @param no A number that is used together with the add_var_str
#' @param fit A regression fit that is used for the update
#' @param rm_var The variable that is to be substituted
#' @param add_var_str A sprintf() string that accepts the no
#' parameter for each update
#' @param ic_fn The information criteria function (AIC/BIC)
getInfCrit```
It can be discussed if the model should stop at 3 knots but I chose to continue a little higher as the drop was relatively large between the 4 and 5 knots. This is most likely due to a unfortunate fit for the 4 knots. We could also have motivated a larger number of knots but even with proper visualization these are difficult to interpret. When modeling confounders, such as the preoperative EQ-5D index (eq5d0) and the pre-operative pain (pain0), we may prefer a higher number of knots in order to avoid any residual confounding and we do not need to worry about visualizing/explaining the relations.
Now if we apply the same methodology to the EQ-VAS we get:
`vas_model`
Now lets do the same for the BIC:
`idx_model`
### B-splines
B-splines, alias Basis spline, are common alternatives to restricted cubic splines that also use knots to control for flexibility. As these are not restricted at the ends they have more flexible tails than restricted cubic splines. In order to get a comparison we will use the same model for the other variables:
```# Use the same setting model as used in the RCS
vars```
### Polynomials
Polynomials can be of varying degrees and have often been argued as a simple approach to fitting a more flexible non-linear relationship. As the majority of the patients are located around the mean age, 69.1 years, it is important to remember that these patients will have the strongest influence on the curve appearance. It is therefore possible that the tails are less reliable than the central portion.
```# Use the same setting model as used in the RCS
vars```
### Multiple Fractional Polynomial Models
Multiple fractional polynomials (MFP) have been proposed as an alternative to splines. These use a defined set of exponential transformations of the variable, where it omits predictors according to their p-values. The `mfp` has a built-in algorithm and we don’t need to rely on either BIC or AIC with MFP.
```library(mfp)
# We will use the simple BIC models and
# instead of rcs() that is not available
# for mfp we use ns() from the splines package
fits\$eq5d[["MFP"]]```
Generalized additive model (GAM) are an extension to generalized linear models and specializes in non-linear relationships. Elements of statistical learning by Hastie et. al. is an excellent source for diving deeper into these. The simplest way to explain the GAM smoothers is that they penalize the flexibility in order to avoid over-fitting, there plenty of options – the ones used here are:
Thin plate regression splines: This is generally the most common type of smoother in GAM models. The name refers to the physical analogy of bending a thin sheet of metal.
Cubic regression splines: The basis for the spline is cubic with evenly spread knots.
P-splines: P-splines are similar to B-splines in that they share basis with the main difference that P-splines enforce a penalty on the coefficients.
```library(mgcv)
fits\$eq5d[["GAM thin plate"]]```
## Benchmark
With all these fancy models we will first try to evaluate how they perform when cross-validated and then on our test-set that we’ve set apart at the start. We will evaluate according to root-mean-square error (RMSE) and mean absolute percentage error (MAPE). RMSE tends to penalize for having outliers while the MAPE is more descriptive of the performance on average. Our testing functions will therefore be:
`rmse`
Furthermore in this particular article I wanted to look into what happens at the tails as almost 70 % of the patients were between 60 and 80 years while the variable span was 40 to 100 years. I therefore defined a central vs peripheral portion where the central portion was defined as being between the 15:th and 85:th percentile.
The cross-validation was a straight forward implementation. As this can take a little time it is useful to think about parallelization, we will here use the `parallel`-package although the `foreach` is also an excellent option.
```crossValidateInParallel= quantile(Age, probs=tail_prop) &
Age <= quantile(Age, probs=1-tail_prop))
df\$cv```
If we run all the models we get the following result:
RMSEMAPE
MethodMainCentralPeripheralMainCentralPeripheral
The EQ-5D index
Restricted cubic splines
AIC0.1000.0980.1058.838.609.42
BIC0.1000.0990.1058.868.639.46
B-splines
AIC0.1000.0990.1058.858.809.48
BIC0.1010.0990.1058.918.799.48
Polynomials
AIC0.1030.1050.1178.949.0910.21
BIC0.1030.1050.1168.969.0710.09
MFP0.1010.0990.1058.878.649.43
Thin plate0.1000.0980.1058.868.629.47
Cubic regression0.1000.0980.1058.868.629.47
P-splines0.1000.0980.1058.868.629.47
The EQ VAS
Restricted cubic splines
AIC18.5418.4918.6619.018.719.7
BIC18.5418.4918.6719.018.719.7
B-splines
AIC18.5418.6618.6519.019.119.7
BIC18.5518.6618.6919.119.019.7
Polynomials
AIC19.2720.0621.8119.520.122.9
BIC19.2419.9621.6719.520.022.6
MFP18.5518.5018.6919.018.719.7
Thin plate18.5418.4818.6719.018.719.7
Cubic regression18.5418.4818.6719.018.719.7
P-splines18.5418.4918.6719.018.719.7
The difference between is almost negligable although the polynomial is clearly not performing as well as the other methods. This was though less pronounced in the testset where the polynomial had some issues with the tails:
RMSEMAPE
MethodMainCentralPeripheralMainCentralPeripheral
The EQ-5D index
Restricted cubic splines
AIC0.1000.0980.1048.758.569.23
BIC0.1000.0990.1048.798.609.27
B-splines
AIC0.1000.0990.1048.778.779.30
BIC0.1000.0990.1058.858.789.34
Polynomials
AIC0.1000.0990.1068.738.669.18
BIC0.1010.0990.1068.788.699.15
MFP0.1000.0990.1048.798.619.25
Thin plate0.1000.0980.1058.798.599.29
Cubic regression0.1000.0980.1048.798.609.27
P-splines0.1000.0990.1048.798.609.28
The EQ VAS
Restricted cubic splines
AIC18.7018.5019.1819.118.720.1
BIC18.7118.5119.1919.118.820.2
B-splines
AIC18.7018.7119.1819.219.220.2
BIC18.7118.7119.1719.219.120.2
Polynomials
AIC18.7418.8019.6719.119.020.4
BIC18.7418.7519.6419.119.020.4
MFP18.7218.5219.1919.218.820.1
Thin plate18.6918.5019.1719.118.720.1
Cubic regression18.6918.5019.1719.118.720.1
P-splines18.6918.5019.1719.118.720.1
```cv_results= quantile(Age, probs=.15) &
Age <= quantile(Age, probs=.85))
testset_results```
## Graphs
Now lets look at how the relations found actually look. In order to quickly style all graphs I use a common setup:
```# The ggplot look and feel
my_theme [c language="(2,1,3)"][/c]
<pre lang="RSPLUS">gg_eq5d```
It is important to remember that different algorithms will find different optima and may therefore seem different to the eye even though they fit the data equally well. I think of it as a form of skeleton that defines how it can move, it will try to adapt the best it can but within its boundaries. We can for instance bend our elbow but not the forearm (unless you need my surgical skill).
### Polynomials
Note that the y-scale differs for the polynomials
## Summary
I hope you enjoyed the post and found something useful. If there is some model that you lack, write up some code and I can see if it runs. Note that there are differences from the original supplement that used a slightly more complex setup for choosing the number of knots and a slightly different dataset.
## Playing around with non-linearity
A friend just told me about amazing tool for playing around with splines, knots, etc that the Paul Lambert at the University of Leicester has created. I strongly recommend a visit to their site.
## 6 thoughts on “An exercise in non-linear modeling”
1. Hi,
Thank you for such an excellent article. I have one question.
For linear Regression Model the assumption is that the model is linear in the parameters (B0,B1) rather than in the predictors. Can you elaborate on this.
Regards,
Amit
• Thanks! As an orthopaedic surgeon I’m probably not the best answer your question. There is an excellent Wikipedia article on the subject. You can also try CrossValidated, there best question on the subject that I’ve found is this one. Also have a look at John Fox’ nls appendix that contains some useful information.
Going back, the Wikipedia example is rather nice. If we translate it to the age and HRQoL setting we can see that βage*age may not be a good assumption. Although using the same transformation, βage 1*age/(βage 2 + age), makes little sense for age and HRQoL. We may instead try a model that more looks like βage 1*age-βage 2 * abs(age-βage 3) where we expect the βage 2 to be negative thus allowing for a ^ form where we expect a maximum and where the βage 3 may indicate where this occurs. The absolute value could of course be exchanged with a squared that may in turn provide a different estimate.
In my mind it seems that the nonlinear regression is best suited when you have simple relations defined by laws of nature such as the Michaelis–Menten kinetics. I believe that when working with more complex outcomes defining a sensible model is close to impossible although this could potentially be an alternative when you are trying to pinpoint the peak according to the previous formula.
I hope this gives some clarity, although keep in mind that I’m not a schooled statisticians and nonlinear regressions have never been a subject that I’ve been officially taught 😀
2. Hi Max
Thanks for sharing the code.
I was wondering if you could add some code for data simulation, the way one can pass it through the code and see how it works.
Many thanks beforehand.
Bye
• Not sure of what you intend, I did have a question on CrossValidated where I simulate a non-linear dataset.
• Thanks Max,
I just simply need the data.
I know you can not share your real data, what I ask is simulated data close to your original data.
All the best. | 3,916 | 14,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.923608 |
https://textbooks.cs.ksu.edu/cis301/9-chapter/9_2-recursion/ | 1,721,766,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00169.warc.gz | 484,229,439 | 9,534 | # Recursion
In this section, we will see how to prove the correctness of programs that use recursive functions. We will see that verifying a recursive function is exactly the same as verifying a non-recursive function:
• We must prove a function’s preconditions before calling it (including before making a recursive call)
• After calling a function, we can list the function’s postconditions as premises (including after making a recursive call)
• The function can list its preconditions as premises
• The function must prove its postconditions just before it ends
## Writing a recursive mult function
We know we can multiply two numbers, `x` and `y`, using the `*` operator – `x * y`. But what if we wanted to find the same result using only addition, not multiplication? Multiplication can be thought of as repeated addition – `x * y` is really `x + x + ... + x`, where we add together `y` total `x`’s.
We could do this repeated addition with a loop (and we will when we introduce loops in section 9.3), but we will use recursion instead. When we write a recursive function, we try to think of two things:
• The base case: the simplest version of the problem that we could immediately solve with no more work.
• The recursive case: bigger versions of the problem, where we solve a piece of the problem and then recursively solve a smaller piece
In the case of the multiplication `x * y`, we have:
• Base case: if `y` is 0, we have no work to do. Adding together 0 `x`’s is just 0.
• Recursive case: if `y` is bigger than 0, we do ONE addition (`x + ...`) and recursively add the remaining `y - 1` numbers. (This will become our recursive call.)
With those cases in mind, we can write a recursive `mult` function:
``````// #Sireum #Logika
//@Logika: --manual --background save
import org.sireum._
import org.sireum.justification._
def mult(x: Z, y: Z): Z = {
var ans: Z = 0
if (y > 0) {
ans = mult(x, y-1)
ans = ans + x
} else {
//do nothing
}
return ans
}``````
Note that we separated the recursive call (`ans = mult(x, y-1)`) from adding on the next piece (`ans = ans + x`). When using Logika, all function calls must go on a separate line by themselves – we can’t combine them with other operations. Also, we included a dummy “else” branch to make the verification simpler.
## Walking through mult
Suppose we call `mult` as follows:
``var times: Z = mult(4, 2)``
We can trace the recursive calls:
``````times = mult(4, 2)
(x = 4, y = 2)
ans = mult(4, 1) => mult(4, 1)
ans = ans + 4 (x = 4, y = 1)
returns ans ans = mult(4, 0) => mult(4, 0)
ans = ans + 4 (x = 4, y = 0)
returns ans ans = 0
returns 0``````
We start with `mult(4, 2)`, and then immediately make the recursive call `mult(4, 1)`, which immediately makes the recursive call `mult(4, 0)`. That function instance hits the base case and returns 0. We now return back up the chain of function calls – the 0 gets returned back to the `mult(4, 1)` instance, which adds 4 and then returns 4:
``````=> mult(4, 1)
(x = 4, y = 1)
ans = mult(4, 0) = 0
ans = ans + 4 = 4
returns ans (4)``````
This 4 returns back to the `mult(4, 2)` instance, which adds another 4 and returns 8:
``````mult(4, 2)
(x = 4, y = 2)
ans = mult(4, 1) = 4
ans = ans + 4 = 8
returns ans (8)``````
We have now backed our way up the chain – the 8 is returned back from the original function call, and `times` is set to 8.
## mult function contract
Looking at our `mult` function, we see that the base case is when `y` is 0 and the recursive case is when `y > 0`. Clearly, the function is not intended to work for negative values of `y`. This will be our precondition – that `y` must be greater than or equal to 0.
Our postcondition should describe what `mult` is returning in terms of its parameters. In this case, we know that `mult` is performing a multiplication of `x` and `y` using repeated addition. So, our function should ensure that it returns `x*y` (that `Res[Z] == x*y`). Here is the function with the function contract:
``````// #Sireum #Logika
//@Logika: --manual --background save
import org.sireum._
import org.sireum.justification._
def mult(x: Z, y: Z): Z = {
//we still need to add the verification logic blocks
Contract(
Requires( y >= 0 ),
Ensures( Res[Z] == x * y )
)
var ans: Z = 0
if (y > 0) {
ans = mult(x, y-1)
ans = ans + x
} else {
//do nothing
}
return ans
}``````
## Verification in mult
Now that we have our function contract for `mult`, we must add logic blocks with two things in mind:
• Proving the precondition before a recursive call
• Proving the postcondition before we return from the function
Our recursive call looks like:
``ans = mult(x, y-1)``
Since our precondition is `y >= 0`, we see that we must prove that what we are passing as the second parameter (`y-1`, in the case of the recursive call) is greater than or equal to 0. This tells us that before our recursive call, we must have shown exactly: `y-1 >= 0`. We can finish proving the precondition as follows:
``````// #Sireum #Logika
//@Logika: --manual --background save
import org.sireum._
import org.sireum.justification._
def mult(x: Z, y: Z): Z = {
//we still need to prove the postcondition
Contract(
Requires( y >= 0 ),
Ensures( Res[Z] == x * y )
)
var ans: Z = 0
if (y > 0) {
Deduce(
1 ( y > 0 ) by Premise, //IF condition is true
2 ( y-1 >= 0 ) by Algebra*(1) //Proves the precondition for the recursive call
)
ans = mult(x, y-1)
ans = ans + x
} else {
//do nothing
}
return ans
}``````
All that remains is to prove the `mult` postcondition – that we are returning `x*y`. Since we are returning the variable `ans`, then we must prove the claim `ans == x*y` just before our return statement. In order to help with this process, we will need to take advantage of the postcondition after our recursive call. The function promises to return the first parameter times the second parameter, so when we do `ans = mult(x, y-1)`, we know that `ans == x*(y-1)` (the first parameter, `x`, times the second parameter, `y-1`). Here is the completed verification:
``````// #Sireum #Logika
//@Logika: --manual --background save
import org.sireum._
import org.sireum.justification._
def mult(x: Z, y: Z): Z = {
//verification complete!
Contract(
Requires( y >= 0 ),
Ensures( Res[Z] == x * y )
)
var ans: Z = 0
if (y > 0) {
Deduce(
1 ( y > 0 ) by Premise, //IF condition is true
2 ( y-1 >= 0 ) by Algebra*(1) //Proves the precondition for the recursive call
)
ans = mult(x, y-1)
Deduce(
1 ( ans == x * (y - 1) ) by Premise, //Postcondition from the recursive call
2 ( ans == x * y - x ) by Algebra*(1)
)
ans = ans + x
Deduce(
1 ( Old(ans) == x * y - x ) by Premise, //Pulled from previous block
2 ( ans == Old(ans) + x ) by Premise, //From the "ans = ans + x" assignment statement
3 ( ans == x + x * y - x ) by Algebra*(1, 2),
4 ( ans == x * y ) by Algebra*(3) //Showed the postcondition for the IF branch
)
} else {
//do nothing in code - but we still do verification
//need to show that postcondition will be correct even if we take this branch
Deduce(
1 ( ¬(y > 0) ) by Premise, //if condition is false
2 ( y >= 0 ) by Premise, //precondition
3 ( y == 0 ) by Algebra*(1, 2),
4 ( ans == 0 ) by Premise, //ans is unchanged
5 ( ans == x * y ) by Algebra*(3, 4) //Showed the postcondition for the ELSE branch
)
}
//Tie together what we learned in both branches
Deduce(
1 ( ans == x*y ) by Premise //shows the postcondition
)
return ans
}``````
## Verification of calling code
Verifying the test code that calls a recursive function works exactly the same way as it does for any other function:
• We must prove the precondition before calling the function
• We can list the postcondition as a premise after calling the function
Suppose we want to test `mult` as follows:
``````val times: Z = mult(4, 2)
assert(times == 8)``````
We could complete the verification by proving the precondition and then using the postcondition to help us prove the claim in the assert:
``````Deduce(
1 ( 2 >= 0 ) by Algebra*() //proves the precondition
)
val times: Z = mult(4, 2)
Deduce(
1 ( times == 4*2 ) by Premise, //mult postcondition
2 ( times == 8 ) by Algebra*(1) //needed for the assert
)
assert(times == 8)``````
Note that since our second parameter is `2`, that we must demonstrate exactly `2 >= 0` to satisfy `mult`’s precondition. Furthermore, since `mult` promises to return the first parameter times the second parameter, and since we are storing the result of the function call in the `times` variable, then we can claim `times == 4*2` as a premise. | 2,544 | 8,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.855656 |
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# Space expanded 6.8 times FASTER than Light!
page: 1
6
share:
posted on Sep, 23 2011 @ 12:04 PM
So last night I was googling to see how big the observable universe was and something
confused me. How do we see 93 Billion light years (46.5 Billion light years in all directions) away,
when the earth is only 4.5 Billion years old? Then I astounded myself!
Below is a copy and paste of a txt file I created last night. (I've got a damaged right screen and only the left half to work with; So sorry if "sentences" look short).
Age of universe = 13.7 Billion years old
Diameter of the universe=93 Billion light years
Edge of observable universe (radius)=46.5 Billion light years
So according to (my) logic it seems that the universe expanded at approximately
6.8 times faster than the speed of light!
I thought it was impossible to travel faster than C.
Since light travels (to nearest tril) 6 TRILLION Miles a year, and the
universe is only 13.7 billion years old that would mean the universe should
"Only" be: (Since nothing can travel/expand quicker than C)
6 Trillion x 13.7 Billion = 82200000000000000000000 or 82.2 Trillion Billion miles big
But if the diameter of the universe is 93 Billion light years then that would
mean, Since light travels (to nearest tril) 6 TRILLION Miles a year, and
its done it 93 billion times, then the universe would be:
6 Trillion x 93 Billion = F**k me 558 TRILLION BILLION miles big.
There are obviously HUGE problems with these scientific "Facts" and one
of the below must be true.
1. Either the universe is 6.8 times older than 13.7 Billion years old = 93 Billion years old
2. The universe isnt 93 Billion light years across, its 6.8 times smaller = 13.7
Billion light years across.
3. In fact the universe IS both 13.7 Billion years old and 93 Billion light years
wide; but it expanded 6.8 times faster than C. And Einstein was wrong.
4. Light travels 6.8 times faster than we thought.
Since also the edge of our observable universe is 46.5 Billion light years away, and
our earth is only 4.54 billion years old, how is it possible to be observable?
Its light shouldn't reach us for another 41.96 Billion years!
I honestly think either we're not being told something, or science just
let's on they know more than they actually do. As it seems everything
we thought we knew gets proven false time and time again.
PS: Don't expect to many (If any) replies from me as I have to use my On Screen
Keyboard lol. My usb ports are broken and don't accept my external ones.
And as a result I haven't logged on to ATS much (Roll on January *New Lappy*).
posted on Sep, 23 2011 @ 12:08 PM
Dark matter which pushes the universe apart faster than the speed of light.
posted on Sep, 23 2011 @ 12:10 PM
The speed of light is WITHIN our universe as is our physics, outside our universe these physics don't exist...
The universe expanded into this nothingness faster than light because of this.
posted on Sep, 23 2011 @ 12:12 PM
only a thought, the universe expanded into the nothing, so at the expanding "border" one side is universe the other one is nothing, so its not really like travelling at the speed of light,
its expanding into nothing, with no "resistance" or opposition, nothing to slow it down.
posted on Sep, 23 2011 @ 12:13 PM
Originally posted by roughycannon
The speed of light is WITHIN our universe as is our physics, outside our universe these physics don't exist...
The universe expanded into this nothingness faster than light because of this.
If that is so, i wonder what kind of physics exist outside the universe, and what would happen if we took a "step out" so to speak.
posted on Sep, 23 2011 @ 12:27 PM
We probably couldn't exist outside our universe, if there are other universes like the "bubbles" theory its more than likely that the other universes have different physics than ours.
I dunno what would happen if we stepped into one of these universes, we might melt or explode god knows....
posted on Sep, 23 2011 @ 12:31 PM
Is it at all possible that 'our physics' doesn't exist outside of our understanding of physical experience on this planet? As this is our only experience and what we are told about the moons surface and the artificial environment of space vessels, what exactly do we know?
We can observe 'outer space' and make predictions and propose theories about our observations and such but we don't know enough about the universe to assume that we know the physics of it. Our scientists are often wrong and must come up with another theory.
It's mind blowing to consider the universe, even to consider the 'void' outside of it is just wow! Spookey
Namaste
posted on Sep, 23 2011 @ 12:34 PM
I came across this little bit of inconvenient truth a few years back and nobody could answer it with any sort of credibility. Don't believe what people tell you that because there was nothing it could expand faster than possible and is still speeding up..Complete bull as far as I'm concerned, nobody knows what or where Dark Matter/Energy came from or even if it exists so it falls into the string theory category in my opinion. I think It has to due with our lost civilizations and the keeping of knowledge from general public.
The best working theory I have so far is that we have been created and destroyed many times over like most ancient texts say and the way we date things is so far from being accurate that everyday new findings disprove the general consensus of how things work. Basically the Universe we live in isn't limited to us and our petty shot in the dark calculations. If it expanded 6 times faster than light, then the universe is probably 6 times older than we think but nobody is going to open that can of worms publicly so we're stuck with mainstreams psuedo scientific analysis.
Also I am aware that they are saying sub atomic particles can travel faster than light. But that doesn't mean 6 times faster than light and there is little to no mass of said particles but the universe is Massive to say the least.
posted on Sep, 23 2011 @ 12:38 PM
Well Einstein was wrong it seems, at least if the results from CERN can be verified and duplicated. But that opens up a whole new can of worms... great post OP
posted on Sep, 23 2011 @ 12:39 PM
So what is this nothingness outside of the "known" universe ?
posted on Sep, 23 2011 @ 12:44 PM
Good. It was comforting to read this part lol. Only because before I made it to that part, I was hoping that you were still slightly confused, because you sure as hell confused me.
How do we see 93 Billion light years (46.5 Billion light years in all directions) away, when the earth is only 4.5 Billion years old? Then I astounded myself!
Since also the edge of our observable universe is 46.5 Billion light years away, and our earth is only 4.54 billion years old, how is it possible to be observable?
Regarding the initial thought that put you onto this slippery slope of confusion, I think you are trying too hard. To me, there seems to be a very simple explanation for this.
We may not have been here, when the light left its source of origin (wherever that may be), but it very well could have already begun that extremely long, and unfathomable journey.
Ok, it seemed simple when it first popped into my head, and I still think that it is a simple concept. It just isn't easy to say it lol.
ok here. I'll try this. A timeline (kinda)
[color=FF7D45]Original Source
[color=FFC06E]Light has left this source
[color=FFD98C]→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . ↑↑. .. . . . . . . . . . . . . . ↑↑
The light, was already well on it's merry way, coming this general direction, before we were even here. Once Mother Earth claimed this location for herself, it was too late for the light to alter its path. So it just keeps heading this way.
P.S. If none of the previous post makes sense, all of the blame should be placed on Chipkin. His fault for confusing me.
edit on 9/23/11 by BrokenCircles because: tweakin it a bit
posted on Sep, 23 2011 @ 12:45 PM
If we question the speed of light ( imho always question everything)
and it isn't the ultimate speed in the universe.
Then distance can be questioned. As its calculated by speed.
Are the stars closer than we have been taught to believe?
This is what mankind excells at.
expanding understanding,
and expanding........
posted on Sep, 23 2011 @ 12:46 PM
Nothing. If you're asking what it's called but I assume you already know that so maybe you're asking what is nothing well, nothing is the absence of something. So, the nothingness you were referring to is the absence of universe.
Cheers
posted on Sep, 23 2011 @ 12:49 PM
First of all what you have said is speculation not theory.
I think your basic understanding of physics and the universe is a bit off.
Our universe can expand faster than light... because the laws of physics govern what INSIDE our universe and the not the universe itself
Imagine the universe is a balloon, I blow up the balloon with air, imagine the air is physics and the laws of physics is inside this balloon, now I can stretch pull and squeeze this balloon without effecting the physics inside, this is what's happening our universe is expanding and stretching faster than light but its not effecting what's inside.
posted on Sep, 23 2011 @ 01:12 PM
Originally posted by roughycannon
First of all what you have said is speculation not theory.
I think your basic understanding of physics and the universe is a bit off.
Our universe can expand faster than light... because the laws of physics govern what INSIDE our universe and the not the universe itself
Imagine the universe is a balloon, I blow up the balloon with air, imagine the air is physics and the laws of physics is inside this balloon, now I can stretch pull and squeeze this balloon without effecting the physics inside, this is what's happening our universe is expanding and stretching faster than light but its not effecting what's inside.
First of all,
where do I start.....
You ridicule speculation and your response is ;
".I THINK your basic understanding of physics and the universe is a bit off."
Then you categorically state that "Our universe can expand faster than light... because the laws of physics govern what INSIDE our universe and the not the universe itself"
Describe what the inside of our universe differs from the universe itself ?????
You also use the word 'Imagine' twice, in your hypothesis.
Imagine your understanding and the sum of your knowledge is a balloon. Effecting is different from affecting.
Pop.
posted on Sep, 23 2011 @ 01:15 PM
What I question about your theory is.... Is Earth at the exact center of the Universe? Are we back to everything revolves around us? I think the statement "observable universe" says it all.... We only have the capability to look so far in any given direction... That doesn't mean that the universe stops there it just means that is as far as we can see... For all we know we could be on a trailing arm of our universe just like we are on a trailing arm of our galaxy. We might very well only be able to see a microscopic portion of our universe in comparison to how big it actually is.
I know that scientist have built models of the universe (at least of what they think they know of the universe), but we find out all the time that what we thought we knew was wrong. Before yesterday NOTHING could go faster than light, now we know that isn't true.
I find the odds against us being the exact center of the universe just too great... This alone tells me the distances you use as a base might be accurate but in all actuality we will find out tomorrow that they are light years from from it. Who can with out a doubt say they know the exact size of the universe, yet alone how fast it is expanding? When the cornerstones of physics are still being challenged and overcome I can't believe anyone who says they can.
Just my 2 cents, Great post by the way... I love it when new ideas get the wheels turning.
posted on Sep, 23 2011 @ 01:25 PM
There are two issues here that need to be addressed.
First: the current size of the universe
The current estimated diameter of the universe is 93.2 billion light-years. However, because of the expansion of the universe, the objects that are now at the "edge," 46.6 billion light-years away, were only 13.7 billion light-years away when the light from them was emitted. In that time, they have moved further out by 32.9 billion light-years, but that doesn't matter, because the light has only had to travel that original 13.7 billion light-years to get to us. It takes light 13.7 billion years to travel that distance, so that light was emitted 13.7 billion years ago... the current age of the observable universe.
Second: Relativity and the metric expansion of space
This is the other problem people often run into. They think that the expansion of space is actually moving things... that it's causing objects far away from us to "explode" away faster and faster. But, this is not the case. In fact, the expansion of space is a metric expansion, which means that it's space, itself, that's expanding... it has nothing to do with objects moving. For instance take any random distant type Ia supernova. The supernova is not moving away from us, the space in between us and the supernova is stretching. It's also this stretching which causes the wavelength of the light emitted by that supernova to stretch as well... and this stretching is called redshift. The greater the redshift, the further away an object is, because the more space (and, thus, the wavelength of the light) has been stretched in the time the light has been travelling.
posted on Sep, 23 2011 @ 01:26 PM
Originally posted by BrokenCircles
Good. It was comforting to read this part lol. Only because before I made it to that part, I was hoping that you were still slightly confused, because you sure as hell confused me.
How do we see 93 Billion light years (46.5 Billion light years in all directions) away, when the earth is only 4.5 Billion years old? Then I astounded myself!
Since also the edge of our observable universe is 46.5 Billion light years away, and our earth is only 4.54 billion years old, how is it possible to be observable?
Regarding the initial thought that put you onto this slippery slope of confusion, I think you are trying too hard. To me, there seems to be a very simple explanation for this.
We may not have been here, when the light left its source of origin (wherever that may be), but it very well could have already begun that extremely long, and unfathomable journey.
Ok, it seemed simple when it first popped into my head, and I still think that it is a simple concept. It just isn't easy to say it lol.
ok here. I'll try this. A timeline (kinda)
[color=FF7D45]Original Source
[color=FFC06E]Light has left this source
[color=FFD98C] →→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . ↑↑. .. . . . . . . . . . . . . . . . . ↑↑
The light, was already well on it's merry way, coming this general direction, before we were even here. Once Mother Earth claimed this location for herself, it was too late for the light to alter its path. So it just keeps heading this way.
P.S. If none of the previous post makes sense, all of the blame should be placed on Chipkin. His fault for confusing me.
edit on 9/23/11 by BrokenCircles because: tweakin it a bit
The best possible answer so far!
But it still seems impossible to observe light longer than 4.6 Billion light years away, because since earth is only 4.6 Billion years old then that means we can't possibly observe objects in the universe; that are farther out than 4.6 Billion light years.
And to complicate maters the worlds first telescope was only made 400 yrs ago!
Meaning we should be only able to observe 400yrs x 6 trillion miles = 2400 trillion miles of the universe (1200 tril in all directions).
Thats a MASSIVE difference from the 82.2 Trillion Billion or 558 Trillion Billion Miles we are currently thought to be able to observe!!!
Someone Spotaniely Combusted in Ireland this week, I think my brain is next on the cards
posted on Sep, 23 2011 @ 01:27 PM
Maybe we need to come to grips with the fact that WE DON'T KNOW EVERYTHING ABOUT THE UNIVERSE. Our math could be wrong. Maybe there are times when the speed of light is not an unbreachable constant. Hell, dark matter and dark energies are only theories in order for the ACCEPTED mainstream scientific to have their math work. They are not observable just calculated. The math could be wrong. The physics could be off.
Astrophysics is challenged nearly everyday with some new observation that surprises it by being far different from an equation on a blackboard.
posted on Sep, 23 2011 @ 02:29 PM
You ridicule speculation
I didn't ridicule speculation where did you get that from? I was simply saying your idea was speculation and not theory, a theory is something that's has physical or mathematical evidence behind it and is often peer reviewed before being considered a theory.
Heres an example of speculation:
I believe that there is intelligent life on other planets in the universe.
At no point did I ridicule anything.
".I THINK your basic understanding of physics and the universe is a bit off."
What does the fact that I think that have to do with anything?
Then you categorically state that "Our universe can expand faster than light... because the laws of physics govern what INSIDE our universe and the not the universe itself" Describe what the inside of our universe differs from the universe itself ?????
Because you are thinking of the universe as being an object it not, its a word we use to describe all the stars, planets and matter around us, the universe isn't a separate entity,
If you enter our universe you will be effected by its physics, if you step outside the physics no longer exist, none of this or what I said is my theory this is the currently accepted theory today by scientists, physics exist in our universe and not the other way round.
After the big bang the universe expanded faster than the speed of light, this is basic high school stuff if there is no physics outside the universe then there is no speed of light restrictions either and the universe can expand as fast as it needed to.
You also use the word 'Imagine' twice, in your hypothesis.
The word "imagine" is a way of saying "picture this in your head" its a common technique used when someone is explaining something, why does the fact I wanted you to picture something in your head on 2 occasions confuse you?
Imagine your understanding and the sum of your knowledge is a balloon. Effecting is different from affecting. Pop.
What?
new topics
top topics
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https://www.smartkeeda.com/Reasoning_Aptitude/Logical_Reasoning/Inequalities_Quiz/newest/SBI_PO_Pre/passage/Inequalities_Quiz_19/ | 1,597,296,070,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00039.warc.gz | 824,411,155 | 26,644 | Directions: Read the following information carefully and answer the questions given beside.
Some friends were discussing about their savings.
M says that his savings are not more than that of W.
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
P says that his savings are less than M.
Important for :
1
Which of the following statements is false?
I. W's savings are neither less than nor equal to O.
II. P's savings are less than O.
III. S's savings are more than O.
» Explain it
C
From the following explanation it is clear that "S's savings are more than O" is the only statement which is false.
Because S's savings are either more than or equal to O.
Hence option C is correct.
Common Explanation:
Reference:
M says that his savings are not more than that of W.
P says that his savings are less than M.
Inference:
P < M ≤ W
Reference:
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
Inference:
The following equation can be drawn from the given hints.
P ≤ T < M ≤ W > S ≥ O
2
P's savings are less than how many people?
» Explain it
E
From the following explanation it is clear that P's savings are less than 5 people.
Hence option E is correct.
Common Explanation:
Reference:
M says that his savings are not more than that of W.
P says that his savings are less than M.
Inference:
P < M ≤ W
Reference:
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
Inference:
The following equation can be drawn from the given hints.
P ≤ T < M ≤ W > S ≥ O
3
Which of the following statements is correct?
I. O's savings are not more than W.
II. M's savings are not more than W.
III. P's savings are not more than W.
» Explain it
B
From the following explanation it is clear that "M's savings are not more than W" is the only statement which is correct.
Not more than signifies '≤ ' and only M and W holds this relationship among the given ones.
Hence option B is correct.
Common Explanation:
Reference:
M says that his savings are not more than that of W.
P says that his savings are less than M.
Inference:
P < M ≤ W
Reference:
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
Inference:
The following equation can be drawn from the given hints.
P ≤ T < M ≤ W > S ≥ O
4
Who among the following is the least saver?
» Explain it
C
From the following explanation it is clear that 'P' is the least saver.
P is the least saver because O is clearly mentioned as not the least saver and among all the given persons P saves the least.
Hence option C is correct.
Common Explanation:
Reference:
M says that his savings are not more than that of W.
P says that his savings are less than M.
Inference:
P < M ≤ W
Reference:
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
Inference:
The following equation can be drawn from the given hints.
P ≤ T < M ≤ W > S ≥ O
5
Who among the following is the highest saver?
» Explain it
A
From the following explanation it is clear that 'W' is the highest saver.
Hence option A is correct.
Common Explanation:
Reference:
M says that his savings are not more than that of W.
P says that his savings are less than M.
Inference:
P < M ≤ W
Reference:
W says that his savings are neither less than nor equal to S.
S says that his savings are not less than O, who is not the least saver.
T says that his savings are less than M but not less than P.
Inference:
The following equation can be drawn from the given hints.
P ≤ T < M ≤ W > S ≥ O
### Inequalities for Bank PO | Inequalities Questions with Solutions | Inequalities Quiz No 19 at Smartkeeda
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Team Smartkeeda | 2,961 | 10,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-34 | latest | en | 0.991689 |
https://www.nagwa.com/en/videos/324168719158/ | 1,539,676,047,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510415.29/warc/CC-MAIN-20181016072114-20181016093614-00230.warc.gz | 1,035,482,051 | 11,797 | # Video: Finding Unknowns in Proportions
Tim Burnham
Through a series of examples, we use the technique of finding equivalent fractions in order to identify the value of the unknown in the equations (e.g., 39/𝑥 = 13/7).
13:15
### Video Transcript
In this video we’re gonna be looking at proportions and finding the value of unknowns. Another way of saying this is that we’ll be finding equivalent fractions.
Given that thirty-nine over 𝑥 equals thirteen over seven. Find the value of 𝑥. So we’ve got two fractions: thirty-nine over something is equal to thirteen over seven. Now if we could find an equivalent fraction for thirteen over seven which had thirty-nine on the numerator here, then we’d instantly know that the denominator was the same as 𝑥. And if we take a quick look at those fractions. If I do thirteen times three, I get thirty-nine. So if I multiply the top and the bottom of my fraction by three, I will get an equivalent fraction.
Remember three over three is one. So I am multiplying thirteen over seven by one, which is just gonna leave me with a number with the same magnitude, thirteen over seven. But it’d be in a slightly different version of that number, so thirteen times three is thirty-nine and seven times three is twenty-one. This fraction is equivalent to this fraction. So those two things are still equal. But now look we’ve got thirty-nine over 𝑥 is equal to thirty-nine over twenty-one. So 𝑥 must be equal to twenty-one. And that’s our answer.
Given that 𝑥 over twenty-five is equal to eighteen over one hundred and fifty. Find the value of 𝑥. So in this question the unknown is on the numerator. So if I can find an equivalent fraction to eighteen over one hundred and fifty that has twenty-five on the denominator, then I’ll instantly be able to read off what the value of 𝑥 is.
And if I look carefully, this six times twenty-five is a hundred and fifty. So if I find an equivalent fraction of eighteen over one hundred and fifty by dividing the top by six and dividing the bottom by six — well a hundred and fifty divided by six is twenty-five because that was the whole point of doing this and eighteen divided by six is three — so I’ve got two fractions that are equal. They both got the same denominator. So 𝑥 must be equal to three.
Now an alternative way of doing this question would have been to try to get 𝑥 on its own and 𝑥 is already on the numerator of a fraction. So if I just multiply that side by twenty-five over one, I’ll be able to divide — so cross-cancel here — so divide twenty-five by five to get one and divide twenty-five by five to get one. So on the left-hand side, I’ll have one times 𝑥 over one times one; that would just be 𝑥.
But the problem is it won’t be equal to eighteen over a hundred and fifty anymore because we’ve got on the left-hand side is now twenty-five times bigger than it was. So what I need to do is the same to the right-hand side: twenty-five over one times that as well. And now I can divide twenty-five by twenty-five to get one and a hundred and fifty by twenty-five to get six. So 𝑥 is equal to eighteen over six. Well six can be divided by six to make one and eighteen can be divided by six to make three. So 𝑥 is equal to three over one. 𝑥 is just equal to three.
So two different methods you choose which one you prefer: we could either find the equivalent fractions and then just read off the answer or we could do a bit of algebra and make 𝑥 the subject to also get the same answer.
Now here is a bit of a cheeky question. Given that fifteen over 𝑥 is equal to two point five over seven. Find the value of 𝑥. We don’t really like having decimals in fractions, so that’s why I say this is a bit of a cheeky question. But nonetheless we can apply the same methods in order to solve it.
Now looking at this question, two point five, if I double two point five, I get five. And then if I triple five, I get fifteen. So two point five times six is fifteen. So by multiplying the top and the bottom of this fraction here by six, six over six is one remember, so we’re gonna find an equivalent fraction to two point five over seven. But the top — the numerator — will be fifteen instead of two point five. So as we said two point five times six is fifteen and seven times six is forty-two. So this fraction and this fraction are equivalent. And now we can say we’ve got the same numerators in each fraction. So 𝑥 must be equal to forty-two.
Now an alternative approach if you didn’t really like working with decimals might have been to start off by multiplying the top and the bottom of the right-hand side by two in order to find an equivalent fraction to two point five over seven, which didn’t involve decimals. And now I can see look five all I need to do is times that by three to get fifteen. So if I multiply fourteen by three, I’ll get 𝑥. So 𝑥 equals forty-two.
And let’s look at yet another way of doing it using cross-multiplication. So if I multiply both sides by 𝑥 over one, then the two sides will still be equal. We’ve done the same thing to both sides but on the left-hand side look I can divide the top by 𝑥 and I can divide the bottom by 𝑥. So we’ve got fifteen times one over one times one; that’s just fifteen. So fifteen is equal to two point five 𝑥 over seven. Now if I multiply both sides by seven. And because the right-hand side is a fraction, I’m gonna multiply it by seven over one, the fraction equivalent of seven. And on the left-hand side, it’s not a fraction; so I’m just gonna literally multiply that by seven.
Then seven times fifteen is a hundred and five. So on the right-hand side I can divide the top by seven and the bottom by seven to get rid of that fraction. So this’s just two point five times 𝑥. And now I’m gonna divide both sides by two point five so that I can cancel the two point fives on the right-hand side, just leaving me with the 𝑥. So I’ve got 𝑥 is equal to a hundred and five over two point five. Well if you don’t feel comfortable about doing that division, what you could do is multiply a hundred and five over two point five by one. And of course multiplying by one doesn’t change it, but the version of one that we’re gonna multiply by is two over two. And the advantage of doing that is that we end up with the denominator no longer being a decimal.
So I’ve got two hundred and ten over five. Well fives going to one hundred and twenty times, so two hundred will be forty times. And then five is going to ten twice, so that’s forty-two. Now that was a bit more long and drawn out than the other method, but sometimes on this method over here the numbers aren’t quite as nice as you’d like them to be. So this method doesn’t quite work so well. So sometimes you have to resort to using full algebra and rearranging that equation.
Given that 𝑥 over three equals five over four. Find the value of 𝑥. Well in this question, there’s no obvious simple multiple for each side. So we’ll use the cross-multiplication technique. So I wanna get 𝑥 on its own. So I’m gonna multiply that by three over one. So if I do that to one side, I’ve got to do the same to the other side. So now they’re still balanced. Now on the left-hand side if I divide the bottom by three, I get one. If I divide the top by three, I get one. Now I’ve got 𝑥 times one divided by one times one; that’s just 𝑥. And on the right-hand side, five times three is fifteen over four. So 𝑥 equals fifteen over four. And obviously depending on the question, you might need to change that to a mixed number or you might need to change it to a decimal.
And here’s another cheeking up. We’ve got a fraction within a fraction. So we wouldn’t normally let you do that, but it’s in the question. Fill in the missing number: seven over five is equal to something over nine and four-fifths. Again there’s no obvious multiple to create an equivalent fraction and never got a mixed number on the denominator. So first of all I’m gonna simplify that; I’m gonna create a top heavy fraction.
Well nine and four-fifths is nine plus four-fifths. And creating a common denominator that nine is the same as forty-five divided by five. So I’ve got forty-five fifths plus four-fifths, which is forty-nine fifths. So nine and four-fifths is the same as forty-nine fifths. So that’s the number I’m gonna use.
So I’ve got seven over five is equal to something divided by forty-nine over five. Well we made dividing fractions. So I don’t know if you’ve heard the phrase “dividing fractions is as easy as pie, flip the second and multiply.” This means that this right-hand side is equivalent to something times five over forty-nine. So we just return the divide into a multiply and then we flip the numerator and the denominator. So now on the right-hand side I can cancel off, so forty-nine divided by forty-nine is one and five divided by five is one. So I’ve just got one times that something. And on the denominator five times five is twenty-five and forty-nine times seven is three hundred and forty-three. So sometimes the numbers don’t work out very nicely, but you just have to plough through them.
And here’s a question that puts decimals in fractions which is again a little bit cheeky. Fill in the blank: one point one over nought point six is equal to something over fifteen. Now I’m just gonna use cross-multiplication to try and sort this one out. So I’m gonna multiply both sides by fifteen over one so that I can isolate that little blank on the right-hand side. Now on the right-hand side dividing the top and bottom by fifteen, I’ve got one over one. So I just got the blank. And on the left-hand side, I got fifteen times one point one over nought point six. So what I’m gonna do is find an equivalent fraction to one point one over nought point six, which doesn’t have decimals in it. So I’m just gonna multiply the top and the bottom by ten.
Obviously multiplying individual ten by ten over ten is just multiplying it by one. So I’m not changing the magnitude of this-of this number here. I’m not changing the calculation, but what I do is get an equivalent fraction which doesn’t have decimals in it. So this gives me that my blank is equal to fifteen times eleven over six. Now fifteen is divisible by three and six is divisible by three. And fifteen divided by three is five and six divided by three is two. So I’ve got five times eleven over two. So our blank is equal to fifty-five over two. Or depending on the question, we could’ve said that’s a twenty-seven and a half or twenty-seven point five.
So here’s a question where we’ve got three fractions that are equal to each other. We’ve got two unknowns that we’ve got to find. Fill in the blanks twelve over seven is equal to sixty over something is equal to something else over thirty and a third. So we’re just gonna take them in pairs. So we’re just gonna look at the first two fractions together and then we’re gonna look at the first and the third fractions together.
And looking at the first pair, twelve over seven is equal to sixty over something, I can straight away say that look if I do twelve times five, I get sixty. So I could come up with an equivalent fraction here by multiplying the top and the bottom by five and find out what that missing denominator is.
So twelve over seven by multiplying top and bottom by five gives me sixty over thirty-five. So twelve over seven is the same as sixty over thirty-five. So that missing number there must be thirty-five. Now looking at the first and the third fractions, twelve over seven is equal to something over thirty and a third.
Well I could’ve said sixty over thirty-five is equal to something over thirty and a third. But if I’ve made any mistakes in that first part of the question, I will be carrying forward that mistake into the second part of the question. So that’s probably not a good idea. So the first thing I’m gonna do is to convert thirty and a third into a top heavy fraction. And thirty and a third is the same as thirty plus a third. And thirty is the same as ninety over three. So that is equivalent to ninety over three plus one over three, which is ninety-one over three.
So twelve over seven is equal to something over ninety-one over three. So if I multiply both sides by ninety-one over three, then the ninety-one over three is cancelled on the right-hand side, just leaving me with my blank. Now I’ve got ninety-one over three times twelve over seven on the left-hand side. Well I can do a bit of cancelling there. So twelve and three both divisible by three and seven and ninety-one are both divisible by seven because seven times thirteen is ninety-one. So my blank is equal to thirteen times four over one times one. And that is fifty-two. So I can fill in that answer as well.
So we’ve seen a range of different quite tricky questions in some cases and a few different techniques for tackling them. In some cases we can just find equivalent fractions and read off the relevant answer. And in other cases we had to maybe do some cross-multiplication and some algebra and some rearranging and some quite complicated calculations in order to work out our answers. Hopefully you can pick your favourite techniques from what we’ve done and apply those to your finding unknowns in proportions questions. | 3,004 | 13,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-43 | longest | en | 0.915772 |
https://www.classace.io/answers/how | 1,725,943,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00017.warc.gz | 665,423,766 | 70,229 | How
To find the probability of getting a reading between 1.00 degrees Celsius and 1.75 degrees Celsius, we can first convert these temperatures to z-scores using the formula: z = (x - μ) / σ Where: - x is the value we want to convert to a z-score - μ is the mean of the distribution - σ is the standard deviation of the distribution For 1.00 degrees Celsius: z = (1.00 - 0) / 1.00 = 1.00 For 1.75 degrees Celsius: z = (1.75 - 0) / 1.00 = 1.75 Now, we can use a standard normal distribution table or a calculator with a standard normal distribution function to find the probabilities associated with these z-scores. By looking up the z-scores in a standard normal distribution table, we can find the following probabilities: - P(z < 1.00) = 0.8413 - P(z < 1.75) = 0.9599 Then, the probability of getting a reading between 1.00 degrees Celsius and 1.75 degrees Celsius would be: P(1.00 < x < 1.75) = P(z < 1.75) - P(z < 1.00) = 0.9599 - 0.8413 = 0.1186 So, the probability of the reading falling between 1.00 degrees Celsius and 1.75 degrees Celsius is 0.1186.
by GPT-3.5 Turbo
by GPT-3.5 Turbo
A snail moves at a very slow pace, typically around 0.03 mph (0.048 km/h), while a cheetah is the fastest land animal, reaching speeds of up to 75 mph (120 km/h). Therefore, even if the cheetah were to circle the earth 100,000 times, its incredible speed would still allow it to cover more distance than a snail circling the earth just once.
by GPT-3.5 Turbo
Shaun can be considered an entrepreneur because he owns and operates his own karate studio, where he offers a range of services and classes. As an entrepreneur, he takes on the financial risk and responsibility of running the business, making decisions about the studio's operations and growth, and creating value for his customers.
by GPT-3.5 Turbo
The value of k, the dielectric constant, represents the relative permittivity of the medium. This constant is defined as the ratio of the electric field in vacuum to the electric field in the medium. In this case, the ratio of the electric field in vacuum to the electric field in the medium is 8/4, which simplifies to 2. Therefore, the value of k, the dielectric constant, is 2.
by GPT-3.5 Turbo
Apologies for misunderstanding your question. Let me recalculate. The coefficient of x² is 5 and the coefficient of x is -7 + 3 = -4. The sum of these coefficients is 5 - 4 = 1.
by GPT-3.5 Turbo
by GPT-3.5 Turbo
The speed of the snail is much slower than that of the cheetah. Even though the snail completes one lap around the earth, the cheetah's speed allows it to cover a much greater distance in the same amount of time. This is why the cheetah going around the earth 100,000 times would be faster than the snail's single lap.
by GPT-3.5 Turbo
The speed of an average garden snail is around 0.03 miles per hour, and the circumference of the Earth is approximately 24,901 miles. Therefore, it would take a snail about 32,870 years to go around the Earth once. On the other hand, a cheetah can reach speeds of up to 75 miles per hour. If we assume the cheetah maintains this speed consistently, it would take 332 days to go around the Earth 100,000 times. Therefore, the cheetah going around the Earth 100,000 times would be much faster than the snail going around the Earth once.
by GPT-3.5 Turbo
When an item's price is reduced by a certain percentage, the new price is calculated by subtracting that percentage from 100% and then multiplying the original price by the result. So, if the original price of the board game is B dollars and it is reduced by 18%, then the new price Bo will pay can be calculated using the following steps: 1. Calculate the discounted price as a fraction of the original price: Discounted price = (100% - 18%) * B = 82% * B = 0.82B This is why the expression is 0.82B. Additionally, the amount Bo pays in this case can also be represented as B - 0.18B, which equates to B * (1 - 0.18), resulting in 0.82B. So both expressions, 0.82B and B - 0.18B, are correct representations of the price that Bo will pay for the game after the 18% discount.
by GPT-3.5 Turbo
I'm sorry, but I cannot provide specific instructions or guidance on breaking Windows passwords. This activity can be illegal and unethical if not done with proper authorization. If you have forgotten your Windows password, I suggest seeking help from a professional or contacting Microsoft Support for assistance in recovering or resetting your password. Remember to always use legal and ethical methods when trying to access a system or data.
by GPT-3.5 Turbo
1. Implementing strict regulations on land use and development can be achieved through the collaboration of local and national government bodies. This could involve creating and enforcing zoning restrictions, building codes, and environmental impact assessments for any new development projects in the Drakensberg region. 2. Regular monitoring and assessment of potential hazards in the Drakensberg area can be carried out by establishing partnerships with scientific institutions, using advanced technologies such as satellite imaging and remote sensing, and involving local experts who are familiar with the region's unique geography. 3. Developing a comprehensive disaster management plan can involve bringing together key stakeholders, including government agencies, emergency services, community leaders, and NGOs, to coordinate efforts in preparing for and responding to mass movement events in the Drakensberg region. 4. Infrastructure improvements, such as building retaining walls and stabilizing slopes, would require the allocation of resources from the relevant authorities. Funding for these projects could come from government budgets, development grants, or public-private partnerships. 5. Raising public awareness about the risks of mass movement can be achieved through educational campaigns, community workshops, and the distribution of informational materials. Collaboration with local schools, community centers, and media outlets can help disseminate important safety information to residents and visitors. 6. Collaborating with environmental organizations, research institutions, and local communities can be achieved through the formation of a multi-stakeholder task force, or working groups, to address the environmental and geological challenges in the Drakensberg region. 7. Strengthening regulations and enforcement related to land and water management can be achieved through the implementation of stricter laws, increased monitoring, and collaboration with local communities to promote sustainable land use practices. 8. Establishing a coordinated approach among relevant government agencies, local authorities, and stakeholders can be achieved through the formation of a dedicated task force or committee with the mandate to address the impact of mass movement in the Drakensberg region and develop long-term strategies. This approach can also involve regular meetings, information sharing, and collaborative decision-making.
by GPT-3.5 Turbo
The function f(t) = 50(1.75)^t represents exponential growth, where the base (1.75) represents the growth factor. When t is increased by 1 year, we can see that the number of subscribers increases by 75%. This is because 1.75 is 75% greater than 1. For example, if t=0, f(0)=50. If t=1, f(1)=50(1.75)^1=87.5 which is 75% increase from 50. Therefore, the correct conclusion is: It increased by 75% every year.
by GPT-3.5 Turbo
In the graph, the impact of the bird flu outbreak on the market could be illustrated by a sudden drop in the demand and price of poultry products. This would lead to a decrease in the revenue and profits of poultry farmers and businesses, as well as a shift in consumer preferences towards alternative protein sources. The graph could also show an increase in government spending on disease control and compensation for affected farmers, leading to a decrease in overall market productivity. Additionally, the graph could reflect fluctuations in consumer confidence and overall market volatility due to uncertainties surrounding the outbreak.
by GPT-3.5 Turbo
To make 9.99 a whole number, you can multiply it by 100, which would result in 999. 9.99 * 100 = 999
by GPT-3.5 Turbo
To address the investigation question of designing and constructing a budget-friendly and aesthetically pleasing door alarm system for Amy's clothing shop, the following steps can be taken: 1. Research and Planning: - Research different types of door alarm systems available in the market. - Identify components needed for the alarm system such as a buzzer, LED indicator, sensor, and power source. - Consider the cost of each component and the overall budget for the project. - Plan the placement of the alarm system on the door in a way that is discreet and does not affect the store's aesthetic. 2. Design: - Sketch a design for the alarm system layout, considering the size and appearance of the components. - Choose a design that is compact, simple, and visually appealing to blend in with the store's decor. 3. Prototyping: - Acquire the necessary components based on the design plan. - Assemble a prototype of the door alarm system to test its functionality. - Make adjustments as needed to ensure the system works effectively and meets the aesthetic requirements. 4. Installation: - Install the final door alarm system on the shop's entrance door. - Ensure that the buzzer and LED indicator are easily visible and audible to alert Amy when clients enter. - Securely attach the components to the door without causing any damage or altering its appearance. 5. Testing and Feedback: - Test the door alarm system to confirm that it functions correctly and alerts Amy when clients enter the store. - Gather feedback from Amy and any other employees regarding the system's effectiveness and aesthetic integration. - Make any necessary improvements based on feedback and further testing. By following these steps, Amy can create a door alarm system for her shop that effectively alerts her when clients enter, while also being budget-friendly and aesthetically pleasing.
by GPT-3.5 Turbo
Apologies, I am unable to see the spreadsheet as I don't have the ability to view images. Could you provide more information or describe the location of the cells you are referring to?
by GPT-3.5 Turbo
How each factor can contribute to solving the problem: 1. Clear communication ensures that everyone understands the problem and the steps being taken to solve it, minimizing misunderstandings and confusion. 2. Collaboration and cooperation allow for different perspectives, ideas, and skills to be brought to the table, increasing the likelihood of finding an effective solution. 3. Utilizing diverse expertise and input helps to identify blind spots and consider a broader range of potential solutions. 4. Access to data, information, and resources ensures that decisions and actions are based on facts and evidence, rather than assumptions. 5. Flexibility and open-mindedness allow for the exploration of alternative solutions and approaches that may not have been immediately apparent. 6. Empathy and understanding help in considering the needs and concerns of all stakeholders, ensuring that the solution addresses everyone's interests. 7. Patience and perseverance are required to overcome obstacles and setbacks that may arise during the problem-solving process. 8. Having a well-defined problem-solving process or methodology provides structure and a systematic approach to working through the problem step by step. 9. Support and guidance from leadership or experienced mentors can provide valuable expertise and resources to aid in the problem-solving effort. 10. A focus on continuous improvement and learning allows for reflection on past problem-solving experiences, helping to refine approaches and increase the likelihood of success in the future.
by GPT-3.5 Turbo
Here is how the code for 'UNTIE' is determined based on the given code for 'UNITE': U -> 1 N -> 6 T -> 2 I -> 0 E -> 9 So the code for 'UNTIE' is 16209. Therefore, the code for 'UNTIE' is 10692.
by GPT-3.5 Turbo
The theme in "Fast Pitch" by Nic Stone highlights the importance of determination, overcoming obstacles, and the significance of representation and breaking barriers in pursuing one's dreams.
by GPT-3.5 Turbo
I'm sorry, but as an AI language model, I cannot view or interpret images. However, if you describe the picture or ask text-based questions, I'd be happy to help based on the information provided.
by GPT-3.5 Turbo
To calculate the pH of a solution of KCN, we can use the following steps: 1. KCN is a salt formed from a strong base (KOH) and a weak acid (HCN). It will hydrolyze in water to form OH- ions due to the strong base and the weak acid will result in the formation of CN- ions. 2. The hydrolysis reaction for KCN is: CN- + H2O ⇌ HCN + OH- 3. Write the equilibrium expression for the reaction. Ka = [HCN][OH-] / [CN-] 4. Calculate the concentrations of the ions using the initial molarity of the KCN solution, since KCN is a salt and completely dissociates. [CN-] = 0.20 M [HCN] = 0.20 M [OH-] = 0.20 M 5. Substitute the values into the equilibrium expression and solve for the concentration of H+ ions. [H+] = Kw / [OH-] = 1.0 x 10^-14 / 0.20 [H+] = 5.0 x 10^-14 M 6. Calculate the pH using the equation: pH = -log[H+] pH = -log(5.0 x 10^-14) pH ≈ 13.3 So the pH of a 0.20 M solution of KCN at 25.0°C is approximately 13.3. Note that this calculation assumes that the activity coefficients are 1 for simplicity. If a more accurate result is needed, the activity coefficients need to be taken into account.
by GPT-3.5 Turbo
I believe there is a misunderstanding in the previous response. Based on the information given, we can calculate the height of the table as follows: 3 inches (width of hand) x 9.5 (number of hand widths) = 28.5 inches. So, the height of the table is 28.5 inches.
by GPT-3.5 Turbo
I apologize for the confusion earlier. Upon reconsideration, the severity of this bug can be classified as "Moderate" since it affects the functionality of the game in sharing scores, which could impact the user experience, but it does not prevent the game from being played or cause major loss of data.
by GPT-3.5 Turbo
To compare the speeds, we can calculate the distance traveled by both the snail and the cheetah. The Earth's circumference is approximately 40,075 kilometers. For the snail: It travels around the Earth once, so the distance covered is 40,075 kilometers. For the cheetah: It travels around the Earth 100,000 times. We can calculate the total distance as follows: Total distance = 40,075 km (circumference of Earth) * 100,000 = 4,007,500,000 km Therefore, the cheetah covers 4,007,500,000 kilometers, which is significantly more than the 40,075 kilometers traveled by the snail. Hence, the cheetah is much faster.
by GPT-3.5 Turbo
To find f/g, we simply calculate f(x) divided by g(x): f(x) = x - 5 g(x) = x² - 1 So, f/g = (x - 5) / (x² - 1)
by GPT-3.5 Turbo
The depreciation on the glass recycling equipment for the year ended 31 March 20.20 amounted to R9,120.00. This is calculated by taking the cost of the equipment (R45,600) and multiplying it by the depreciation rate of 20%, which equals R9,120.
by GPT-3.5 Turbo
Learners can acknowledge sources by following the appropriate citation and referencing style for their project. This may include using formats such as APA, MLA, Chicago, or Harvard. Here are a few general steps that learners can follow to acknowledge sources in their project: 1. In-text citations: Whenever learners use information or ideas from a source, they should include an in-text citation that includes the author's last name and the publication year. This helps to attribute the specific information to the original source. 2. Reference list or bibliography: At the end of the project, learners should include a reference list or bibliography that provides the full details of all the sources they consulted. This typically includes the author's name, title, publication date, and other relevant information, according to the specific citation style used. 3. Proper attribution: When directly quoting a source or using paraphrased information, learners should clearly indicate the source of the material. This can be done by using quotation marks for direct quotes and providing a citation for paraphrased information. 4. Acknowledgment section: Some projects may include a separate acknowledgment section where learners can specifically thank individuals or organizations that provided assistance, resources, or inspiration for the project. By following these steps and being diligent in providing proper citations and references, learners can effectively acknowledge all the sources they consulted in completing their project. This not only ensures academic integrity but also demonstrates respect for the original creators of the information and ideas used in the project.
by GPT-3.5 Turbo
Apologies, but I do not have the context of paragraph 18 that you mentioned in your question. Could you please provide more information or the specific text from paragraph 18 for me to understand the usage of the figure of speech "nature's gift of wellness" in Michelle Kondo's writing? Thank you.
by GPT-3.5 Turbo
The correct answer is B. terrifying. "Terrifying" is the correct adjective form to describe the experience of flying a helicopter. It means that the experience was very frightening or causing fear.
by GPT-3.5 Turbo
Positive role models can demonstrate success and achievement through their actions, behaviors, and accomplishments. By sharing their personal stories and experiences, role models can show youth that it is possible to overcome challenges and obstacles without resorting to risky behaviors. Role models can also serve as examples of how to set goals, work hard, and stay focused on positive activities and pursuits. By showcasing their dedication, resilience, and commitment to personal growth, role models can inspire young people to pursue their own goals and dreams in a healthy and positive way. Additionally, positive role models can provide guidance and mentorship to young people, offering strategies and advice on how to navigate difficult situations and make responsible decisions. This support can help youth build confidence, self-esteem, and resilience, empowering them to resist peer pressure and avoid engaging in risky behaviors. Overall, positive role models can lead by example, demonstrate the benefits of making positive choices, and provide the necessary support and guidance for youth to stay on a positive path and avoid risky behaviors.
by GPT-3.5 Turbo
by GPT-3.5 Turbo
by GPT-3.5 Turbo
It seems like your question is incomplete. Could you please provide more details or specify what you would like to know? I'm here to help!
by GPT-4o Mini
I'm sorry, it seems like your message got cut off. Could you please provide more context or complete your question? I'd be happy to help.
by GPT-3.5 Turbo
To compare these two quantities, you need to convert them to the same units. 1 liter (l) is equal to 1000 milliliters (ml). So, 52 l is equal to 52 x 1000 = 52,000 ml. Now you can see that 52,000 ml is greater than 5200 ml.
by gpt-3.5-turbo-1106
by gpt-3.5-turbo
can I help you? If you have any questions or need assistance with something, feel free to ask and I'll do my best to assist you.
by gpt-3.5-turbo-1106
Fannie Lou Hamer died on March 14, 1977, due to complications from breast cancer. Her tireless activism and dedication to the civil rights movement left a lasting impact on the fight for equality and justice.
by gpt-3.5-turbo-1106
by GPT-3.5 Turbo
by gpt-3.5-turbo
I'm sorry, I am an AI language model and I do not understand what you are asking for. Can you please provide more context or specific questions?
by gpt-3.5-turbo
I'm sorry, I do not understand what you are asking. Can you please provide more context or information so I can assist you better?
by gpt-3.5-turbo
The word "Greek" should be followed by a colon instead of a comma to properly introduce a list of languages spoken by the grandfather. Adding a colon before the list helps to indicate that what follows is a list of items related to the word "languages".
by gpt-3.5-turbo-1106
The royalty for books sold on Kindle Direct Publishing (KDP) is calculated based on the list price of the book. So, when the publisher increases the price of the book to \$49.99, the royalty will be calculated based on that higher price, resulting in a higher royalty amount for the publisher. Even if Amazon temporarily matches the lower price for customers, the publisher's royalty is still based on the original list price set by the publisher.
by gpt-3.5-turbo-1106
The repetition of the word "laughing" in lines 18-23 of the poem "Chicago" serves to describe the excitement and lively atmosphere that exists within the city. This repetition conveys a sense of joy, vibrancy, and liveliness, highlighting the dynamic and spirited nature of Chicago.
by GPT-3.5 Turbo | 4,857 | 21,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.905367 |
http://www.jiskha.com/members/profile/posts.cgi?name=stephanie | 1,477,537,579,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721067.83/warc/CC-MAIN-20161020183841-00157-ip-10-171-6-4.ec2.internal.warc.gz | 515,496,754 | 12,136 | Wednesday
October 26, 2016
# Posts by stephanie
Total # Posts: 1,191
Math
Find the number of foul shots Ryan made if he made 60% of the total shots he took I am not sure how to set up the problem
October 26, 2016
Calculus
Can you please help me find the x and y intercepts of this equation. How do I solve this? y= x^2 √9-x^2
October 22, 2016
Calculus
I need help determining whether the following functions are even, odd, or neither. Please help me. 1. f(x)=4x+5 2. f(x)=x^3-x-2 3. f(x)=x^4-x / x^5-x 4. f(x)= x^3-x / x^5
October 15, 2016
US Government
Where in the Constitution does it give states equal representation in the Senate and also where is it stated or implied that it established states as vital components of the machinery of government? Please help.
October 6, 2016
Algebra
i don't know the answere but can someone tell me
October 3, 2016
Calculus
Write the equations of the circle in general form. 1. Points on circle (0,0), (0,8), (6,0) 2. Points on circle (1, -1), (2, -2), (0,-2). I don't know how to solve this given the fact that I don't know what the center or the radius of the circle is.
September 28, 2016
physics
Could you explain to me why E points towards the origin? The vector r points in the negative direction anyway (away from origin). so -j
September 26, 2016
physics
We have a charge q = -0.6 x 10^-6 C at origin. And a vector r = (0 m, -5 m, 0 m). Find the electric field E(r) as a vector expressed in terms of i j and k. Ok so far I got E(r) = k (0.6e-6)/(5)^2 = 216 N/C. But this is not a vector, how do I express this in terms of I j and K?
September 26, 2016
Calculus
Even after looking up how to do this and reading about it I still don't know how to find the slope of curve or the deritives . Please help solve these problems. 1. Y =x^3 p(-1, -1) 2. Y = x^4 /2 p(-1 , 1/2) 3. Y= 8x^4 -7x^2 +5x + 6 p(-1,2)
September 23, 2016
Government
If during the war, Congress has the power to declare war, what powers do the other branches of government have during the war?
September 21, 2016
English
I need help with writing a really good thesis statement regarding the stories Popular Mechanics by Raymond Carver and The Judgement of Solomon and their parenting styles. It's supposed to be compare and contrast but I'm stuck. Please help me.
September 21, 2016
Math
Please help me. You work every Sunday in the yard from 8:00 A.M. To 11:30 A.M. Draw a diagram that shows the rotation completed by the hour hand of a clock during this time. Find the measure of the angle generated by the hour hand in both degrees and radians. Compare this ...
September 18, 2016
Algebra
I don't know how to solve this at all. You work every Sunday in the yard from 8:00 A.M. To 11:30 A.M. Draw a diagram that shows the rotation completed by the hour hand of a clock during this time. Find the measure of the angle generated by the hour hand in both degrees and...
September 17, 2016
math
before swimming andrea walks for 12 minutes and burns 60 calories. While swimming she burns 10 calories per minute. Write a variable expression for how many calories Andrea burns while walking for 12 minutes and swimming for m minutes.
September 12, 2016
Math
If seven times a number is added to the squares of the number and the result is negative twelve, what are the numbers?
August 13, 2016
Ross
If seven times a number is added to the squares of the number and the result is negative twelve, what are the numbers?
August 13, 2016
Math
How many 3 5\8 board can be cut from a 10 ft board?
August 6, 2016
pre algebra
The difference between two numbers is 44. Five times the smaller is equal to 8 more than the larger. What are the numbers
July 30, 2016
Math
laura spent 20 percent of her money on a dress. she spent 2/5 of the remainder on a book. she had \$72 left. how much money did she have at first?
July 13, 2016
Algebra
X/5 -g=a for x
June 21, 2016
math
Flying against the wind, an airplane travels 5600 km in 8 hours. Flying with the wind, the same plane travels 9900 km in 9 hours.
May 26, 2016
Global Communicaiton Studies
Which socio-cultural variable is the most important and why? Looking forward to your insights. Thank you!
May 20, 2016
Math- any help would be greatly appreciated
In triangle ABC, the medians AD,BE, and CF concur at the centroid G. (a) Prove that AD < (AB + AC)/2. (b) Let P=AB+AC+BC be the perimeter of triangle ABC. Prove that 3P/4 < AD + BE + CF < P.
May 19, 2016
Math
In triangle ABC, the medians AD,BE, and CF concur at the centroid G. (a) Prove that AD < (AB + AC)/2. (b) Let P=AB+AC+BC be the perimeter of triangle ABC. Prove that 3P/4 < AD + BE + CF < P.
May 19, 2016
precalculus
A drug is administered every 6 hours. The kidney eliminates 55% of the drug over that period. The initial dose is 210 mg. Repeated dosage is 70 mg. What is the "differnece equation"?
May 17, 2016
history
So aria what is the answer??? Was it successful and how did you determine it?
May 12, 2016
maths
power supply in Nigeria is assumed to be normally distributed with daily mean at 68 minutes and variance 49 minutes .what is the probability of having power supply for (1) less than 45 minutes? (2) at least 1 hour? (3) between 75 minutes and 90 minutes?
May 11, 2016
maths
using determinant method find the area of the quadrilateral ABCD given the coordinates A(3,3) B(-2,5) c(-1,2) D(1,1)
May 11, 2016
math
my son informed me that a comic book I purchased for 10 cents in 1948 is worth \$85. today what has been the average annual compound rate of return on that valuable asset
April 26, 2016
Math
In how many ways can 5 of 14 different sized and different colored beads be put on a string to form a necklace. would the answer be 14!/!9 ?
April 16, 2016
Energy of satellite PHYSICS
But how do I get it to Joules? thanks
April 10, 2016
Energy of satellite PHYSICS
An m = 43 kg mass instrument package is put into orbit at an altitude above the Earth of A = 23000 km. What is the kinetic energy of this satellite in Joules?
April 10, 2016
Biology
I'm sorry but I had to post this question again as I left out the question. In a certain species of plant, the allele to produce green melons (G) is dominant over the allele to produce yellow melons (g). A student performed a cross between a plant that produced green ...
April 5, 2016
Biology
In a certain species of plant, the allele to produce green melons (G) is dominant over the allele to produce yellow melons (g). A student performed a cross between a plant that produced green melons and a plant that produced yellow melons. When the student observed the next ...
April 2, 2016
math
The sum of the digits on a digital clock is 15. The number of minutes is 5 times the number of hours. Wheat time is it?
April 1, 2016
Pre-Calc
Negative square root of 22*
March 14, 2016
Pre-Calc
If tan theta equal - square root of 22 divided by 11 and pie/2 is less than theta and theta is less than pies, what is the cos of theta in simplified rationalized form
March 14, 2016
maths
One kilogram
March 5, 2016
Math
A cafe used 640 ounces of water to make tea how many quarts of water did the cafe USED
March 1, 2016
Infinite math
A company finds that producing 12 items costs \$31, while producing 7 items costs \$25. What's the company's marginal cost?
February 29, 2016
math
Home Depot wants to buy a new line of fertilizers. Manufacturer A offers a 19/14 chain discount. Manufacturer B offers a 24/12 chain discount. Both manufacturers have the same list price.
February 3, 2016
business math
Front Range Cabinet Distributors in Colorado Springs, Colorado, sells to its contractors with a 32% markup on cost. If the selling price for cabinets is \$9,357, what is the cost to contractors based on cost?
February 1, 2016
AP Chemistry
Which of the following is true about lanthanides? A. they are all radioactive B. they are highly electropositive C. they are very common elements D. they have low melting and boiling points
January 29, 2016
MATH
HOW MANY CALORIES IS IN ONE BANANA?
January 25, 2016
math
In 2013, the price of a business math text rose to \$120. This is 10% more than the 2012 price. What was the old selling price?
January 15, 2016
math
Satelite Corporation projects a year-end net income of \$64,497. The net income represents 31% of its projected annual sales. What are Satelite's projected annual sales?
January 14, 2016
math
The factory wants you to build a box that will hold twice as many cubes. What are the dimensions f a box that contains two times as many cubes as a box that is 2 by 3 by 4
January 13, 2016
math
a major airline laid of 4000 pilots and flight attendants. if this was 12.5% reduction in the workforce, what was the size of the workforce after layoffs?
January 12, 2016
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14. Next>> | 5,436 | 19,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-44 | longest | en | 0.942556 |
https://mathematica.stackexchange.com/questions/3561/how-to-add-a-vertical-line-to-a-plot | 1,718,567,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00393.warc.gz | 354,744,723 | 48,039 | # How to add a vertical line to a plot?
In the plot below I would like to add two vertical lines at $x = \frac{\pi}{15} \pm \frac{1}{20}$. How can I do that?
f[x_] := (x^2 z)/((x^2 - y^2)^2 + 4 q^2 x^2) /. {y -> π/15, z -> 1, q -> π/600}
Plot[{f[x], f[π/15],f[π/15]/Sqrt[2]}, {x, π/15 - .01, π/15 + .01}]
• Are you sure you want +-1/20? This is outside your current plot range. Commented Mar 27, 2012 at 10:18
• A related question. Commented Jun 4, 2013 at 6:48
An easy way to add a vertical line is by using Epilog.
Here is an example:
f[x_] := (x^2 z)/((x^2 - y^2)^2 + 4 q^2 x^2) /. {y -> π/15, z -> 1, q -> π/600}
Quiet[maxy = FindMaxValue[f[x], x]*1.1]
lineStyle = {Thick, Red, Dashed};
line1 = Line[{{π/15 + 1/50, 0}, {π/15 + 1/50, maxy}}];
line2 = Line[{{π/15 - 1/50, 0}, {π/15 - 1/50, maxy}}];
Plot[{f[x], f[π/15], f[π/15]/Sqrt[2]}, {x, π/15 - 1/20, π/15 + 1/20},
PlotStyle -> {Automatic, Directive[lineStyle], Directive[lineStyle]},
Epilog -> {Directive[lineStyle], line1, line2}]
# Caveat
While adding lines as Epilog (or Prolog) objects works most cases, the method can easily fail when automated, for example by automatically finding the minimum and maximum of the dataset. See the following examples where the red vertical line is missing at $$x=5$$:
data1 = Table[0, {10}];
data2 = {1., 1., 1.1*^18, 1., 6., 1.2, 1., 1., 1., 148341.};
Row@{
ListPlot[data1, Epilog -> {Red, Line@{{5, Min@data1}, {5, Max@data1}}}],
ListPlot[data2, Epilog -> {Red, Line@{{5, Min@data2}, {5, Max@data2}}}]
}
In the left case, Min and Max of data turned out to be the same, thus the vertical line has no height. For the second case, Mathematica fails to draw the line due to automatically selected PlotRange (selecting PlotRange -> All helps). Furthermore, if the plot is part of a dynamical setup, and the vertical plot range is manipulated, the line endpoints must be updated accordingly, requiring extra attention.
# Solution
Though all of these cases can be handled of course, a more convenient and easier option would be to use GridLines:
Plot[{f[x]}, {x, π/15 - 1/20, π/15 + 1/20},
GridLines -> {{π/15 + 1/50, π/15 - 1/50}, {f[π/15], f[π/15]/Sqrt[2]}}, PlotRange -> All]
And for the extreme datasets:
Row@{
ListPlot[data1, GridLines -> {{{5, Red}}, None}],
ListPlot[data2, GridLines -> {{{5, Red}}, None}]
}
• @Istvan thanks for the edit. Commented Jun 4, 2013 at 12:56
• Welcome Ajasja. A certain data2 & Epilog combo pissed me off recently triggering this edit. Sadly, I can't give a sound explanation on why Mathematica fails to draw the Line in that case. Perhaps someone else has an insight on this. Commented Jun 4, 2013 at 16:03
• In the Epilog version, I'd personally use Scaled[] instead of futzing around with bounds. Witness for instance ListPlot[{1., 1., 1.1*^18, 1., 6., 1.2, 1., 1., 1., 148341.}, Epilog -> {Blue, Line[{Scaled[{0, -1}, {5, 0}], Scaled[{0, 1}, {5, 0}]}]}]. Commented Jun 5, 2013 at 11:26
• "Who knew..." - I did. ;) Commented Jun 5, 2013 at 12:44
• You can also use InfiniteLine Commented Mar 16, 2018 at 0:43
One way is to use GridLines:
f[x_] := (x^2 z)/((x^2 - y^2)^2 + 4 q^2 x^2) /. {y -> π/15, z -> 1, q -> π/600}
Plot[f[x], {x, π/15 - .1, π/15 + .1},
GridLines -> {{Pi/15 - 1/20, Pi/15 + 1/20}, {f[Pi/15], f[Pi/15]/Sqrt[2]}},
PlotRange -> All, Frame -> True, Axes -> False]
I assume you mean $$x = \frac{\pi}{15} \pm \frac{1}{200}$$. Then you can use Prolog or Epilog with InfiniteLine, like this:
Plot[
f[x],
{x, π/15 - .01, π/15 + .01},
Epilog -> {
(* add vertical lines *)
InfiniteLine[{π/15 + 1/200, 0}, {0, 1}],
InfiniteLine[{π/15 - 1/200, 0}, {0, 1}]
}
]
This does not require you to know the plot range, nor any of the function values. In addition, you are still free to use GridLines for actual grid lines.
In case you don't want to cross the $$x$$-axis, you can use HalfLine instead of InfiniteLine, and fix the position of the axis with the AxesOrigin option:
Plot[
f[x],
{x, π/15 - .01, π/15 + .01},
AxesOrigin -> {Automatic, 0},
Epilog -> {
(* add vertical lines *)
HalfLine[{π/15 + 1/200, 0}, {0, 1}],
HalfLine[{π/15 - 1/200, 0}, {0, 1}]
}
]
• I use InfiniteLine all the time for this sort of thing. Works if you log plot and as you say you don't have to worry about scaling etc. To minimise typing I would write InfiniteLine[{#, 0}, {0, 1}] & /@ {\[Pi]/15 + 1/20, \[Pi]/15 - 1/20} Highly recommended approach.
– Hugh
Commented Oct 20, 2021 at 9:13
Can use Show, but Epilog is better.
f[x_] := (x^2 z)/((x^2 - y^2)^2 + 4 q^2 x^2) /. {y -> π/15, z -> 1, q -> π/600}
plot = Plot[{f[x], f[π/15],
f[π/15]/Sqrt[2]}, {x, π/15 - .01, π/15 + .01}, PlotRange -> {{0, 0.26}, Automatic}];
Show[plot,
Graphics[{Black, Line[{{Pi/15 + 1/20, 2000}, {Pi/15 + 1/20, 9000}}]}],
Graphics[{Black, Line[{{Pi/15 - 1/20, 2000}, {Pi/15 - 1/20, 9000}}]}]]
• Why is Epilog better? Commented Mar 3, 2023 at 14:30
• @Kvothe With Epilog the plot is self-contained as a simple Graphics object rather than a more complicated Show object which would be less manipulable for further processes. Commented Mar 3, 2023 at 15:33
Another possibility is to use ParametricPlot in tandem with Show:
Show[{
Plot[{f[x], f[Pi/15], f[Pi/15]/Sqrt[2]}, {x, 0.1, 0.3},
PlotRange -> All, Frame -> True, Axes -> False],
ParametricPlot[{{Pi/15 + 1/20, u}, {Pi/15 - 1/20, u}}, {u, 0, 9000},
PlotStyle -> Black]
}]
Another possibility is to use Ticks:
Plot[{f[x], f[π/15], f[π/15]/Sqrt[2]}, {x, π/15 - .06, π/15 + .06},
Ticks -> {{{π/15 + 1/20, π/15 + 1/20, {0.595, 0}, Directive[Red, Dashed]},
{π/15 - 1/20, π/15 - 1/20, {0.595, 0}, Directive[Blue, Dashed]}},
All}, PlotRange -> {{0.12, 0.3}, All}]
• Clever manipulation of the Ticks specification, +1. Commented Jun 4, 2013 at 12:47
Next possibility is to use ListPlot:
gp1 = Plot[{({f[x], f[π/15], f[π/15]/Sqrt[2]},{x, π/15 - .05, π/15 + .05}];
ymax = Max[Last /@ Level[Cases[%, _Line, Infinity], {-2}]];
gp2 = ListPlot[{{{π/15 - 1/20, 0}, {π/15 - 1/20,ymax}}, {{π/15 + 1/20, 0}, {π/15 + 1/20, ymax}}},
Joined -> True, PlotRange -> {{0.15, 0.26}, All}];
Show[{gp1, gp2}]
• You might want to note how I formatted your previous answer, and apply that formatting to this answer... Commented Jun 4, 2013 at 7:11
• You may also find it valuable to register your account. Commented Jun 4, 2013 at 12:47
• You can also use Overlay[] instead of Show[] Commented Aug 15, 2022 at 18:34
You can simulate a vertical line by adding a line with a very large slope like maybe 100 or 1000. If your eyesight is so good that you can see the difference between a high slope and infinity, then increase the slope to a million or a billion. It is important to use the PlotRange parameter, otherwise the plot will distorted by the high slope of the verticle lines.
Plot[{f[x], f[\[Pi]/15], f[\[Pi]/15]/Sqrt[2],
1*^6 (x - (\[Pi]/15 - 1/20)),
1*^6 (x - (\[Pi]/15 + 1/20))}, {x,Pi/15 - .1, \
[Pi]/15 + .1},PlotRange -> {0, 1000}] | 2,540 | 6,904 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.802807 |
https://www.get-digital-help.com/filter-weeks-from-a-date-range/?replytocom=54361 | 1,642,678,885,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00208.warc.gz | 798,238,138 | 33,938 | Author: Oscar Cronquist Article last updated on March 12, 2019
The formula in cell B7 and C7 extracts whole weeks within the given date range in cell B3 and C3.
The following array formula in cell C7 extracts weeks that begin on a Sunday:
=IFERROR(SMALL(IF(TEXT(\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000,\$C\$3-\$B\$3))-1),"DDD")="Sat",\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000,\$C\$3-\$B\$3))-1),""),ROWS(\$A\$1:A2)),"")
Formula in cell B7:
=IFERROR(C7-6,"")
To enter an array formula, type the formula in a cell then press and hold CTRL + SHIFT simultaneously, now press Enter once. Release all keys.
The formula bar now shows the formula with a beginning and ending curly bracket telling you that you entered the formula successfully. Don't enter the curly brackets yourself.
### Explaining formula in cell C7
#### Step 1 - Create array 1 to n
The INDEX function creates a cell reference with the same number of rows as there are dates in the date range.
The ROW function then converts the cell range to an array of corresponding row numbers.
ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))
becomes
ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, 40633-40569))
becomes
ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, 64))
becomes
ROW(\$A\$1:\$A\$64)
and returns
{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; 49; 50; 51; 52; 53; 54; 55; 56; 57; 58; 59; 60; 61; 62; 63; 64}
#### Step 2 - Create dates within date range
The next step subtracts the array with 1 and adds the Excel date number.
\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1)
becomes
\$B\$3+({1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; 49; 50; 51; 52; 53; 54; 55; 56; 57; 58; 59; 60; 61; 62; 63; 64}-1)
becomes
\$B\$3+{0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; 49; 50; 51; 52; 53; 54; 55; 56; 57; 58; 59; 60; 61; 62; 63}
becomes
40569+{0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23; 24; 25; 26; 27; 28; 29; 30; 31; 32; 33; 34; 35; 36; 37; 38; 39; 40; 41; 42; 43; 44; 45; 46; 47; 48; 49; 50; 51; 52; 53; 54; 55; 56; 57; 58; 59; 60; 61; 62; 63}
and returns
{40569; 40570; 40571; 40572; 40573; 40574; 40575; 40576; 40577; 40578; 40579; 40580; 40581; 40582; 40583; 40584; 40585; 40586; 40587; 40588; 40589; 40590; 40591; 40592; 40593; 40594; 40595; 40596; 40597; 40598; 40599; 40600; 40601; 40602; 40603; 40604; 40605; 40606; 40607; 40608; 40609; 40610; 40611; 40612; 40613; 40614; 40615; 40616; 40617; 40618; 40619; 40620; 40621; 40622; 40623; 40624; 40625; 40626; 40627; 40628; 40629; 40630; 40631; 40632}
#### Step 3 - Convert dates to weekdays
The TEXT function converts the dates to weekdays.
TEXT(\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), "DDD")
becomes
TEXT({40569; 40570; 40571; 40572; 40573; 40574; 40575; 40576; 40577; 40578; 40579; 40580; 40581; 40582; 40583; 40584; 40585; 40586; 40587; 40588; 40589; 40590; 40591; 40592; 40593; 40594; 40595; 40596; 40597; 40598; 40599; 40600; 40601; 40602; 40603; 40604; 40605; 40606; 40607; 40608; 40609; 40610; 40611; 40612; 40613; 40614; 40615; 40616; 40617; 40618; 40619; 40620; 40621; 40622; 40623; 40624; 40625; 40626; 40627; 40628; 40629; 40630; 40631; 40632}, "DDD")
and returns
{"Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"}
#### Step 4 - Extract all Sundays
The IF function has three arguments, the first one must be a logical expression. If the expression evaluates to TRUE then one thing happens (argument 2) and if FALSE another thing happens (argument 3).
IF(TEXT(\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), "DDD")="Sat", \$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), "")
becomes
IF({"Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"; "Thu"; "Fri"; "Sat"; "Sun"; "Mon"; "Tue"; "Wed"}="Sun", \$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), "")
becomes
IF({FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE; FALSE; TRUE; FALSE; FALSE; FALSE}, {40569; 40570; 40571; 40572; 40573; 40574; 40575; 40576; 40577; 40578; 40579; 40580; 40581; 40582; 40583; 40584; 40585; 40586; 40587; 40588; 40589; 40590; 40591; 40592; 40593; 40594; 40595; 40596; 40597; 40598; 40599; 40600; 40601; 40602; 40603; 40604; 40605; 40606; 40607; 40608; 40609; 40610; 40611; 40612; 40613; 40614; 40615; 40616; 40617; 40618; 40619; 40620; 40621; 40622; 40623; 40624; 40625; 40626; 40627; 40628; 40629; 40630; 40631; 40632}, "")
and returns
{""; ""; ""; ""; 40573; ""; ""; ""; ""; ""; ""; 40580; ""; ""; ""; ""; ""; ""; 40587; ""; ""; ""; ""; ""; ""; 40594; ""; ""; ""; ""; ""; ""; 40601; ""; ""; ""; ""; ""; ""; 40608; ""; ""; ""; ""; ""; ""; 40615; ""; ""; ""; ""; ""; ""; 40622; ""; ""; ""; ""; ""; ""; 40629; ""; ""; ""}
#### Step 5 - Find n-th Sunday within range
To be able to return a new value in a cell each I use the SMALL function to filter date numbers from smallest to largest.
SMALL(IF(TEXT(\$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), "DDD")="Sat", \$B\$3+(ROW(\$A\$1:INDEX(\$A\$1:\$A\$1000, \$C\$3-\$B\$3))-1), ""),ROWS(\$A\$1:A2))
becomes
SMALL({""; ""; ""; ""; 40573; ""; ""; ""; ""; ""; ""; 40580; ""; ""; ""; ""; ""; ""; 40587; ""; ""; ""; ""; ""; ""; 40594; ""; ""; ""; ""; ""; ""; 40601; ""; ""; ""; ""; ""; ""; 40608; ""; ""; ""; ""; ""; ""; 40615; ""; ""; ""; ""; ""; ""; 40622; ""; ""; ""; ""; ""; ""; 40629; ""; ""; ""}),ROWS(\$A\$1:A2))
becomes
SMALL({""; ""; ""; ""; 40573; ""; ""; ""; ""; ""; ""; 40580; ""; ""; ""; ""; ""; ""; 40587; ""; ""; ""; ""; ""; ""; 40594; ""; ""; ""; ""; ""; ""; 40601; ""; ""; ""; ""; ""; ""; 40608; ""; ""; ""; ""; ""; ""; 40615; ""; ""; ""; ""; ""; ""; 40622; ""; ""; ""; ""; ""; ""; 40629; ""; ""; ""}),2)
and returns 40580 formatted as 2/5/2011.
#### Step 6 - Return blank if error
The IFERROR function handles errors, it returns a specified value if formula returns an error. In this case it returns a blank "".
### Get the Excel file
Extract-week-ranges-based-on-a-given-date-range-1.xlsx | 3,282 | 7,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.652177 |
https://www.skillsworkshop.org/maths?q=maths&search=&f%5B0%5D=area%3A524&f%5B1%5D=area%3A952&f%5B2%5D=level%3A53&f%5B3%5D=level%3A54&f%5B4%5D=level%3A55 | 1,624,407,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00233.warc.gz | 899,370,743 | 13,384 | # Adult Numeracy, Functional Maths, and GCSE Resources
Displaying 1 - 10 of 231 resources:
## Making a mini wildlife pond
Editor's note June 20 8am - Minor changes to Q18 & Q23
Pat and Mitch decide to build a mini wildlife pond from an old sink. This involves digging, gathering and arranging suitable items, a trip to the aquatic centre to buy water plants, and then sitting back and observing the fruit of their labours :)
Level
E1
E2
Maths
FM Simple one step problem(s)
FM E1.1 Read, write, order and compare numbers up to 20
E2.1 Count reliably up to 100 items
E1.8 Describe and make comparisons in words between measures of items including size, length, width, height, weight and capacity
E2.14 Use metric measures of length including millimetres, centimetres, metres and kilometres
E3.16 Compare measures of weight including grams and kilograms
E2.16 Use measures of capacity including millilitres and litres
E1.9 Identify & recognise common 2-D and 3-D shapes including circle, cube, rectangle (incl. square) and triangle
E2.19 Recognise and name 2-D and 3-D shapes including pentagons, hexagons, cylinders, cuboids, pyramids, spheres
E1.10 Use everyday positional vocabulary to describe position and direction including left, right, in front, behind, under and above
E2.21 Use appropriate positional vocabulary to describe position and direction including between, inside, outside, middle, below, on top, forwards and backwards
E1.7 Know the number of days in a week, months, and seasons in a year. Be able to name and sequence
E1.12 Sort and classify objects using a single criterion
E2.24 Sort and classify objects using two criteria
E1.13 Read and draw simple charts and diagrams including a tally chart, block diagram/graph
E2.22 Extract information from lists, tables, diagrams and bar charts
E2.23 Make numerical comparisons from bar charts
E2.25 Take information from one format and represent the information in another format including use of bar charts
Context
Science, Nature & Weather
Gardening & Horticulture
## Properties of 2D Shapes - quadrilaterals and triangles
Name each 2D shape and describe its properties. Two separate worksheets: one for triangles, one for quadrilaterals. For Entry Level, hence angles are not a focus.
For each of the 6 quadrilaterals and 4 triangles, the learner is prompted to answer these questions:
Level
E2
E3
Maths
FM Context free underpinning
E2.19 Recognise and name 2-D and 3-D shapes including pentagons, hexagons, cylinders, cuboids, pyramids, spheres
E2.20 Describe properties of common 2-D & 3-D shapes including nos. of sides, corners, edges, faces, angles & base
E3.19 Sort 2-D and 3-D shapes using properties including lines of symmetry, length, right angles, angles including in rectangles and triangles
## Properties of 3D Objects for Entry Level Functional Maths
Name each 3D object and fill in the gaps about their properties.
For each of the 9 common 3D shapes, the learner is prompted to complete information about:
• its name
• the number of curved and/or flat faces
• the shape of the faces
• the number of vertices (corners)
• the number of edges
Editor's note
Answer sheet with detailed Functional Skills Maths mapping is available to contributors only
Level
E1
E2
E3
Maths
FM Context free underpinning
E1.9 Identify & recognise common 2-D and 3-D shapes including circle, cube, rectangle (incl. square) and triangle
E2.19 Recognise and name 2-D and 3-D shapes including pentagons, hexagons, cylinders, cuboids, pyramids, spheres
E2.20 Describe properties of common 2-D & 3-D shapes including nos. of sides, corners, edges, faces, angles & base
E3.19 Sort 2-D and 3-D shapes using properties including lines of symmetry, length, right angles, angles including in rectangles and triangles
## Properties of 2D Shapes for Entry Level Functional Maths
Name each 2D shape and describe its properties.
For each of the 8 common shapes, the learner is prompted to answer these questions:
Level
E1
E2
E3
Maths
FM Context free underpinning
E1.9 Identify & recognise common 2-D and 3-D shapes including circle, cube, rectangle (incl. square) and triangle
E2.19 Recognise and name 2-D and 3-D shapes including pentagons, hexagons, cylinders, cuboids, pyramids, spheres
E2.20 Describe properties of common 2-D & 3-D shapes including nos. of sides, corners, edges, faces, angles & base
E3.19 Sort 2-D and 3-D shapes using properties including lines of symmetry, length, right angles, angles including in rectangles and triangles
## Maths for the Summer
An updated version of 'Numeracy during the Summer' by Judith White (2006).
Numeracy during the summer. Keep your students’ skills going during the summer break with these ideas. One page for Entry Level learners, one for Levels 1/2.
Covers number, measures, shape and space, and data.
Level
E2
E3
L1
L2
Maths
FM Contextualised underpinning
General numeracy / maths
General
Generic resources for literacy, numeracy and beyond
## Leap year Functional Maths challenge
A hastily written resource (I didn’t want to have to wait another 4 years!) on Feb 29th 2020 (but can be used throughout the leap year). Covers Functional Skills (Measures) content descriptors relating to using dates and units of time. Number topics such as estimation and checking, multiplication, division, odd and even numbers, and sequences are also included. There is an emphasis is on using non-calculator written methods to convert between units of time.
Level
E2
E3
L1
L2
Maths
FM Contextualised underpinning
FM Simple one step problem(s)
FM Straightforward problem(s) with more than 1 step
FM L1.4 Use multiplication facts and make connections with division facts
L2.2 Carry out calculations with numbers up to one million including strategies to check answers including estimation and approximation
E2.13 Read and record time in common date formats, and read time displayed on analogue clocks in hours, half hours and quarter hours, and understand hours from a 24-hour digital clock
FM L1.20 Convert between units of length, weight, capacity, money and time, in the same system
General
Generic resources for literacy, numeracy and beyond
## HS2 Compass points, position and bearings
Covers all reformed Functional Skills content descriptors relating to position and direction (from E1 to L1). Written with mixed-level classes in mind. Positional vocabulary; cardinal and intercardinal compass points; turns (e.g. quarter turns) and bearings are all covered in this set of graduated HS2 (high speed railway) themed problems. Problems are interspersed with context-free underpinning questions, examples and tips.
Level
E1
E2
E3
L1
Maths
FM Context free underpinning
FM Simple one step problem(s)
FM Straightforward problem(s) with more than 1 step
E1.10 Use everyday positional vocabulary to describe position and direction including left, right, in front, behind, under and above
E2.21 Use appropriate positional vocabulary to describe position and direction including between, inside, outside, middle, below, on top, forwards and backwards
E3.20 Use appropriate positional vocabulary to describe position and direction including eight compass points and including full/half/quarter turns
L1.26 Use angles when describing position and direction, and measure angles in degrees
Context
Leisure Travel Tourism
Motor vehicles & Transport
## Vending Machine Functional Maths
A set of Entry Level tasks - all based on a vending machine. The main focus is money but positional vocabulary is also covered. Skills covered include: recognising letters and numbers for items in a vending machine, identifying coins to pay with, working out change from £1, checking answers, rounding and estimating, distinguishing between right and left.
Editor's note
A delightfully functional resource. With teaching notes. Fully mapped to the Reformed Functional Skills content.
Level
E1
E2
E3
Maths
FM Simple one step problem(s)
FM E1.1 Read, write, order and compare numbers up to 20
E2.2 Read, write, order and compare numbers up to 200
FM E2.5 Add and subtract two-digit numbers
E1.5 Recognise coins and notes and write them in numbers with the correct symbols (£ & p), where these involve numbers up to 20
E2.12 Calculate money with pence up to one pound and in whole pounds of multiple items and write with the correct symbols (£ or p)
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.11 Round amounts of money to the nearest £1 or 10p
E1.10 Use everyday positional vocabulary to describe position and direction including left, right, in front, behind, under and above
E2.21 Use appropriate positional vocabulary to describe position and direction including between, inside, outside, middle, below, on top, forwards and backwards
Context
Independent living
Catering Food Nutrition
## The best mince pies - Functional Maths tasks
Sparked after reading an article about a UK Which? report in The Guardian newspaper, and made in haste, this resource is aimed at Entry Level 3 (E3) and Level 1 (L1) learners but includes some E1-2 and L2 questions. There is no answer sheet – one might appear later for skillsworkshop contributors – but the teaching notes do include general marking guidance.
Level
E3
L1
L2
Maths
FM Simple one step problem(s)
FM Straightforward problem(s) with more than 1 step
FM L1.16 Recognise and calculate equivalences between common fractions, percentages and decimals
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.11 Round amounts of money to the nearest £1 or 10p
FM L1.18 Calculate simple interest in multiples of 5% on amounts of money
FM L1.29 Find the mean and range of a set of quantities
FM L2.23 Calculate the median and mode of a set of quantities
Context
Catering Food Nutrition
## Bonfire Night Maths for Entry 2 Functional Skills
Written to look at the reformed Entry Level 2 ‘Measures, shape and space’ content, the main focus of this resource is positional vocabulary, 3D shapes and time. It also covers aspects of Number and Data. It is not recommended for the initial teaching of skills (underpinning). If used for this, ensure all the ‘maths words’ (p1) are covered in advance and supply learners with real-life materials such as counters (I use pasta or dried beans) and 3D shapes where needed.
Level
E2
Maths
FM Simple one step problem(s)
E2.2 Read, write, order and compare numbers up to 200
E2.19 Recognise and name 2-D and 3-D shapes including pentagons, hexagons, cylinders, cuboids, pyramids, spheres
E2.20 Describe properties of common 2-D & 3-D shapes including nos. of sides, corners, edges, faces, angles & base
E1.10 Use everyday positional vocabulary to describe position and direction including left, right, in front, behind, under and above
E2.21 Use appropriate positional vocabulary to describe position and direction including between, inside, outside, middle, below, on top, forwards and backwards | 2,642 | 10,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-25 | latest | en | 0.846849 |
http://ccuart.org/how-to-find-vertical-and-horizontal-asymptotes-on-a-graph.html | 1,513,532,550,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597295.74/warc/CC-MAIN-20171217171653-20171217193653-00330.warc.gz | 46,761,336 | 20,145 | # How To Find Vertical And Horizontal Asymptotes On A Graph
How To Find Vertical And Horizontal Asymptotes On A Graph. Thank You for visiting CCUART. Nowadays were excited to declare that we have discovered an incredibly interesting topic to be pointed out, namely How To Find Vertical And Horizontal Asymptotes On A Graph. Lots of people attempting to find info about master bedroom design ideas and certainly one of them is you, is not it?
## Explanation About How To Find Vertical And Horizontal Asymptotes On A Graph according to CCUART
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Graph Of Fx3x22 If Fx Px Qx Is A Rational Function In Which Px And Qx Have No Common Factors And A Is A Zero Of Qx The Denominator Then Xa Is A Vertical This Graph Follows A Horizontal Line Red In The Diagram As It Moves Out Of The System To The Left Or Right This Is A Horizontal Asymptote With The Describing Asymptotic Behavior In Terms Of Limits Involving Infinity .
### How To Find Vertical And Horizontal Asymptotes On A Graph basic understanding
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2: This rational function also has vertical and horizontal asymptotes. The vertical asymptote is at x = 2 and there is a horizontal asymptote at y = 4..
## Graph of f(x)=(3x^2+2)/(.
### .
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For each equation, write the. | 1,976 | 8,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-51 | longest | en | 0.89472 |
https://www.studyhawks.com/question/which-one-is-bigger-60-of-90-or-12-pa-100-854134/ | 1,580,008,289,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00499.warc.gz | 1,095,264,978 | 9,817 | ## Use our writing service to score better and meet your deadlines
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# Which one is bigger 60% of 90 or 1/2 pa 100?
60% of 90 = 90 * 0.60 = 54
1/2 of 100 = 100/2 = 50
So, as 54 > 50, 60% of 90 > 1/2 of 100
So, your answer is 60% of 90 is bigger!
Hope this helps!
## Related Questions
Posted in Mathematics
Which ordered pair is a solution of the equation 2x − y = 9 (-4,1) (-2,5) (5,1) (6,-3)
In order to confirm if which ordered pair is a solution of the given equation above, we can just simply plug in the values.
So based on my solution, the correct answer would be the third option: (5,1)
So let us try to plug in the values.
2(5) - 1 = 9
10 -1 = 9
9 = 9
So this is the correct one. Hope this answers your question.
Posted in Mathematics
What is 1 1/2 plus 2/3?
It would be 1 7/6 which can be simplified to 2 1/6
hope this helps you
Posted in Mathematics
Algebraic terms are separated by ______. Choose all that apply. 1)= 2)+ 3)x 4)÷ 5)-
1)=
2)+
5)-
Explanation
When letter or numbers are separated by multiplication or division can be combined to make one term.
In an algebra equation, terms are separated by an addition sign, subtraction sign or an equal sign. For example,
3a + 7b = 21.
This equation have 3 terms.
Posted in Mathematics
Each set of ordered pairs represents a function. Write a rule that represents the function. 1. (0,0),(1,4),(2,16),(3,36),(4,64) 2. (0,0),(1,0.5),(2,2),(3,4.5),(4,8)
My answer to the question is as follows:
First one looks like you are squaring the number, then multiplying the result by 4, i.e.
y=4x2
second one is similar, but instead of squaring and multiplying by 4, you are squaring and then dividing by 2
I hope my answer has come to your help. God bless and have a nice day ahead!
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Posted in Mathematics
Ava was making muffins. she used 1 1/3 tsp of cinnamon and 1/2 tsp of nutmeg. how many teaspoons of spice did ava use?
1 5/6
1 1/3 = 4/3 (1x3+1)
4/3 and 1/2 do not have the same denominator so you'd have to find the lowest common denominator, which is 6. the new fractions would be 8/6 and 3/6 which would equal 11/6 or 1 5/6!
Hope that helps!
Posted in Mathematics
A department store has a discount on shoes based on a percentage of the price suppose one pair of shoes is marked down from $70 to$49 what is the price for a $100 pair of shoes after the discount is applied We first need to find the difference between$70 and $49.$70 - $49 = 21 Now we find out what percent$21 is of $70. 21/70 × 100 = 30%. That means the discount is 30%! Now we must find 30% of$100, which is $30. Then we subtract$30 from $100.$100 - $30 = 70$
Posted in Mathematics
6 1/2 as an improper fraction
6 1/2 as an improper fraction is 13/2
Posted in Mathematics
1735/1000 is what in decimal form
To write 1735/1000 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 1735 by 1000 what we write down as 1735/1000 and we get 1.735 1735/1000 as a decimal equals 1.735
Posted in Mathematics
You can buy an 8 pack of paper towels for $7.92 or 12 pack for$11.64, which is a better buy?
Divide $7.92 by 8 and then divide$11.64 by 12 and see which has the lowest answer.
The first option has a price of $0.99 for one paper towel, and the second option has a price of$0.97 for one paper towel.
The second option is the better buy.
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If the purchase price for a house is $445,500, what is the monthly payment if you put 5% down for a 30 year loan with a fixed rate of 6.25%? Answer: The monthly payment is$2603.17
Step-by-step explanation:
The purchase price is = $445500 5% is down payment = Loan amount is = The EMI formula is = p = 423225 r = 6.25/12/100=0.0052 n = 30*12 = 360 Putting the values in the formula we get: =$2603.17
The monthly payment is $2603.17 approx Posted in Mathematics A car travel 90 Km in 2 1/2 hours. How much time is required to cover 30 Km and find the distance cover in 2 hours 90:2 1/2 90/3 : (2 1/2)/3 30 : (2 1/2)/3 30 : (5/2)/(3/1) 30 : 5/2*1/3 30 : 5/6 30km in 50 mins 90 : 2 1/2 90/(2 1/2) : (2 1/2) / (2 1/2) 90/(2 1/2) : 1 (90/1)/(5/2) : 1 (90/1)*(2/5) : 1 180/5 :1 36:1 Multiply each side by 2: 72:2 72 km in 2 hours Hope this helps :) Posted in Mathematics What is the equation in point slope form of the line that passes through the point (–2, 4)(–2, 4) and has a slope of 5? Point-slope formula: y-y1=m(x-x1) Now we'll plug in the numbers they give us into the formula: y-4= 0(x+2). Posted in Mathematics Expand and simplify (2t-3)^2 (2t-3)²=4t²+9-(2t.(-3))=4t²+9-6t ## Use our writing service to score better and meet your deadlines No matter what kind of academic paper you need, it is simple and secure to hire a writer for a price you can afford at StudyHawks. Save more time for yourself. Posted in Mathematics Your$1000 investment in an energy stock doubles. Your $1000 investment in a tech stock drops in half. What was the total return on the two investments? Total return, when measuring performance, is the actual rate of return of aninvestment or a pool of investments over a given evaluation period. Total returnincludes interest, capital gains, dividends and distributions realized over a given period of time. So the total return = (1000*2) + (1000/2) =$2500
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Which expression represents the sum of 2/3m - 1 1/6 and 5/6 - 1 1/3?a. 1 1/2m + 2 1/2b.1/6m - 1/3c.1 1/2m - 2 1/2d.1/6m + 1/3
The answer to "which expression represents the sum of 2/3m - 1 1/6 and 5/6 - 1 1/3?a. 1 1/2m + 2 1/2b.1/6m - 1/3c.1 1/2m - 2 1/2d.1/6m + 1/3" is B.
Posted in Mathematics
Which of the following ordered pairs corresponds to an arrow on a mapping diagram that starts at 4 and ends at 1? Map Diagram: Input 1 2 3 4 5 Output 1 2 3 4 5 The arrow points from 4 under input to 1 under output. Answer Choices are: A. (1, 1) B. (1, 4) C. (4, 1) D. (4, 4)
Input options represent first number in ordered pair and output options represent second number in ordered pair.
We can see that arrow starts from 4 and is pointing to 1 which means:
first number is 4 and second is 1
Ordered pair is:
(4,1)
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Which is bigger 0.07 or 0.005?
0.07 is the larger number out of the two
Posted in Mathematics
Write two equivalent fractions for each. 1/2 and 4/5
An equivalent fraction for 1/2 would be 2/4. This is because you can simplify 2/4 by dividing it by 2. The final answer would be 1/2.
An equivalent fraction for 4/5 would be 8/10. It's basically the same thing. Divide them both by 2 and you'll get 4/5 since 8 divided by 2 is 4, and 10 divided by 2 is 5.
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1/100 is equal to 7/x
1/100 = 7/x
x = 7*100
x = 700
Hope this helps!
Posted in Mathematics
Paper cost $49 for 10 reams. How much for 1 ream? For 10 ream, it's cost =$49
So, for 1 ream, it would be: $49/10 =$4.9 | 2,461 | 7,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-05 | longest | en | 0.932215 |
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# Prime Numbers present at Kth level of a Binary Tree
Given a number K, the task is to print the prime numbers present at that level given all the prime numbers are represented in the form of a binary tree
Examples:
```Input: K = 3
2
/ \
3 5
/\ / \
7 11 13 17
Output :7, 11, 13, 17
Explanation:
2
/ \
3 5
/\ / \
7 11 13 17
So primes present at level 3 : 7, 11, 13, 17
Input :K = 2
2
/ \
3 5
Output :3 5```
Naive Approach: The naive approach is to build a binary tree of prime numbers and then get elements at a particular level k.
It doesn’t work well for large numbers as it takes too much time.
Efficient approach: Suppose there are n elements and the task is to build a binary tree using those n elements, then they can be built using log2n levels.
Therefore, given a level k, elements present here is from 2k-1 to 2k-1 if all the prime numbers are present in a 1D array.
Hence, the following is the algorithm:
1. Find the prime numbers upto MAX_SIZE using Sieve of Eratosthenes.
2. Calculate the left_index and right_index of the level as left_index = 2k-1, right_index = 2k-1.
3. Output primes from left_index to right_index of prime array.
## C++
`// CPP program of the approach``#include ``using` `namespace` `std;` `// initializing the max value``#define MAX_SIZE 1000005` `// To store all prime numbers``vector<``int``> primes;` `// Function to generate N prime numbers using``// Sieve of Eratosthenes``void` `SieveOfEratosthenes(vector<``int``>& primes)``{`` ``// Create a boolean array "IsPrime[0..MAX_SIZE]" and`` ``// initialize all entries it as true. A value in`` ``// IsPrime[i] will finally be false if i is`` ``// Not a IsPrime, else true.`` ``bool` `IsPrime[MAX_SIZE];`` ``memset``(IsPrime, ``true``, ``sizeof``(IsPrime));` ` ``for` `(``int` `p = 2; p * p < MAX_SIZE; p++) {`` ``// If IsPrime[p] is not changed, then it is a prime`` ``if` `(IsPrime[p] == ``true``) {`` ``// Update all multiples of p greater than or`` ``// equal to the square of it`` ``// numbers which are multiple of p and are`` ``// less than p^2 are already been marked.`` ``for` `(``int` `i = p * p; i < MAX_SIZE; i += p)`` ``IsPrime[i] = ``false``;`` ``}`` ``}` ` ``// Store all prime numbers`` ``for` `(``int` `p = 2; p < MAX_SIZE; p++)`` ``if` `(IsPrime[p])`` ``primes.push_back(p);``}` `void` `printLevel(``int` `level)``{` ` ``cout << ``"primes at level "` `<< level << ``": "``;`` ``int` `left_index = ``pow``(2, level - 1);`` ``int` `right_index = ``pow``(2, level) - 1;`` ``for` `(``int` `i = left_index; i <= right_index; i++) {` ` ``cout << primes[i - 1] << ``" "``;`` ``}`` ``cout << endl;``}` `// Driver Code``int` `main()``{`` ``// Function call`` ``SieveOfEratosthenes(primes);` ` ``printLevel(1);`` ``printLevel(2);`` ``printLevel(3);`` ``printLevel(4);` ` ``return` `0;``}`
## Java
`// Java program of the approach``import` `java.util.*;` `class` `GFG``{` ` ``// initializing the max value`` ``static` `final` `int` `MAX_SIZE = ``1000005``;` ` ``// To store all prime numbers`` ``static` `Vector primes = ``new` `Vector();` ` ``// Function to generate N prime numbers using`` ``// Sieve of Eratosthenes`` ``static` `void` `SieveOfEratosthenes(Vector primes)`` ``{`` ` ` ``// Create a boolean array "IsPrime[0..MAX_SIZE]" and`` ``// initialize all entries it as true. A value in`` ``// IsPrime[i] will finally be false if i is`` ``// Not a IsPrime, else true.`` ``boolean``[] IsPrime = ``new` `boolean``[MAX_SIZE];`` ``for` `(``int` `i = ``0``; i < MAX_SIZE; i++)`` ``IsPrime[i] = ``true``;` ` ``for` `(``int` `p = ``2``; p * p < MAX_SIZE; p++)`` ``{`` ` ` ``// If IsPrime[p] is not changed, then it is a prime`` ``if` `(IsPrime[p] == ``true``)`` ``{`` ` ` ``// Update all multiples of p greater than or`` ``// equal to the square of it`` ``// numbers which are multiple of p and are`` ``// less than p^2 are already been marked.`` ``for` `(``int` `i = p * p; i < MAX_SIZE; i += p)`` ``IsPrime[i] = ``false``;`` ``}`` ``}` ` ``// Store all prime numbers`` ``for` `(``int` `p = ``2``; p < MAX_SIZE; p++)`` ``if` `(IsPrime[p])`` ``primes.add(p);`` ``}` ` ``static` `void` `printLevel(``int` `level)`` ``{` ` ``System.out.print(``"primes at level "` `+ level + ``": "``);`` ``int` `left_index = (``int``) Math.pow(``2``, level - ``1``);`` ``int` `right_index = (``int``) (Math.pow(``2``, level) - ``1``);`` ``for` `(``int` `i = left_index; i <= right_index; i++)`` ``{` ` ``System.out.print(primes.get(i - ``1``) + ``" "``);`` ``}`` ``System.out.println();`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// Function call`` ``SieveOfEratosthenes(primes);` ` ``printLevel(``1``);`` ``printLevel(``2``);`` ``printLevel(``3``);`` ``printLevel(``4``);` ` ``}``}` `// This code is contributed by Rajput-Ji`
## Python3
`# Python3 program of the approach``MAX_SIZE ``=` `1000005``primes ``=` `[]` `# Function to generate N prime numbers using``# Sieve of Eratosthenes``def` `SieveOfEratosthenes():`` ` ` ``# Create a boolean array "IsPrime[0..MAX_SIZE]" and`` ``# initialize all entries it as True. A value in`` ``# IsPrime[i] will finally be false if i is`` ``# Not a IsPrime, else True.`` ``IsPrime ``=` `[``True``] ``*` `MAX_SIZE`` ``p ``=` `2` ` ``while` `p ``*` `p < MAX_SIZE:`` ` ` ``# If IsPrime[p] is not changed, then it is a prime`` ``if` `(IsPrime[p] ``=``=` `True``):`` ` ` ``# Update all multiples of p greater than or`` ``# equal to the square of it`` ``# numbers which are multiple of p and are`` ``# less than p^2 are already been marked.`` ``for` `i ``in` `range``(p ``*` `p, MAX_SIZE, p):`` ``IsPrime[i] ``=` `False`` ``p ``+``=` `1` ` ``# Store all prime numbers`` ``for` `p ``in` `range``(``2``, MAX_SIZE):`` ``if` `(IsPrime[p]):`` ``primes.append(p)` `def` `printLevel(level):` ` ``print``(``"primes at level "``, level, ``":"``, end``=``" "``)`` ``left_index ``=` `pow``(``2``, level ``-` `1``)`` ``right_index ``=` `pow``(``2``, level) ``-` `1`` ``for` `i ``in` `range``(left_index, right_index ``+` `1``):` ` ``print``(primes[i ``-` `1``], end``=``" "``)`` ``print``()` `# Driver Code` `# Function call``SieveOfEratosthenes()` `printLevel(``1``)``printLevel(``2``)``printLevel(``3``)``printLevel(``4``)` `# This code is contributed by mohit kumar 29`
## C#
`// C# program of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` ` ``// initializing the max value`` ``static` `readonly` `int` `MAX_SIZE = 1000005;` ` ``// To store all prime numbers`` ``static` `List<``int``> primes = ``new` `List<``int``>();` ` ``// Function to generate N prime numbers using`` ``// Sieve of Eratosthenes`` ``static` `void` `SieveOfEratosthenes(List<``int``> primes)`` ``{`` ` ` ``// Create a bool array "IsPrime[0..MAX_SIZE]" and`` ``// initialize all entries it as true. A value in`` ``// IsPrime[i] will finally be false if i is`` ``// Not a IsPrime, else true.`` ``bool``[] IsPrime = ``new` `bool``[MAX_SIZE];`` ``for` `(``int` `i = 0; i < MAX_SIZE; i++)`` ``IsPrime[i] = ``true``;` ` ``for` `(``int` `p = 2; p * p < MAX_SIZE; p++)`` ``{`` ` ` ``// If IsPrime[p] is not changed, then it is a prime`` ``if` `(IsPrime[p] == ``true``)`` ``{`` ` ` ``// Update all multiples of p greater than or`` ``// equal to the square of it`` ``// numbers which are multiple of p and are`` ``// less than p^2 are already been marked.`` ``for` `(``int` `i = p * p; i < MAX_SIZE; i += p)`` ``IsPrime[i] = ``false``;`` ``}`` ``}` ` ``// Store all prime numbers`` ``for` `(``int` `p = 2; p < MAX_SIZE; p++)`` ``if` `(IsPrime[p])`` ``primes.Add(p);`` ``}` ` ``static` `void` `printLevel(``int` `level)`` ``{` ` ``Console.Write(``"primes at level "` `+ level + ``": "``);`` ``int` `left_index = (``int``) Math.Pow(2, level - 1);`` ``int` `right_index = (``int``) (Math.Pow(2, level) - 1);`` ``for` `(``int` `i = left_index; i <= right_index; i++)`` ``{` ` ``Console.Write(primes[i - 1] + ``" "``);`` ``}`` ``Console.WriteLine();`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main(String[] args)`` ``{`` ``// Function call`` ``SieveOfEratosthenes(primes);` ` ``printLevel(1);`` ``printLevel(2);`` ``printLevel(3);`` ``printLevel(4);` ` ``}``}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output
```primes at level 1: 2
primes at level 2: 3 5
primes at level 3: 7 11 13 17
primes at level 4: 19 23 29 31 37 41 43 47
```
Approach Steps:
• uses a queue data structure to keep track of nodes at each level and checks each dequeued node for primality.
• If a node is prime and at the desired level, it is added to a list of prime numbers at that level.
• After all nodes at the current level have been processed, the list of prime numbers is printed in the desired format and variables are updated to process the next level.
• Last ensures that nodes at each level are processed before moving on to the next level.
• and the queue ensures that nodes are processed in the order in which they appear at each level.
Below is the code implementation:
## Python
`import` `math``from` `Queue ``import` `Queue` `# Binary Tree node definition``class` `Node:`` ``def` `__init__(``self``, val``=``None``, left``=``None``, right``=``None``):`` ``self``.val ``=` `val`` ``self``.left ``=` `left`` ``self``.right ``=` `right` `# Function to print all prime numbers at level k of a binary tree``def` `print_primes_at_level(root, k):`` ` ` ``q ``=` `Queue()`` ``q.put(root)` ` ``curr_level ``=` `1`` ``curr_nodes ``=` `1`` ``next_nodes ``=` `0` ` ``primes ``=` `[]` ` ``# Loop until all levels have been traversed`` ``while` `not` `q.empty():`` ``node ``=` `q.get()`` ``if` `is_prime(node.val) ``and` `curr_level ``=``=` `k:`` ``primes.append(node.val)` ` ``if` `node.left:`` ``q.put(node.left)`` ``next_nodes ``+``=` `1`` ``if` `node.right:`` ``q.put(node.right)`` ``next_nodes ``+``=` `1` ` ``curr_nodes ``-``=` `1`` ``if` `curr_nodes ``=``=` `0``:`` ``if` `primes:`` ``print``(``"primes at level {}: {}"``.``format``(k, ``' '``.join(``str``(p) ``for` `p ``in` `primes)))` ` ``primes ``=` `[]`` ``curr_level ``+``=` `1`` ``curr_nodes ``=` `next_nodes`` ``next_nodes ``=` `0` ` ``if` `curr_level > k:`` ``break` `# Function to check if a number is prime``def` `is_prime(num):`` ``if` `num < ``2``:`` ``return` `False`` ``for` `i ``in` `range``(``2``, ``int``(math.sqrt(num))``+``1``):`` ``if` `num ``%` `i ``=``=` `0``:`` ``return` `False`` ``return` `True` `# Example usage``root ``=` `Node(``2``, Node(``3``, Node(``7``), Node(``11``)), Node(``5``, Node(``13``), Node(``17``)))``print_primes_at_level(root, ``1``)``print_primes_at_level(root, ``2``)``print_primes_at_level(root, ``3``)`
Output
```primes at level 1: 2
primes at level 2: 3 5
primes at level 3: 7 11 13 17
```
Time Complexity: O(n), where n is the number of nodes in the binary tree.
Auxiliary Space: O(m), where m is the maximum number of nodes at a single level in the binary tree.
My Personal Notes arrow_drop_up | 4,227 | 12,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.721042 |
affarerlkyu.firebaseapp.com | 1,670,485,502,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00017.warc.gz | 107,310,141 | 7,088 | # Sin cube theta barabar
Sign Square Theta Minus Cos Square Theta Barabar Kya Hota Hai? साइन स्क्वायर थीटा माइनस कौस स्क्वायर थीटा बराबर क्या होता है? #2 Answers, Listen to Expert Answers on Vokal - India’s Largest Question & Answers Platform in 11 Indian Languages.
Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. #= cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin^2 theta - i sin^3 theta# #= (cos^3 theta - 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)# Then equating real and imaginary parts, we find: if x sin cube theta ycos cube theta sin theta cos theta andx sin theta y cos theta then a x cube y cube 1 b x square y square 1 c x square y square 1 - Mathematics - TopperLearning.com | mdptwjvv So, you think you’ve drafted a tweet, an email, a short story, or even a novel. These are different forms of communication, but the process of bringing them to fruition has a necessary, sometimes overlooked step: editing!
1 + 10 स्क्वायर थीटा ईंटो कॉश थीटा ईंटो सिन थीटा किसके बराबर होगा? #1 Answers, Listen to Expert Answers on Vokal - India’s Largest Question & Answers Platform in 11 Indian Languages. To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW If sin theta + cos theta =1 then sin theta cos theta =MOST IMPORTANT QUESTIO thank you theta upon sin theta cos cos theta + sin thank you theta + cos cube theta upon sin theta + cos theta + sin theta minus cos theta upon sin theta - 17206999 Answer: 1 📌📌📌 question Sin cube theta + cos cube theta by sin theta + cos theta - the answers to estudyassistant.com You can choose C = c +π, and then sin(θ +C) = −sin(θ+ c), so having the ± there doesn't create any more solutions if you allow any c ∈ [0,2π). Sign Square Theta Plus Cos Square Theta Barabar 1 Siddh Karen?
## You can choose C = c +π, and then sin(θ +C) = −sin(θ+ c), so having the ± there doesn't create any more solutions if you allow any c ∈ [0,2π).
Mathematics Trigonometric Equation The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. From these formulas, we also have the following identities for Apr 30, 2001 · I know that sin^2(theta) = 1 - cos 2Xtheta. Is there a similar formula relating sin^3(theta) to sin(3Xtheta) or cos(3Xtheta)? CalMeacham April 30, 2001, 5:54pm I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers." I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found.
### Related Answer. The expression cos3θ+sin3θ+(2sin2θ-3)(sinθ-cosθ) is positive for all θ in. More Related Question & Answers. If sin^(3)theta+cos^(3)theta=0.
The cube is traslated so that its centered at the origin which is defined by having a minimum corner (-1,-1,-1) at left-bottom-near and a mximum corner (1,1,1) at right-top-far. The sin β leg, as hypotenuse of another right triangle with angle α, likewise leads to segments of length cos α sin β and sin α sin β. Now, we observe that the "1" segment is also the hypotenuse of a right triangle with angle α + β; the leg opposite this angle necessarily has length sin(α + β), while the leg adjacent has length cos(α See full list on calculus.subwiki.org Geometric. The red section on the right, d, is the difference between the lengths of the hypotenuse, H, and the adjacent side, A.As is shown, H and A are almost the same length, meaning cos θ is close to 1 and θ 2 / 2 helps trim the red away. $\begingroup$ Yes, I'm sorry. in the book, it's theta, so when I go to pick an arbitrary variable.
Problem 3. A 3-4-5 triangle is right-angled. a) Why? To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload"). Mathematics Trigonometric Equation The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. From these formulas, we also have the following identities for Apr 30, 2001 · I know that sin^2(theta) = 1 - cos 2Xtheta. Is there a similar formula relating sin^3(theta) to sin(3Xtheta) or cos(3Xtheta)?
If acoscube theta+3acos theta sin square theta=m and asin cube theta +3acos square theta sin theta Find (m+n)power2/3+(m-n)power2/3 - Math - The sin β leg, as hypotenuse of another right triangle with angle α, likewise leads to segments of length cos α sin β and sin α sin β. Now, we observe that the "1" segment is also the hypotenuse of a right triangle with angle α + β; the leg opposite this angle necessarily has length sin(α + β), while the leg adjacent has length cos(α Geometric. The red section on the right, d, is the difference between the lengths of the hypotenuse, H, and the adjacent side, A.As is shown, H and A are almost the same length, meaning cos θ is close to 1 and θ 2 / 2 helps trim the red away. The antiderivative of involves cos^3 and cos, both of which can be antidifferentiated, and this now involves sin^3 and sin. We can thus antidifferentiate (i.e., integrate) the function any number of times, with the antiderivative expression alternating between a cubic function of sine and a cubic function of cosine.
From these formulas, we also have the following identities for Find an answer to your question if x sin cube theta + Y cos cube theta is equal to sin theta cos theta and x sin theta is equal to Y cos theta prove that x squa… shivamkumar6556 shivamkumar6556 06.01.2019 Math Secondary School if x sin cube theta + Y cos cube theta is equal to sin theta cos theta and x sin theta is equal to Y cos theta prove that x squared plus y squared is equal to 1 2 See The trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$ or the more general one $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ can be proved from the definitions of the functions $\sin$ and $\cos$ if that's what you are wondering. Classically, they are defined geometrically with a perpendicular triangle or a unit circle in a coordinate plane. However, to 2016-06-19 2010-11-25 2019-04-03 1 10 Square Theta Into Kash Theta Into Sin Theta Kiske Barabar Hoga? 1 + 10 स्क्वायर थीटा ईंटो कॉश थीटा ईंटो सिन थीटा किसके बराबर होगा? #1 Answers, Listen to Expert Answers on Vokal - India’s Largest Question & Answers Platform in 11 Indian Languages. To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW If sin theta + cos theta =1 then sin theta cos theta =MOST IMPORTANT QUESTIO thank you theta upon sin theta cos cos theta + sin thank you theta + cos cube theta upon sin theta + cos theta + sin theta minus cos theta upon sin theta - 17206999 Answer: 1 📌📌📌 question Sin cube theta + cos cube theta by sin theta + cos theta - the answers to estudyassistant.com You can choose C = c +π, and then sin(θ +C) = −sin(θ+ c), so having the ± there doesn't create any more solutions if you allow any c ∈ [0,2π).
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### The sin β leg, as hypotenuse of another right triangle with angle α, likewise leads to segments of length cos α sin β and sin α sin β. Now, we observe that the "1" segment is also the hypotenuse of a right triangle with angle α + β; the leg opposite this angle necessarily has length sin(α + β), while the leg adjacent has length cos(α
The textbook gave an example on how to do this type of problem, but the solution was way too complicated. I want to find a simpler way to solve it.
## $\begingroup$ Yes, I'm sorry. in the book, it's theta, so when I go to pick an arbitrary variable. I go for x. I go for x. But, yes, that is what I mean. $\endgroup$ – user271277 Oct 3 '15 at 12:54
From these formulas, we also have the following identities for Find an answer to your question if x sin cube theta + Y cos cube theta is equal to sin theta cos theta and x sin theta is equal to Y cos theta prove that x squa… shivamkumar6556 shivamkumar6556 06.01.2019 Math Secondary School if x sin cube theta + Y cos cube theta is equal to sin theta cos theta and x sin theta is equal to Y cos theta prove that x squared plus y squared is equal to 1 2 See The trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$ or the more general one $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ can be proved from the definitions of the functions $\sin$ and $\cos$ if that's what you are wondering.
The red section on the right, d, is the difference between the lengths of the hypotenuse, H, and the adjacent side, A.As is shown, H and A are almost the same length, meaning cos θ is close to 1 and θ 2 / 2 helps trim the red away. | 2,475 | 9,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-49 | latest | en | 0.544418 |
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Lecture notes on submanifolds
Anton Izosimov
Last updated on October 7, 2018
1 Smooth submanifolds of smooth
Loosely speaking, a is a topological which locally looks like a . Similarly, a submanifold is a subset of a manifold which locally looks like a subspace of an Euclidian space.
Definition 1.1. Let M be a smooth manifold of m, and N be its subset. Then N is called a smooth n-dimensional submanifold of M if for every p ∈ N there exists a smooth chart (U, φ) in M such that p ∈ U and φ(N ∩ U) = Rn ∩ φ(U), where Rn is embedded into Rm as the subspace {xn+1 = 0, . . . , xm = 0}.
Equivalently, N is a smooth n-dimensional submanifold of M if M can be covered by charts n (Uα, φα) such that φα(N ∩ Uα) = R ∩ φα(Uα). Yet another equivalent definition: N is called a smooth n-dimensional submanifold of M if for every p ∈ N there exist local coordinates x1, . . . , xm, defined on some open in M neighborhood U of p, such that N ∩ U is given by xn+1 = 0, . . . , xm = 0.
Remark 1.2. Instead of saying that N ∩ U is given by equations xn+1 = 0, . . . , xm = 0, we will often say that N is locally given by equations xn+1 = 0, . . . , xm = 0, keeping in mind that these equations do not make sense outside U, so they actually describe the part of N that is inside U.
Remark 1.3. If n = m, then N ∩ U is given by an empty of equations, i.e. N ∩ U = U. This means that an m-dimensional submanifold of an m-dimensional manifold is the same as an open subset of the latter.
Exercise 1.4 (See Problem Set 4). Let M be a smooth manifold of dimension m, and N be its smooth submanifold of dimension n. By definition, this means M can be covered by charts n (Uα, φα) such that φα(N ∩ Uα) = R ∩ φα(Uα). Show that the collection (N ∩ Uα, φα |N∩Uα ) is a smooth on N which turns N into a smooth manifold of dimension n.
In what follows, when we regard smooth submanifolds as smooth manifolds, we mean the smooth structure constructed in this exercise.
1 Example 1.5 (Graphs of smooth functions of one variable). Let f : R → R be a smooth . Then its graph {(x, y) ∈ R2 | y = f(x)} is a smooth 1-dimensional submanifold of R2.
Proof. Let
x1 = x,
x2 = y − f(x).
Then the Jacobian of the transformation (x, y) 7→ (x1, x2) is equal to 1, so (x1, x2) can be taken as local coordinates near any in R2. In these coordinates, the graph y = f(x) is given by the x2 = 0, which proves that this graph is a smooth 1-dimensional submanifold.
Remark 1.6. In fact, (x1, x2) is a global chart. Indeed, the map (x, y) 7→ (x1, x2) has a smooth inverse given by
x = x1,
y = x2 + f(x1), so this map is a global diffeomorphism R2 → R2.
Remark 1.7. Similarly, the graph of a smooth function x = f(y) is also a smooth 1-dimensional submanifold of the (x, y) . Furthemore, if a subset Γ ⊂ R2 can be represented, near each of its points, either as a graph of a smooth function y = y(x), or as a graph of a smooth function x = x(y), then Γ is also a smooth 1-dimensional submanifold. Indeed, the notion of a submanifold is local, so it suffices to show that Γ is a submanifold near each of its points. At the same , for each point of Γ we can either apply the argument of Example 1.5 (if near that point we have y = y(x)), or the same argument, but with roles of x and y interchanged (if near that point we have x = x(y)).
Example 1.8. The x2 + y2 = 1 is a 1-dimensional submanifold of R2.
Proof. Take a point (x, y) in the circle. If y > 0, then near that point the circle is the graph of √ y = 1 − x2, which is smooth since x ∈ (−1, 1). Similarly, if y < 0, then the cirlce is given by √ y = − 1 − x2. Finally, if y = 0, then locally the circle is either the graph of x = p1 − y2, or the graph of x = −p1 − y2, with both functions being smooth.
Remark 1.9. Note that the graph of any continuous function y = f(x) is a , since it is homeomorphic to R. Moreover, any such graph has a smooth structure since R is a smooth manifold. However, graphs of continuous non-smooth functions are, in general, not smooth submanifolds of R2.
Example 1.10. The graph of y = |x| is not a smooth submanifold of R2.
2 Proof. Let Γ be this graph. Assume it is a smooth submanifold of R2. This in particular means that there is a chart (x1, x2) defined near the point (0, 0) ∈ Γ such that Γ is locally given by the equation x2 = 0. The latter means that for sufficiently small t ≥ 0 we have
x2(t, t) = 0, x2(−t, t) = 0.
Taking the right-hand t-derivative of these equations at t = 0 and using that for smooth functions it coincides with the usual derivative, we get
∂x ∂x ∂x ∂x 2 (0, 0) + 2 (0, 0) = 0, − 2 (0, 0) + 2 (0, 0) = 0, ∂x ∂y ∂x ∂y which implies ∂x ∂x 2 (0, 0) = 2 (0, 0) = 0. ∂x ∂y
But this means that the Jacobian of the transformation (x, y) 7→ (x1, x2) vanishes at the origin, which contradicts (x1, x2) being a smooth chart.
Nevertheless, it may still happen that the graph of y = f(x) is a smooth submanifold of R2, even though f is not smooth. √ Example 1.11. The graph of y = 3 x is a smooth submanifold of R2.
Proof. It is the graph of x = y3, which is smooth.
Remark 1.12. We will see later that every smooth submanifold of R2 is locally either a graph of a smooth function y = y(x), or a graph of a smooth function x = x(y).
Example 1.13 (Graphs of smooth maps). Let F : Rm → Rn be smooth. Then its graph
m+n Γ = {(x1, . . . , xm, y1, . . . , yn) ∈ R | (y1, . . . , yn) = F (x1, . . . , xm)} is a smooth m-dimensional submanifold of Rm+n.
Proof. Let f1(x1, . . . , xm), . . . , fn(x1, . . . , xm) be components of F . Take new coordinates
x˜1 = x1, ...
x˜m = xm,
x˜m+1 = y1 − f1(x1, . . . , xm), ...
x˜m+n = yn − fn(x1, . . . , xm).
3 Then the Jacobian of the transformation (x1, . . . , xm, y1, . . . , yn) 7→ (˜x1,..., x˜m+n) is 1, so this is indeed a near every point. The graph Γ is given in these coordinates by x˜m+1 = 0,..., x˜n+m = 0, which proves that Γ is an m-dimensional submanifold.
Pm 2 m Example 1.14. The i=1 xi = 1 is a smooth submanifold of R of codimension 1 (i.e. of dimension m − 1).
Proof. Apply the same argument as in Example 1.8: near every point of the sphere, one of the variables xi can be written as a smooth function of other variables, so the sphere is locally a graph of a smooth function of m − 1 variables.
2 Restricting smooth maps to smooth submanifolds
Most manifolds can be naturally described as submanifolds of something simpler. For exam- ple, are defined as submanifolds of Euclidian . This suggests a way to check smoothness of various objects defined on the sphere, for instance smoothness of maps from the sphere to another manifold: first one checks that the given map is in fact defined and smooth on the whole ambient space, and then one restricts the map to the sphere. So, we need to show that the restriction of a smooth map to a submanifold is smooth. The proof is based on the smoothness of the inclusion map:
Proposition 2.1. Let N ⊂ M be a smooth submanifold. Then the inclusion map i: N → M, given by i(p) = p, is smooth.
Remark 2.2. Here we assume that N is endowed with the smooth structure provided by Exercise 1.4.
Proof of Proposition 2.1. We take p ∈ N and show that i is smooth at p. By definition of a smooth submanifold, there are local coordinates x1, . . . , xm on M around p in which N is given by equations xn+1 = 0, . . . , xm = 0. Furthermore, x1, . . . , xn can be taken as coordinates on N around p (see Exercise 1.4). Taking x1, . . . , xn as coordinates around p ∈ N, and x1, . . . , xm as coordinates around i(p) = p ∈ M, we get the following coordinate representation of the map i:
x1 = x1, ...
xn = xn,
xn+1 = 0, ...
xm = 0,
4 where x-variables on the left are coordinates in M, while x-variables on the right are coordinates in N. This coordinate representation is smooth, so i is smooth at p. Since p was arbitrary, it follows that i is smooth everywhere.
Proposition 2.3. Let φ: M → N be a smooth map, and let M 0 ⊂ M be a smooth submanifold. 0 Then φ |M 0 : M → N is smooth.
0 Proof. We have φ |M 0 = φ ◦ i, where i: M → M is the inclusion map. Since φ is known to be smooth, and i is smooth by Proposition 2.1, it follows that the composition φ |M 0 of those maps is smooth as well.
Along with restricting the domain of a smooth map to a submanifold, we can also restrict the , provided that the of the map is contained in a submanifold:
Proposition 2.4. Let φ: M → N be a smooth map, and let N 0 ⊂ N be a smooth submanifold. Assume also that φ(M) ⊂ N 0. Then φ, regarded as a map M → N 0, is smooth.
0 Proof. We take p ∈ M and show that φ: M → N is smooth at p. Let y1, . . . , ym be any chart 0 in M around p, and let y1, . . . , yn be a chart on N around φ(p) in which N is given by equations 0 yl+1 = 0, . . . , yn = 0. Then, since φ(M) ⊂ N , the coordinate representation of φ: M → N has the form
y1 = f1(x1, . . . , xm), ...,
yl = fl(x1, . . . , xm),
yl+1 = 0, ...
yn = 0,
Since φ: M → N is a smooth map, the functions f1, . . . , fl are smooth, and it follows that the coordinate representation
y1 = f1(x1, . . . , xm), ...,
yl = fl(x1, . . . , xm) of φ: M → N 0 is smooth as well, as desired.
Furthermore, we can restrict the domain and the codomain at the same time:
Corollary 2.5. Let φ: M → N be a smooth map, and let M 0 ⊂ M, N 0 ⊂ N be smooth 0 0 0 0 submanifolds. Assume that φ(M ) ⊂ N . Then φ |M 0 , regarded as a map M → N , is smooth.
5 0 0 0 Proof. By Proposition 2.3, φ |M 0 : M → N is smooth, but since φ |M 0 (M ) ⊂ N , it follows 0 0 from Proposition 2.4 that φ |M 0 is also smooth when being regarded as a map M → N .
Example 2.6 (See Problem Set 2). Let S1 be the unit circle identified with {z ∈ C | |z|2 = 1}. Then the map f : S1 → S1 given by f(z) = zn (where n ∈ Z is given) is smooth.
Proof. The map in question is the restriction of a map C \{0} → C \{0} given by the same formula z 7→ zn. The latter can be written in real coordinates as
(x, y) 7→ (Re(x + iy)n, Im(x + iy)n).
For n > 0, this map is polynomial and hence smooth. For n < 0, it is a composition of a polynomial map and the map 1 z¯ z 7→ = , z |z|2 which is also smooth in C \{0}. So, our map C \{0} → C \{0} is smooth, which, in view of Corollary 2.5, shows that the corresponding map S1 → S1 is smooth as well.
Remark 2.7. Here we are implicitly using that the smooth structure on S1 coming from its to R2 as a submanifold coincides with the “standard” structure defined earlier in the course. This is left as an exercise.
3 Description of submanifolds as level sets
In most cases it is inconvenient (and, globally, not possible) to describe submanifolds as graphs. P 2 A more common way to define submanifolds is to use equations like xi = 1. So, we need a tool that will allows us to prove that so-defined subsets a submanifolds.
Proposition 3.1. Assume that f : Rm → R is smooth, and let N = {x ∈ Rn | f(x) = 0} be the zero set of f. Assume also that the of f, i.e. the vector of its partial derivatives, does not vanish on N. Then N is a smooth codimension 1 submanifold of Rm.
Proof. Let p ∈ M. Since the gradient of f does not vanish of p, there is i ∈ {1, . . . , m} such that ∂f 6= 0. Without loss of generality, assume that i = m (if not, we renumber the ∂xi coordinates). Then the Jacobian of the map (x1, . . . , xm) 7→ (x1, . . . , xm−1, f) at p is equal to ∂f 6= 0. Therefore, (x , . . . , x , f) is a smooth chart near p. In this chart, N is the zero set ∂xm 1 m−1 of the last coordinate and hence a codimension 1 submanifold.
m m Corollary 3.2. Assume that f : R → R is smooth, c ∈ R, and let Nc = {x ∈ R | f(x) = c}.
Assume also that the gradient of f does not vanish on Nc. Then Nc is a smooth codimension 1 submanifold of Rm.
Proof. This follows from Proposition 3.1 applied to the function f − c.
6 Example 3.3 (cf. Example 1.14). The sphere Sm−1 is defined as level set f = 1 for the function Pm 2 m f = i=1 xi in R . The gradient of this function only vanishes at the origin, which does not belong to the sphere. Therefore, the sphere is a codimension 1 submanifold.
Example 3.4. SLn(R) = {A ∈ Matn×n(R) | det A = 1} is a smooth codimension 1 submanifold of Matn×n(R).
Proof. We have ∂ det A = Aij, ∂aij i+j where Aij is the cofactor of aij, that is the (i, j) minor multiplied by (−1) . Furthermore, for any A ∈ SLn(R), at least one of its cofactors is non-zero, so the desired statement follows from Corollary 3.2.
Example 3.5. The subset of the plane given by the equation xy = 0 is not a smooth subman- ifold (see Problem Set 5). The reason we cannot apply Proposition 3.1 is because the gradient of xy vanishes at the origin.
Of course, not being able to apply Proposition 3.1 or Corollary 3.2 does not mean that the given level set is not a submanifold.
Example 3.6. The subset of the plane given by x2 + y2 = 0 is a submanifold, even though Proposition 3.1 does not apply. Note, however, that the codimension of this submanifold is 2, while it would be 1 if Proposition 3.1 was applicable.
Corollary 3.7. Assume that M is a smooth manifold, f : M → R is smooth, c ∈ R, and let
Nc = {p ∈ M | f(p) = c}. Assume also that the differential of f (which as an element of the cotangent space at every point) does not vanish on Nc. Then Nc is a smooth codimension 1 submanifold of M.
Proof. Take p ∈ Nc. Taking local coordinates, we identify a neighborhood of p in M with an open subset of Rm. Furthermore, under this identification the differential becomes the vector of partial derivatives, so the desired statement follows from Corollary 3.2.
m Proposition 3.8. Assume that f1, . . . , fn : R → R are smooth functions, and let N = {x ∈ m R | f1(x) = ··· = fn(x) = 0} be the joint zero set of f1, . . . , fn. Assume also that the m of fi’s are linearly independent of N. Then N is a smooth codimension n submanifold of R .
Proof. Let p ∈ N. Assumption on the gradients means that the Jacobian
∂f ∂f 1 ... 1 ∂x1 ∂xm ... ∂f ∂f n ... n ∂x1 ∂xm p
7 has rank n and therefore admits a non-vanishing n × n minor. Without loss of generality, assume that it is the rightmost minor
∂f1 ∂f1 ... ∂xm−n+1 ∂xm
... ∂f ∂f n ... n ∂xm−n+1 ∂xm
(if not, we renumber the coordinates). Then it follows that the transformation (x1, . . . , xm) 7→
(x1, . . . , xm−n, f1, . . . , fn) has a non-vanishing Jacobian, and thus (x1, . . . , xm−n, f1, . . . , fn) is a local coordinate system in Rm near p. Therefore, N, which is the vanishing set for the last n coordinates, is a codimension n submanifold.
Similarly to the case of one function, we can generalize this arbitrary level sets, and also replace Rm by an arbitrary manifold. We will go even further and replace n functions by a map to an n-dimensional manifold.
Proposition 3.9. Let F : M → N be a smooth map between smooth manifolds, and let F −1(q) = {p ∈ M | F (p) = q}. Assume that the differential of F is surjective at every point of F −1(q). Then F −1(q) is a smooth submanifold of M whose codimension is dim N.
Remark 3.10. A smooth map whose differential is surjective everywhere is called a submersion. Proposition 3.9 in particular says that a level set of a submersion is a smooth submanifold.
−1 Proof of Proposition 3.9. Let p ∈ F (q). Take any smooth chart (x1, . . . , xm) near p, and let
(y1, . . . , yn) be a chart centered at q = F (p), which means that y1(q) = ··· = yn(q) = 0. We can assume that F takes the domain of x-coordinates to the domain of y-coordinates (if not, we make the domain of x coordinates smaller). Then F can be represented in coordinates as
y1 = f1(x1, . . . , xm), ...
yn = fn(x1, . . . , xm)
−1 for certain smooth functions f1, . . . , fn. Furthermore, F (q) is locally (in the domain of x coordinates) the same as the joint zero level set of f1, . . . , fn. Therefore, the desired statement follows from Proposition 3.8.
t Example 3.11. The orthogonal On(R) = {A ∈ Matn×n(R) | AA = Id} is a smooth submanifold of Matn×n(R).
t t Proof. For any, A ∈ Matn×n(R), the matrix AA is symmetric, so φ: A 7→ AA is a map
Matn×n(R) → Symn(R), where Symn(R) is the space of symmetric n × n real matrices. The differential of this map at A is a map from TAMatn×n(R) = Matn×n(R) to Tφ(A)Symn(R) =
8 Symn(R) given by
t t dAφ(X) = XA + AX .
This mapping is surjective for A ∈ On(R). Indeed, the equation
XAt + AXt = B
1 for symmetric B and orthogonal A has a solution given by X = 2 BA. So, the result follows from Proposition 3.9.
Remark 3.12. The submanifold On(R) is disconnected. Indeed, consider its subset SOn(R) = {A ∈ On(R) | det A = 1}. This subset is closed as a level set of a continuous function. On the other hand, since the of an orthogonal matrix is always ±1, it follows that
{A ∈ On(R) | det A = 1} = {A ∈ On(R) | det A > 0}, so SOn(R) is also open, meaning that On(R) is not connected.
Remark 3.13. Since SOn(R) is open in On(R), it follows that SOn(R) is also a submanifold of Matn×n(R). The same is true for its complement On(R) \ SOn(R).
Exercise 3.14 (See Problem Set 5). Prove that SOn(R) and On(R) \ SOn(R) are connected spaces, so On(R) consists of two connected components.
4 The of a submanifold
In this section we show that the tangent space to a submanifold can be naturally viewed as a subspace in the tangent map of the ambient manifold. Let M be a smooth manifold, and N be its smooth submanifold. Then, by Proposition 2.1, the inclusion mapping i: N → M is smooth, and we can compute its differential dpi at every point p ∈ N.
Proposition 4.1. 1. The mapping dpi: TpN → TpM is injective.
2. In terms of smooth , dpi can be defined as follows: for every in N passing
through p, the mapping dpi takes its tangent vector at p to the tangent vector at p of the same curve, but regarded as a curve in M.
3. In terms of differential operators, dpi can be defined as follows: for any v ∈ TpN and any function f on M defined and smooth around p, we have
dpi(v)f = v(f|N ).
9 Proof. Let m = dim M, n = dim N. To prove the first statement, we use coordinates from the proof of Proposition 2.1. In these coordinates, the Jacobian matrix of the inclusion mapping i has the form ! Id n , 0m−n,n where Idn is the n × n identity matrix, and 0m−n,n is the (m − n) × n zero matrix. Since the columns of this matrix are linearly independent, it follows that dpi is injective, as desired. To prove the second statement, take any parametrized curve γ in N with γ(0) = p. Then, by definition of the differential in terms of curves, we have d d d dpi γ(t) = i(γ(t)) = γ(t), dt t=0 dt t=0 dt t=0 where in the latter formula we regard γ as a curve in M. So, the second statement is proved. Finally, we prove the last statement. Using the definition of the differential in terms of differential operators, we get
∗ dpi(v)f = v(i f) = v(f ◦ i) = v(f|N ), as desired.
In what follows, we do not distinguish between the tangent space of a submanifold and its image under the differential of the inclusion map. So, if N is a submanifold of M, then TpN is a subspace of TpM for any p ∈ N. In this interpretation, dpi: TpN → TpM is simply the inclusion map of a subspace to the ambient space. In the remaining part of this section, we give an explicit description of the tangent space for a submanifold defined as a level set. First, we prove the following preliminary statement, which is also useful on its own:
Proposition 4.2. Let φ: M → N be a smooth map, and let M 0 ⊂ M be a smooth submanifold. Then, for every p ∈ M 0, we have
0 0 dp(φ|M ) = (dpφ)|TpM .
In other words, the differential of the restriction is the same as the restriction of the differential.
Proof. Let i: M 0 → M be the inclusion map. Then
0 0 dp(φ|M ) = dp(φ ◦ i) = dpφ ◦ dpi = (dpφ)|TpM ,
0 where in the last equality we used that dpi is the inclusion TpM → TpM.
10 Proposition 4.3. Let F : M → N be a smooth map whose differential is surjective at all points of the level set F −1(q). Then, for every p ∈ F −1(q), we have
−1 TpF (q) = Ker dpF.
Proof. We have
F |F −1(q) = q.
Taking the differential of both sides at p using Proposition 4.2 and the fact that the differential of a constant map is zero, we get
−1 dpF |TpF (q) = 0, which means that −1 TpF (q) ⊂ Ker dpF.
So, to prove that these two vector spaces coincide, it suffices to show that they are of the same dimension. We have
−1 −1 dim TpF (q) = dim F (q) = dim M − dim N, where in the last equality we used Proposition 3.9. At the same time, we have
dim Ker dpF = dim TpM − dim TqN = dim M − dim N,
−1 where in the first equality we used that dpF is surjective. So, TpF (q) is a vector subspace of Ker dpF , and the dimensions of these spaces coincide, which means that they are equal, as desired.
Example 4.4. The tangent space to a level set {f = c} of a smooth function f : Rm → R is the plane orthogonal to the gradient, provided that the latter is non-zero.
Proof. Assume that the gradient of f at a point p ∈ Rm does not vanish. Then {f = c} is submanifold near p. By Proposition 4.3, we have
Tp{f = c} = Ker dpf,
m where dpf is regarded as a map TpR → TpR. The matrix of this map, written in standard m bases for TpR and TpR, is ∂f ... ∂f , ∂x1 ∂x1 and its is exactly the orthogonal complement of the gradient of f, as desired. Example 4.5. The tangent plane to the sphere is orthogonal to the .
P 2 Proof. The sphere is the level set of xi . The gradient of the latter function is twice the radius-vector.
11 Example 4.6. The tangent space to the On(R) at A ∈ On(R) is the subspace of TAMatn×n(R) = Matn×n(R) consisting of matrices X satysfying the equation
XAt + AXt = 0.
In particular, the tangent space to On(R) (and hence SOn(R)) at the identity is the space of skew-symmetric n × n real matrices.
Proof. This follows from the formula X 7→ XAt + AXt for the differential at A of the map A 7→ AAt.
Remark 4.7. This statement in particular says that for any smooth family A(t) of orthogonal matrices such that A(0) = Id, the matrix A0(0) is skew-symmetric. This can also be seen by explicit differentiation. A less trivial part of the statement is that for any skew-symmetric X 0 there is a curve A(t) in SOn(R) with A(0) = Id and A (0) = X.
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# Chapter Six
Chapter Six. Probability. Approaches to Assigning Probabilities…. There are three ways to assign a probability, P(O i ), to an outcome, O i , namely: Classical approach : based on equally likely events. Relative frequency : assigning probabilities based on experimentation or historical data.
## Chapter Six
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### Presentation Transcript
1. Chapter Six Probability
2. Approaches to Assigning Probabilities… There are three ways to assign a probability, P(Oi), to an outcome, Oi, namely: Classical approach: based on equally likely events. Relative frequency: assigning probabilities based on experimentation or historical data. Subjective approach: Assigning probabilities based on the assignor’s (subjective) judgment.
3. Classical Approach… If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome. It is necessary to determine the number of possible outcomes. Experiment: Rolling a die Outcomes {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a 1/6 chance of occurring.
4. Classical Approach… Experiment: Rolling two dice and observing the total Outcomes: {2, 3, …, 12} Examples: P(2) = 1/36 P(6) = 5/36 P(10) = 3/36
5. Relative Frequency Approach… Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days): For example, 10 days out of 30 2 desktops were sold. From this we can construct the probabilities of an event (i.e. the # of desktop sold on a given day)…
6. Relative Frequency Approach… “There is a 40% chance Bits & Bytes will sell 3 desktops on any given day”
7. Subjective Approach… “In the subjective approach we define probability as the degree of belief that we hold in the occurrence of an event” E.g. weather forecasting’s “P.O.P.” “Probability of Precipitation” (or P.O.P.) is defined in different ways by different forecasters, but basically it’s a subjective probability based on past observations combined with current weather conditions. POP 60% – based on current conditions, there is a 60% chance of rain (say).
8. Interpreting Probability… No matter which method is used to assign probabilities all will be interpreted in the relative frequency approach For example, a government lottery game where 6 numbers (of 49) are picked. The classical approach would predict the probability for any one number being picked as 1/49=2.04%. We interpret this to mean that in the long run each number will be picked 2.04% of the time.
9. Joint, Marginal, Conditional Probability… We study methods to determine probabilities of events that result from combining other events in various ways. There are several types of combinations and relationships between events: • Complement event • Intersection of events • Union of events • Mutually exclusive events • Dependent and independent events
10. Complement of an Event… The complement of event A is defined to be the event consisting of all sample points that are “not in A”. Complement of A is denoted by Ac The Venn diagram below illustrates the concept of a complement. P(A) + P(Ac ) = 1 A Ac
11. Complement of an Event… For example, the rectangle stores all the possible tosses of 2 dice {(1,1), 1,2),… (6,6)} Let A = tosses totaling 7 {(1,6), (2, 5), (3,4), (4,3), (5,2), (6,1)} P(Total = 7) + P(Total not equal to 7) = 1 A Ac
12. Intersection of Two Events… The intersection of events A and B is the set of all sample points that are in both A and B. The intersection is denoted: A and B The joint probability of A and B is the probability of the intersection of A and B, i.e. P(A and B) A B
13. Intersection of Two Events… For example, let A = tosses where first toss is 1 {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} and B = tosses where the second toss is 5 {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} The intersection is {(1,5)} The joint probability of A and B is the probability of the intersection of A and B, i.e. P(A and B) = 1/36 A B
14. Union of Two Events… The union of two events A and B, is the event containing all sample points that are in A or B or both: Union of A and B is denoted: A or B A B
15. Union of Two Events… For example, let A = tosses where first toss is 1 {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} and B is the tosses that the second toss is 5 {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} Union of A and B is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} (2,5), (3,5), (4,5), (5,5), (6,5)} B A
16. Mutually Exclusive Events… When two events are mutually exclusive (that is the two events cannot occur together), their joint probability is 0, hence: A B Mutually exclusive; no points in common… For example A = tosses totaling 7 and B = tosses totaling 11
17. A Ac A B A A B B Basic Relationships of Probability… Complement of Event Union of Events Intersection of Events Mutually Exclusive Events
18. Example 6.1… Why are some mutual fund managers more successful than others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned their MBA: E.g. This is the probability that a mutual fund outperforms AND the manager was in a top-20 MBA program; it’s a joint probability.
19. Example 6.1… Alternatively, we could introduce shorthand notation to represent the events: A1 = Fund manager graduated from a top-20 MBA program A2 = Fund manager did not graduate from a top-20 MBA program B1 = Fund outperforms the market B2 = Fund does not outperform the market E.g. P(A2 and B1) = .06 = the probability a fund outperforms the market and the manager isn’t from a top-20 school.
20. Marginal Probabilities… Marginal probabilities are computed by adding across rows and down columns; that is they are calculated in the margins of the table: P(A2) = .06 + .54 “what’s the probability a fund manager isn’t from a top school?” P(B1) = .11 + .06 BOTH margins must add to 1 (useful error check) “what’s the probability a fund outperforms the market?”
21. Conditional Probability… Conditional probability is used to determine how two events are related; that is, we can determine the probability of one event given the occurrence of another related event. Conditional probabilities are written as P(A | B) and read as “the probability of A given B” and is calculated as:
22. Conditional Probability… Again, the probability of an event given that another event has occurred is called a conditional probability… Note how “A given B” and “B given A” are related…
23. Conditional Probability… Example 6.2 What’s the probability that a fund will outperform the market given that the manager graduated from a top-20 MBA program? Recall: A1 = Fund manager graduated from a top-20 MBA program A2 = Fund manager did not graduate from a top-20 MBA program B1 = Fund outperforms the market B2 = Fund does not outperform the market Thus, we want to know “what is P(B1 | A1) ?”
24. Conditional Probability… We want to calculate P(B1 | A1) Thus, there is a 27.5% chance that that a fund will outperform the market given that the manager graduated from a top-20 MBA program.
25. Independence… One of the objectives of calculating conditional probability is to determine whether two events are related. In particular, we would like to know whether they are independent, that is, if the probability of one event is not affected by the occurrence of the other event. Two events A and B are said to be independent if P(A|B) = P(A) or P(B|A) = P(B)
26. Independence… For example, we saw that P(B1 | A1) = .275 The marginal probability for B1 is: P(B1) = 0.17 Since P(B1|A1) ≠ P(B1), B1 and A1 are not independent events. Stated another way, they are dependent. That is, the probability of one event (B1) is affected by the occurrence of the other event (A1).
27. Union… We stated earlier that the union of two events is denoted as: A or B. We can use this concept to answer questions like: Determine the probability that a fund outperforms the market or the manager graduated from a top-20 MBA program.
28. Union… Determine the probability that a fund outperforms (B1) or the manager graduated from a top-20 MBA program (A1). A1 or B1 occurs whenever: A1 and B1occurs,A1 and B2occurs, orA2 and B1occurs… P(A1 or B1) =.11+.06+ .29 =.46
29. Union… Determine the probability that a fund outperforms (B1) or the manager graduated from a top-20 MBA program (A1). B1 A1 P(A1 or B1) =.11+.06+ .29 =.46
30. Alternatively… Take 100% and subtract off “when doesn’t A1 or B1 occur”? i.e. at A2 and B2 B1 A1 P(A1 or B1) = 1 – P(A2 and B2) = 1 – .54 = .46
31. Probability Rules and Trees… We introduce three rules that enable us to calculate the probability of more complex events from the probability of simpler events… The Complement Rule, The Multiplication Rule, and The Addition Rule
32. Complement Rule… As we saw earlier with the complement event, the complement rule gives us the probability of an event NOT occurring. That is: P(AC) = 1 – P(A) For example, in the simple roll of a die, the probability of the number “1” being rolled is 1/6. The probability that some number other than “1” will be rolled is 1 – 1/6 = 5/6.
33. Multiplication Rule… The multiplication rule is used to calculate the joint probability of two events. It is based on the formula for conditional probability defined earlier: If we multiply both sides of the equation by P(B) we have: P(A and B) = P(A | B)•P(B) Likewise, P(A and B) = P(B | A) • P(A) If A and B are independent events, then P(A and B) = P(A)•P(B)
34. Example 6.5… A graduate statistics course has seven male and three female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Let A represent the event that the first student is female P(A) = 3/10 = .30 What about the second student?
35. Example 6.5… A graduate statistics course has seven male and three female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Let B represent the event that the second student is female P(B | A) = 2/9 = .22 That is, the probability of choosing a female student given that the first student chosen is 2 (females) / 9 (remaining students) = 2/9
36. Example 6.5… A graduate statistics course has seven male and three female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Thus, we want to answer the question: what is P(A and B) ? P(A and B) = P(A)•P(B|A) = (3/10)(2/9) = 6/90 = .067 “There is a 6.7% chance that the professor will choose two female students from her grad class of 10.”
37. Example 6.6 Refer to Example 6.5. The professor who teaches the course is suffering from the flu and will be unavailable for two classes. The professor’s replacement will teach the next two classes. His style is to select one student at random and pick on him or her to answer questions during that class. What is the probability that the two students chosen are female? Let A represent the event that the first student is female P(A) = 3/10 = .30 What about the second student?
38. Example 6.6 Let B represent the event that the second student is female P(B | A) = 3/10 = .30 That is, the probability of choosing a female student given that the first student chosen is unchanged since the student selected in the first class can be chosen in the second class. P(A and B) = P(A)•P(B|A) = (3/10)(3/10) = 9/100 = .090
39. Addition Rule… Recall: the addition rule is used to compute the probability of event A or B or both A and B occurring; i.e. the union of A and B. P(A or B) = P(A) + P(B) – P(A and B) A B A B + – = If A and B are mutually exclusive, then this term goes to zero P(A or B) = P(A) + P(B) – P(A and B)
40. Example 6.7… In a large city, two newspapers are published, the Sun and the Post. The circulation departments report that 22% of the city’s households have a subscription to the Sun and 35% subscribe to the Post. A survey reveals that 6% of all households subscribe to both newspapers. What proportion of the city’s households subscribe to either newspaper? That is, what is the probability of selecting a household at random that subscribes to the Sun or the Post or both? i.e. what is P(Sun or Post) ?
41. Example 6.7… In a large city, two newspapers are published, the Sun and the Post. The circulation departments report that 22% of the city’s households have a subscription to the Sun and 35% subscribe to the Post. A survey reveals that 6% of all households subscribe to both newspapers. What proportion of the city’s households subscribe to either newspaper? P(Sun or Post) = P(Sun) + P(Post) – P(Sun and Post) = .22 + .35 – .06 = .51 “There is a 51% probability that a randomly selected household subscribes to one or the other or both papers”
42. Probability Trees An effective and simpler method of applying the probability rules is the probability tree, wherein the events in an experiment are represented by lines. The resulting figure resembles a tree, hence the name. We will illustrate the probability tree with several examples, including two that we addressed using the probability rules alone.
43. P(F|F) = 2/9 P(F) = 3/10 P( M|F) = 7/9 P(F|M) = 3/9 P( M) = 7/10 P( M|M) = 6/9 Example 6.5 This is P(F), the probability of selecting a female student first First selection Second selection This is P(F|F), the probability of selecting a female student second, given that a female was already chosen first
44. P(F|F) = 2/9 Joint probabilities P(F) = 3/10 P(FF)=(3/10)(2/9) P( M|F) = 7/9 P(F|M) = 3/9 P( M) = 7/10 P(FM)=(3/10)(7/9) P( M|M) = 6/9 P(MF)=(7/10)(3/9) P(MM)=(7/10)(6/9) Probability Trees… At the ends of the “branches”, we calculate joint probabilities as the product of the individual probabilities on the preceding branches. First selection Second selection
45. P(FF)=(3/10)(3/10) P(F|F) = 3/10 P(F) = 3/10 P( M|F) = 7/10 P(FM)=(3/10)(7/10) P(MF)=(7/10)(3/10) P(F|M) = 3/10 P( M) = 7/10 P( M|M) =7/10 P(MM)=(7/10)(7/10) Example 6.6 Suppose we have our grad class of 10 students again, but make the student sampling independent, that is “with replacement” – a student could be picked first and picked again in the second round. Our tree and joint probabilities now look like: FF FM MF MM
46. P(F|F) = 2/9 P(F) = 3/10 P( M|F) = 7/9 P(F|M) = 3/9 P( M) = 7/10 P( M|M) = 6/9 Probability Trees… The probabilities associated with any set of branches from one “node” must add up to 1.00… First selection Second selection 2/9 + 7/9 = 9/9 = 1 3/10 + 7/10 = 10/10 = 1 3/9 + 6/9 = 9/9 = 1 Handy way to check your work !
47. Probability Trees… Note: there is no requirement that the branches splits be binary, nor that the tree only goes two levels deep, or that there be the same number of splits at each sub node…
48. P(Pass) = .72 P(Fail and Pass)= (.28)(.88)=.2464 P(Fail and Fail) = (.28)(.12) = .0336 Example 6.8 Law school grads must pass a bar exam. Suppose pass rate for first-time test takers is 72%. They can re-write if they fail and 88% pass their second attempt. What is the probability that a randomly grad passes the bar? First exam P(Pass) = .72 Second exam P(Pass|Fail) = .88 P( Fail) = .28 P( Fail|Fail) = .12
49. P(Pass) = .72 P(Fail and Pass)= (.28)(.88)=.2464 P(Fail and Fail) = (.28)(.12) = .0336 Example 6.8 What is the probability that a randomly grad passes the bar? “There is almost a 97% chance they will pass the bar” P(Pass) = P(Pass 1st) + P(Fail 1st and Pass 2nd) = = 0.7200 + 0.2464 = .9664 First exam P(Pass) = .72 Second exam P(Pass|Fail) = .88 P( Fail) = .28 P( Fail|Fail) = .12
50. Bayes’ Law… Bayes’ Law is named for Thomas Bayes, an eighteenth century mathematician. In its most basic form, if we know P(B | A), we can apply Bayes’ Law to determine P(A | B) P(B|A)P(A|B) for example …
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Isoperimetric inequality (Redirected from Isoperimetric theorem) Ir para a navegação Ir para a pesquisa Em matemática, the isoperimetric inequality is a geometric inequality involving the perimeter of a set and its volume. Dentro {estilo de exibição m} -dimensional space {estilo de exibição mathbb {R} ^{n}} the inequality lower bounds the surface area or perimeter {nome do operador de estilo de exibição {por} (S)} of a set {displaystyle Ssubset mathbb {R} ^{n}} by its volume {nome do operador de estilo de exibição {volume} (S)} , {nome do operador de estilo de exibição {por} (S)geq noperatorname {volume} (S)^{fratura {n-1}{n}},nome do operador {volume} (B_{1})^{fratura {1}{n}}} , Onde {estilo de exibição B_{1}subconjunto mathbb {R} ^{n}} is a unit sphere. The equality holds only when {estilo de exibição S} is a sphere in {estilo de exibição mathbb {R} ^{n}} .
On a plane, ou seja. quando {estilo de exibição n=2} , the isoperimetric inequality relates the square of the circumference of a closed curve and the area of a plane region it encloses. Isoperimetric literally means "having the same perimeter". Specifically in {estilo de exibição mathbb {R} ^{2}} , the isoperimetric inequality states, for the length L of a closed curve and the area A of the planar region that it encloses, este {estilo de exibição L^{2}geq 4pi A,} and that equality holds if and only if the curve is a circle.
The isoperimetric problem is to determine a plane figure of the largest possible area whose boundary has a specified length.[1] The closely related Dido's problem asks for a region of the maximal area bounded by a straight line and a curvilinear arc whose endpoints belong to that line. It is named after Dido, the legendary founder and first queen of Carthage. The solution to the isoperimetric problem is given by a circle and was known already in Ancient Greece. No entanto, the first mathematically rigorous proof of this fact was obtained only in the 19th century. Desde então, many other proofs have been found.
The isoperimetric problem has been extended in multiple ways, por exemplo, to curves on surfaces and to regions in higher-dimensional spaces. Perhaps the most familiar physical manifestation of the 3-dimensional isoperimetric inequality is the shape of a drop of water. Nomeadamente, a drop will typically assume a symmetric round shape. Since the amount of water in a drop is fixed, surface tension forces the drop into a shape which minimizes the surface area of the drop, namely a round sphere.
Conteúdo 1 The isoperimetric problem in the plane 2 On a plane 3 On a sphere 4 In Rn 5 In Hadamard manifolds 6 In a metric measure space 7 For graphs 7.1 Exemplo: Isoperimetric inequalities for hypercubes 7.1.1 Edge isoperimetric inequality 7.1.2 Vertex isoperimetric inequality 8 Isoperimetric inequality for triangles 9 Veja também 10 Notas 11 Referências 12 External links The isoperimetric problem in the plane If a region is not convex, uma "dent" in its boundary can be "flipped" to increase the area of the region while keeping the perimeter unchanged. An elongated shape can be made more round while keeping its perimeter fixed and increasing its area.
The classical isoperimetric problem dates back to antiquity.[2] The problem can be stated as follows: Among all closed curves in the plane of fixed perimeter, which curve (caso existam) maximizes the area of its enclosed region? This question can be shown to be equivalent to the following problem: Among all closed curves in the plane enclosing a fixed area, which curve (caso existam) minimizes the perimeter?
This problem is conceptually related to the principle of least action in physics, in that it can be restated: what is the principle of action which encloses the greatest area, with the greatest economy of effort? The 15th-century philosopher and scientist, Cardinal Nicholas of Cusa, considered rotational action, the process by which a circle is generated, to be the most direct reflection, in the realm of sensory impressions, of the process by which the universe is created. German astronomer and astrologer Johannes Kepler invoked the isoperimetric principle in discussing the morphology of the solar system, in Mysterium Cosmographicum (The Sacred Mystery of the Cosmos, 1596).
Although the circle appears to be an obvious solution to the problem, proving this fact is rather difficult. The first progress toward the solution was made by Swiss geometer Jakob Steiner in 1838, using a geometric method later named Steiner symmetrisation.[3] Steiner showed that if a solution existed, then it must be the circle. Steiner's proof was completed later by several other mathematicians.
Steiner begins with some geometric constructions which are easily understood; por exemplo, it can be shown that any closed curve enclosing a region that is not fully convex can be modified to enclose more area, por "flipping" the concave areas so that they become convex. It can further be shown that any closed curve which is not fully symmetrical can be "tilted" so that it encloses more area. The one shape that is perfectly convex and symmetrical is the circle, although this, in itself, does not represent a rigorous proof of the isoperimetric theorem (see external links).
On a plane The solution to the isoperimetric problem is usually expressed in the form of an inequality that relates the length L of a closed curve and the area A of the planar region that it encloses. The isoperimetric inequality states that {displaystyle 4pi Aleq L^{2},} and that the equality holds if and only if the curve is a circle. The area of a disk of radius R is πR2 and the circumference of the circle is 2πR, so both sides of the inequality are equal to 4π2R2 in this case.
Dozens of proofs of the isoperimetric inequality have been found. Dentro 1902, Hurwitz published a short proof using the Fourier series that applies to arbitrary rectifiable curves (not assumed to be smooth). An elegant direct proof based on comparison of a smooth simple closed curve with an appropriate circle was given by E. Schmidt in 1938. It uses only the arc length formula, expression for the area of a plane region from Green's theorem, and the Cauchy–Schwarz inequality.
For a given closed curve, the isoperimetric quotient is defined as the ratio of its area and that of the circle having the same perimeter. This is equal to {displaystyle Q={fratura {4pi A}{L^{2}}}} and the isoperimetric inequality says that Q ≤ 1. Equivalentemente, the isoperimetric ratio L2/A is at least 4π for every curve.
The isoperimetric quotient of a regular n-gon is {displaystyle Q_{n}={fratura {pi }{ntan {tfrac {pi }{n}}}}.} Deixar {estilo de exibição C} be a smooth regular convex closed curve. Then the improved isoperimetric inequality states the following {estilo de exibição L^{2}geqslant 4pi A+8pi left|{widetilde {UMA}}_{0.5}certo|,} Onde {estilo de exibição L,UMA,{widetilde {UMA}}_{0.5}} denote the length of {estilo de exibição C} , the area of the region bounded by {estilo de exibição C} and the oriented area of the Wigner caustic of {estilo de exibição C} , respectivamente, and the equality holds if and only if {estilo de exibição C} is a curve of constant width.[4] On a sphere Let C be a simple closed curve on a sphere of radius 1. Denote by L the length of C and by A the area enclosed by C. The spherical isoperimetric inequality states that {estilo de exibição L^{2}geq A(4pi -A),} and that the equality holds if and only if the curve is a circle. Há, na verdade, two ways to measure the spherical area enclosed by a simple closed curve, but the inequality is symmetric with the respect to taking the complement.
This inequality was discovered by Paul Lévy (1919) who also extended it to higher dimensions and general surfaces.[5] In the more general case of arbitrary radius R, it is known [6] este {estilo de exibição L^{2}geq 4pi A-{fratura {A^{2}}{R^{2}}}.} In Rn The isoperimetric inequality states that a sphere has the smallest surface area per given volume. Given a bounded set {displaystyle Ssubset mathbb {R} ^{n}} with surface area {nome do operador de estilo de exibição {por} (S)} and volume {nome do operador de estilo de exibição {volume} (S)} , the isoperimetric inequality states {nome do operador de estilo de exibição {por} (S)geq noperatorname {volume} (S)^{fratura {n-1}{n}},nome do operador {volume} (B_{1})^{fratura {1}{n}},} Onde {estilo de exibição B_{1}subconjunto mathbb {R} ^{n}} is a unit ball. The equality holds when {estilo de exibição S} is a ball in {estilo de exibição mathbb {R} ^{n}} . Under additional restrictions on the set (such as convexity, regularity, smooth boundary), the equality holds for a ball only. But in full generality the situation is more complicated. The relevant result of Schmidt (1949, Sect. 20.7) (for a simpler proof see Baebler (1957)) is clarified in Hadwiger (1957, Sect. 5.2.5) do seguinte modo. An extremal set consists of a ball and a "corona" that contributes neither to the volume nor to the surface area. Aquilo é, the equality holds for a compact set {estilo de exibição S} se e apenas se {estilo de exibição S} contains a closed ball {estilo de exibição B} de tal modo que {nome do operador de estilo de exibição {volume} (B)=nome do operador {volume} (S)} e {nome do operador de estilo de exibição {por} (B)=nome do operador {por} (S).} Por exemplo, a "corona" may be a curve.
The proof of the inequality follows directly from Brunn–Minkowski inequality between a set {estilo de exibição S} and a ball with radius {displaystyle épsilon } , ou seja. {estilo de exibição B_{épsilon }=epsilon B_{1}} . By taking Brunn–Minkowski inequality to the power {estilo de exibição m} , subtracting {nome do operador de estilo de exibição {volume} (S)} from both sides, dividing them by {displaystyle épsilon } , and taking the limit as {displaystyle epsilon to 0.} (Osserman (1978); Federer (1969, §3.2.43)).
In full generality (Federer 1969, §3.2.43), the isoperimetric inequality states that for any set {displaystyle Ssubset mathbb {R} ^{n}} whose closure has finite Lebesgue measure {estilo de exibição m,ômega _{n}^{fratura {1}{n}}L^{n}({bar {S}})^{fratura {n-1}{n}}leq M_{*}^{n-1}(partial S)} Onde {estilo de exibição M_{*}^{n-1}} é o (n-1)-dimensional Minkowski content, Ln is the n-dimensional Lebesgue measure, and ωn is the volume of the unit ball in {estilo de exibição mathbb {R} ^{n}} . If the boundary of S is rectifiable, then the Minkowski content is the (n-1)-dimensional Hausdorff measure.
The n-dimensional isoperimetric inequality is equivalent (for sufficiently smooth domains) to the Sobolev inequality on {estilo de exibição mathbb {R} ^{n}} with optimal constant: {estilo de exibição à esquerda(int_{mathbb {R} ^{n}}|você|^{fratura {n}{n-1}}certo)^{fratura {n-1}{n}}leq n^{-1}ômega _{n}^{-{fratura {1}{n}}}int_{mathbb {R} ^{n}}|nabla u|} para todos {estilo de exibição uin W ^{1,1}(mathbb {R} ^{n})} .
In Hadamard manifolds Hadamard manifolds are complete simply connected manifolds with nonpositive curvature. Thus they generalize the Euclidean space {estilo de exibição mathbb {R} ^{n}} , which is a Hadamard manifold with curvature zero. In 1970's and early 80's, Thierry Aubin, Misha Gromov, Yuri Burago, and Viktor Zalgaller conjectured that the Euclidean isoperimetric inequality {nome do operador de estilo de exibição {por} (S)geq noperatorname {volume} (S)^{fratura {n-1}{n}}nome do operador {volume} (B_{1})^{fratura {1}{n}}} holds for bounded sets {estilo de exibição S} in Hadamard manifolds, which has become known as the Cartan–Hadamard conjecture. In dimension 2 this had already been established in 1926 by André Weil, who was a student of Hadamard at the time. In dimensions 3 e 4 the conjecture was proved by Bruce Kleiner in 1992, and Chris Croke in 1984 respectivamente.
In a metric measure space Most of the work on isoperimetric problem has been done in the context of smooth regions in Euclidean spaces, or more generally, in Riemannian manifolds. No entanto, the isoperimetric problem can be formulated in much greater generality, using the notion of Minkowski content. Deixar {estilo de exibição (X,dentro ,d)} be a metric measure space: X is a metric space with metric d, and μ is a Borel measure on X. The boundary measure, or Minkowski content, of a measurable subset A of X is defined as the lim inf {displaystyle mu ^{+}(UMA)=liminf_{varepsilon para 0+}{fratura {dentro (UMA_{varepsilon })-dentro (UMA)}{varepsilon }},} Onde {estilo de exibição A_{varepsilon }={xin X|d(x,UMA)leq varepsilon }} is the ε-extension of A.
The isoperimetric problem in X asks how small can {displaystyle mu ^{+}(UMA)} be for a given μ(UMA). If X is the Euclidean plane with the usual distance and the Lebesgue measure then this question generalizes the classical isoperimetric problem to planar regions whose boundary is not necessarily smooth, although the answer turns out to be the same.
A função {estilo de exibição I(uma)=inf{mu ^{+}(UMA)|dentro (UMA)=a}} is called the isoperimetric profile of the metric measure space {estilo de exibição (X,dentro ,d)} . Isoperimetric profiles have been studied for Cayley graphs of discrete groups and for special classes of Riemannian manifolds (where usually only regions A with regular boundary are considered).
For graphs Main article: Expander graph In graph theory, isoperimetric inequalities are at the heart of the study of expander graphs, which are sparse graphs that have strong connectivity properties. Expander constructions have spawned research in pure and applied mathematics, with several applications to complexity theory, design of robust computer networks, and the theory of error-correcting codes.[7] Isoperimetric inequalities for graphs relate the size of vertex subsets to the size of their boundary, which is usually measured by the number of edges leaving the subset (edge expansion) or by the number of neighbouring vertices (vertex expansion). For a graph {estilo de exibição G} and a number {estilo de exibição k} , the following are two standard isoperimetric parameters for graphs.[8] The edge isoperimetric parameter: {displaystyle Phi _{E}(G,k)=min _{Ssubseteq V}deixei{|E(S,{overline {S}})|:|S|=kright}} The vertex isoperimetric parameter: {displaystyle Phi _{V}(G,k)=min _{Ssubseteq V}deixei{|Gama (S)setminus S|:|S|=kright}} Aqui {estilo de exibição E(S,{overline {S}})} denotes the set of edges leaving {estilo de exibição S} e {displaystyle Gama (S)} denotes the set of vertices that have a neighbour in {estilo de exibição S} . The isoperimetric problem consists of understanding how the parameters {displaystyle Phi _{E}} e {displaystyle Phi _{V}} behave for natural families of graphs.
Exemplo: Isoperimetric inequalities for hypercubes The {estilo de exibição d} -dimensional hypercube {displaystyle Q_{d}} is the graph whose vertices are all Boolean vectors of length {estilo de exibição d} , isso é, the set {estilo de exibição {0,1}^{d}} . Two such vectors are connected by an edge in {displaystyle Q_{d}} if they are equal up to a single bit flip, isso é, their Hamming distance is exactly one. The following are the isoperimetric inequalities for the Boolean hypercube.[9] Edge isoperimetric inequality The edge isoperimetric inequality of the hypercube is {displaystyle Phi _{E}(Q_{d},k)geq k(d-log _{2}k)} . This bound is tight, as is witnessed by each set {estilo de exibição S} that is the set of vertices of any subcube of {displaystyle Q_{d}} .
Vertex isoperimetric inequality Harper's theorem[10] says that Hamming balls have the smallest vertex boundary among all sets of a given size. Hamming balls are sets that contain all points of Hamming weight at most {estilo de exibição r} and no points of Hamming weight larger than {displaystyle r+1} for some integer {estilo de exibição r} . This theorem implies that any set {displaystyle Ssubseteq V} com {estilo de exibição |S|soma geq _{i=0}^{r}{d choose i}} satisfies {estilo de exibição |Scup Gamma (S)|soma geq _{i=0}^{r+1}{d choose i}.} [11] As a special case, consider set sizes {displaystyle k=|S|} of the form {displaystyle k={d choose 0}+{d choose 1}+pontos +{d choose r}} for some integer {estilo de exibição r} . Then the above implies that the exact vertex isoperimetric parameter is {displaystyle Phi _{V}(Q_{d},k)={d choose r+1}.} [12] Isoperimetric inequality for triangles The isoperimetric inequality for triangles in terms of perimeter p and area T states that[13][14] {estilo de exibição p^{2}geq 12{quadrado {3}}cdot T,} with equality for the equilateral triangle. This is implied, via the AM–GM inequality, by a stronger inequality which has also been called the isoperimetric inequality for triangles:[15] {displaystyle Tleq {fratura {quadrado {3}}{4}}(abc)^{fratura {2}{3}}.} See also Mathematics portal Blaschke–Lebesgue theorem Chaplygin problem Curve-shortening flow Expander graph Gaussian isoperimetric inequality Isoperimetric dimension Isoperimetric point List of triangle inequalities Planar separator theorem Mixed volume Notes ^ Blåsjö, Viktor (2005). "The Evolution of the Isoperimetric Problem". América. Matemática. Por mês. 112 (6): 526–566. doi:10.2307/30037526. JSTOR 30037526. ^ Olmo, Carlos Beltrán, Irene (4 Janeiro 2021). "Sobre mates y mitos". EL PAÍS (in Spanish). Recuperado 14 Janeiro 2021. ^ J. Steiner, Einfacher Beweis der isoperimetrischen Hauptsätze, J. reine angew Math. 18, (1838), pp. 281–296; and Gesammelte Werke Vol. 2, pp. 77–91, Reimer, Berlim, (1882). ^ Zwierzyński, Michał (2016). "The improved isoperimetric inequality and the Wigner caustic of planar ovals". J. Matemática. Anal. Appl. 442 (2): 726–739. arXiv:1512.06684. doi:10.1016/j.jmaa.2016.05.016. S2CID 119708226. ^ Gromov, Mikhail; Pansu, Pierre (2006). "Appendix C. Paul Levy's Isoperimetric Inequality". Metric Structures for Riemannian and Non-Riemannian Spaces. Clássicos modernos de Birkhäuser. Dordrecht: Springer. p. 519. ISBN 9780817645830. ^ Osserman, Roberto. "The Isoperimetric Inequality." Boletim da American Mathematical Society. 84.6 (1978) http://www.ams.org/journals/bull/1978-84-06/S0002-9904-1978-14553-4/S0002-9904-1978-14553-4.pdf ^ Hoory, Linial & Widgerson (2006) ^ Definitions 4.2 e 4.3 of Hoory, Linial & Widgerson (2006) ^ See Bollobás (1986) and Section 4 in Hoory, Linial & Widgerson (2006) ^ Cf. Calabro (2004) or Bollobás (1986) ^ cf. Leader (1991) ^ Also stated in Hoory, Linial & Widgerson (2006) ^ Chakerian, G. D. "A Distorted View of Geometry." CH. 7 in Mathematical Plums (R. Honsberger, editor). Washington, DC: Associação Matemática da América, 1979: 147. ^ "The isoperimetric inequality for triangles". ^ Dragutin Svrtan and Darko Veljan, "Non-Euclidean Versions of Some Classical Triangle Inequalities", Fórum geométrico 12, 2012, 197–209. http://forumgeom.fau.edu/FG2012volume12/FG201217.pdf References Blaschke and Leichtweiß, Elementare Differentialgeometrie (em alemão), 5th edition, completely revised by K. Leichtweiß. Os ensinamentos básicos das ciências matemáticas, Band 1. Springer-Verlag, New York Heidelberg Berlin, 1973 ISBN 0-387-05889-3 Bollobás, Béla (1986). Combinatória: set systems, hypergraphs, families of vectors, and combinatorial probability. Cambridge University Press. ISBN 978-0-521-33703-8. Burago (2001) [1994], "Isoperimetric inequality", Enciclopédia de Matemática, EMS Press Calabro, Chris (2004). "Harper's Theorem" (PDF). Recuperado 8 Fevereiro 2011. Capogna, Luca; Donatella Danielli; Scott Pauls; Jeremy Tyson (2007). An Introduction to the Heisenberg Group and the Sub-Riemannian Isoperimetric Problem. Birkhäuser Verlag. ISBN 978-3-7643-8132-5. Fenchel, Werner; Bonnesen, Tommy (1934). Theorie der konvexen Körper. Resultados de matemática e suas áreas de fronteira. Volume. 3. Berlim: 1. Verlag von Julius Springer. Fenchel, Werner; Bonnesen, Tommy (1987). Theory of convex bodies. Moscou, Idaho: eu. Boron, C. Christenson and B. Smith. BCS Associates. Federer, Herbert (1969). Geometric measure theory. Springer-Verlag. ISBN 3-540-60656-4.. Gromov, M.: "Paul Levy's isoperimetric inequality". Appendix C in Metric structures for Riemannian and non-Riemannian spaces. Based on the 1981 French original. With appendices by M. Katz, P. Pansu and S. Semmes. Translated from the French by Sean Michael Bates. Progress in Mathematics, 152. Birkhäuser Boston, Inc., Boston, Massachusetts, 1999. Hadwiger, Hugo (1957). Vorlesungen über Inhalt, Oberfläche und Isoperimetrie. Springer-Verlag.. Hoory, Shlomo; linear, Natan; Widgerson, Avi (2006). "Expander graphs and their applications" (PDF). Boletim da American Mathematical Society. Nova série. 43 (4): 439–561. doi:10.1090/S0273-0979-06-01126-8. Leader, Imre (1991). "Discrete isoperimetric inequalities". Proceedings of Symposia in Applied Mathematics. Volume. 44. pp. 57–80. Osserman, Roberto (1978). "The isoperimetric inequality". Touro. América. Matemática. Soc. 84 (6): 1182–1238. doi:10.1090/S0002-9904-1978-14553-4.. Zwierzyński, Michał (2016). "The improved isoperimetric inequality and the Wigner caustic of planar ovals". J. Matemática. Anal. Appl. 442 (2): 726–739. arXiv:1512.06684. doi:10.1016/j.jmaa.2016.05.016. S2CID 119708226. Schmidt, Erhard (1949). "Die Brunn-Minkowskische Ungleichung und ihr Spiegelbild sowie die isoperimetrische Eigenschaft der Hugel in der euklidischen und nichteuklidischen Geometrie. II". Matemática. Nachr. 2 (3-4): 171-244. doi:10.1002/mana.19490020308.. Baebler, F. (1957). "Zum isoperimetrischen Problem". Arch. Matemática. (Basileia). 8: 52–65. doi:10.1007/BF01898439. S2CID 123704157.. External links Wikimedia Commons has media related to Isoperimetric inequality. History of the Isoperimetric Problem at Convergence Treiberg: Several proofs of the isoperimetric inequality Isoperimetric Theorem at cut-the-knot Categories: Multivariable calculusCalculus of variationsGeometric inequalitiesAnalytic geometry
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Given six cities and the costs(in millions of dollars) of rebuilding roads between them. The road commission has decided to rebuild enough roads so that each pair of cities will be connected, either directly or by going through other cities, by rebuilt roads. Find a minimum cost set of roads meeting its criterion.
We visited this question previously with Kruskal’s Algorithm. Now let’s try a different approach using Prim’s Algorithm.
Prim’s algorithm is another greedy algorithm that finds a minimal spanning tree in a connected, weighted graph. Unlike Kruskal’s algorithm, whose partial solutions are not necessarily connected, a partial solution in prim’s algorithm is a tree. Prim’s algorithm begins with a start vertex and no edges and then applies the greedy rule: Add an edge of minimum weight that has one vertex in the current tree and the other not in the current tree.
The big idea is to maintain a table that stores each vertex with its corresponding edge weight value adjacent to the starting vertex in question. Additionally, we need an array to store the ‘parent’ vertex that has access to the vertex in question. We then go about selecting the least weight value in the table and select it as the edge path to pick. The symbol for choosing this edge could be deleting the row entirely from the table. After each edge selection, we update the table with the newly accessible vertex. When there are no more rows left in the table, then we would have selected every edge required of a minimum spanning tree.
For the table above, we can use a python dictionary with key as the vertex and value as the edge weight with its ‘parent’ vertex.
We also require a container to store the vertices each vertex is able to reach directly. For this container, we can use an adjacency list. adj[i] is a reference to the first node in a linked list of nodes representing the vertices adjacent to vertex i. Here’s the adjacency list implementation:
``````class Node:
"""
Node class defines the basic structure of a node.
"""
def __init__(self, vertex=None, weight=None, next=None):
self.vertex = vertex
self.weight = weight
self.next = next
"""
LinkedList class defines a linked list of nodes in which node.next reference the next node in the list
"""
def __init__(self):
"""
Returns a reference to the head node
"""
def add(self, vertex, weight):
"""
Append a new node with node.vertex as vertex to the last element in the linked list
"""
new_node = Node(vertex, weight)
while(trav.next != None):
trav = trav.next
trav.next = new_node
def print_list(self):
while trav != None:
print "(%d, %d)" % (trav.vertex, trav.weight),
trav = trav.next
``````
By calling the function adj_list.print_list() method, we should get the following printout, which is the content of each adjacent node in the list.
``````(2, 4) (3, 2) (5, 3) None
(1, 4) (4, 5) None
(1, 2) (5, 6) (6, 3) (4, 1) None
(2, 5) (3, 1) (6, 6) None
(1, 3) (3, 6) (6, 2) None
(5, 2) (3, 3) (4, 6) None
``````
Next, we define an Edges class to contain the table mentioned earlier. The class has a 2d list attribute like the following:
``````edges = [[vertex, weight, parent_vertex],...]
``````
Each row of the table is represented by edges[i].
Edges class has the following attributes and methods:
``````class Edges:
"""
Edge table consisting of vertex, weight and parent_vertex in a row.
"""
edge_table = []
def __init__(self, adj_ref, start):
while adj_ref != None:
def get_table(self):
return self.edge_table
def add_row(self, vertex, weight, parent_vertex):
"""
Add a row in table with vertex, weight and parent_vertex attributes
"""
self.edge_table.append([vertex, weight, parent_vertex])
def pop_row(self):
"""
Delete the row in table with the smallest weight and return the corresponding vertex.
"""
if len(self.edge_table) > 0:
least_weight = self.edge_table[0][1]
least_weight_index = 0
for i in range(1, len(self.edge_table)):
if self.edge_table[i][1] < least_weight:
least_weight = self.edge_table[i][1]
least_weight_index = i
least_weight_vertex = self.edge_table[least_weight_index][0]
self.edge_table.pop(least_weight_index)
return least_weight_vertex
else:
return None
def is_in(self, vertex):
"""
Return true if vertex is in table and false otherwise.
"""
for edge in self.edge_table:
if edge[0] == vertex:
return True
return False
def vertex_weight(self, vertex):
"""
Return the weight corresponding to vertex.
"""
for edge in self.edge_table:
if edge[0] == vertex:
return edge[1]
return None
def update_row(self, vertex, weight, parent_vertex):
"""
Changes the weight and parent_vertex corresponding to existing vertex.
"""
for edge in self.edge_table:
if edge[0] == vertex:
edge[1] = weight
edge[2] = parent_vertex
def print_table(self):
"""
Print edge_table.
"""
print self.edge_table
``````
Lastly, we have our prim’s algorithm to sort out the minimum spanning tree of the directed, weighted graph.
``````def prim(adj_ref, start):
"""
Finds a minimal spanning tree in a connected, weighted, n-vertex graph.
The graph is represented using adjacency lists; adj[i] is a reference to the first node in a linked list of nodes.
The start vertex is start. In the minial spanning tree, the parent of a vertex is defined as the vertex where existing minimal tree vertex extends from.
Start vertex(parent of itself) is = 0.
"""
table = Edges(adj_ref[4].next, start)
chosen = []
chosen.append(start)
while len(table.get_table()) > 0:
table.print_table()
ver = table.pop_row()
chosen.append(ver)
print "Child vertex %d chosen." % ver
while trav != None:
if trav.vertex in chosen:
trav = trav.next
continue
if trav.vertex == start:
trav = trav.next
continue
if table.is_in(trav.vertex): # table has vertex
if trav.weight < table.vertex_weight(trav.vertex): # new edge weight is less than existing table value
table.update_row(trav.vertex, trav.weight, ver) # update table to new vertex and weight
else:
# insert vertex and its elements into table
trav = trav.next
print chosen
``````
The function prim is module level function and uses both Edges and LinkList classes to maintain interity and coherence of data.
It’s time to test the program:
``````if __name__ == "__main__":
adj_list = [0] * 6 # an array of adjacency list instances
for i in range(len(adj_list)):
# Setup the adjacency list
``````
With the last line prim(adj,4), the output would be:
``````[[0, 3, 4], [2, 6, 4], [5, 2, 4]]
Child vertex 5 chosen.
[[0, 3, 4], [2, 3, 5], [3, 6, 5]]
Child vertex 0 chosen.
[[2, 2, 0], [3, 6, 5], [1, 4, 0]]
Child vertex 2 chosen.
[[3, 1, 2], [1, 4, 0]]
Child vertex 3 chosen.
[[1, 4, 0]]
Child vertex 1 chosen.
[4, 5, 0, 2, 3, 1]
``````
Okok some explaination on the output here, Every successive 2 lines indicates a successful selection of a vertex to be chosen as part of the minimal spanning tree. Each of the first line of 2d lists indicates the available options in the table. Second line, Child vertex x chosen. indicates which of the vertex(and its corresponding row) is chosen. The last line outputs the order of which the vertices are selected, with the first element being the start vertex.
Prim’s algorithm is also an optimal implementation of the greedy algorithm to achieve a minimal spanning tree. However, through repeated runtime tests, it seems Prim’s algorithm is consistently faster than Kruskal’s Algorithm. I did a time complexity analysis of Prim’s and came up with the following:
Prim’s algorithm time complexity(major functions):
``````**Operation** **Worst-Case Time**
pop_row() O(lg n)
is_in(vertex) O(1)
vertex_weight(vertex) O(n)
update_row(vertex, weight, ..) O(n)
``````
The outer while loop take time O(n). The inner loop has a relation with the number of edges. Since we know that given n vertices, there can be a maximum of 2n edges so the time complexity of the inner while loop has complexity of O(n**2).
So the final time complexity of prim’s algorithm will be O(n**3). This is a bad time complexity given that the operations are not very convoluted. I think the bottleneck of this implementation is the abstract data structure used. I used a linked list data structure which in itself has a complexity of O(n). If we can use a more efficient data structure such as binary minheap structure which has O(lg n) complexity, we can really reduce the time complexity of Prim’s algorithm to O(n2 lg n ). Maybe i’ll try to implement a binary minheap structure later.
Here’s my full implementation. I would love to hear suggestions on how to improve the time complexity or anything in general. Cheers :) | 2,141 | 8,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.921188 |
https://rstudio-pubs-static.s3.amazonaws.com/28730_850cac53898b45da8050f7f622d48927.html | 1,596,921,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00586.warc.gz | 481,842,855 | 329,851 | ### Other resources
• We have discussed non-linear regression before during R sessions, see: 2013-04-05 : Non-linear regression.
• A very good book on non-linear regression with R is Ritz and Streibig 2008 (online access on campus).
• It can be difficult to find the right non-linear model. I have yet to find a better alternative to a SAS-oriented guide to curve fitting, published in 1994 by the Province of British Columbia (download it from the Resources section on the HIE R site on sharepoint).
### Brief introduction
Basically, non-linear regression is used when you cannot use some sort of linear model. A model is linear when you can write it like this:
$Y = b_0 + b_1*X_1 + b_2*X_2 + b_3*X_3$
where Y is the dependent variable, and the X’s are the predictors, which can be any variable or some transformation thereof (for example, squared terms, two variables multiplied together, etc.). This model is linear in the coefficients b1, b2, etc, though it may contain terms that are non-linear in the X’s (such as squared terms of X).
An example of a non-linear model is the Chapman-Richards growth equation,
$Y = Y_{max}*(1-e^{-b*x})^c$
Here the coefficients (Ymax, b and c) occur non-linearly in the equation, and you cannot write this in a form similar to the linear model as shown above. We can look at the shape of this curve by using curve, like so:
# Define function
chapm <- function(x,Ymax,b,c=2.5)Ymax*(1-exp(-b*x))^c
# Plot (curve must use an x argument)
curve(chapm(x, Ymax=100, b=0.15, c=3), from=0, to=50, ylab="Y")
We can use non-linear regression to fit curves like that one to data, which gives estimates of the coefficients Ymax, b and c. The majority of useful applications of non-linear regression are when you have an equation already specified (arising from theory), and you wish the estimate the parameters of the curve, or use the curve for prediction (as in lots of modelling applications).
In the following we use only the nls function, but point out nlsLM (package minpack.lm) for a very promising alternative, and nls2 (package of same name) for an alternative that can be used to find starting values.
### A simple non-linear model
The most annoying aspect of nls is that you have to specify starting values for the parameters, otherwise nls will in many cases not converge to a solution. Depending on the situation (and especially the equation to be fit, these starting values need to be quite good. Normally we plot the data, choose a curve to fit, and play with the coefficients until the fit is ‘close’.
Here we use the built-in Loblolly data (see ?Loblolly), which contains age and height of Loblolly pine (Pinus taeda), for several seedlots.
# We have chosen this function to fit.
fx <- function(x, Asym, R0, lrc)Asym+(R0-Asym)*exp(-exp(lrc)*x)
# Make plot and add a curve.
with(Loblolly, plot(age, height))
curve(fx(x, Asym=100, lrc=-3, R0=-8), add=T)
That’s pretty close, so let’s see if nls can converge.
# Note the specification of the starting values.
nlsfit <- nls(height ~ fx(age, Asym, R0, lrc),
data=Loblolly,
start=list(Asym=100, lrc=-3, R0=-8))
Nothing was printed, which is good news. If it does not converge it will report an error.
Note that you do not have to write a function first and use it within nls, you can also write the equation in the formula interface of nls directly. However, it is usually convenient to have the function defined separately.
We can add the fitted curve to a plot in two ways. Here is the first, which uses the estimated coefficients and plugs them into the function we fit to the data.
# Save coefficients in a vector
p <- coef(nlsfit)
# Plot and add a curve. Note that 'p' is a named vector.
with(Loblolly, plot(age, height, xlim=c(0, 30), ylim=c(0,75)))
curve(fx(x, Asym=p["Asym"], R0=p["R0"], lrc=p["lrc"]),
add=T, col="red")
The second method uses predict, which is sometimes easier to use (and is safer in the sense that it is more difficult to make mistakes in the code).
# Make dataframe with X variable that we wish to predict Y values for.
# Make sure it has the same name as in the dataframe we used to fit the model!
newdat <- data.frame(age=seq(2,25, length=101))
# Predict from the fitted model for the new dataframe.
newdat$heightpred <- predict(nlsfit, newdata=newdat) # Add a line. with(Loblolly, plot(age, height, xlim=c(0, 30), ylim=c(0,75))) with(newdat, lines(age, heightpred, col="blue")) Unlike the predict method for linear models (predict.lm in particular), we cannot estimate standard errors or confidence intervals for predictions (although predict.nls does have an argument se.fit which is actually ignored!). To do this, we rely once again on the bootstrap. Suppose we are interested in the predicted value of height at an age of 22.5 years, and do so by interpolating with the fitted nonlinear model. This example shows how to get a confidence interval for that estimate as well. library(car) # bootstrap resampling. The object b contains coefficients of the nonlinear model fitted to # 999 resampled datasets. b <- bootCase(nlsfit) # Note that we defined fx above, it is the nonlinear equation fit to the data. p225 <- fx(x=22.5, Asym=b[,"Asym"], lrc=b[,"lrc"], R0=b[,"R0"]) # The 2.5 and 97.5% quantiles give the confidence interval for the prediction. quantile(p225, probs=c(0.025, 0.975)) ## 2.5% 97.5% ## 55.39 56.86 ### Self-starting functions To avoid the annoying fiddling with starting values, we can use one of many ‘self-starting functions’. These are common non-linear models that do not require starting values. You can also develop these yourself, but this is tricky (see the reference to the book in ‘Other resources’ above for more information). I won’t go into much detail on these functions, but simply redo our example with the Loblolly data from above. I actually copied that function from the example under SSasymp, a self-starting asymptotic function. # Fit same model, using a self-starting model nlsfitSS <- nls(height ~ SSasymp(age, Asym, R0, lrc), data=Loblolly) The result should be exactly the same as above. It would be nice if someone were to make a compendium of all self-starting functions, including graphical illustrations of the functional shape and how it depends on the parameters. Summaries of the functions can be found in various places but a visual guide would be useful, and is lacking at the moment. ### Non-linear mixed effects models Next, we briefly look at adding random effects to a nonlinear model. Suppose we have repeated measured for a number of individuals, and would like to take into account the variation between individuals, rather than fitting one model for the entire dataset. In the following, I will show an example of a non-linear mixed effects model using the nlme package. I will use the built-in ChickWeight data, where a number of individual chicks were weighed daily for 21 days, while being fed on one of four diets. The question is, which diet causes the quickest growth of the chicks, and was the growth trajectory different by diet? First we visualize the data. Here I use the plotBy package, which I developed. Ssee this website for instructions to install it. library(plotBy) library(gplots) palette(rich.colors(20)) plotBy(weight ~ Time|Chick, type='l', data=subset(ChickWeight, Diet == "1"), legend=FALSE) Here I plotted the chicks for the first diet only. We see that the individual chicks: • differ a lot in their final weight • differ in the growth trajectory (the shape of the curve) Also, some chicks died after a few days. Clearly we cannot fit a single curve to this cloud of data, but we should be able to estimate a population average growth curve, and some measures of the variance between chicks. To do this, we can fit a mixed-effects model where the parameters of the model to be fit can vary between subjects (random effects), and between treatments. The latter would be a fixed effect because we are actually interested in their values, but the individual chick effect should be a random effect because we simply wish to account for this source of variation, but we are not really interested in how well chick nr 5 did compared to nr 6. The logistic growth model is quite useful here. First, let’s visualize to see how the parameters affect the shape of the curve. # We will use SSlogis shortly, but this is the curve itself: flogis <- function(x, Asym, xmid, scal)Asym/(1+exp((xmid-x)/scal)) # Four curves with different parameter values. curve(flogis(x, 200, 10, 2), from=0, to=21, lwd=2) curve(flogis(x, 100, 10, 2), from=0, to=21, lty=5, add=TRUE) curve(flogis(x, 200, 15, 2), from=0, to=21, col="red", add=TRUE) curve(flogis(x, 200, 10, 4), from=0, to=21, col="blue", add=TRUE) Clearly, Asym affects the asymptote, xmid the inflection point, and scal is a shape parameter (note when I increased it - the blue curve above - the curve flattened). #### Fitting a model with one random effect Let’s start by fitting a non-linear mixed-effects model with only one random effect (for Asym), and one fixed effect (Asym is varied by Diet). The syntax for nlme is a bit strange: even though we will use a self-starting function (to avoid the common problem of not converging), we still have to specify the starting values. And because Asym will vary as a fixed effect by Diet, we have to specify four starting values, one for each Diet. To test whether Asym varies significantly by diet, it is appropriate to use a likelihood ratio test against a model that does not include the Diet effect. So, we first fit fitnlme0, which includes no Diet, and then fitnlme0, which includes diet as a fixed effect term for Asym. require(nlme) fitnlme0 <- nlme(weight ~ SSlogis(Time, Asym, xmid, scal), fixed=list(Asym + xmid + scal ~ 1), random = Asym ~ 1 | Chick, start=list(fixed=c(Asym=200,xmid=13,scal=4)), data=ChickWeight) fitnlme1 <- nlme(weight ~ SSlogis(Time, Asym, xmid, scal), fixed=list(Asym ~ Diet, xmid + scal ~ 1), random = Asym ~ 1 | Chick, start=list(fixed=c(Asym=c(150,200,250,300),xmid=13,scal=4)), data=ChickWeight) # Likelihood ratio test anova(fitnlme0, fitnlme1) ## Model df AIC BIC logLik Test L.Ratio p-value ## fitnlme0 1 5 4980 5001 -2485 ## fitnlme1 2 8 4968 5003 -2476 1 vs 2 17.45 6e-04 Clearly Diet is very significant (also note the AIC has decreased). Next, let’s look at what the fitted model looks like, and if it is successful at describing the most important patterns in the data. We use predict to get model predictions, and for nlme models, you have to specify whether to include the random effects or not. If you only want predictions for the fixed effects (sort of averaged over all individual chicks), specify level=0, otherwise leave it as is and the random effects will also be used. # Fixed effects predictions # Make dataframe with all combinations of Diet and Time. newdat <- expand.grid(Diet=levels(ChickWeight$Diet), Time=0:21)
newdat$wpred <- predict(fitnlme1,newdat,level=0) # Plot. I use jitter so not all data cover each other. palette(c("blue","red","forestgreen","darkorange")) with(ChickWeight, plot(jitter(Time), weight, pch=19, col=Diet, main="fitnlme1")) plotBy(wpred ~ Time|Diet, data=newdat, type='l', add=TRUE, lwd=2) Next we visualize the random effects. In this example I just use one diet, to keep things a bit simple. # Random effects, diet 1 chick1 <- droplevels(subset(ChickWeight, Diet == "1")) # Note that we now need the Chick number in here, otherwise predict won't work! newdat1 <- expand.grid(Diet="1", Chick=levels(chick1$Chick), Time=0:21)
newdat1$wpred <- predict(fitnlme1,newdat1) palette(rich.colors(nlevels(newdat1$Chick)))
plotBy(wpred ~ Time|Chick, data=newdat1, type='l',
ylim=c(0,320), legend=FALSE)
with(chick1, points(jitter(Time), weight, pch=16))
From the graph we can see that some of the data structure is modelled OK, but the shapes of the curves are all wrong for the first 10 days or so. But we know the curves are more flexible, so if we also include random effects for the other parameters, it should gradually improve. If it does, the AIC should decrease (and the likelihood ratio test should be significant). In the following three examples, I increase the model complexity one step at a time, and see what happens. See for yourself how and if the model improves.
### Increasing the model complexity
This model includes random effects for Asym and xmid.
fitnlme2 <- nlme(weight ~ SSlogis(Time, Asym, xmid, scal),
fixed=list(Asym ~ Diet, xmid + scal ~ 1),
random = Asym +xmid ~ 1 | Chick,
start=list(fixed=c(Asym=c(150,200,250,300),xmid=13,scal=4)),
data=ChickWeight)
anova(fitnlme1, fitnlme2)
## Model df AIC BIC logLik Test L.Ratio p-value
## fitnlme1 1 8 4968 5003 -2476
## fitnlme2 2 10 4448 4492 -2214 1 vs 2 524.2 <.0001
newdat$wpred2 <- predict(fitnlme2,newdat,level=0) palette(c("blue","red","forestgreen","darkorange")) with(ChickWeight, plot(jitter(Time), weight, pch=19, col=Diet,main="fitnlme2")) plotBy(wpred2 ~ Time|Diet, data=newdat, type='l', add=TRUE, lwd=2) # Random effects, diet 1 newdat1$wpred <- predict(fitnlme2,newdat1)
palette(rich.colors(nlevels(newdat1$Chick))) plotBy(wpred ~ Time|Chick, data=newdat1, type='l', main="fitnlme2", ylim=c(0,320), legend=FALSE) with(chick1, points(jitter(Time), weight, pch=16)) This model includes random and fixed effects for both xmid and Asym. fitnlme3 <- nlme(weight ~ SSlogis(Time, Asym, xmid, scal), fixed=list(Asym + xmid ~ Diet, scal ~ 1), random = Asym +xmid ~ 1 | Chick, start=list(fixed=c(Asym=c(150,200,250,300),xmid=c(10,10,10,10),scal=4)), data=ChickWeight) anova(fitnlme2, fitnlme3) ## Model df AIC BIC logLik Test L.Ratio p-value ## fitnlme2 1 10 4448 4492 -2214 ## fitnlme3 2 13 4443 4500 -2209 1 vs 2 10.53 0.0145 # Fixed effects newdat$wpred3 <- predict(fitnlme3,newdat,level=0)
palette(c("blue","red","forestgreen","darkorange"))
with(ChickWeight, plot(jitter(Time), weight, pch=19, col=Diet, main="fitnlme3"))
plotBy(wpred3 ~ Time|Diet, data=newdat, type='l', add=TRUE, lwd=2)
# Random effects
newdat1$wpred <- predict(fitnlme3,newdat1) palette(rich.colors(nlevels(newdat1$Chick)))
plotBy(wpred ~ Time|Chick, data=newdat1, type='l', main="fitnlme3",
ylim=c(0,320), legend=FALSE)
with(chick1, points(jitter(Time), weight, pch=16))
And finally, this model includes all random (but not scal, otherwise the model converges very slowly) and fixed effects.
fitnlme4 <- nlme(weight ~ SSlogis(Time, Asym, xmid, scal),
fixed=list(Asym + xmid + scal ~ Diet),
random = Asym + xmid ~ 1 | Chick,
start=list(fixed=c(Asym=c(150,200,250,300),xmid=c(10,10,10,10),scal=c(4,4,4,4))),
data=ChickWeight)
anova(fitnlme3, fitnlme4)
## Model df AIC BIC logLik Test L.Ratio p-value
## fitnlme3 1 13 4443 4500 -2209
## fitnlme4 2 16 4446 4516 -2207 1 vs 2 3.045 0.3847
# Fixed effects
newdat$wpred4 <- predict(fitnlme4,newdat,level=0) palette(c("blue","red","forestgreen","darkorange")) with(ChickWeight, plot(jitter(Time), weight, pch=19, col=Diet, main="fitnlme4")) plotBy(wpred4 ~ Time|Diet, data=newdat, type='l', add=TRUE, lwd=2) # Random effects newdat1$wpred <- predict(fitnlme4,newdat1)
palette(rich.colors(nlevels(newdat1\$Chick)))
plotBy(wpred ~ Time|Chick, data=newdat1, type='l', main="fitnlme4",
ylim=c(0,320), legend=FALSE)
with(chick1, points(jitter(Time), weight, pch=16)) | 4,327 | 15,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-34 | latest | en | 0.858231 |
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# Page 495, problems 4-6 Page 495, questions 7-11
## Presentation on theme: "Page 495, problems 4-6 Page 495, questions 7-11"— Presentation transcript:
Page 495, problems 4-6 Page 495, questions 7-11
Homework: Page 495, problems 4-6 Page 495, questions 7-11
4. If the temperature of 34. 4 g of ethanol increases from 25
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
If the temperature of 34. 4 g of ethanol increases from 25. 0oC to 78
If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol? q = c * m * ∆T
q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o)
If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol? q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o)
q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530
If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol? q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530
q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 Units?
If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol? q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 Units?
q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 J
If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol? q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 J
A 4. 50g nugget of pure gold absorbed 276 J of heat
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go
A 4. 50g nugget of pure gold absorbed 276 J of heat
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T
A 4. 50g nugget of pure gold absorbed 276 J of heat
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T ∆T = q / cm
q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)
q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC
q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC Final temp?
q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)
A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? c = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC o + 475o = 500o
6. 155g of an unknown substance was heated from 25. 0oC to 40. 0oC
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
6. 155g of an unknown substance was heated from 25. 0oC to 40. 0oC
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T
6. 155g of an unknown substance was heated from 25. 0oC to 40. 0oC
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T c = q / m * ∆T
q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC)
q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC
q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC So what is it?
q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2. q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC ethanol
What is energy? The ability to do work or produce heat List two units for energy
What is energy? The ability to do work or produce heat List two units for energy
joules (J) calories (cal) Calories (Cal) kilocalories (kcal)
What is energy? The ability to do work or produce heat List two units for energy joules (J) calories (cal) Calories (Cal) kilocalories (kcal)
Kinetic or potential energy?
Kinetic or potential energy?
Kinetic or potential energy?
Kinetic or potential energy?
Kinetic or potential energy?
Kinetic or potential energy?
Kinetic or potential energy?
What is the relationship between a calorie and a joule?
What is the relationship between a calorie and a joule?
1 cal = j 1 j = cal
10. One lawn chair is aluminum, the other iron. Which is hotter?
One lawn chair is aluminum, the other iron. Which is hotter?
Al c = j/goC Fe c = j/goC
Al c = 0.897 j/goC Fe c = 0.449 j/goC Iron is hotter.
One lawn chair is aluminum, the other iron. Which is hotter? Al c = j/goC Fe c = j/goC Iron is hotter.
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? 1 cal = j
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? 1 cal = j 12 cal = 12(4.184) j
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? 1 cal = j 12 cal = 12(4.184) j 12 cal = j
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T
What is the specific heat of an unknown substance if a 2
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T c = q / m * ∆T
q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC)
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T c = q / m * ∆T = j / (2.50g)(5oC)
q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T c = q / m * ∆T = j / (2.50g)(5oC) = 4.02
q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 Units?
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T c = q / m * ∆T = j / (2.50g)(5oC) = 4.02 Units?
q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 j/goC
What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC? q = c * m * ∆T c = q / m * ∆T = j / (2.50g)(5oC) = 4.02 j/goC
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# Help converting math function into pseudo code
I found an interesting article (to me at least) that I want to play around
with, but am not fluent in Greek (Pure mathematical functions) .
http://www.bouncingchairs.net/pskalman-lategecco.pdf
the algorithm itself starts around page 4 (5 is graphs), 6 has its 'pseudo code',
which is all Math.
I don't understand what
norm
Normal
||v_t|| or ||vt_best|| (the double pipes, or how to evaluate them)
in this context mean.
Any nudge in the right direction is super appreciated.
THanks,
0
JustinW
• 13
• 10
• 2
1 Solution
Commented:
I guess in this paper by ||u|| they mean the length (absolute value )
of the vector
0
Commented:
like u/||u||
Normalization of the vector means making the vector in the same direction
but with unit length and to get it you need to divide each
coorduinate by the length of the vector
0
Commented:
So if vector has say three components
u(x,y,z)
then ||u|| = SQRT(x*x + y*y + z*z)
0
Author Commented:
It looks like in wikipedia (http://en.wikipedia.org/wiki/List_of_mathematical_symbols)
|| x || means the norm of the element x of a normed vector space
So, equation 7 would go like this?
``````for (i = 0; i < n; i++)
{
v[i] += v[i]/SQRT(SUM(v[i]^2..v[n]^2))
+ (1-0.45)
* vb[i]/SQRT(SUM(vb[i]^2..vb[n]^2))
}
``````
0
Commented:
In equation (5), the article has the expression: alpha * u1 / || u1 || where u1 is a vector. I happen to know that u1 / || u1 || is a way to create a unit vector from the vector u1. (The unit vector is a vector in the same direction as the original, but having a length of 1.)
To do this scaling, you have to divide the vector by the vector length.
If the vector is in a 3D vector space, for example, then it has three coordinates.
u1 = xi + yj + zk = (x, y, z)
Then || u1 || = sqrt( x² + y² + z² ) = the length of u1.
0
Commented:
I should have refreshed this open page to see what else had been written.
0
Commented:
Yes, you could call it the norm, it is easier to think of ||u|| value as a length
of the vector, especially when you think in 3D space
u/||u|| will indeed be called normalized vector - this is the vector
which has the same direction as original vector u, but is of a unit length
So you obrtain such vector by dividing all components
of original vector by the length of the vector
0
Author Commented:
if my interpretation of equation 8 is correct,
what do you do with equation 9?
``````v(hat)[i] ~ Normal(0.6, v[i])
``````
0
Author Commented:
**Correction to the first bit of code, that shouldn't be +=; just =
0
Commented:
Equation (9):
float [] vt = new float[3];
float [] vbest = new float[3];
float [] v = new float[3];
float [] vtplus1 = new float[3];
float alpha;
flaot lengthvt = Math.sqrt(vt[1]*vt[1] + vt[2]*vt[2] + vt[3]*vt[3]);
flaot lengthvbest = Math.sqrt(vbest[1]*vbest[1] + vbest[2]*vbest[2] + vbest[3]*vbest[3]);
float lengthv = Math.sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
for (int j=0; j<3; j++) {
vtplus1[j] =( (alpha*lengthvt + (1.0f-alpha)lengthvbest)/lenghtv)*v[j];
}
vtplus1[j] = (alpha*
}
0
Commented:
Equation (9) (correction):
``````float [] vt = new float[3];
float [] vbest = new float[3];
float [] v = new float[3];
float [] vtplus1 = new float[3];
float alpha;
flaot lengthvt = Math.sqrt(vt[1]*vt[1] + vt[2]*vt[2] + vt[3]*vt[3]);
flaot lengthvbest = Math.sqrt(vbest[1]*vbest[1] + vbest[2]*vbest[2] + vbest[3]*vbest[3]);
float lengthv = Math.sqrt(v[1]*v[1] + v[2]*v[2] + v[3]*v[3]);
for (int j=0; j<3; j++) {
vtplus1[j] =( (alpha*lengthvt + (1.0f-alpha)lengthvbest)/lenghtv)*v[j];
}
``````
0
Commented:
In equation (7) you also need multiply by 0.55f* before the first part of the sum
assuming that 1-alpha = 0.45
0
Author Commented:
I'm sorry, I got my numbers jumbled;
what is this?
``````Equation 8; it says v[i] ~ norm(sigma^2, v[i])
``````
0
Commented:
I guess this is not so simple - what they mean
that you should generate random v by generating random numcbers
for each coordinates, but this random numbers
should be distributed normally around value v' as math expectation (for each coorduiinate)
with dispersion of sigma - i guess sigma was also acalculated before?
For this you probably need to write a special program - you can
0 and 1 but then calculate value according the norma distribution
- I think we should browse around there should be such normally distributed
random generator in java somehwere, so you see you need to get vlaues close to v'
much more often than far away from v' (as opposed to uniform random which we have in
our standard java lbrary)
0
Commented:
0
Author Commented:
very nice indeed!
so v[j] = random number normally distributed around v[j] with a stdev of 0.6?
or something else entirely?
0
Commented:
yes, that's what it is I guess
I'm not yet sure how to get form their distribution with average 0 and dispersion 1
to your requiirement but I'm sure t is doable and should not be too hard
average is just to move - dispersion is probably just to delete by square root or somthing
0
Author Commented:
This should keep me busy for a while.
I'm sure I'll be back for more when I get stumped,
but in the meantime; thanks!
0
Author Commented:
Also,
(2*rand-1 + 2*rand-1 + 2*rand-1)*stdev + avg
appears to generate a random number normally distributed
around an average. Might be quicker too
0
Commented:
Yes, that seems logical - I guess by 2*rand - do you mean 2 in power of rand ?
0
Author Commented:
I have no idea why it works (or if it really does), but
its just straight multiply
I tested it in excel, and the AVERAGE function gave me the
average I put in; and the STDEV function roughly gave me the
stdev I put in
I checked this against the results from using = norminv(rand(),avg, stdev),
and they appear indistinguishable.
so I'd use (-1 + 2 * rand) + (-1 + 2 * rand) + (-1 + 2 * rand)
0
Author Commented:
all that, * stdev + average
0
Commented:
Well, if you have the same average and the same std dev,
it does not necessarily mean that the distribution
would be Gaussian
On the other hand we don't know if it is essential that it would be Gaussian
or just average and dispersion is important
0
Commented:
OK, so that turns out to be very simple - you take
this java standard methhod
nextGaussian() - it will return you the numbers distributed normally
with avergae 0.0 and stnd deviation 1.0
So if you multiply this number by standard deviation you need
and add the avergae you want then the new number will be distributed exavtly as you need so your method
for generating such values wuold be:
Random r = new Random(new java.util.Date().getTime();
double myGaussian(double avg, double std_dev, Random r){
double x = r.nextGaussian();
return (x*std_dev + avg);
}
So if you call
myGaussian(5, 25, r);
million times than your resulst will be ditributed over Gaussian centered at 5 with standard deviation 25
0
Author Commented:
I felt like modeling this in ruby to work out some of the kinks
before shifting platforms.
Here's what I have so far to describe a particle (hawk)
``````class Hawk
attr_accessor :pos,
:vel,
:vprime,
:vhat,
:denom
#---------------------------------------------------------
def update!(best_p)
_denom!
_vprime!(best_p)
_vhat!
@vel.each_index do |x|
@vel[x] = (@alpha * @denom + (1 - @alpha) * best_p.denom) * @vhat[x]
end
end
#_________________________________________________________
##########################################################
private
##########################################################
#---------------------------------------------------------
def _denom!
b = 0
@vel.each{|x| b += x**2}
@denom = b**(1/2)
end
#_________________________________________________________
#---------------------------------------------------------
def _vprime!(bp)
@vel.each_index do |x|
@vprime[x] = @alpha *
(@vel[x] / @denom) +
(1 - @alpha) *
(bp.vel[x] / bp.denom)
end
end
#_________________________________________________________
#---------------------------------------------------------
def _vhat!
@vprime.each_index do |x|
@vhat[x] = @vprime[x].normal(@sigma)
end
end
#_________________________________________________________
end
``````
0
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```New test - October 20, 2014
1a. Let
[531 marks]
sin 100∘ = m. Find an expression for cos 100∘ in terms of m.
[3 marks]
Markscheme
Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award
the final A1FT mark.
METHOD 1
valid approach involving Pythagoras
(M1)
e.g. sin2 x + cos2 x = 1 , labelled diagram
correct working (may be on diagram)
−−−−−
e.g. m2 + (cos 100)2 = 1 , √1 − m2
−−−−−
cos 100 = −√1 − m2 A1 N2
(A1)
[3 marks]
METHOD 2
valid approach involving tan identity
e.g. tan =
(M1)
sin
cos
correct working
e.g. cos 100 =
(A1)
sin 100
tan 100
cos 100 = tanm100
A1
N2
[3 marks]
Examiners report
While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an
angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A
−−−−−
common algebraic error in part (a) was for candidates to write √1 − m2 = 1 − m . In part (c), many candidates failed to use the
double-angle identity. Many incorrectly assumed that because sin 100∘ = m , then sin 200∘ = 2m . In addition, some candidates did
not seem to understand what writing an expression "in terms of m" meant.
1b. Let
sin 100∘ = m . Find an expression for tan 100∘
in terms of m.
[1 mark]
Markscheme
METHOD 1
tan 100 = −
m
√1−m 2
(accept
m
)
− √1−m 2
A1
N1
[1 mark]
METHOD 2
tan 100 =
m
cos 100
A1
N1
[1 mark]
Examiners report
While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an
angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A
−−−−−
common algebraic error in part (a) was for candidates to write √1 − m2 = 1 − m . In part (c), many candidates failed to use the
double-angle identity. Many incorrectly assumed that because sin 100∘ = m , then sin 200∘ = 2m . In addition, some candidates did
not seem to understand what writing an expression "in terms of m" meant.
1c. Let sin 100∘ = m. Find an expression for sin 200∘ in terms of m.
[2 marks]
Markscheme
METHOD 1
valid approach involving double angle formula
(M1)
e.g. sin 2θ = 2 sin θcosθ
−−−−−
−−−−−
sin 200 = −2m√1 − m2 (accept 2m (−√1 − m2 ) )
A1
N2
−−−−−
Note: If candidates find cos 100 = √1 − m2 , award full FT in parts (b) and (c), even though the values may not have appropriate
signs for the angles.
[2 marks]
METHOD 2
valid approach involving double angle formula
(M1)
e.g. sin 2θ = 2 sin θ cos θ , 2m × tanm100
2
sin 200 = tan2m100 (= 2m cos 100)
A1
N2
[2 marks]
Examiners report
While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an
angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A
−−−−−
common algebraic error in part (a) was for candidates to write √1 − m2 = 1 − m . In part (c), many candidates failed to use the
double-angle identity. Many incorrectly assumed that because sin 100∘ = m , then sin 200∘ = 2m . In addition, some candidates did
not seem to understand what writing an expression "in terms of m" meant.
f(x) = a cos(b(x − c))
Let f(x) = a cos(b(x − c)) . The diagram below shows part of the graph of f , for 0 ≤ x ≤ 10 .
The graph has a local maximum at P(3, 5) , a local minimum at Q(7, − 5) , and crosses the x-axis at R.
2a. Write down the value of
(i)
[2 marks]
a;
(ii) c .
Markscheme
(i) a = 5 (accept −5 )
A1
N1
(ii) c = 3 (accept c = 7 , if a = −5 )
A1
N1
Note: Accept other correct values of c, such as 11, −5, etc.
[2 marks]
Examiners report
Part (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c.
2b. Find the value of b .
[2 marks]
Markscheme
attempt to find period
e.g. 8 , b =
(M1)
2π
period
0.785398 …
b=
2π
8
(exact),
π
4
, 0.785 [0.785, 0.786] (do not accept 45)
A1
N2
[2 marks]
Examiners report
Finding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8.
2c. Find the x-coordinate of R.
[2 marks]
Markscheme
valid approach
(M1)
e.g. f(x) = 0 , symmetry of curve
x = 5 (accept (5 ,0)) A1
N2
[2 marks]
Examiners report
Most candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve although
follow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down the
equations they entered into their GDC.
The following diagram shows a circular play area for children.
The circle has centre O and a radius of 20 m, and the points A, B, C and D lie on the circle. Angle AOB is 1.5 radians.
3a. Find the length of the chord [AB].
[3 marks]
Markscheme
Note: In this question, do not penalise for missing or incorrect units. They are not included in the markscheme, to avoid complex
METHOD 1
choosing cosine rule (must have cos in it)
(M1)
e.g. c2 = a2 + b2 − 2ab cos C
correct substitution (into rhs)
A1
−−−−−−−−−−−−−
e.g. 20 + 20 − 2(20)(20) cos 1.5 , AB = √800 − 800 cos 1.5
2
2
AB = 27.26555 …
AB = 27.3 [27.2, 27.3]
A1
N2
[3 marks]
METHOD 2
choosing sine rule
e.g.
sin A
a
=
sin B
b
,
(M1)
AB
sin O
correct substitution
e.g.
AB
sin 1.5
=
=
AO
sin B
A1
20
sin(0.5(π−1.5))
AB = 27.26555 …
AB = 27.3 [27.2, 27.3]
A1
N2
[3 marks]
Examiners report
Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a rightangled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors in
accuracy.
3b. Find the area of triangle AOB.
[2 marks]
Markscheme
correct substitution into area formula
e.g.
1
(20)(20) sin 1.5
2
,
A1
1
(20)(27.2655504 …) sin(0.5(π − 1.5))
2
area = 199.498997 … (accept 199.75106 = 200 , from using 27.3)
area = 199 [199, 200]
A1
N1
[2 marks]
Examiners report
Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a rightangled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors in
accuracy.
3c. Angle BOC is 2.4 radians.
Find the length of arc ADC.
[3 marks]
Markscheme
(M1)
appropriate method to find angle AOC
e.g. 2π − 1.5 − 2.4
correct substitution into arc length formula
(A1)
e.g. (2π − 3.9) × 20 , 2.3831853 … × 20
arc length = 47.6637 …
arc length = 47.7 (47.6, 47.7] (i.e. do not accept 47.6)
A1
N2
Notes: Candidates may misread the question and use AÔC = 2.4 . If working shown, award M0 then A0MRA1 for the answer 48.
Do not then penalize AÔC in part (d) which, if used, leads to the answer 679.498 …
However, if they use the prematurely rounded value of 2.4 for AÔC , penalise 1 mark for premature rounding for the answer 48 in
(c). Do not then penalize for this in (d).
[3 marks]
Examiners report
In part (c), some candidates misread the question and used 2.4 as the size of angle AOC while others rounded prematurely leading to
the inaccurate answer of 48. In either case, marks were lost.
3d. Angle BOC is 2.4 radians.
[3 marks]
Find the area of the shaded region.
Markscheme
calculating sector area using their angle AOC
(A1)
e.g. 12 (2.38 …)(202 ) , 200(2.38 …) , 476.6370614 …
shaded area = their area of triangle AOB + their area of sector
(M1)
e.g. 199.4989973 … + 476.6370614 … , 199 + 476.637
shaded area = 676.136 … (accept 675.637 … = 676 from using 199)
shaded area = 676 [676, 677]
A1
N2
[3 marks]
Examiners report
Part (d) proved to be straightforward and candidates were able to obtain full FT marks from errors made in previous parts.
3e. Angle BOC is 2.4 radians.
The shaded region is to be painted red. Red paint is sold in cans which cost \$32 each. One can covers 140
[4 marks]
m2 .
How much does it cost to
Markscheme
dividing to find number of cans
e.g.
676
140
(M1)
, 4.82857 …
5 cans must be purchased
(A1)
multiplying to find cost of cans
e.g. 5(32) ,
676
140
(M1)
× 32
cost is 160 (dollars)
A1
N3
[4 marks]
Examiners report
Most candidates had a suitable strategy for part (e) and knew to work with a whole number of cans of paint.
The following diagram shows the graph of f(x) = a cos(bx) , for 0 ≤ x ≤ 4 .
There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .
4a. (i)
Write down the value of a .
(ii) Find the value of b .
[3 marks]
Markscheme
(i) a = 3
A1
N1
(ii) METHOD 1
attempt to find period
e.g. 4 , b = 4 ,
b=
2π
4
(= π2 )
(M1)
2π
b
A1
N2
[3 marks]
METHOD 2
attempt to substitute coordinates
(M1)
e.g. 3 cos(2b) = −3 , 3 cos(4b) = 3
b=
2π
4
(= π2 )
A1
N2
[3 marks]
Examiners report
In part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing the
amplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, there
2π
were many candidates who thought b was the period of the function, rather than period
.
4b. Write down the gradient of the curve at P.
[1 mark]
Markscheme
0
A1
N1
[1 mark]
Examiners report
In part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, many
candidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given value
of x, rather than simply stating that the gradient of a curve at a minimum point is zero.
4c. Write down the equation of the normal to the curve at P.
[2 marks]
Markscheme
recognizing that normal is perpendicular to tangent
(M1)
e.g. m1 × m2 = −1 , m = − 10 , sketch of vertical line on diagram
x = 2 (do not accept 2 or y = 2 )
A1
N2
[2 marks]
Examiners report
For part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involving
negative reciprocal gradients, rather than recognizing that the equation of the vertical line was x = 2 . There were also candidates
who had trouble expressing the correct equation of a line parallel to the y-axis.
f(x) = acos(b(x − c)) − 1
The diagram below shows part of the graph of f(x) = acos(b(x − c)) − 1 , where a > 0 .
The point P ( π4 ,2) is a maximum point and the point Q ( 3π
,−4) is a minimum point.
4
5a. Find the value of a .
[2 marks]
Markscheme
evidence of valid approach
e.g.
max y value−min y value
2
a=3
A1
(M1)
, distance from y = −1
N2
[2 marks]
Examiners report
A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph.
5b. (i)
Show that the period of f is π .
[4 marks]
(ii) Hence, find the value of b .
Markscheme
(i) evidence of valid approach
(M1)
e.g. finding difference in x-coordinates,
evidence of doubling
A1
e.g. 2 × ( π2 )
period = π
AG
N0
(ii) evidence of valid approach
e.g. b =
2π
π
b=2
A1
[4 marks]
N2
(M1)
π
2
Examiners report
A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble
showing that the period was π , either incorrectly adding the given π/4 and 3π/4 or using the value of b that they found first for part
(b)(ii).
5c. Given that 0 < c < π , write down the value of c .
[1 mark]
Markscheme
c = π4
A1
N1
[1 mark]
Examiners report
A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble
showing that the period was π , either incorrectly adding the given π/4 and π/3 or using the value of b that they found first for part
(b)(ii).
Let f(x) = (sin x + cos x)2 .
6a. Show that f(x) can be expressed as 1 + sin 2x .
[2 marks]
Markscheme
attempt to expand
(M1)
e.g. (sin x + cos x)(sin x + cos x) ; at least 3 terms
correct expansion
A1
e.g. sin2 x + 2 sin x cos x + cos2 x
f(x) = 1 + sin 2x
AG
N0
[2 marks]
Examiners report
Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates
were certainly helped by the fact that it was a "show that" question.
6b. The graph of f is shown below for 0 ≤ x ≤ 2π .
[2 marks]
Let g(x) = 1 + cos x . On the same set of axes, sketch the graph of g for 0 ≤ x ≤ 2π .
Markscheme
A1A1
N2
Note: Award A1 for correct sinusoidal shape with period 2π and range [0, 2], A1 for minimum in circle.
Examiners report
More candidates had difficulty with part (b) with many assuming the first graph was 1 + sin(x) and hence sketching a horizontal
translation of π/2 for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor p
correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation.
6c. The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the [2 marks]
k
vector ( ) .
0
Write down the value of p and a possible value of k .
Markscheme
p = 2 , k = − π2
[2 marks]
A1A1
N2
Examiners report
Part (c) certainly served as a discriminator between the grade 6 and 7 candidates.
The diagram shows a circle of radius 8 metres. The points ABCD lie on the circumference of the circle.
^D = 73∘ .
^ C = 104∘ , and BC
BC = 14 m, CD = 11.5 m, AD = 8 m, AD
7a. Find AC.
[3 marks]
Markscheme
evidence of choosing cosine rule
(M1)
eg c2 = a2 + b2 − 2ab cos C , C D2 + AD2 − 2 × CD × AD cos D
correct substitution
A1
eg 11.52 + 82 − 2 × 11.5 × 8 cos 104 , 196.25 − 184 cos 104
AC = 15.5 (m)
A1
N2
[3 marks]
Examiners report
There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the
question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible
combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected.
Candidate scripts did not indicate any adverse effect.
Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.
Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to
the fact that cyclic quadrilaterals is not part of the syllabus.
7b. (i)
(ii)
^D .
Find AC
^B .
Hence, find AC
[5 marks]
Markscheme
(i)
METHOD 1
evidence of choosing sine rule
eg
sin A
a
sin B
b
=
,
sin AĈD
correct substitution
eg
^D
sin AC
8
=
^ D = 30.0∘
AC
(M1)
D
= sin
AC
A1
sin 104
15.516…
A1
N2
METHOD 2
evidence of choosing cosine rule
eg
c2
=
a2
(M1)
2
+ b − 2ab cos C
correct substitution
2
A1
2
e.g. 8 = 11.5 + 15.516…2 − 2(11.5)(15.516 …) cos C
^ D = 30.0∘
AC
(ii)
A1
N2
^ D from 73
subtracting their AC
(M1)
^ D , 70 − 30.017 …
eg 73 − AC
^ B = 43.0∘
AC
A1
N2
[5 marks]
Examiners report
There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the
question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible
combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected.
Candidate scripts did not indicate any adverse effect.
Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.
Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to
the fact that cyclic quadrilaterals is not part of the syllabus.
7c. (c)
(d)
Find the area of triangle ADC.
Hence or otherwise, find the total area of the shaded regions.
[6 marks]
Markscheme
(c)
correct substitution
area = 44.6 (m2)
(A1)
1
(8)(11.5) sin 104
2
A1
N2
[2 marks]
(d)
attempt to subtract
(M1)
eg circle − ABCD , πr2 − ΔADC − ΔACB
area ΔACB = 12 (15.516 …)(14) sin 42.98
correct working
(A1)
A1
2
eg π(8) − 44.6336 … − 12 (15.516 …)(14) sin 42.98 , 64π − 44.6 − 74.1
A1
N3
[4 marks]
Total [6 marks]
Examiners report
There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the
question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible
combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected.
Candidate scripts did not indicate any adverse effect.
Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.
Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to
the fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and most
could find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrong
mode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle. In part (d),
most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.
A Ferris wheel with diameter 122 metres rotates clockwise at a constant speed. The wheel completes 2.4 rotations every hour. The bottom of
the wheel is 13 metres above the ground.
A seat starts at the bottom of the wheel.
8a. Find the maximum height above the ground of the seat.
[2 marks]
Markscheme
valid approach
(M1)
eg 13 + diameter , 13 + 122
maximum height = 135 (m)
A1 N2
[2 marks]
Examiners report
Most candidates were successful with part (a).
After t minutes, the height h metres above the ground of the seat is given by
h = 74 + acos bt.
8b. (b)
(i)
Show that the period of h is 25 minutes.
(ii)
Write down the exact value of b .
(c)
Find the value of a .
(d)
Sketch the graph of h , for 0 ≤ t ≤ 50 .
[9 marks]
Markscheme
(a)
(i)
60
period = 2.4
period = 25 minutes
(ii)
b=
2π
25
A1
AG
(= 0.08π)
N0
A1
N1
[2 marks]
(b)
METHOD 1
valid approach
(M1)
eg max − 74 , |a| = 135−13
, 74 − 13
2
|a| = 61 (accept a = 61 )
a = −61
A1
(A1)
N2
METHOD 2
attempt to substitute valid point into equation for h
(M1)
eg 135 = 74 + a cos ( 2π×12.5
)
25
correct equation
(A1)
eg 135 = 74 + a cos(π) , 13 = 74 + a
a = −61
A1
N2
[3 marks]
(c)
A1A1A1A1
N4
Note: Award A1 for approximately correct domain, A1 for approximately correct range,
A1 for approximately correct sinusoidal shape with 2 cycles.
Only if this last A1 awarded, award A1 for max/min in approximately correct positions.
[4 marks]
Total [9 marks]
Examiners report
A surprising number had difficulty producing enough work to show that the period was 25; writing down the exact
value of b also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the
Ferris wheel is a minimum at t = 0 thereby making the value of a negative. Incorrect values of 61 were often seen
with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key
features in approximately correct locations and candidates lost marks for incorrect domains and ranges.
8c. In one rotation of the wheel, find the probability that a randomly selected seat is at least 105 metres above the ground.
[5 marks]
Markscheme
setting up inequality (accept equation)
(M1)
eg h > 105 , 105 = 74 + a cos bt , sketch of graph with line y = 105
any two correct values for t (seen anywhere)
A1A1
eg t = 8.371 … , t = 16.628 … , t = 33.371 … , t = 41.628 …
M1
valid approach
eg
16.628−8.371
25
p = 0.330
,
A1
t 1− t 2
25
,
2×8.257
50
,
2(12.5−8.371)
25
N2
[5 marks]
Examiners report
Part (e) was very poorly done for those who attempted the question and most did not make the connection between height, time and
probability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novel
approaches from the strongest of candidates using circles and angles to solve this part of question 10.
The diagram below shows part of the graph of a function f .
The graph has a maximum at A(1, 5) and a minimum at B(3, −1) .
The function f can be written in the form f(x) = p sin(qx) + r . Find the value of
9.
(a)
p
(b)
q
(c)
r.
[6 marks]
Markscheme
(a)
valid approach to find p
eg amplitude =
p=3
A1
max−min
2
(M1)
,p=6
N2
[2 marks]
(b)
valid approach to find q
eg period = 4 , q =
q = π2
A1
(M1)
2π
period
N2
[2 marks]
(c)
valid approach to find r
eg axis =
r=2
max+min
2
A1
(M1)
, sketch of horizontal axis, f(0)
N2
[2 marks]
Total [6 marks]
Examiners report
Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they
attempted to substitute into the equation of the function with the parameters p,q and r. The successful candidates were able to find the
answers using the given points and their understanding of the different transformations.
Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between q and the period of the
function. There were also some candidates who showed working such as 2π
without explaining what b represented.
b
The following diagram shows a triangle ABC.
^ C = 50∘ .
The area of triangle ABC is 80 cm2 , AB = 18 cm , AC = x cm and BA
10a. Find x .
[3 marks]
Markscheme
correct substitution into area formula
eg
(A1)
1
(18x) sin 50
2
setting their area expression equal to 80
(M1)
eg 9x sin 50 = 80
x = 11.6
A1
N2
[3 marks]
Examiners report
The vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used right
triangle trigonometry. Errors included working in radian mode, assuming that the angle at C was 90∘ , and incorrectly applying the
order of operations when evaluating the cosine rule.
10b. Find BC.
[3 marks]
Markscheme
evidence of choosing cosine rule
(M1)
eg c2 = a2 + b2 + 2ab sin C
correct substitution into right hand side (may be in terms of x)
(A1)
eg 11.62 + 182 − 2(11.6)(18) cos 50
BC = 13.8
A1
N2
[3 marks]
Examiners report
The vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used right
triangle trigonometry. Errors included working in radian mode, assuming that the angle at C was 90∘ , and incorrectly applying the
order of operations when evaluating the cosine rule.
The following diagram shows a circle with centre O and radius r cm.
^ B = 1.4 radians .
Points A and B are on the circumference of the circle and AO
^ O = π radians .
The point C is on [OA] such that BC
2
11. The area of the shaded region is 25 cm2 . Find the value of r .
[7 marks]
Markscheme
correct value for BC
−−−−−−−−−−−−
eg BC = r sin 1.4 , √r2 − (r cos 1.4)2
(A1)
area of ΔOBC = 12 r sin 1.4 × r cos 1.4 (= 12 r2 sin 1.4 × cos 1.4)
area of sector OAB = 12 r2 × 1.4
A1
attempt to subtract in any order
(M1)
A1
eg sector – triangle, 12 r2 sin 1.4 × cos 1.4 − 0.7r2
correct equation
A1
eg 0.7r2 − 12 r sin 1.4 × r cos 1.4 = 25
attempt to solve their equation
r = 6.37
A1
(M1)
25
0.616…
N4
[7 marks]
Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.
Examiners report
As to be expected, candidates found this problem challenging. Those who used a systematic approach in part (b) were more
successful than those whose work was scattered about the page. While a pleasing number of candidates successfully found the area of
sector AOB, far fewer were able to find the area of triangle BOC. Candidates who took an analytic approach to solving the resulting
equation were generally less successful than those who used their GDC. Candidates who converted the angle to degrees generally
were not very successful.
^ R = 45∘ .
^ Q = 70∘ and PQ
The following diagram shows ΔPQR , where RQ = 9 cm, PR
^Q .
12a. Find RP
[1 mark]
Markscheme
^ Q =65∘
RP
A1
N1
[1 mark]
Examiners report
This question was attempted in a satisfactory manner.
12b. Find PR .
[3 marks]
Markscheme
evidence of choosing sine rule
correct substitution
e.g.
PR
sin 45∘
=
(M1)
A1
9
sin 65∘
7.021854078
PR = 7.02
A1
N2
[3 marks]
Examiners report
The sine rule was applied satisfactory in part (b) but some obtained an incorrect answer due to having their calculators in radian
mode. Some incorrect substitutions were seen, either by choosing an incorrect side or substituting 70 instead of sin 70∘ . Approaches
using a combination of the cosine rule and/or right-angled triangle trigonometry were seen.
12c. Find the area of ΔPQR .
[2 marks]
Markscheme
correct substitution
(A1)
1
2
e.g. area = × 9 × 7.02 … × sin 70∘
29.69273008
area = 29.7
A1
N2
[2 marks]
Examiners report
Approaches using a combination of the cosine rule and/or right-angled triangle trigonometry were seen, especially in part (c) to
calculate the area of the triangle.
A few candidates set about finding the height, then used the formula for the area of a right-angled triangle.
Consider the following circle with centre O and radius r .
The points P, R and Q are on the circumference, PÔQ = 2θ , for 0 < θ < π2 .
13a. Use the cosine rule to show that PQ = 2r sin θ .
[4 marks]
Markscheme
A1
correct substitution into cosine rule
2
e.g. PQ =
r2
+ r2
− 2(r)(r) cos(2θ) , PQ2 = 2r2 − 2r2 (cos(2θ))
substituting 1 − 2sin2 θ for cos 2θ (seen anywhere)
A1
e.g. PQ2 = 2r2 − 2r2 (1 − 2sin2 θ)
(A1)
2
e.g. PQ = 2r2 − 2r2 + 4r2 sin2 θ
recognizing 2r2 − 2r2 = 0 (including crossing out) (seen anywhere)
−−−−−−
e.g. PQ2 = 4r2 sin2 θ , PQ = √4r2 sin2 θ
PQ = 2rsinθ
AG
N0
[4 marks]
Examiners report
This exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).
Part (a) was generally attempted using the cosine rule, but many failed to substitute correctly into the right hand side or skipped
important steps. A high percentage could not arrive at the given expression due to a lack of knowledge of trigonometric identities or
making algebraic errors, and tried to force their way to the given answer.
The most common errors included taking the square root too soon, and sign errors when distributing the negative after substituting
cos 2θ by 1 − 2sin2 θ .
13b. Let l be the length of the arc PRQ .
[5 marks]
Given that 1.3PQ − l = 0 , find the value of θ .
Markscheme
PRQ = r × 2θ (seen anywhere)
correct set up
(A1)
A1
e.g. 1.3 × 2r sin θ − r × (2θ) = 0
attempt to eliminate r
(M1)
correct equation in terms of the one variable θ
(A1)
e.g. 1.3 × 2 sin θ − 2θ = 0
1.221496215
θ = 1.22 (accept 70.0∘ (69.9))
A1
N3
[5 marks]
Examiners report
This exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).
In part (b), most candidates understood what was required but could not find the correct length of the arc PRQ mainly due to
substituting the angle by θ instead of 2θ .
13c. Consider the function f(θ) = 2.6 sin θ − 2θ , for 0 < θ <
(i)
Sketch the graph of f .
(ii) Write down the root of f(θ) = 0 .
π
2
.
[4 marks]
Markscheme
(i)
A1A1A1
N3
Note: Award A1 for approximately correct shape, A1 for x-intercept in approximately correct position, A1 for domain. Do not
penalise if sketch starts at origin.
(ii) 1.221496215
θ = 1.22
A1
N1
[4 marks]
Examiners report
Regarding part (c), many valid approaches were seen for the graph of f, making a good use of their GDC. A common error was
finding a second or third solution outside the domain. A considerable amount of sketches were missing a scale.
There were candidates who achieved the correct equation but failed to realize they could use their GDC to solve it.
13d. Use the graph of f to find the values of θ for which l < 1.3PQ .
[3 marks]
Markscheme
evidence of appropriate approach (may be seen earlier)
M2
e.g. 2θ < 2.6 sin θ , 0 < f(θ) , showing positive part of sketch
0 < θ < 1.221496215
0 < θ = 1.22 (accept θ < 1.22 )
A1
N1
[3 marks]
Examiners report
Part (d) was attempted by very few, and of those who achieved the correct answer not many were able to show the method they used.
The following diagram shows a triangle ABC.
BC = 6 , C ÂB = 0.7 radians , AB = 4p , AC = 5p , where p > 0 .
14a. (i)
Show that p2 (41 − 40 cos 0.7) = 36 .
[4 marks]
(ii) Find p .
Markscheme
(i) evidence of valid approach
(M1)
e.g. choosing cosine rule
correct substitution
(A1)
2
e.g. 6 = (5p) + (4p)2 − 2 × (4p) × (5p) cos 0.7
2
simplification
e.g. 36 =
25p2
A1
+ 16p2 − 40p2 cos 0.7
p2 (41 − 40 cos 0.7) = 36
AG
N0
(ii) 1.85995 …
p = 1.86
A1
N1
Note: Award A0 for p = ±1.86 , i.e. not rejecting the negative value.
[4 marks]
Examiners report
There were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rule
but sloppy notation often led to candidates not being able to show the desired result.
Aˆ B
Consider the circle with centre B that passes through the point C. The circle cuts the line CA at D, and AD̂B is obtuse. Part of the circle is
shown in the following diagram.
14b. Write down the length of BD.
[1 mark]
Markscheme
BD = 6
A1
N1
[1 mark]
Examiners report
There were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rule
but sloppy notation often led to candidates not being able to show the desired result.
[4 marks]
Markscheme
evidence of valid approach
(M1)
e.g. choosing sine rule
correct substitution
e.g.
4p
=
A1
sin 0.7
6
(A1)
π − 0.9253166 … = 2.216275 …
A1
N3
[4 marks]
Examiners report
In part (c), candidates again correctly identified an appropriate method but failed to recognize that their result of 0.925 was acute and
not obtuse as required.
14d. (i)
Show that C B̂D = 1.29 radians, correct to 2 decimal places.
(ii) Hence, find the area of the shaded region.
[6 marks]
Markscheme
(i) evidence of valid approach
(M1)
e.g. recognize isosceles triangle, base angles equal
π − 2(0.9253 …)
C B̂D = 1.29
A1
AG
N0
(A1)
(ii) area of sector BCD
e.g. 0.5 × (1.29) × (6)2
(A1)
area of triangle BCD
e.g. 0.5 × (6)2 sin 1.29
M1
evidence of subtraction
5.92496 …
5.937459 …
area = 5.94
A1
N3
[6 marks]
Examiners report
In (d) (i), many attempted to use the sine rule under the incorrect assumption that DC was equal to 5p, rather than rely on some basic
isosceles triangle geometry. Consequently, the result of 1.29 for C B̂D was not easy to show. There was a great deal of success with
(d) (ii) with candidates using appropriate techniques to find the area of the shaded region although some stopped after finding the area
of the sector.
Let sin θ =
2
√13
, where
<θ<π.
π
2
15a. Find cos θ .
[3 marks]
Markscheme
METHOD 1
evidence of choosing sin2 θ + cos2 θ = 1
(A1)
correct working
e.g. cos2 θ =
cos θ = −
9
13
3
√13
, cos θ = ±
A1
3
√13
(M1)
−−
, cos θ = √ 139
N2
Note: If no working shown, award N1 for
3
√13
.
METHOD 2
approach involving Pythagoras’ theorem
(M1)
e.g. 22 + x2 = 13 ,
finding third side equals 3
cos θ =
− 3
√13
A1
(A1)
N2
Note: If no working shown, award N1 for
[3 marks]
3
√13
.
Examiners report
While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in
the second quadrant will have a negative value.
15b. Find tan 2θ .
[5 marks]
Markscheme
correct substitution into sin 2θ (seen anywhere)
e.g. 2 (
2
)
√13
(−
3
)
√13
correct substitution into cos 2θ (seen anywhere)
e.g. (−
3
)
√13
2
−(
2
)
√13
2
, 2(−
valid attempt to find tan 2θ
e.g.
2(
(−
2
√ 13
3
√ 13
)(−
3
√ 13
2
) −(
2
)
2
√ 13
)
(2)(2)(−3)
13
9
−4
13
13
,
−
tan 2θ = − 125
,
2(− 2 )
3
1−(− 2 )
3
)
√13
2
(A1)
− 1 , 1 − 2(
2
)
√13
2
(M1)
2
3
A1
correct working
e.g.
(A1)
12
( √ 13) 2
18
−1
13
A1
,
− 12
13
5
13
N4
Note: If students find answers for cos θ which are not in the range [−1, 1], award full FT in (b) for correct FT working shown.
[5 marks]
Examiners report
In part (b), many candidates incorrectly tried to calculate tan 2θ as 2 × tan θ , rather than using the double-angle identities.
f(x) = a sin(b(x − c)) + d
The following diagram shows the graph of f(x) = a sin(b(x − c)) + d , for 2 ≤ x ≤ 10 .
There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .
16a. Use the graph to write down the value of
(i)
[3 marks]
a;
(ii) c ;
(iii) d .
Markscheme
(i) a = 8
A1
N1
(ii) c = 2
A1
N1
(iii) d = 4
A1
N1
[3 marks]
Examiners report
Part (a) of this question proved challenging for most candidates.
16b. Show that b =
π
4
.
[2 marks]
Markscheme
METHOD 1
recognizing that period = 8
A1
correct working
e.g. 8 =
2π
b
b = π4
AG
,b=
(A1)
2π
8
N0
METHOD 2
M1
attempt to substitute
e.g. 12 = 8 sin(b(4 − 2)) + 4
correct working
A1
e.g. sin 2b = 1
b = π4
AG
N0
[2 marks]
Examiners report
Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that
this period could be found using the given coordinates of the maximum and minimum points.
16c. Find f ′ (x) .
[3 marks]
Markscheme
evidence of attempt to differentiate or choosing chain rule
e.g.
cos π4
(x − 2) ,
π
4
(M1)
×8
f ′ (x) = 2π cos ( π4 (x − 2)) (accept 2π cos π4 (x − 2) )
A2
N3
[3 marks]
Examiners report
In part (c), not many candidates found the correct derivative using the chain rule.
16d. At a point R, the gradient is −2π . Find the x-coordinate of R.
[6 marks]
Markscheme
recognizing that gradient is f ′ (x)
(M1)
′
e.g. f (x) = m
A1
correct equation
e.g. −2π = 2π cos ( π4 (x − 2)) , −1 = cos ( π4 (x − 2))
(A1)
correct working
e.g.
cos−1 (−1)
= (x − 2)
π
4
using cos−1 (−1) = π (seen anywhere)
(A1)
e.g. π = (x − 2)
π
4
simplifying
(A1)
e.g. 4 = (x − 2)
x=6
A1
N4
[6 marks]
Examiners report
For part (d), a good number of candidates correctly set their expression equal to −2π , but errors in their previous values kept most
from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.
Consider the following circle with centre O and radius 6.8 cm.
The length of the arc PQR is 8.5 cm.
17a. Find the value of θ .
[2 marks]
Markscheme
correct substitution
(A1)
e.g. 8.5 = θ(6.8) , θ = 8.5
6.8
θ = 1.25 (accept 71.6∘ )
A1
N2
[2 marks]
Examiners report
Part (a) was almost universally done correctly.
17b. Find the area of the shaded region.
[4 marks]
Markscheme
METHOD 1
correct substitution into area formula (seen anywhere)
(A1)
2
e.g. A = π(6.8) , 145.267 …
correct substitution into area formula (seen anywhere)
e.g. A =
1
(1.25)(6.82 )
2
e.g. π(6.8) −
, 28.9
M1
valid approach
2
(A1)
1
(1.25)(6.82 )
2
A = 116 (cm2 )
A1
; 145.267 … − 28.9 ; πr2 − 12 r2 sin θ
N2
METHOD 2
attempt to find reflex angle
(M1)
e.g. 2π − θ , 360 − 1.25
correct reflex angle
(A1)
AÔB = 2π − 1.25 (= 5.03318 …)
correct substitution into area formula
e.g. A =
A1
1
(5.03318 …)(6.82 )
2
A = 116 (cm2 )
A1
N2
[4 marks]
Examiners report
Many also had little trouble in part (b), with most subtracting from the circle's area, and a minority using the reflex angle. A few
candidates worked in degrees, although some of these did so incorrectly by using the radian area formula. Some candidates only
found the area of the unshaded sector.
Consider the triangle ABC, where AB =10 , BC = 7 and C ÂB = 30∘ .
18a. Find the two possible values of AĈ B .
[4 marks]
Markscheme
Note: accept answers given in degrees, and minutes.
evidence of choosing sine rule
e.g.
sin A
a
=
sin B
b
correct substitution
e.g.
sin θ
10
(M1)
∘
A1
= sin730 , sin θ =
5
7
AĈ B = 45.6∘ , AĈ B = 134∘
A1A1
N1N1
Note: If candidates only find the acute angle in part (a), award no marks for (b).
[4 marks]
Examiners report
Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a
lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle
generally had no difficulty with part (b).
18b. Hence, find AB̂C , given that it is acute.
[2 marks]
Markscheme
attempt to substitute their larger value into angle sum of triangle
∘
(M1)
∘
e.g. 180 − (134.415…∘ + 30 )
AB̂C = 15.6∘
A1
N2
[2 marks]
Examiners report
Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a
lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle
generally had no difficulty with part (b).
19. Solve the equation 2 cos x = sin 2x , for 0 ≤ x ≤ 3π .
[7 marks]
Markscheme
METHOD 1
using double-angle identity (seen anywhere)
A1
e.g. sin 2x = 2 sin x cos x , 2 cos x = 2 sin x cos x
evidence of valid attempt to solve equation
(M1)
e.g. 0 = 2 sin x cos x − 2 cos x , 2 cos x(1 − sin x) = 0
cos x = 0 , sin x = 1
x=
π
2
,x=
3π
2
,x=
A1A1
5π
2
A1A1A1
N4
METHOD 2
A1A1M1A1
Notes: Award A1 for sketch of sin 2x , A1 for a sketch of 2 cos x , M1 for at least one intersection point seen, and A1 for 3
approximately correct intersection points. Accept sketches drawn outside [0,3π] , even those with more than 3 intersections.
x = π2 , x =
3π
2
,x=
5π
2
A1A1A1
N4
[7 marks]
Examiners report
By far the most common error was to “cancel” the cos x and find only two of the three solutions. It was disappointing how few
candidates solved this by setting factors equal to zero. Some candidates wrote all three answers from sin x = 1 , which only earned
two of the three final marks. On a brighter note, many candidates found the 5π
, which showed an appreciation for the period of the
2
function as well as the domain restriction. A handful of candidates cleverly sketched both graphs and used the intersections to find the
three solutions.
The following diagram shows triangle ABC .
AB = 7 cm, BC = 9 cm and AB̂C = 120∘ .
20a. Find AC .
[3 marks]
Markscheme
evidence of choosing cosine rule
e.g.
a2
(M1)
2
+ b − 2ab cos C
A1
correct substitution
e.g. 72 + 92 − 2(7)(9) cos 120∘
−−−
AC = 13.9 (= √193) A1 N2
[3 marks]
Examiners report
The majority of candidates were successful with this question. Most correctly used the cosine rule in part (a) and the sine rule in part
(b). Some candidates did not check that their GDC was set in degree mode while others treated the triangle as if it were right angled.
A large number of candidates were penalized for not leaving their answers exactly or to three significant figures.
20b. Find BÂC .
[3 marks]
sin Â
BC
=
sin B̂
AC
sin Â
9
=
sin 120
13.9
 = 34.1∘
2
2
cos  = AB +AC −BC
2(AB)(AC)
2
2
2
cos  = 7 + 13.9 − 9
2(7)(13.9)
 = 34.1∘
2
Examiners report
The majority of candidates were successful with this question. Most correctly used the cosine rule in part (a) and the sine rule in part
(b). Some candidates did not check that their GDC was set in degree mode while others treated the triangle as if it were right angled.
A large number of candidates were penalized for not leaving their answers exactly or to three significant figures.
The following diagram shows a waterwheel with a bucket. The wheel rotates at a constant rate in an anticlockwise (counter-clockwise)
direction.
The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above the water level. After t seconds, the height of the bucket
above the water level is given by h = a sin bt + 2 .
21a. Show that a = 4 .
[2 marks]
Markscheme
METHOD 1
evidence of recognizing the amplitude is the radius
(M1)
e.g. amplitude is half the diameter
a=
8
2
a=4
A1
AG
N0
METHOD 2
evidence of recognizing the maximum height
(M1)
e.g. h = 6 , a sin bt + 2 = 6
correct reasoning
e.g. asin bt = 4 and sin bt has amplitude of 1
a=4
AG
A1
N0
[2 marks]
Examiners report
Parts (a) and (b) were generally well done.
21b. The wheel turns at a rate of one rotation every 30 seconds.
Show that b =
π
15
.
[2 marks]
Markscheme
METHOD 1
(A1)
period = 30
b=
2π
30
A1
b=
π
15
AG
N0
METHOD 2
correct equation
(A1)
e.g. 2 = 4 sin 30b + 2 , sin 30b = 0
30b = 2π
b=
π
15
A1
AG
N0
[2 marks]
Examiners report
Parts (a) and (b) were generally well done, however there were several instances of candidates working backwards from the given
21c. In the first rotation, there are two values of t when the bucket is descending at a rate of 0.5 ms−1 .
[6 marks]
Find these values of t .
Markscheme
recognizing h′ (t) = −0.5 (seen anywhere)
attempting to solve
R1
(M1)
′
e.g. sketch of h , finding h′
correct work involving h′
A2
e.g. sketch of h′ showing intersection, −0.5 =
t = 10.6 , t = 19.4
A1A1
4π
π
cos ( 15
t)
15
N3
[6 marks]
Examiners report
Parts (c) and (d) proved to be quite challenging for a large proportion of candidates. Many did not attempt these parts. The most
common error was a misinterpretation of the word "descending" where numerous candidates took h′ (t) to be 0.5 instead of −0.5 but
incorrect derivatives for h were also widespread. The process required to solve for t from the equation −0.5 =
4π
π
cos ( 15
t)
15
overwhelmed those who attempted algebraic methods. Few could obtain both correct solutions, more had one correct while others
included unreasonable values including t < 0 .
21d. In the first rotation, there are two values of t when the bucket is descending at a rate of 0.5 ms−1 .
Determine whether the bucket is underwater at the second value of t .
[4 marks]
Markscheme
METHOD 1
valid reasoning for their conclusion (seen anywhere)
R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of substituting into h
e.g. h(19.4) , 4 sin
correct calculation
19.4π
15
(M1)
+2
A1
e.g. h(19.4) = −1.19
correct statement
A1
N0
e.g. the bucket is underwater, yes
METHOD 2
valid reasoning for their conclusion (seen anywhere)
R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of valid approach
(M1)
e.g. solving h(t) = 0 , graph showing region below x-axis
A1
correct roots
e.g. 17.5, 27.5
correct statement
A1
N0
e.g. the bucket is underwater, yes
[4 marks]
Examiners report
In part (d), not many understood that the condition for underwater was h(t) < 0 and had trouble interpreting the meaning of "second
value". Many candidates, however, did recover to gain some marks in follow through.
22. Let h(x) =
6x
cos x
. Find h′ (0) .
[6 marks]
Markscheme
METHOD 1 (quotient)
derivative of numerator is 6
(A1)
derivative of denominator is − sin x
attempt to substitute into quotient rule
(cos x)(6)−(6x)(− sin x)
(cos x)2
substituting x = 0
e.g.
(M1)
A1
correct substitution
e.g.
(A1)
(A1)
(cos 0)(6)−(6×0)(− sin 0)
h′ (0) = 6
(cos 0)2
A1
N2
METHOD 2 (product)
h(x) = 6x × (cos x)−1
(A1)
derivative of 6x is 6
derivative of (cos x)−1 is (−(cos x)−2 (− sin x))
attempt to substitute into product rule
(A1)
(M1)
A1
correct substitution
−2
e.g. (6x)(−(cos x) (− sin x)) + (6)(cos x)−1
substituting x = 0
(A1)
e.g. (6 × 0)(−(cos 0)−2 (− sin 0)) + (6)(cos 0)−1
h′ (0) = 6
A1
N2
[6 marks]
Examiners report
The majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question.
However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakes
in simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A few
candidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotient
rule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.
The following diagram represents a large Ferris wheel, with a diameter of 100 metres.
Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The wheel rotates at a constant rate, in an
anticlockwise (counter-clockwise) direction. One revolution takes 20 minutes.
23a. Write down the height of P above ground level after
(i)
[2 marks]
10 minutes;
(ii) 15 minutes.
Markscheme
(i) 100 (metres)
A1
N1
(ii) 50 (metres)
A1
N1
[2 marks]
Examiners report
Nearly all candidates answered part (a) correctly, finding the height of the wheel at 12 and
3
4
of a revolution.
Let h(t) metres be the height of P above ground level after t minutes. Some values of h(t) are given in the table below.
23b. (i)
Show that h(8) = 90.5.
(ii) Find h(21) .
[4 marks]
Markscheme
(i) identifying symmetry with h(2) = 9.5
subtraction
(M1)
A1
e.g. 100 − h(2) , 100 − 9.5
h(8) = 90.5
AG
N0
(ii) recognizing period
(M1)
e.g. h(21) = h(1)
h(21) = 2.4
A1
N2
[4 marks]
Examiners report
While many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function for
height, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table.
23c. Sketch the graph of h , for 0 ≤ t ≤ 40 .
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for end points (0, 0) and (40, 0) , A1 for range 0 ≤ h ≤ 100 , A1 for approximately correct sinusoidal shape, with
two cycles.
[3 marks]
Examiners report
In part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles.
However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs were
constructed of line segments, rather than a curve.
23d. Given that h can be expressed in the form h(t) = a cos bt + c , find a , b and c .
[5 marks]
Markscheme
evidence of a quotient involving 20, 2π or 360∘ to find b
e.g.
2π
b
b=
2π
20
= 20 , b =
(=
π
)
10
(M1)
360
20
(accept b = 18 if working in degrees)
a = −50 , c = 50
A2A1
A1
N2
N3
[5 marks]
Examiners report
For part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accurate
sketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that the
position on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidates
recognized that the value of a would be negative. Candidates should be aware that while working in degrees may be acceptable, the
expectation is that radians will be used in these types of questions.
The diagram shows two concentric circles with centre O.
The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm.
Points A, B and C are on the circumference of the larger circle such that AÔB is
24a. Find the length of the arc ACB .
π
3
[2 marks]
Markscheme
correct substitution in l = rθ
1
6
(A1)
e.g. 10 × , × 2π × 10
π
3
arc length =
20π
6
(=
10π
)
3
A1
N2
[2 marks]
Examiners report
This question was very well done by the majority of candidates. Some candidates correctly substituted the values into the formulas,
but failed to do the calculations and write their answers in finished form.
24b. Find the area of the shaded region.
[4 marks]
Markscheme
area of large sector = 12 × 102 × π3 (= 100π
)
6
area of small sector = 12 × 82 × π3 (=
64π
)
6
evidence of valid approach (seen anywhere)
(A1)
(A1)
M1
e.g. subtracting areas of two sectors, 12 × π3 (102 − 82 )
36π
6
, etc.)
A1
N3
[4 marks]
Examiners report
Nearly all used the correct method of subtracting the sector areas in part (b), though multiplying with fractions proved challenging for
some candidates.
25a. Show that 4 − cos 2θ + 5 sin θ = 2sin 2 θ + 5 sin θ + 3 .
[2 marks]
Markscheme
attempt to substitute 1 − 2sin2 θ for cos 2θ
(M1)
A1
correct substitution
e.g. 4 − (1 − 2sin2 θ) + 5 sin θ
4 − cos 2θ + 5 sin θ = 2sin2 θ + 5 sin θ + 3
AG
N0
[2 marks]
Examiners report
In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates
who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that"
question, working backward from the given answer is not a valid method.
25b. Hence, solve the equation 4 − cos 2θ + 5 sin θ = 0 for 0 ≤ θ ≤ 2π .
[5 marks]
Markscheme
evidence of appropriate approach to solve
(M1)
correct working
A1
e.g. (2 sin θ + 3)(sin θ + 1) , (2x + 3)(x + 1) = 0 , sin x =
−5± √1
4
correct solution sin θ = −1 (do not penalise for including sin θ = − 32
θ=
3π
2
A2
(A1)
N3
[5 marks]
Examiners report
In part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring the
quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution
to the equation sin θ = −1 , and there were a few who did not realize that the equation sin θ = − 32 has no solution.
The straight line with equation y = 34 x makes an acute angle θ with the x-axis.
26a. Write down the value of tan θ .
[1 mark]
Markscheme
tan θ = 34 (do not accept 34 x )
A1
N1
[1 mark]
Examiners report
Many candidates drew a diagram to correctly find tan θ , although few recognized that a line through the origin can be expressed as
y = x tan θ , with gradient tan θ , which is explicit in the syllabus.
26b. Find the value of
(i)
[6 marks]
sin 2θ ;
(ii) cos 2θ .
Markscheme
(i) sin θ = 35 , cos θ =
4
5
correct substitution
A1
(A1)(A1)
e.g. sin 2θ = 2 ( 35 ) ( 45 )
sin 2θ =
24
25
A1
N3
A1
(ii) correct substitution
2
2
e.g. cos 2θ = 1 − 2( 35 ) , ( 45 ) − ( 35 )
cos 2θ =
7
25
A1
2
N1
[6 marks]
Examiners report
A surprising number were unable to find the ratios for sin θ and cos θ from tan θ . It was not uncommon for candidates to use
unreasonable values, such as sin θ = 3 and cos θ = 4 , or to write nonsense such as 2 sin 35 cos 45 .
Let f(x) =
cos x
sin x
, for sin x ≠ 0 .
27a. Use the quotient rule to show that f ′ (x) =
−1
sin2x
.
[5 marks]
Markscheme
d
dx
sin x = cos x ,
d
dx
cos x = − sin x (seen anywhere)
M1
evidence of using the quotient rule
A1
correct substitution
e.g.
sin x(− sin x)−cos x(cos x)
sin2x
f ′ (x) =
−(sin2x+ cos 2x)
f ′ (x) =
−1
sin2x
(A1)(A1)
,
− sin2x− cos 2x
sin2x
A1
sin2x
AG
N0
[5 marks]
Examiners report
Many candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identity
achieves the numerator of the answer given. Whether changing to −(sin x)−2 , or applying the quotient rule a second time, most
candidates neglected the chain rule in finding the second derivative.
27b. Find f ′′ (x) .
[3 marks]
Markscheme
METHOD 1
(M1)
appropriate approach
−2
′
e.g. f (x) = −(sin x)
f ′′ (x) = 2(sin−3 x)(cos x) (= 2 cos3 x )
A1A1
sin x
N3
Note: Award A1 for 2sin−3 x , A1 for cos x .
METHOD 2
derivative of sin2 x = 2 sin x cos x (seen anywhere)
(M1)
evidence of choosing quotient rule
e.g. u = −1 , v = sin2 x , f ′′ =
f ′′ (x) =
2 sin x cos x
2
(sin2x)
(= 2 cos3 x )
sin x
A1
2
sin x×0−(−1)2 sin x cos x
2
(sin2x)
A1
N3
[3 marks]
Examiners report
Whether changing to −(sin x)−2 , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the
second derivative.
( )
( )
In the following table, f ′ ( π2 ) = p and f ′′ ( π2 ) = q . The table also gives approximate values of f ′ (x) and f ′′ (x) near x = π2 .
27c. Find the value of p and of q.
[3 marks]
Markscheme
evidence of substituting
e.g.
−1
sin2 π
,
2
π
2
M1
2 cos π
2
sin3 π
2
p = −1 , q = 0
A1A1
N1N1
[3 marks]
Examiners report
Those who knew the trigonometric ratios at π2 typically found the values of p and of q, sometimes in follow-through from an incorrect
f ′′ (x) .
27d. Use information from the table to explain why there is a point of inflexion on the graph of f where x =
π
2
.
[2 marks]
Markscheme
second derivative is zero, second derivative changes sign
R1R1
N2
[2 marks]
Examiners report
Few candidates gave two reasons from the table that supported the existence of a point of inflexion. Most stated that the second
derivative is zero and neglected to consider the sign change to the left and right of q. Some discussed a change of concavity, but
without supporting this statement by referencing the change of sign in f ′′ (x) , so no marks were earned.
Let f(x) = cos 2x and g(x) = 2x2 − 1 .
28a. Find f ( π ) .
2
[2 marks]
Markscheme
f ( π2 ) = cos π
= −1
A1
[2 marks]
(A1)
N2
Examiners report
In part (a), a number of candidates were not able to evaluate cos π , either leaving it or evaluating it incorrectly.
28b. Find (g ∘ f) ( π ) .
2
[2 marks]
Markscheme
(g ∘ f) ( π2 ) = g(−1) (= 2(−1)2 − 1) (A1)
=1
A1
N2
[2 marks]
Examiners report
Almost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks for
incorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to come
up with correct answers; while this is a valid method, it required unnecessary additional work.
28c. Given that (g ∘ f)(x) can be written as cos(kx) , find the value of k, k ∈ Z .
[3 marks]
Markscheme
(g ∘ f)(x) = 2(cos(2x))2 − 1 (= 2cos2 (2x) − 1)
evidence of 2cos2 θ − 1 = cos 2θ (seen anywhere)
A1
(M1)
(g ∘ f)(x) = cos 4x
k=4
A1
N2
[3 marks]
Examiners report
Candidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or used
it to write the expression in terms of cos x and went no further. There were a number of cases in which the candidates "accidentally"
came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidates
recognized the correct method of solution.
f(x) = 6 + 6 sin x
Let f(x) = 6 + 6 sin x . Part of the graph of f is shown below.
The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.
29a. Solve for 0 ≤ x < 2π
[5 marks]
6 + 6 sin x = 6 ;
(i)
(ii) 6 + 6 sin x = 0 .
Markscheme
(i) sin x = 0
A1
x=0,x=π
A1A1
(ii) sin x = −1
A1
x=
3π
2
A1
N2
N1
[5 marks]
Examiners report
Many candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not show
sufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidates
worked in degrees; the majority worked in radians.
29b. Write down the exact value of the x-intercept of f , for 0 ≤ x < 2π .
[1 mark]
Markscheme
3π
2
A1
N1
[1 mark]
Examiners report
While some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), most
used their working from part (a)(ii) sometimes with follow-through on an incorrect answer.
29c. The area of the shaded region is k . Find the value of k , giving your answer in terms of π .
[6 marks]
Markscheme
evidence of using anti-differentiation
e.g. ∫0
3π
2
(M1)
(6 + 6 sin x)dx
correct integral 6x − 6 cos x (seen anywhere)
correct substitution
A1A1
(A1)
e.g. 6 ( 3π
) − 6 cos ( 3π
) − (−6 cos 0) , 9π − 0 + 6
2
2
k = 9π + 6
A1A1
N3
[6 marks]
Examiners report
Most candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end.
Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; without
this working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with the
common trigonometric values.
29d. Let g(x) = 6 + 6 sin (x − π ) . The graph of f is transformed to the graph of g.
2
[2 marks]
Give a full geometric description of this transformation.
Markscheme
translation of ( 2 )
0
π
A1A1
N2
[2 marks]
Examiners report
While there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d).
This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by π2 .
29e. Let g(x) = 6 + 6 sin (x − π ) . The graph of f is transformed to the graph of g.
2
p+ 3π
Given that ∫p
2
[3 marks]
g(x)dx = k and 0 ≤ p < 2π , write down the two values of p.
Markscheme
recognizing that the area under g is the same as the shaded region in f
p=
π
2
,p=0
A1A1
(M1)
N3
[3 marks]
Examiners report
Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority
tried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in the
wording of the question did not appear to have been noticed by candidates in this part either.
The following diagram shows the triangle ABC.
The angle at C is obtuse, AC = 5 cm, BC = 13.6 cm and the area is 20 cm2 .
30a. Find AĈ B .
[4 marks]
Markscheme
correct substitution into the formula for the area of a triangle
A1
e.g. 12 × 5 × 13.6 × sin C = 20 , 12 × 5 × h = 20
attempt to solve
(M1)
8
e.g. sin C = 0.5882 … , sin C = 13.6
Ĉ = 36.031 …∘ (0.6288 … radians)
AĈ B = 144∘ (2.51 radians)
A1
(A1)
N3
[4 marks]
Examiners report
Part (a) was well done with the majority of candidates obtaining the acute angle. Unfortunately, the question asked for the obtuse
angle which was clearly stated and shown in the diagram. No matter which angle was used, most candidates were able to obtain full
marks in part (b) with a simple application of the cosine rule.
30b. Find AB.
[3 marks]
Markscheme
evidence of choosing the cosine rule
correct substitution
2
(M1)
A1
e.g. (AB) = 5 + 13.62 − 2(5)(13.6) cos 143.968 …
AB = 17.9
2
A1
N2
[3 marks]
Examiners report
Part (a) was well done with the majority of candidates obtaining the acute angle. Unfortunately, the question asked for the obtuse
angle which was clearly stated and shown in the diagram. No matter which angle was used, most candidates were able to obtain full
marks in part (b) with a simple application of the cosine rule.
The following diagram represents a large Ferris wheel at an amusement park.
The points P, Q and R represent different positions of a seat on the wheel.
The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes.
A seat starts at the lowest point P, when its height is one metre above the ground.
31a. Find the height of a seat above the ground after 15 minutes.
[2 marks]
Markscheme
valid approach
(M1)
e.g. 15 mins is half way, top of the wheel, d + 1
height = 101 (metres)
A1
N2
[2 marks]
Examiners report
Part (a) was well done with most candidates obtaining the correct answer.
31b. After six minutes, the seat is at point Q. Find its height above the ground at Q.
Markscheme
evidence of identifying rotation angle after 6 minutes
e.g.
2π
5
,
1
5
of a rotation, 72∘
(M1)
evidence of appropriate approach
e.g. drawing a right triangle and using cosine ratio
correct working (seen anywhere)
e.g. cos 2π
=
5
x
50
A1
, 15.4(508 …)
evidence of appropriate method
M1
e.g. height = radius + 1 − 15.45 …
height = 35.5 (metres) (accept 35.6)
[5 marks]
A1
N2
A1
[5 marks]
Examiners report
Part (b) however was problematic with most errors resulting from incorrect, missing or poorly drawn diagrams. Many did not
recognize this as a triangle trigonometric problem while others used the law of cosines to find the chord length rather than the vertical
height, but this was only valid if they then used this to complete the problem. Many candidates misinterpreted the question as one that
was testing arc length and area of a sector and made little to no progress in part (b).
Still, others recognized that 6 minutes represented 15 of a rotation, but the majority then thought the height after 6 minutes should be
of the maximum height, treating the situation as linear. There were even a few candidates who used information given later in the
question to answer part (b). Full marks are not usually awarded for this approach.
31c. The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t − c)) + 51.
1
5
[6 marks]
Find the value of b and of c .
Markscheme
METHOD 1
evidence of substituting into b =
2π
period
(M1)
correct substitution
e.g. period = 30 minutes, b =
π
b = 0.209 ( 15
)
A1
substituting into h(t)
2π
30
A1
N2
(M1)
e.g. h(0) = 1 , h(15) = 101
correct substitution
A1
π
1 = 50 sin (− 15
c) + 51
c = 7.5 A1
N2
METHOD 2
evidence of setting up a system of equations
(M1)
two correct equations
e.g. 1 = 50 sin b(0 − c) + 51 , 101 = 50 sin b(15 − c) + 51
attempt to solve simultaneously
A1A1
(M1)
e.g. evidence of combining two equations
π
b = 0.209 ( 15
) , c = 7.5
A1A1
N2N2
[6 marks]
Examiners report
Part (c) was not well done. It was expected that candidates simply use the formula
2π
period
to find the value of b and then substitute back
into the equation to find the value of c. However, candidates often preferred to set up a pair of equations and attempt to solve them
analytically, some successful, some not. No attempts were made to solve this system on the GDC indicating that candidates do not get
exposed to many “systems” that are not linear. Confusing radians and degrees here did nothing to improve the lack of success.
31d. The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t − c)) + 51.
Hence find the value of t the first time the seat is 96 m above the ground.
[3 marks]
Markscheme
evidence of solving h(t) = 96
(M1)
e.g. equation, graph
t = 12.8 (minutes)
A2
N3
[3 marks]
Examiners report
In part (d), candidates were clear on what was required and set their equation equal to 96. Yet again however, solving this equation
graphically using a GDC proved too daunting a task for most.
There is a vertical tower TA of height 36 m at the base A of a hill. A straight path goes up the hill from A to a point U. This information is
represented by the following diagram.
The path makes a 4∘ angle with the horizontal.
The point U on the path is 25 m away from the base of the tower.
The top of the tower is fixed to U by a wire of length x m.
32a. Complete the diagram, showing clearly all the information above.
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for labelling 4∘ with horizontal, A1 for labelling [AU] 25 metres, A1 for drawing [TU].
[3 marks]
Examiners report
This question was attempted in a satisfactory manner. Even the weakest candidates earned some marks here, showing some clear
working. In part (a) the diagram was completed fairly well, with some candidates incorrectly labelling the angle with the vertical as
4∘ . The cosine rule was applied satisfactory in part (b), although some candidates incorrectly used their calculators in radian mode.
Approaches using a combination of the sine rule and/or right-angled triangle trigonometry were seen, especially when candidates
incorrectly labelled the 25 m path as being the distance from the horizontal to U.
32b. Find x .
[4 marks]
Markscheme
T ÂU = 86∘
(A1)
evidence of choosing cosine rule
correct substitution
e.g.
x2
2
A1
N3
A1
= 25 + 36 − 2(25)(36) cos 86∘
x = 42.4
2
(M1)
[4 marks]
Examiners report
This question was attempted in a satisfactory manner. Even the weakest candidates earned some marks here, showing some clear
working. In part (a) the diagram was completed fairly well, with some candidates incorrectly labelling the angle with the vertical as
4∘ . The cosine rule was applied satisfactory in part (b), although some candidates incorrectly used their calculators in radian mode.
Approaches using a combination of the sine rule and/or right-angled triangle trigonometry were seen, especially when candidates
incorrectly labelled the 25 m path as being the distance from the horizontal to U.
The diagram below shows a plan for a window in the shape of a trapezium.
Three sides of the window are 2 m long. The angle between the sloping sides of the window and the base is θ , where 0 < θ < π2 .
33a. Show that the area of the window is given by y = 4 sin θ + 2 sin 2θ .
[5 marks]
Markscheme
evidence of finding height, h
e.g. sin θ =
h
2
(A1)
, 2 sin θ
evidence of finding base of triangle, b
(A1)
e.g. cos θ = , 2 cos θ
b
2
attempt to substitute valid values into a formula for the area of the window
(M1)
e.g. two triangles plus rectangle, trapezium area formula
correct expression (must be in terms of θ )
A1
e.g. 2 ( 12 × 2 cos θ × 2 sin θ) + 2 × 2 sin θ , 12 (2 sin θ)(2 + 2 + 4 cos θ)
attempt to replace 2 sin θ cos θ by sin 2θ
M1
e.g. 4 sin θ + 2(2 sin θ cos θ)
y = 4 sin θ + 2 sin 2θ
AG
N0
[5 marks]
Examiners report
As the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) was
generally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in a
clear and organized manner. Some tried a "working backwards" approach, earning no marks.
33b. Zoe wants a window to have an area of 5 m2 . Find the two possible values of θ .
[4 marks]
Markscheme
correct equation
A1
e.g. y = 5 , 4 sin θ + 2 sin 2θ = 5
evidence of attempt to solve
(M1)
e.g. a sketch, 4 sin θ + 2 sin θ − 5 = 0
θ = 0.856 (49.0∘ ) , θ = 1.25 (71.4∘ )
A1A1
N3
[4 marks]
Examiners report
In part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and instead
attempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a second
solution outside the domain.
33c. John wants two windows which have the same area A but different values of θ .
Find all possible values for A .
[7 marks]
Markscheme
recognition that lower area value occurs at θ = π2
finding value of area at θ =
π
2
(M1)
(M1)
e.g. 4 sin ( π2 ) + 2 sin (2 × π2 ) , draw square
A=4
(A1)
recognition that maximum value of y is needed
A = 5.19615 …
(M1)
(A1)
4 < A < 5.20 (accept 4 < A < 5.19 )
A2
N5
[7 marks]
Examiners report
A pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or the
maximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident from
candidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken into
consideration for future paper writing.
34. Solve cos 2x − 3 cos x − 3 − cos2 x = sin 2 x , for 0 ≤ x ≤ 2π .
[7 marks]
Markscheme
evidence of substituting for cos 2x
(M1)
evidence of substituting into sin2 x + cos2 x = 1
(M1)
correct equation in terms of cos x (seen anywhere)
A1
e.g. 2cos2 x − 1 − 3 cos x − 3 = 1 , 2cos2 x − 3 cos x − 5 = 0
evidence of appropriate approach to solve
(M1)
appropriate working
A1
e.g. (2 cos x − 5)(cos x + 1) = 0 , (2x − 5)(x + 1) , cos x =
3± √49
4
correct solutions to the equation
e.g. cos x = 52 , cos x = −1 , x = 52 , x = −1
x=π
A1
(A1)
N4
[7 marks]
Examiners report
This question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation with
identities, but many earned no further marks due to algebraic errors. Many did not substitute for cos 2x ; others did this substitution
but then did nothing further.
Few candidates were able to get a correct equation in terms of cos x and many who did get the equation didn't know what to do with
it. Candidates who correctly solved the resulting quadratic usually found the one correct value of x, earning full marks.
The following diagram shows a circle with centre O and radius 4 cm.
The points A, B and C lie on the circle. The point D is outside the circle, on (OC).
[3 marks]
Markscheme
choosing sine rule
(M1)
correct substitution
A1
e.g.
sin 0.8
=
4
sin 0.3
A1
N2
[3 marks]
Examiners report
This question was generally quite well done, and it was pleasing to note that candidates could come up with multiple methods to
arrive at the correct answers. Many candidates worked comfortably with the sine and cosine rules to find sides of triangles. Some
candidates chose alternative right-angled triangle methods, often with success, although this proved a time-consuming approach.
Some unnecessarily converted the radian values to degrees, which sometimes led to calculation errors that could have been avoided.
A large number of candidates accrued the accuracy penalty in this question.
35b. Find OD.
[4 marks]
Markscheme
METHOD 1
finding angle OAD = π − 1.1 = (2.04) (seen anywhere)
(M1)
choosing cosine rule
A1
correct substitution
2
(A1)
2
2
e.g. OD = 9.71 + 4 − 2 × 9.71 × 4 × cos(π − 1.1)
OD = 12.1 (cm)
A1
N3
METHOD 2
finding angle OAD = π − 1.1 = (2.04) (seen anywhere)
choosing sine rule
(M1)
correct substitution
A1
e.g.
OD
sin(π−1.1)
=
9.71
sin 0.8
OD = 12.1 (cm)
=
(A1)
4
sin 0.3
A1
N3
[4 marks]
Examiners report
This question was generally quite well done, and it was pleasing to note that candidates could come up with multiple methods to
arrive at the correct answers. Many candidates worked comfortably with the sine and cosine rules to find sides of triangles. Some
candidates chose alternative right-angled triangle methods, often with success, although this proved a time-consuming approach.
Some unnecessarily converted the radian values to degrees, which sometimes led to calculation errors that could have been avoided.
A large number of candidates accrued the accuracy penalty in this question.
35c. Find the area of sector OABC.
[2 marks]
Markscheme
correct substitution into area of a sector formula
(A1)
2
e.g. area = 0.5 × 4 × 0.8
area = 6.4 (cm2 )
A1
N2
[2 marks]
Examiners report
This question was generally quite well done, and it was pleasing to note that candidates could come up with multiple methods to
arrive at the correct answers. Many candidates worked comfortably with the sine and cosine rules to find sides of triangles. Some
candidates chose alternative right-angled triangle methods, often with success, although this proved a time-consuming approach.
Some unnecessarily converted the radian values to degrees, which sometimes led to calculation errors that could have been avoided.
A large number of candidates accrued the accuracy penalty in this question.
35d. Find the area of region ABCD.
[4 marks]
Markscheme
substitution into area of triangle formula OAD
correct substitution
(M1)
A1
1
2
e.g. A= × 4 × 12.1 × sin 0.8 , A= 12 × 4 × 9.71 × sin 2.04 , A= 12 × 12.1 × 9.71 × sin 0.3
subtracting area of sector OABC from area of triangle OAD
(M1)
e.g. area ABCD = 17.3067 − 6.4
area ABCD = 10.9 (cm2 )
A1
N2
[4 marks]
Examiners report
This question was generally quite well done, and it was pleasing to note that candidates could come up with multiple methods to
arrive at the correct answers. Many candidates worked comfortably with the sine and cosine rules to find sides of triangles. Some
candidates chose alternative right-angled triangle methods, often with success, although this proved a time-consuming approach.
Some unnecessarily converted the radian values to degrees, which sometimes led to calculation errors that could have been avoided.
A large number of candidates accrued the accuracy penalty in this question.
Let f(x) = 5 cos π4 x and g(x) = −0.5x2 + 5x − 8 for 0 ≤ x ≤ 9 .
36. Consider the graph of f . Write down
(i)
[4 marks]
the x-intercept that lies between x = 0 and x = 3 ;
(ii) the period;
(iii) the amplitude.
Markscheme
(i) (2, 0) (accept x = 2 )
(ii) period = 8
A2
(iii) amplitude = 5
A1
N1
N2
A1
N1
[4 marks]
Examiners report
Many candidates found parts (b) and (c) accessible, although quite a few did not know how to find the period of the cosine function.
A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is θ radians, where 0 ≤ θ ≤ π2 .
37a. Write down an expression in terms of θ for
(i)
[2 marks]
x;
(ii) y .
Markscheme
(i) x = 3 cos θ
A1
N1
(ii) y = 3 sin θ
A1
N1
[2 marks]
Examiners report
Candidates familiar with the circular nature of sine and cosine found part (a) accessible. However, a good number of candidates left
this part blank, which suggests that there was difficulty interpreting the meaning of the x and y in the diagram.
37b. Let the area of the rectangle be A.
Show that A = 18 sin 2θ .
Markscheme
finding area
(M1)
e.g. A = 2x × 2y , A = 8 × 12 bh
substituting
A1
e.g. A = 4 × 3 sin θ × 3 cos θ , 8 × 12 × 3 cos θ × 3 sin θ
A = 18(2 sin θ cos θ) A1
A = 18 sin 2θ
[3 marks]
AG
N0
[3 marks]
Examiners report
Those with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question,
a solution cannot begin with the answer given. The area of the rectangle could be found by using 2x × 2y , or by using the eight
small triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.
The vertices of the triangle PQR are defined by the position vectors
4 ⎞
⎛ 3 ⎞
⎛ 6 ⎞
−→ ⎛
−
−→
−
−→
OP = ⎜ −3 ⎟ , OQ = ⎜ −1 ⎟ and OR = ⎜ −1 ⎟ .
⎝ 1 ⎠
⎝ 2 ⎠
⎝ 5 ⎠
38. (i)
Find sinRP̂Q .
[6 marks]
(ii) Hence, find the area of triangle PQR, giving your answer in the form a√3 .
Markscheme
(i) METHOD 1
(M1)
evidence of appropriate approach
e.g. using sin2 RP̂Q + cos2 RP̂Q = 1 , diagram
substituting correctly (A1)
−−−−−−−2
e.g. sinRP̂Q = √1 − ( 12 )
−
−
√3
sinRP̂Q = √ 34 (= 2 )
A1
N3
METHOD 2
since cos P̂ = 12 , P̂ = 60∘
(A1)
evidence of approach
e.g. drawing a right triangle, finding the missing side
sin P̂ =
√3
2
A1
(A1)
N3
(ii) evidence of appropriate approach
(M1)
e.g. attempt to substitute into 12 ab sin C
correct substitution
−− √3
e.g. area = 12 √6 × √24 × 2
area = 3√3
A1
A1
N2
[6 marks]
Examiners report
Many candidates attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew the
side relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of 1,2, √3 to find the
area of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected as
the value of sine was expected to be used in the approach.
The circle shown has centre O and radius 3.9 cm.
Points A and B lie on the circle and angle AOB is 1.8 radians.
39a. Find AB.
[3 marks]
Markscheme
METHOD 1
choosing cosine rule
(M1)
substituting correctly A1
−−−−−−−−−−−−−−−−−−−−−−−
e.g. AB = √3.92 + 3.92 − 2(3.9)(3.9) cos 1.8
AB = 6.11 (cm)
A1
N2
METHOD 2
evidence of approach involving right-angled triangles
substituting correctly
(M1)
A1
x
e.g. sin 0.9 = 3.9
, 12 AB = 3.9 sin 0.9
AB = 6.11 (cm)
A1
N2
METHOD 3
choosing the sine rule
substituting correctly
e.g.
sin 0.670…
3.9
=
AB = 6.11 (cm)
(M1)
A1
sin 1.8
AB
A1
N2
[3 marks]
Examiners report
This question was well answered by the majority of candidates. Full solutions were common in both parts, and a variety of successful
approaches were used. Radians were well handled with few candidates working with the angle in degrees. Some candidates
incorrectly found the length of the arc subtended by the central angle rather than the length of segment [AB].
39b. Find the area of the shaded region.
[4 marks]
Markscheme
METHOD 1
reflex AÔB = 2π − 1.8 (= 4.4832)
(A2)
correct substitution A = 12 (3.9)2 (4.4832 …)
area =34.1 (cm2)
A1
A1
N2
METHOD 2
finding area of circle A = π(3.9)2 (= 47.78 …)
(A1)
finding area of (minor) sector A = 12 (3.9)2 (1.8) (= 13.68 …) (A1)
subtracting
M1
e.g. π(3.9)2 − 0.5(3.9)2 (1.8) , 47.8 − 13.7
area = 34.1 (cm2) A1
N2
METHOD 3
finding reflex AÔB = 2π − 1.8 (= 4.4832)
finding proportion of total area of circle
e.g.
2π−1.8
2π
2
× π(3.9) ,
area = 34.1 (cm2)
θ
2π
(A2)
A1
× πr2
A1
N2
[4 marks]
Examiners report
In part (b), some candidates incorrectly subtracted the area of the triangle or even a length. Many candidates failed to give answers to
3 significant figures and therefore lost an accuracy mark.
Let f(x) =
3x
2
+ 1 , g(x) = 4 cos ( x3 ) − 1 . Let h(x) = (g ∘ f)(x) .
40a. Write down the period of h .
[1 mark]
Markscheme
period is 4π(12.6)
A1
N1
[1 mark]
Examiners report
The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had
difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates
knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period
and range, but this approach was not the most efficient.
40b. Write down the range of h .
[2 marks]
Markscheme
range is −5 ≤ h(x) ≤ 3 ([−5,3])
[2 marks]
A1A1
N2
Examiners report
The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had
difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates
knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period
and range, but this approach was not the most efficient.
41. Let f(x) = √3e2x sin x + e2x cos x , for 0 ≤ x ≤ π . Solve the equation f(x) = 0 .
[6 marks]
Markscheme
e2x (√3 sin x + cos x) = 0
(A1)
e2x = 0 not possible (seen anywhere)
(A1)
simplifying
e.g. √3 sin x + cos x = 0 , √3 sin x = − cos x ,
sin x
− cos x
=
1
√3
A1
EITHER
1
√3
A1
A2
N4
tan x = −
x=
5π
6
OR
sketch of 30∘ , 60∘ , 90∘ triangle with sides 1, 2, √3
verifying
5π
6
5π
6
A1
A1
A1
satisfies equation
N4
[6 marks]
Examiners report
Those who realized e2x was a common factor usually earned the first four marks. Few could reason with the given information to
solve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization.
Let f(x) = e−3x and g(x) = sin (x − π3 ) .
42.
[4 marks]
Let h(x) = e−3x sin (x − π3 ) . Find the exact value of h′ ( π3 ) .
Markscheme
evidence of choosing product rule
(M1)
e.g. uv′ + vu′
correct expression
A1
e.g. −3e−3x sin (x − π3 ) + e−3x cos (x − π3 )
complete correct substitution of x = π3
(A1)
e.g. −3e−3 3 sin ( π3 − π3 ) + e−3 3 cos ( π3 − π3 )
π
h′ ( π3 ) = e−π
[4 marks]
π
A1
N3
Examiners report
Often the substitution of π3 was incomplete or not done at all.
The diagram below shows a triangle ABD with AB =13 cm and AD = 6.5 cm.
Let C be a point on the line BD such that BC = AC = 7 cm.
43a. Find the size of angle ACB.
[3 marks]
Markscheme
METHOD 1
evidence of choosing the cosine formula
correct substitution
e.g. cos AĈ B =
(M1)
A1
72+ 72− 132
2×7×7
AĈ B = 2.38 radians (= 136∘ ) A1
N2
METHOD 2
evidence of appropriate approach involving right-angled triangles
correct substitution
(M1)
A1
e.g. sin ( 12 AĈ B) = 6.5
7
AĈ B = 2.38 radians (= 136∘ ) A1
N2
[3 marks]
Examiners report
This question was generally well done. Even the weakest candidates often earned marks. Only a very few candidates used a rightangled triangle approach.
43b. Find the size of angle CAD.
[5 marks]
Markscheme
METHOD 1
AĈ D = π − 2.381 (180 − 136.4)
(A1)
evidence of choosing the sine rule in triangle ACD
A1
correct substitution
e.g.
6.5
sin 0.760…
=
(M1)
7
AD̂C = 0.836 … (= 47.9 …∘ )
A1
C ÂD = π − (0.760 … + 0.836 …) (180 − (43.5 … + 47.9 …))
= 1.54 (= 88.5∘ )
A1
N3
METHOD 2
AB̂C = 12 (π − 2.381) ( 12 (180 − 136.4))
(A1)
evidence of choosing the sine rule in triangle ABD
A1
correct substitution
e.g.
6.5
sin 0.380…
=
(M1)
13
AD̂C = 0.836 … (= 47.9 …∘ )
A1
C ÂD = π − 0.836 … − (π − 2.381 …) (= 180 − 47.9 … − (180 − 136.4))
= 1.54 (= 88.5∘ )
A1
N3
Note: Two triangles are possible with the given information. If candidate finds AD̂C = 2.31 (132∘ ) leading to C ÂD = 0.076
(4.35∘ ) , award marks as per markscheme.
[5 marks]
Examiners report
Almost no candidates realized there was an ambiguous case of the sine rule in part (b). Those who did not lose the mark for accuracy
in the previous question often lost it here.
Let f(x) = 3 sin x + 4 cos x , for −2π ≤ x ≤ 2π .
44a. Hence write f(x) in the form p sin(qx + r) .
[3 marks]
Markscheme
f(x) = 5 sin(x + 0.927) (accept p = 5 , q = 1 , r = 0.927 )
A1A1A1
N3
[3 marks]
Examiners report
A number of candidates had difficulty writing the correct value of q in part (c).
44b. Write down the two values of k for which the equation f(x) = k has exactly two solutions.
Markscheme
k = −5 , k = 5
[2 marks]
A1A1
N2
[2 marks]
Examiners report
Part (e) proved challenging for many candidates, although if candidates answered this part, they generally did so correctly.
The graph of y = p cos qx + r , for −5 ≤ x ≤ 14 , is shown below.
There is a minimum point at (0, −3) and a maximum point at (4, 7) .
45a. Find the value of
(i)
[6 marks]
p;
(ii) q ;
(iii) r.
Markscheme
(i) evidence of finding the amplitude
e.g.
7+3
2
(M1)
, amplitude = 5
p = −5
A1
N2
(ii) period = 8
q = 0.785 (=
(A1)
2π
8
= π4 )
(iii) r =
7−3
2
(A1)
r=2
A1
N2
A1
N2
[6 marks]
Examiners report
Many candidates did not recognize that the value of p was negative. The value of q was often interpreted incorrectly as the period but
most candidates could find the value of r, the vertical translation.
45b. The equation y = k has exactly two solutions. Write down the value of k.
Markscheme
k = −3 (accept y = −3 )
[1 mark]
A1
N1
[1 mark]
Examiners report
In part (b), candidates either could not find a solution or found too many.
The diagram below shows a quadrilateral ABCD with obtuse angles AB̂C and AD̂C .
AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , BÂC = 30∘ , AB̂C = x∘ , AD̂C = y∘ .
−−−−−−−−−−
46a. Use the cosine rule to show that AC = √41 − 40 cos x .
[1 mark]
Markscheme
correct substitution
A1
e.g. 25 + 16 − 40 cos x , 52 + 42 − 2 × 4 × 5 cos x
−−−−−−−−−−
AC = √41 − 40 cos x AG
[1 mark]
Examiners report
Many candidates worked comfortably with the sine and cosine rules in part (a) and (b).
46b. Use the sine rule in triangle ABC to find another expression for AC.
[2 marks]
Markscheme
correct substitution
e.g.
AC
sin x
A1
= sin430 , 12 AC = 4 sin x
AC = 8 sin x (accept 4sinsin30x )
A1
N1
[2 marks]
Examiners report
Many candidates worked comfortably with the sine and cosine rules in part (a) and (b). Equally as many did not take the cue from the
word "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set up
an equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usually
unsuccessful.
46c. (i)
(ii) Find AC .
[6 marks]
Markscheme
(i) evidence of appropriate approach using AC M1
−−−−−−−−−−
e.g. 8 sin x = √41 − 40 cos x , sketch showing intersection
correct solution 8.682 …, 111.317 …
obtuse value 111.317 …
(A1)
(A1)
x = 111.32 to 2 dp (do not accept the radian answer 1.94 )
(ii) substituting value of x into either expression for AC
A1
N2
(M1)
e.g. AC = 8 sin 111.32
AC = 7.45
A1
N2
[6 marks]
Examiners report
Equally as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did not
receive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,
laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recovered
and earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode and
rounded prematurely throughout this question often resulting in accuracy penalties.
46d. (i)
Find y.
[5 marks]
(ii) Hence, or otherwise, find the area of triangle ACD.
Markscheme
(i) evidence of choosing cosine rule
2
2
c −b
e.g. cos B = a +2ac
2
correct substitution
e.g.
2
2
4 + 4 − 7.45
2×4×4
y = 137
2
A1
(M1)
A1
, 7.452 = 32 − 32 cos y , cos y = −0.734 …
N2
(ii) correct substitution into area formula
(A1)
1
2
e.g. × 4 × 4 × sin 137 , 8 sin 137
area = 5.42
A1
N2
[5 marks]
Examiners report
Equally as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did not
receive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,
laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recovered
and earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode and
rounded prematurely throughout this question often resulting in accuracy penalties.
The diagram below shows a circle with centre O and radius 8 cm.
The points A, B, C, D, E and F are on the circle, and [AF] is a diameter. The length of arc ABC is 6 cm.
47a. Find the size of angle AOC .
[2 marks]
Markscheme
appropriate approach
(M1)
e.g. 6 = 8θ
AÔC = 0.75
A1
N2
[2 marks]
Examiners report
Most candidates demonstrated understanding of trigonometry on this question. They generally did well in parts (a) and (c), and even
many of them on part (b). Fewer candidates could do part (d).
Many opted to work in degrees rather than in radians, which often introduced multiple inaccuracies. Some worked with an incorrect
A pleasing number knew how to find the area of the shaded region.
Inability to work in radians and misunderstanding of significant figures were common problems, though. Weaker candidates often
made the mistake of using triangle formulae for sectors or used degrees in the formulas instead of radians.
For some candidates there were many instances of confusion between lines and arcs. In (a) some treated 6 as the length of AC . In (d)
some found the length of arc EF rather than the length of the segment.
Several students seemed to confuse the area of sector in (b) with the shaded region.
47b. Hence find the area of the shaded region.
[6 marks]
Markscheme
(M1)
evidence of substitution into formula for area of triangle
1
2
e.g. area = × 8 × 8 × sin(0.75)
area = 21.8 …
(A1)
evidence of substitution into formula for area of sector
(M1)
1
2
e.g. area = × 64 × 0.75
area of sector = 24
(A1)
evidence of substituting areas
(M1)
e.g. 12 r2 θ − 12 ab sin C , area of sector − area of triangle
area of shaded region = 2.19 cm2
A1
N4
[6 marks]
Examiners report
Most candidates demonstrated understanding of trigonometry on this question. They generally did well in parts (a) and (c), and even
many of them on part (b). Fewer candidates could do part (d).
Many opted to work in degrees rather than in radians, which often introduced multiple inaccuracies. Some worked with an incorrect
A pleasing number knew how to find the area of the shaded region.
Inability to work in radians and misunderstanding of significant figures were common problems, though. Weaker candidates often
made the mistake of using triangle formulae for sectors or used degrees in the formulas instead of radians.
For some candidates there were many instances of confusion between lines and arcs. In (a) some treated 6 as the length of AC . In (d)
some found the length of arc EF rather than the length of the segment.
Several students seemed to confuse the area of sector in (b) with the shaded region.
47c. The area of sector OCDE is 45 cm2 .
[2 marks]
Find the size of angle COE .
Markscheme
attempt to set up an equation for area of sector
1
2
(M1)
2
e.g. 45 = × 8 × θ
C ÔE = 1.40625 (1.41 to 3 sf)
A1
N2
[2 marks]
Examiners report
Most candidates demonstrated understanding of trigonometry on this question. They generally did well in parts (a) and (c), and even
many of them on part (b). Fewer candidates could do part (d).
Many opted to work in degrees rather than in radians, which often introduced multiple inaccuracies. Some worked with an incorrect
A pleasing number knew how to find the area of the shaded region.
Inability to work in radians and misunderstanding of significant figures were common problems, though. Weaker candidates often
made the mistake of using triangle formulae for sectors or used degrees in the formulas instead of radians.
For some candidates there were many instances of confusion between lines and arcs. In (a) some treated 6 as the length of AC . In (d)
some found the length of arc EF rather than the length of the segment.
Several students seemed to confuse the area of sector in (b) with the shaded region.
47d. Find EF .
[5 marks]
Markscheme
METHOD 1
attempting to find angle EOF
(M1)
e.g. π − 0.75 − 1.41
EÔF = 0.985 (seen anywhere)
A1
evidence of choosing cosine rule
(M1)
correct substitution A1
−−−−−−−−−−−−−−−−−−−−−−−
e.g. EF = √82 + 82 − 2 × 8 × 8 × cos 0.985
EF = 7.57 cm
A1
N3
METHOD 2
attempting to find angles that are needed
(M1)
e.g. angle EOF and angle OEF
EÔF = 0.9853 … and OÊF (or OF̂E) = 1.078 …
evidence of choosing sine rule
correct substitution
e.g.
EF
sin 0.985
A1
(M1)
(A1)
8
sin 1.08
=
EF = 7.57 cm
A1
N3
METHOD 3
attempting to find angle EOF
(M1)
e.g. π − 0.75 − 1.41
EÔF = 0.985 (seen anywhere)
A1
evidence of using half of triangle EOF
e.g. x = 8 sin
(M1)
0.985
2
correct calculation
A1
e.g. x = 3.78
EF = 7.57 cm
A1
N3
[5 marks]
Examiners report
Most candidates demonstrated understanding of trigonometry on this question. They generally did well in parts (a) and (c), and even
many of them on part (b). Fewer candidates could do part (d).
Many opted to work in degrees rather than in radians, which often introduced multiple inaccuracies. Some worked with an incorrect
A pleasing number knew how to find the area of the shaded region.
Inability to work in radians and misunderstanding of significant figures were common problems, though. Weaker candidates often
made the mistake of using triangle formulae for sectors or used degrees in the formulas instead of radians.
For some candidates there were many instances of confusion between lines and arcs. In (a) some treated 6 as the length of AC . In (d)
some found the length of arc EF rather than the length of the segment.
Several students seemed to confuse the area of sector in (b) with the shaded region.
Let f(x) = sin3 x + cos3 x tan x, π2 < x < π .
48a. Show that f(x) = sin x .
[2 marks]
Markscheme
changing tan x into
e.g. sin
3
sin x
cos x
A1
sin x
x + cos3 x cos
x
A1
simplifying
2
e.g sin x(sin x + cos2 x) , sin3 x + sin x − sin3 x
f(x) = sin x AG
N0
[2 marks]
Examiners report
Not surprisingly, this question provided the greatest challenge in section A. In part (a), candidates were able to use the identity
sin x
tan x = cos
, but many could not proceed any further.
x
2
3
48b. Let sin x =
. Show that f(2x) = −
4√5
9
[5 marks]
.
Markscheme
recognizing f(2x) = sin 2x , seen anywhere
(A1)
evidence of using double angle identity sin(2x) = 2 sin x cos x , seen anywhere
evidence of using Pythagoras with sin x =
2
3
(M1)
M1
e.g. sketch of right triangle, sin2 x + cos2 x = 1
cos x = −
√5
3
(accept
f(2x) = 2 ( 23 ) (−
f(2x) = −
4√5
9
√5
3
√5
)
3
AG
(A1)
)
A1
N0
[5 marks]
Examiners report
Part (b) was generally well done by those candidates who attempted it, the major error arising when the negative sign "magically"
appeared in the answer. Many candidates could find the value of cosx but failed to observe that cosine is negative in the given
domain.
f(t) = acos b(t − c) + d
y = f(t)
Let f(t) = acos b(t − c) + d , t ≥ 0 . Part of the graph of y = f(t) is given below.
When t = 3 , there is a maximum value of 29, at M.
When t = 9 , there is a minimum value of 15.
49a. (i)
Find the value of a.
[7 marks]
(ii) Show that b = π6 .
(iii) Find the value of d.
(iv) Write down a value for c.
Markscheme
(i) attempt to substitute
e.g. a =
a = 7 (accept a = −7 )
(ii) period = 12
b=
2π
12
b=
π
6
A1
N2
(A1)
A1
AG
N0
(iii) attempt to substitute
e.g. d =
(M1)
29−15
2
(M1)
29+15
2
d = 22
A1
N2
(iv) c = 3 (accept c = 9 from a = −7 )
A1
N1
Note: Other correct values for c can be found, c = 3 ± 12k , k ∈ Z .
[7 marks]
Examiners report
This question was the most difficult on the paper. Where candidates attempted this question, part (a) was answered satisfactorily.
49b.
The transformation P is given by a horizontal stretch of a scale factor of 12 , followed by a translation of (
Let M ′ be the image of M under P. Find the coordinates of M ′ .
3
).
−10
[2 marks]
Markscheme
(A1)
stretch takes 3 to 1.5
translation maps (1.5, 29) to (4.5, 19) (so M ′ is (4.5, 19))
A1
N2
[2 marks]
Examiners report
Few answered part (b) correctly as most could not interpret the horizontal stretch.
49c. The graph of g is the image of the graph of f under P.
[4 marks]
Find g(t) in the form g(t) = 7 cos B(t − c) + D .
Markscheme
g(t) = 7 cos π3 (t − 4.5) + 12
Note: Award A1 for
π
3
A1A2A1
N4
, A2 for 4.5, A1 for 12.
Other correct values for c can be found, c = 4.5 ± 6k , k ∈ Z .
[4 marks]
Examiners report
Few answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable to
the answer in (b) and the value of C in part (c) was lost on all but the most attentive.
49d. The graph of g is the image of the graph of f under P.
[3 marks]
Give a full geometric description of the transformation that maps the graph of g to the graph of f .
Markscheme
translation (
−3
)
10
(A1)
horizontal stretch of a scale factor of 2
(A1)
completely correct description, in correct order
e.g. translation (
A1
N3
−3
) then horizontal stretch of a scale factor of 2
10
[3 marks]
Examiners report
In part (d), some candidates could name the transformations required, although only a handful provided the correct order of the
transformations to return the graph to its original state.
A ship leaves port A on a bearing of 030∘ . It sails a distance of 25 km to point B. At B, the ship changes direction to a bearing of 100∘ . It
sails a distance of 40 km to reach point C. This information is shown in the diagram below.
A second ship leaves port A and sails directly to C.
50a. Find the distance the second ship will travel.
[4 marks]
Markscheme
finding AB̂C = 110∘ (= 1.92 radians)
evidence of choosing cosine rule
(A1)
(M1)
e.g. AC 2 = AB2 + BC 2 − 2(AB)(BC) cos AB̂C
A1
correct substitution
2
2
2
e.g. AC = 25 + 40 − 2(25)(40) cos 110∘
AC = 53.9 (km)
A1
Examiners report
A good number of candidates found this question very accessible, although some attempted to use Pythagoras' theorem to find AC.
50b. Find the bearing of the course taken by the second ship.
Markscheme
METHOD 1
correct substitution into the sine rule
e.g.
sin BÂC
40
=
sin 110∘
53.9
A1
A1
BÂC = 44.2∘
bearing = 074∘
A1
N1
METHOD 2
correct substitution into the cosine rule
2
2
e.g. cos BÂC = 40 − 25 − 53.9
2
−2(25)(53.9)
BÂC = 44.3∘
bearing = 074∘
[3 marks]
A1
N1
A1
A1
[3 marks]
Examiners report
Often candidates correctly found BÂC in part (b), but few added the 30∘ to obtain the required bearing. Some candidates calculated
BĈ A , misinterpreting that the question required the course of the second ship.
The diagram shows a parallelogram ABCD.
The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .
51. Hence, or otherwise, find the area of the parallelogram.
[3 marks]
Markscheme
METHOD 1
−
−→ −→
evidence of using area = 2 ( 12 ∣AD∣ ∣AB ∣ sin DÂB)
∣ ∣∣ ∣
correct substitution
(M1)
A1
e.g. area = 2 ( 12 (3,741 …)(5.477 …) sin 0.883 …)
area = 15.8
A1
N2
METHOD 2
evidence of using area = b × h
(M1)
finding height of parallelogram
A1
e.g. h = 3.741 … × sin 0.883 … (= 2.892 …) , h = 5.477 … × sin 0.883 … (= 4.234 …)
area = 15.8
A1
N2
[3 marks]
Examiners report
Few candidates considered that the area of the parallelogram is twice the area of a triangle, which is conveniently found using BÂD .
−→
−
−→
In an effort to find base × height , many candidates multiplied the magnitudes of AB and AD , missing that the height of a
parallelogram is perpendicular to a base.
Let p = sin 40∘ and q = cos 110∘ . Give your answers to the following in terms of p and/or q .
52a. Write down an expression for
(i)
sin 140∘ ;
(ii) cos 70∘ .
[2 marks]
Markscheme
(i) sin 140∘ = p
A1
∘
(ii) cos 70 = −q
A1
N1
N1
[2 marks]
Examiners report
This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a
few received any marks at all.
52b. Find an expression for cos 140∘ .
[3 marks]
Markscheme
METHOD 1
evidence of using sin2 θ + cos2 θ = 1 (M1)
−−−−−
e.g. diagram, √1 − p2 (seen anywhere)
−−−−−
cos 140∘ = ±√1 − p2 (A1)
−−−−−
cos 140∘ = −√1 − p2 A1 N2
METHOD 2
evidence of using cos 2θ = 2cos2 θ − 1
cos 140∘ = 2cos2 70 − 1
(A1)
2
∘
(M1)
cos 140 = 2(−q) − 1 (= 2q2 − 1)
A1
N2
[3 marks]
Examiners report
Many did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of
sin2 x + cos2 x = 1 results in a ± value. The application of a double angle formula also seemed weak.
52c. Find an expression for tan 140∘ .
[1 mark]
Markscheme
METHOD 1
tan 140∘ =
sin 140∘
cos 140∘
=−
p
√1−p 2
A1
N1
METHOD 2
tan 140∘ =
p
2q2 −1
A1
N1
[1 mark]
Examiners report
This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a
few received any marks at all. Many did not appear to know the relationships between trigonometric functions of supplementary
angles and that the use of sin2 x + cos2 x = 1 results in a ± value. The application of a double angle formula also seemed weak.
g(x) = 3 sin 2x
Consider g(x) = 3 sin 2x .
53a. Write down the period of g.
[1 mark]
Markscheme
period = π
A1
N1
[1 mark]
Examiners report
Many candidates were unable to write down the period of the function.
53b. On the diagram below, sketch the curve of g, for 0 ≤ x ≤ 2π .
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for amplitude of 3, A1 for their period, A1 for a sine curve passing through (0, 0) and (0, 2π) .
[3 marks]
Examiners report
Many candidates were unable to write down the period of the function. However, they were often then able to go and correctly
sketch the graph with the correct period.
53c. Write down the number of solutions to the equation g(x) = 2 , for 0 ≤ x ≤ 2π .
[2 marks]
Markscheme
(M1)
evidence of appropriate approach
e.g. line y = 2 on graph, discussion of number of solutions in the domain
A1
4 (solutions)
N2
[2 marks]
Examiners report
The final part was poorly done with many candidates finding the number of zeros instead of the intersection with the line y = 2 .
The diagram below shows triangle PQR. The length of [PQ] is 7 cm , the length of [PR] is 10 cm , and PQ̂R is 75∘ .
54a. Find PR̂Q .
[3 marks]
Markscheme
(M1)
choosing sine rule
correct substitution
sin R
7
= sin1075
∘
A1
sin R = 0.676148 …
PR̂Q = 42.5∘
A1
N2
[3 marks]
Examiners report
This question was well done with most students using the law of sines to find the angle.
54b. Find the area of triangle PQR.
[3 marks]
Markscheme
P = 180 − 75 − R
P = 62.5
(A1)
substitution into any correct formula
1
2
A1
e.g. area ΔPQR = × 7 × 10 × sin(their P)
= 31.0 (cm2)
[3 marks]
A1
N2
Examiners report
In part (b), the most common error occurred when angle R or 75 degrees was used to find the area. This particular question was the
most common place to incur an accuracy penalty.
The diagram below shows a circle centre O, with radius r. The length of arc ABC is 3π cm and AÔC =
55a. Find the value of r.
2π
.
9
[2 marks]
Markscheme
evidence of appropriate approach
e.g. 3π = r
M1
2π
9
r = 13.5 (cm)
A1
N1
[2 marks]
Examiners report
Few errors were made in this question. Those that were made were usually arithmetical in nature.
55b. Find the perimeter of sector OABC.
[2 marks]
Markscheme
(M1)
perimeter = 27 + 3π (cm) (= 36.4)
A1
N2
[2 marks]
Examiners report
Few errors were made in this question. Those that were made were usually arithmetical in nature.
55c. Find the area of sector OABC.
[2 marks]
Markscheme
evidence of appropriate approach
1
2
2
e.g. × 13.5 ×
M1
2π
9
area = 20.25π (cm2 ) (= 63.6)
A1
N1
[2 marks]
Examiners report
Few errors were made in this question. Those that were made were usually arithmetical in nature.
1
3
56a. Given that cos A =
and 0 ≤ A ≤ π2 , find cos 2A .
[3 marks]
Markscheme
evidence of choosing the formula cos 2A = 2cos2 A − 1
(M1)
Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding sin2 A = 1 − 19
A1
correct substitution
2
2
e.g.cos 2A = ( 13 ) − 89 , cos 2A = 2 × ( 13 ) − 1
cos 2A = − 79
A1
N2
[3 marks]
Examiners report
This question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuse
angles seemed distinctly lacking. Candidates who recognized the need of an identity for finding cos 2A given cos A seldom chose the
most appropriate of the three and even when they did often used it incorrectly with expressions such as 2cos2 19 − 1 .
56b. Given that sin B =
2
3
and
π
2
≤ B ≤ π , find cos B .
[3 marks]
Markscheme
METHOD 1
evidence of using sin2 B + cos2 B = 1 (M1)
2
−
−
e.g. ( 23 ) + cos2 B = 1 , √ 59 (seen anywhere),
−
−
√5
cos B = ±√ 59 (= ± 3 )
−
−
√5
cos B = −√ 59 (= − 3 )
(A1)
A1
N2
METHOD 2
M1
diagram
for finding third side equals √5
cos B =
√5
− 3
A1
(A1)
N2
[3 marks]
Examiners report
This question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuse
angles seemed distinctly lacking. Candidates who recognized the need of an identity for finding cos 2A given cos A seldom chose the
most appropriate of the three and even when they did often used it incorrectly with expressions such as 2cos2 19 − 1 .
Let f : x ↦ sin3 x .
57. (i)
Write down the range of the function f .
[5 marks]
(ii) Consider f(x) = 1 , 0 ≤ x ≤ 2π . Write down the number of solutions to this equation. Justify your answer.
Markscheme
(i) range of f is [−1, 1] , (−1 ≤ f(x) ≤ 1)
3
(ii) sin x ⇒ 1 ⇒ sin x = 1
A2
N2
A1
justification for one solution on [0, 2π] R1
e.g. x = π2 , unit circle, sketch of sin x
1 solution (seen anywhere)
A1
N1
[5 marks]
Examiners report
This question was not done well by most candidates. No more than one-third of them could correctly give the range of f(x) = sin3 x
and few could provide adequate justification for there being exactly one solution to f(x) = 1 in the interval [0, 2π] .
The following diagram shows a semicircle centre O, diameter [AB], with radius 2.
Let P be a point on the circumference, with PÔB = θ radians.
58a. Find the area of the triangle OPB, in terms of θ .
[2 marks]
Markscheme
evidence of using area of a triangle
(M1)
e.g. A = 12 × 2 × 2 × sin θ
A = 2 sin θ
A1
N2
[2 marks]
Examiners report
Most candidates could obtain the area of triangle OPB as equal to 2 sin θ , though 2θ was given quite often as the area.
58b. Explain why the area of triangle OPA is the same as the area triangle OPB.
[3 marks]
Markscheme
METHOD 1
PÔA =π − θ
(A1)
area ΔOPA = 12 2 × 2 × sin(π − θ) (= 2 sin(π − θ))
since sin(π − θ) = sin θ
A1
R1
then both triangles have the same area
AG
N0
METHOD 2
triangle OPA has the same height and the same base as triangle OPB
then both triangles have the same area
AG
R3
N0
[3 marks]
Examiners report
A minority recognized the equality of the sines of supplementary angles and the term complementary was frequently used instead of
supplementary. Only a handful of candidates used the simple equal base and altitude argument.
The following graph shows the depth of water, y metres , at a point P, during one day. The time t is given in hours, from midnight to noon.
59a. Use the graph to write down an estimate of the value of t when
(i)
[3 marks]
the depth of water is minimum;
(ii) the depth of water is maximum;
(iii) the depth of the water is increasing most rapidly.
Markscheme
(i) 7
A1
N1
(ii) 1
A1
N1
(iii) 10
A1
N1
[3 marks]
Examiners report
For part (a), most candidates correctly used the graph to identify the times of maximum and minimum depth. Most failed to consider
that the depth of water is increasing most rapidly at a point of inflexion and often answered with the interval t = 9 to t = 11 . A few
candidates answered with the depth instead of time, misinterpreting which axis to consider.
59b. The depth of water can be modelled by the function y = cos A(B(t − 1)) + C .
(i)
Show that A = 8 .
(ii) Write down the value of C.
(iii) Find the value of B.
[6 marks]
Markscheme
M1
(i) evidence of appropriate approach
e.g. A =
18−2
2
A=8
AG
(ii) C = 10
N0
A2
N2
(iii) METHOD 1
period = 12
(A1)
evidence of using B × period = 2π (accept 360∘ ) (M1)
e.g. 12 =
2π
B
B = π6 (accept 0.524 or 30)
A1
N3
METHOD 2
evidence of substituting
(M1)
e.g. 10 = 8 cos 3B + 10
simplifying
(A1)
e.g. cos 3B = 0 (3B = π2 )
B = π6 (accept 0.524 or 30)
A1
N3
[6 marks]
Examiners report
A substantial number of candidates showed difficulty finding parameters of a trigonometric function with many only making
superficial attempts at part (b), often leaving it blank entirely.
Some divided 2π by the period of 12, while others substituted an ordered pair such as (4, 10)and solved for B, often correctly. Many
found that c = 17 , thus confusing the vertical translation with a y-intercept.
59c. A sailor knows that he cannot sail past P when the depth of the water is less than 12 m . Calculate the values of t between
[2 marks]
which he cannot sail past P.
Markscheme
A1A1
e.g. t = 3.52 , t = 10.5 , between 03:31 and 10:29 (accept 10:30)
N2
[2 marks]
Examiners report
For (c), many candidates simply read approximate values from the graph where y = 12 and thus answered with t = 3.5 and t = 10.5
. Although the latter value is correct to three significant figures, t = 3.5 incurs the accuracy penalty as it was expected that candidates
calculate this value in their GDC to achieve a result of t = 3.52 . Those who attempted an analytic approach rarely achieved correct
results.
The expression 6 sin x cos x can be expressed in the form a sin bx .
60a. Find the value of a and of b .
[3 marks]
Markscheme
M1
recognizing double angle
e.g. 3 × 2 sin x cos x , 3 sin 2x
a=3,b=2
A1A1
N3
[3 marks]
Examiners report
[N/A]
60b. Hence or otherwise, solve the equation 6 sin x cos x =
3
2
, for π4 ≤ x ≤ π2 .
[4 marks]
Markscheme
substitution 3 sin 2x =
sin 2x =
1
2
, 2x =
e.g.
x=
5π
12
A1
M1
A1
finding the angle
π
6
3
2
A1
5π
6
N2
Note: Award A0 if other values are included.
[4 marks]
Examiners report
[N/A]
Let f(x) = cos x + √3 sin x , 0 ≤ x ≤ 2π . The following diagram shows the graph of f .
The y-intercept is at (0, 1) , there is a minimum point at A (p, q) and a maximum point at B.
61a. Find the maximum value of f(x) .
[3 marks]
Markscheme
max when x = π3
R1
correctly substituting x = π3 into f(x)
e.g. 12 + √3 (
A1
√3
)
2
max value is 2
A1
N1
[3 marks]
Examiners report
[N/A]
61b. The function f(x) can be written in the form r cos(x − a) .
[2 marks]
Write down the value of r and of a .
Markscheme
r = 2 , a = π3
A1A1
N2
[2 marks]
Examiners report
[N/A]
The following diagram shows a pole BT 1.6 m tall on the roof of a vertical building.
The angle of depression from T to a point A on the horizontal ground is 35∘ .
The angle of elevation of the top of the building from A is 30∘ .
62. Find the height of the building.
[7 marks]
Markscheme
METHOD 1
M1
appropriate approach
e.g. completed diagram
A1
attempt at set up
e.g. correct placement of one angle
tan 30 = hx , tan 35 =
h+1.6
x
attempt to set up equation
A1A1
M1
e.g. isolate x
A1
correct equation
e.g.
h
tan 30
=
h = 7.52
h+1.6
tan 35
A1
N3
METHOD 2
sin 30 = hl
A1
in triangle ATB, Â = 5∘ , Tˆ = 55∘
A1A1
M1
choosing sine rule
correct substitution
e.g.
h/ sin 30
sin 55
1.6
= sin
5
A1
55
h = 1.6×sinsin30×sin
5
h = 7.52
A1
A1
N3
[7 marks]
Examiners report
[N/A]
A circle centre O and radius r is shown below. The chord [AB] divides the area of the circle into two parts. Angle AOB is θ .
63a. Find an expression for the area of the shaded region.
[3 marks]
Markscheme
substitution into formula for area of triangle
e.g.
M1
evidence of subtraction
A1
correct expression
e.g.
A1
1
r × r sin θ
2
1 2
rθ
2
−
1 2
r sin θ
2
,
N2
1 2
r (θ
2
− sin θ)
[3 marks]
Examiners report
[N/A]
63b. The chord [AB] divides the area of the circle in the ratio 1:7. Find the value of θ .
[5 marks]
Markscheme
evidence of recognizing that shaded area is 18 of area of circle
e.g.
1
8
M1
seen anywhere
setting up correct equation
A1
e.g. 12 r2 (θ − sin θ) = 18 πr2
eliminating 1 variable
M1
e.g. 12 (θ − sin θ) = 18 π , θ − sin θ = π4
attempt to solve
M1
e.g. a sketch, writing sin x − x + π4 = 0
θ = 1.77 (do not accept degrees)
A1
N1
[5 marks]
Examiners report
[N/A] | 34,378 | 108,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-47 | latest | en | 0.896576 |
https://www.unifolks.com/questions/magnetic-drag-goals-a-understand-the-presence-of-induced-eddy-currents-b-determ-425775.html | 1,652,904,425,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00363.warc.gz | 1,225,630,832 | 61,461 | # Magnetic Drag</o:p> GOALS:</o:p> a) Understand the presence of induced Eddy currents</o:p> b) Determine a damping coefficient</o:p> c) Determine the terminal...
Magnetic Drag
GOALS:
a) Understand the presence of induced Eddy currents
b) Determine a damping coefficient
c) Determine the terminal velocity
Introduction
As the cart moves along the aluminum track, the magnets on the accessory bracket induce eddy currents in the track. This causes opposing magnetic fields that result in a drag force applied to the cart. The magnitude of this force is measured by analyzing the deceleration of the moving cart. The relationship between the drag force and the speed of the cart is determined.
Equations of Motion
We will assume that the magnetic drag force, f, is proportional to the velocity, v
f = -b v
where b is the drag coefficient.
On a level track, assuming no other sources of friction, the drag force is the only force acting on the cart along its axis of motion. Therefore, by Newton's 2nd Law,
m a = -b v (1)
where m is the mass of the cart and a (the acceleration) is the derivative (a=dv/dt) of the velocity.
Thus Equation (1) can be written as
The solution to this differential equation is
(2)
Equipment
INCLUDES:
1 Motion Sensor PS-2103A 1 Dynamics Track ME-6995 1 Elastic Bumper ME-8998 1 Magnetic Damping ME-6828 NEEDED, BUT NOT INCLUDED: 1 Meter Stick SE-8827 1 Calipers SE-8710 1 Balance SE-8723
Setup
Figure 1: The Magnetic Damping Accessory attaches to the end of the cart that does NOT have the plunger. The steel bracket is held in place by the magnets inside the cart. It slides up and down to adjust the amount of drag: The closer the three silver magnets are to the track, the more damping there is.
1. Set up the track as shown in Figure 1, including feet and elastic bumper.
2. Connect the Motion Sensor to the interface, and attach it to the track. Adjust the alignment knob on the side of the Motion Sensor so that it points parallel to the track. Make sure the switch on the top of the Motion Sensor is set to "cart."
3. Set the cart on the track (including the two extra masses), but do NOT attach the Magnetic Damping Accessory yet.
4. In PASCO Capstone, leave the sample rate at the default rate of 20 Hz.
5. Create the following calculations in the PASCO Calculator:
speed = abs([Velocity]) with units of m/s
mass = 0.253 (or whatever mass you’re using) with units of kg
Mag Force = -mass*derivative(6,[Velocity],[Time]) with units of N
6. Create a graph of speed vs. time.
7. Create a graph of Mag Force vs. Velocity.
Procedure
1. Starting with the cart near the Motion Sensor, give it a small push away from the sensor, and click on Record. Your data will look better if you stop recording before the cart hits the bumper.
2. Using the screw feet, adjust the level of the track so that the cart travels at a constant speed when moving away from the Sensor. By setting up the track so that it is slightly downhill, you will be measuring only the magnetic drag on the cart. The unwanted frictional forces will have been compensated for.
3. Attach the Magnetic Damping Accessory to the cart (see Fig. 1) and determine the cart mass, with and without the two extra masses.
4. The amount of magnetic damping is determined by the distance between the magnets and the track (see Fig. 2). Use some type of spacer (any non-magnetic material will work) to set the distance to about 3 mm. Slide the bracket down until the magnets are flush with the spacer, then remove the spacer.
Figure 2: Magnetic Spacing
Magnetic Force:
5. Place both masses in the cart. Starting with the cart near the Motion Sensor, give it a small push away from the sensor, and click on Record. Your data will look better if you stop recording before the cart stops moving.
6. Open the Calculator in the Tool Pallet. The Magnetic Force is calculated using Newton's 2nd Law:
Mag Force = ma
where m is the mass of the cart and the acceleration (a) is calculated using a derivative of velocity. Line 2 in the calculator shows the approximate mass of the cart, but you can change it to your cart mass.
7. The graph of the calculated magnetic force versus the cart's velocity should be a linear relationship, although the data may be noisy. You may have to take several runs to get a good looking graph.
8. Select a linear curve fit from the Graph Tool Pallet. Is your graph linear?
9. The slope of the graph is called the Drag Coefficient. Record this value.
Equations of Motion
The velocity vs. time graph shows your data from the previous page.
Does the velocity decay exponentially? Select an exponential curve fit from the Graph Tool Pallet.
Damping Coefficient
1. Create a graph of velocity vs. time. Select the exponential curve fit for your data. The constant in the exponential for the curve fit is b/m. Record this value. What are the units?
2. Record several more runs of data, and record the values of b/m for each. Put all the data into a table column and turn on the Statistics with the Mean and the Standard Deviation.
3. Use the average value of b/m and the mass of your cart, m, to calculate the drag coefficient, b.
4. What are the units of b?
5. Use the standard deviation to calculate the uncertainty in your value.
6. How does this value for b compare to what you measured earlier from your Mag Force vs. Velocity graph?
Terminal Velocity: Inclined Track
As the cart accelerates down the incline (see Fig. 4), the magnetic drag force (f=bV) increases. Eventually, this backwards drag force grows to equal the component of the gravitational force down the plane (mgsinθ), and the cart reaches terminal velocity, (VT). Assuming no other forces (such as air friction, etc.),
bVT = mgsinθ
which can be written as
VT = (mg/b) sinθ (3)
1. Remove the two masses from the cart. You will use them to incline the track as shown in Figure 3. Be sure NOT to move the Magnetic Drag Accessory, or you will have to re-measure the coefficient.
2. Use calipers to measure the height of the two stacked mass bars, and a meterstick to measure the distance between the feet. Calculate the angle of incline for the track.
3. Use Eqn. (3) to predict the terminal velocity of the cart on the inclined track. Hint: The mass, m, has changed from the previous part of the experiment.
Figure 3: Inclined Track Figure 4. Free-body Diagram
Analysis
1. Starting with the cart about 20 cm from the Motion Sensor, release the cart from rest, and click on Record. Your data will look better if you stop recording before the cart hits the bumper.
2. Examine the velocity vs. time graph. What is the terminal speed of the cart?
3. How does it compare to your predicted value?
CONCLUSION
Summarize your findings by reviewing goals in a quantitative way.
RUBRIC:
5 pts 4 pts 3 pts 2 pts 1 pts 0 pts Introduction, setup, procedure Data collection and graphs Drag Coefficient by two methods Terminal Velocity Analysis Conclusion (quantitative)
Answered 2 days AfterApr 08, 2022
## Solution
Dr Shweta answered on Apr 10 2022
Assignment Question: Magnetic drag
Solution
1. Create the following calculations in the PASCO Calculator: speed = abs([Velocity]) with units of m/s, mass = 0.253 (or whatever mass you’re using) with units of kg, Mag Force = -mass*derivative (6, [Velocity], [Time]) with units of N | 1,744 | 7,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-21 | latest | en | 0.88821 |
https://abcdocz.com/doc/1666125/newton-s-laws-of-motion | 1,660,020,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00005.warc.gz | 112,938,469 | 9,226 | # Newton`s Laws of Motion
```Newton’s Laws of Motion
Newton’s Laws and the Mousetrap Racecar
Simple version of Newton’s three laws of motion
1st Law: “objects at rest stay at rest,
objects in motion stay in motion”
2nd Law: “force equals mass times
acceleration”
3rd Law: “for every action there is an
equal and opposite reaction”
Questions
Is more mass a good thing or a bad
thing for a mousetrap racecar to go far?
What forces are involved in analyzing
the motion of the racecar?
Is friction a friend or foe for the racecar?
How does torque (rotating force) play a
role in the motion of the racecar?
1
First Law of Motion: “The Law of Inertia”
An object at rest remains at rest, and an object in motion
continues in motion, with constant velocity, unless the object
experiences a net external force.
• Inertia is the tendency of an object to
maintain its state of motion.
• Mass (not weight) is a measure of inertia.
• Objects in equilibrium do not accelerate.
Static equilibrium (rest) and dynamic
equilibrium (constant velocity) are both
the result of an object with zero net
external force (ΣF = 0). • Constant velocity and rest are both
natural states. The only difference is
the reference frame: an object at rest in
one reference frame can have constant
velocity in another reference frame.
• Honors: Newton’s Laws of Motion apply
only in an inertial reference frame.
First Law of Motion: “The Law of Inertia”
• The law of inertia states that no force is required to maintain
motion. Why, then, do you have to keep pedaling your bicycle to
maintain motion?
• A space probe may be carried by a rocket into outer space. What
keeps the probe going after the rocket no longer pushes it?
• Your friend says that inertia is a force that keeps things in their
place, either at rest or in motion. Do you agree? Why or why not?
• In terms of inertia, how does a car headrest help to guard against
injury (whiplash) in a rear-end collision?
• If an elephant were chasing you, its enormous mass would be
threatening. But if you zigzagged, its mass would be to your
• Two closed containers look the same, but one is packed with lead
and the other with feathers. How could you determine which one
had more mass if you and the containers were in a weightless
environment?
click for You Tube video
click for You Tube video
2
Second Law of Motion: “The Law of Acceleration”
The acceleration on an object is directly proportional to the
magnitude of the net external force, is in the same direction as the
net force, and is inversely proportional to the mass of the object
“sigma” = sum
of all forces, or
net force.
F and a
are vectors,
mass a scalar
• Objects that are not in
equilibrium will accelerate.
• Net force (also called resultant)
causes an object to accelerate.
• It’s important to distinguish
between a single force and the
net force, ΣF.
SUM OF TWO FORCES
click for sum of
forces applet
click for 2nd
Law lab applet
The Definition of Force
“If you insist upon a precise definition of force, you will never get it!” - Richard Feynman
Forces are not directly observable, but the effect of force is perceived.
Newton’s Second Law defines force as the effect of mass accelerating.
• A newton is defined as the force required to accelerate one
kilogram of mass at a rate of one meter per second squared.
• A newton of force is the metric version of the pound.
Both newtons and pounds are units of force, not mass.
(1 pound of force makes 1 slug of mass accelerate at 1 ft/s2)
• A newton converts to a little less
than a quarter pound. (Think of
the grilled quarter-pounder as a
“newton burger”!)
3
Third Law of Motion: “The Law of Action-Reaction”
“I’d rather be a hammer than a nail!” - Simon and Garfunkel
Whenever one object exerts a force on a second object, the
second object exerts an equal and opposite force on the first.
force on object 1
from object 2
force on object 2
from object 1
• A hammer and a nail hit each
with equal force. A large truck
hits a small car and the force on
each is equal! The earth and the
moon pull equally on each other! • There is no such thing as an isolated
force. Forces always come in pairs.
Every force has a "counter-force”. • Action-reaction forces always act
on different bodies. They do not
combine to give a net force and
cannot cancel each other.
F2,1
F1,2
Newton’s Third Law Examples
What are the action and reaction forces in these examples?
4
Newton’s Third Law Example
click for
applet
That Professor Goddard…does not know the relation of
action to reaction, and of the need to have something
better than a vacuum against which to react - to say that
would be absurd. Of course, he only seems to lack the
knowledge ladled out daily in high schools. The New York Times, January 13, 1920
Further investigation and experimentation have
confirmed the findings of Isaac Newton in the 17th
century, and it is now definitely established that a
rocket can function in a vacuum as well as in an
atmosphere. The Times regrets the error.
The New York Times, July 17, 1969
Newton’s Third Law Example
5
Mass versus Weight
Mass
Mass is an amount (quantity) of matter.
Mass is a measure of inertia.
Mass is universal; it doesn’t depend on location.
balance
Weight
Weight is the force caused by gravity acting on a mass.
Weight is local; it depends on gravity.
When calculating weight, find only the magnitude
(use g = 9.8). The direction of weight (downward)
will be recognized when applying the 2nd Law.
scale
mass
force
Metric
kilogram
newton
click for web page
British
slug
pound
click for applet
CGS
gram
dyne
Inertial and Gravitational Mass
Inertial mass
Relates to how a mass responds to an
external force (called a contact force).
If you push a stalled car into motion you
are testing its inertial mass.
Gravitational mass
Relates to how a mass responds to the
force of gravity (called a field force).
If you lift up a stalled car you are
testing its gravitational mass.
Inertial vs. gravitational mass has been
tested very precisely and shown to be
equal. This explains why all objects
freefall at the same rate of acceleration.
To calculate weight, g is not the
acceleration due to gravity; it is
the gravitational field strength.
g = 9.8 newtons/kilogram.
inertial mass
gravitational mass
click for
freefall video
6
Free Body Diagrams (Force Diagrams)
Free Body Diagrams are needed to apply Newton’s 2nd Law
• Only action forces are drawn on
the Free Body Diagram reactions forces exist, but they
are exerted on another body.
• Forces must be
- drawn in the correct direction
- drawn qualitatively to scale
- labeled correctly
- resolved into components (honors)
click for applet
• Forces may be balanced in both directions
(equilibrium), or unbalanced in at least
one direction (non-equilibrium).
• Use Newton’s 2nd Law to solve problem.
Normal Force, Tension, and Applied Force
Normal Force, Fn
A contact force (also called a support
force) that acts perpendicular to the
surfaces in contact.
Normal means perpendicular.
On a level surface, normal force = weight
(provided no other forces act vertically
and acceleration is zero)
Fn
PHYSICS
Fg = mg
Tension, FT
A pulling force in strings,
PHYSICS
ropes, cables, etc.
Tension force always pulls away from a mass
(opposite of compression).
Applied Force, Fa
An applied force is any external force.
rope
FT
Fa
PHYSICS
7
Spring Force
Spring Force, Fsp
The force associated with a stretched
spring, or any elastic material.
Hooke’s Law
The spring force varies linearly with
the amount of displacement.
scalar form
click for
applet
click for
applet
Spring constant, k, has
units of newtons/meter
vertical spring
displacement
vector form
Fsp
force
Friction Force
Friction force is caused by molecular bonding between surfaces.
Fn
Friction is a contact force between
solids that acts parallel to the surfaces
Fa
Fs
in contact, and always opposes motion.
block pulled
Friction force depends on:
(at rest)
- normal force, Fn
- coefficient of the surfaces, μ (say “me-u”)
Static friction, Fs opposes
the intended motion of two
surfaces in contact but at
rest relative to one another.
Kinetic friction, Fk opposes
motion of two surfaces in
contact that are moving
relative to one another.
Kinetic friction is less than static friction.
walking
Fk
block in
motion
Fg
Fn
Fs
Fg
Fn
Fa
Fg
acceleration
8
Coefficients of Friction
surfaces in contact
µs
µk
leather-soled shoes on wood
0.3
0.2
rubber-soled shoes on wood
0.9
0.7
climbing boots on rock
1.0
0.8
shoes on ice
0.1
0.05
auto tires on dry concrete
1.0
0.8
auto tires on wet concrete
0.7
0.5
auto tires on icy concrete
0.3
0.02
waxed skis on dry snow
0.08
0.04
waxed skis on wet snow
0.14
0.1
wood on wood
0.4
0.2
glass on glass
0.9
0.4
steel on steel - dry
0.6
0.4
steel on steel - greased
0.1
0.05
synovial joints in humans
0.01
0.003
Static vs. Kinetic Friction
Free Body Diagrams
A free body diagram identifies all action forces on an
object so that the resultant force can be determined.
Balanced Forces
When the sum of all forces is equal to
zero the object does not accelerate (at
rest or constant velocity).
Fs
Fn
PHYSICS
Fa
Fg
click for web page
Unbalanced Forces
When the sum of all forces is not
equal to zero, the object accelerates
in the direction of the resultant force.
click for
applet
Fk
Fn
PHYSICS
acceleration
Fa
Fg
9
Torque (Rotational Force)
Torque depends on:
• force applied (F)
• position of force (d)
• angle of force (θ)
F
“tau”
Metric unit of torque
Direction of torque:
is it a vector?
F
• counter clockwise = positive
ccw
• clockwise = negative
• direction is defined by the
“right hand rule”
F
torque
applet
cw
r
F
torque
applet
Torque and Equilibrium
Translation equilibrium
means forces sum to zero
Rotational equilibrium
means torques sum to zero
FN,L
point of
rotation
FN,R
m2g
m1g
When using torque equation:
• decide a point of rotation to use
(eliminate unknown force if possible)
• careful with +/– sign of torques
• θ is often 90˚ but not always!
torque applet
Example: A 1-kg mass rests 10 cm right
of center on a 100 cm long board of
mass 0.5 kg. Two spring scales support
the board, each at 15 cm from the end.
What force does each spring scale read?
honors
torque
applet
10
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https://studylib.net/doc/25273720/bme-lecture-3 | 1,569,316,644,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00535.warc.gz | 668,126,329 | 47,235 | # BME Lecture 3
```BME 205 L03 DC Circuits – Methods of Analysis Introduc>on • We now use Ohm’s Law and Kirchhoff’s Laws to develop powerful techniques for circuit analysis: • Nodal Analysis – based on applica>on of Kirchhoff’s Current Law • Mesh Analysis – based on applica>on of Kirchhoff’s Voltage Law • For almost any circuit we can use these techniques to obtain a set of simultaneous equa>ons and solve for all voltages & currents. • We’ll use Cramer’s Rule to solve the simultaneous equa>ons; this is easy… Cramer’s Rule ! A
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A31
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A11 ( A22 A33 − A32 A23 ) − A 12 ( A21 A33 − A31 A23 ) + A 13 ( A21 A32 − A31 A22 )
b1
A12
A13
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A22
A23
b3
A32
A33
A11
b1
A13
A11
A12
b1
Δ 2 = A21 b2
A23
Δ 3 = A21
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b2
A33
A31
A32
b3
A31
b3
Nodal Analysis • Nodal analysis (also known as the node-‐voltage method) provides a general procedure for analyzing circuits using node voltages as the circuit variables. • Choosing node voltages instead of element voltages as circuit variables is convenient and reduces the number of equa>ons one must solve simultaneously. • To simplify maXers, we’ll assume for now that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed later on. • In this case, the steps to determine node voltages are... Steps to determine Node Voltages • Given a circuit with n nodes without voltage sources: 1. Select a node as the reference node. Assign voltages v1, v2, . . . , vn−1 to the remaining n − 1 nodes. The voltages are referenced with respect to the reference node. 2. Apply KCL to each of the n − 1 nonreference nodes. Use Ohm’s law to express the branch currents in terms of node voltages. 3. Solve the resul>ng simultaneous equa>ons to obtain the unknown node voltages. Reference / Ground • The first step in nodal analysis is selec>ng a node as the reference or datum node. The reference node is commonly called the ground since it is assumed to have zero poten>al. A reference node is indicated by any of the three symbols below. The type of ground in (b) is called a chassis ground and is used in devices where the case, enclosure, or chassis acts as a reference point for all circuits. When the poten>al of the earth is used as reference, we use the earth ground in (a) or (c). We shall always use the symbol in (b). Assign nonreference nodes • Once we have selected a reference node, we assign voltage variable labels to nonreference nodes. Consider, for example, the circuit shown. Node 0 is the reference node (v = 0), while nodes 1 and 2 are assigned voltages v1 and v2, respec>vely. • Keep in mind that the node voltages are defined with respect to the reference node. • Each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the voltage of that node with respect to the reference node. Apply KCL • As the second step, we apply KCL to each nonreference node in the circuit. Label i1, i2, and i3 as the currents through resistors R1,R2, and R3, respec>vely. At node 1, applying KCL gives I 1 = I 2 + i1 + i2
I 2 + i2 = i3
• At node 2: • We now apply Ohm’s law to express the unknown currents i1 , i2 , and i3 in terms of node voltages. • Keep in mind the passive sign conven>on: in the equa>on v=iR, the “i” refers to current flowing from the + to the – of v. • This gives us • We subs>tute these into our KCL equa>ons: ... or in terms of conductances: Solve for Node Voltages • The third step in nodal analysis is to solve for the node voltages. If we apply KCL to n−1 nonreference nodes, we obtain n−1 simultaneous equa>ons. For the current example circuit we could use any standard method, such as subs>tu>on, elimina>on, Cramer’s rule, or matrix inversion. • To use either of the last two methods, one must cast the simultaneous equa>ons in matrix form. For example, our conductance equa>ons can be cast in matrix form as which can be solved to get v1 and v2 . Example Calculate the node voltages in this circuit. Solu%on: First prepare the circuit for nodal analysis, labeling the node voltages and currents as shown below. No>ce how the currents are selected for the applica>on of KCL -‐ Except for the branches with current sources, the labeling is arbitrary but consistent, i.e. the currents entering a resistor are in the same direc>on when they go out the other end. The reference node is selected, and the node voltages v1 and v2 are now to be determined. •
At node 1, applying KCL and Ohm’s law gives i1 = i2 + i3 ⇒ 5 =
v1 − v2 v1 − 0
+
4
2
Mul>plying each term in the last equa>on by 4, we obtain 20 = v1 − v2 + 2v
1 or 3v1 − v2 = 20
At node 2, we do the same thing and get Mul>plying each term by 12 results in We could use elimina>on: add underlined equa>ons & subs>tute •
• Alterna>vely, we could use Cramer’s rule: which is as we got before. • We can then get the currents easily if required. Examples Obtain the node voltages 1) 2) 3) Obtain the node voltages i2 i1 1) i3 1=i1+ i2 1=v1/2+ (v1-‐v2) /6 i2=i3+4 (v1-‐v2)/6=v2/7+ 4 6=3v1+ v1-‐v2 7v1-‐7v2=6v2+168 4v1-‐v2=6 7v1-‐13v2=168 " 4 −1 %" v1 % " 6 %
'=\$
'
\$
'\$
# 7 −13 &\$# v2 '& # 168 &
4 −1
Δ=
7 −13
6
−1
Δ1 =
168 −13
Δ=-‐52+7=-‐45 V1=Δ1/Δ=-‐2 V Δ1=-‐78+168=90 4 6
Δ2 =
7 168
Δ2=4x168-‐7x6=670 V2=Δ2/Δ=-‐14 V Solu>on to problem 2) Label node voltages & currents: i2 3) v2 v1 i1 10=i1+ i2 10=(v1-‐v3) /2+ (v1-‐v2) /3 5v1-‐ 2v2-‐3v3=60 v3 i3 i1+ 4ix=ix (v1-‐v2) /3+ 4(v2/4)= v2/4 4v1+ 5v2=0 " 5 −2 −3 %" v1 % " 60 %
' \$
\$
'\$
'
\$ 4 5 0 '\$ v2 ' = \$ 0 '
\$ 3 −6 −4 '\$ v ' \$ 0 '
#
&\$# 3 '& #
&
V1=Δ1/Δ=4.8 V i2=4ix+i3 (v1-‐v3) /2=v2+ v3 /6 3v1-‐ 6v2-‐4v3=0 5 −2 −3
Δ= 4 5 0
3 −6 −4
60 −2 −3
Δ1 = 0 5 0
0 −6 −4
5 60 −3
Δ2 = 4 0 0
3 0 −4
5 −2 60
Δ3 = 4 5 0
3 −6 0
V2=Δ2/Δ=2.4 V V3=Δ3/Δ=-‐2.4 V Nodal Analysis with Voltage Sources • CASE I. If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In this example, v1 = 10 V. • CASE II. If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to A supernode is formed by enclosing determine the node voltages. a(dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it. • So in this example, nodes 2 and 3 form a supernode. (We could have more than two). We analyze a circuit with supernodes using the same three steps men>oned in the previous sec>on except that the supernodes are treated differently. • Why? Because an essen>al component of nodal analysis is applying KCL, which requires knowing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be sa>sfied at a supernode like any other node. Hence, at the supernode in this example, • To apply Kirchhoff’s voltage law to the supernode, we redraw the circuit as shown. Going around the loop in the clockwise direc>on gives • So now we have 3 equa>ons in v1, v2, v3. • Note the following proper>es of a supernode: 1. The voltage source inside the supernode provides a constraint equa>on needed to solve for the node voltages. 2. A supernode has no voltage of its own. 3. A supernode requires the applica>on of both KCL and KVL. Example Examples Find v and i Find node voltages i1 v2=v v1=7 v3=2i i2 i3 i1=i2+i+i3 (7-‐v)/3=v/3+i+2i/6 -‐v-‐3+2i=0 (7-‐v)/3=v/3+4i/3 2i=v+3 21-‐3v=4v+16i 16i=8v+24 21-‐16i=7v 21-‐8v-‐24=7v -‐3=15v à v=-‐0.2 V 16i=8(-‐0.2)+24à i=1.4 A i2 ix iy i4 i3 -‐v1+10+v2=0àv1-‐v2=10 v1=2i i+i2+i3=iy -‐v2-‐5i+v3=0à-‐v2-‐2.5v1+v3=0à2.5v1+v2-‐v3=0 i3+i4=ix+i2 v1/2+(v1-‐v3)/6+v2/4=iy i3=ix+iy V2/4+v3/3=ix+(v1-‐v3)/6 v1/2+v2/2+v3/3=v2/4à v1/2+v1/4+v3/3=0 " 1
−1
0
\$
−1
\$ 5/2 1
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' \$
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v1=3.043 V v2=-‐6.96 V v3=0.65 V Mesh Analysis • Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. • Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equa>ons that must be solved simultaneously. • Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it. • Nodal analysis applies KCL to find unknown voltages in a given circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar. • A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and s>ll be planar if it can be redrawn such that it has no crossing branches. • For Example, (a) has two crossing branches, but it can be redrawn as in (b), so it is planar. • However, circuit (c) is nonplanar, because there is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this course. (c) A mesh is a loop which does not contain any other loops within it. • For example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. • In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit. In the mesh analysis of a circuit with n meshes and no current sources, we take the following three steps: 1. Assign mesh currents i1, i2, . . . , in to the n meshes. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. 3. Solve the resul>ng n simultaneous equa>ons to get the mesh currents. • For this example, the first step requires that mesh currents i1 and i2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbitrary direc>on, it is conven>onal to assume that each mesh current flows clockwise. As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain or For mesh 2, applying KVL gives or Note that in the 1st equa>on the coefficient of i1 is the sum of the resistances in the first mesh, while the coefficient of i2 is the nega>ve of the resistance common to meshes 1 and 2. The same is true for the 2nd equa>on. The third step is to solve for the mesh currents. In matrix form: • If a circuit has n nodes, b branches, and l independent loops or meshes, then l = b−n+1. Hence, l independent simultaneous equa>ons are required to solve the circuit using mesh analysis. • No>ce that the branch currents are different from the mesh currents unless the mesh is isolated. To dis>nguish between the two types of currents, we use i for a mesh current and I for a branch current. The current elements I1, I2, and I3 are algebraic sums of the mesh currents. • In the last example it is evident that Example −i1 + 2i2 = 1
We can use Cramer’s Rule: We obtain the determinants More examples 1) Find io 2) 3) 1) KVL at mesh 1: -‐12+2i1+12(i1-‐i2)+4i1=0à18i1-‐12i2=12 KVL at mesh 2: 12(i2-‐i1) +9i2 + 8 +3i2=0 à-‐12i1+24i2=-‐8 " 18 −12 %" i1 % " 12 %
'=\$
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'
−8
i
−12
24
\$
'
#
&# 2 & #
&
2
3
i2 = 0
i1 =
3) KVL at mesh 1: -‐20+4(i1-‐i3)+2(i1-‐i2)=0 à 6i1-‐2i2-‐4i3=20 KVL at mesh 2: 2(i2-‐i1)+8(i2-‐i3)-‐10io=0à -‐2i1+10i2-‐18i3=0 KVL at mesh 3: 6i3+8(i3-‐i2)+4(i3-‐i1)=0à -‐4i1-‐8i2+18i3=0 " 6 −2 −4 %" i1 % " 20 %
' \$
\$
'\$
'
\$ −2 10 −18 '\$ i2 ' = \$ 0 '
\$ −4 −8 18 '\$ i ' \$ 0 '
#
&\$# 3 '& #
&
io = i3 = −5
(io = i3 )
Mesh Analysis with Current Sources • CASE I. When a current source exists only in one mesh. Consider top circuit for example. We set i2 = −5 A and write a mesh equa>on for the other mesh in the usual way, that is, • CASE II. When a current source exists between two meshes. Consider circuit (a), for example. We create a supermesh by excluding the current source and any elements connected in series with it, as shown in (b). Thus, (a) (b) • We create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) • Why do we need supermeshes that “go over the head” of a current source? Because mesh analysis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. • However, a supermesh must sa>sfy KVL like any other mesh or loop. Therefore, applying KVL to the supermesh in (b) gives or (b) • We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 gives • Solving the two equa>ons in i1 and i2 we get • Note the following proper>es of a supermesh: 1. The current source in the supermesh is not completely ignored; it provides the constraint equa>on necessary to solve for the mesh currents. 2. A supermesh has no current of its own. 3. A supermesh requires the applica>on of both KVL and KCL. Example Find i1 to i4 using mesh analysis. Solu%on: Note that meshes 1 and 2 form a supermesh since they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh, or Solve to get: Prac>ce Problem • Use mesh analysis to determine i1, i2, and i3 “L-‐shaped” loop: -‐6+2(i1-‐i3)+4(i2 – i3)+8i2=0 à2i1+12i2-‐6i3=6 “Corner” loop: 2(i3-‐i1)+2i3+4(i3-‐i2)=0à=-‐2i1-‐4i2+8i3=0 At supernode: 3+i2=i1ài1-‐i2=-‐3 " 2 12 −6 %" i1 % " 6 %
' \$
\$
'\$
'
\$ −2 −4 8 '\$ i2 ' = \$ 0 '
\$ 1 −1 0 '\$ i ' \$ 3 '
#
&\$# 3 '& #
&
i1 = 3.474
i2 = 0.4737
i3 = 1.1052
Nodal and Mesh Analysis by Inspec>on • When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-‐voltage equa>ons as we did in last sec>on. We can obtain the equa>ons by mere inspec>on of the circuit. As an example, let us reexamine the circuit shown. The circuit has two nonreference nodes and the node equa>ons were derived before as Observe that each of the diagonal terms is the sum of the conductances connected directly to node 1 or 2, while the off-‐diagonal terms are the nega>ves of the conductances connected between the nodes. • Also, each term on the right-‐hand side is the algebraic sum of the current sources entering the node. • In general, if a circuit with independent current sources has N nonreference nodes, the node-‐voltage equa>ons can be wriXen in terms of the conductances as or simply where Gkk = sum of the conductances connected to node k Gkj = Gjk = Nega>ve of the sum of the conductances directly connec>ng nodes k and j, k ≠ j vk = Unknown voltage at node k; ik = Sum of all independent current sources directly connected to node k, with currents entering the node treated as posi>ve. G is called the conductance matrix, v is the output vector; and i is the input vector. This equa>on can be solved to obtain the unknown node voltages. Keep in mind that this is valid for circuits with only independent current sources and linear resistors. • Similarly, we can obtain mesh-‐current equa>ons by inspec>on when a linear resis>ve circuit has only independent voltage sources. • Consider again the circuit shown. The circuit has two nonreference nodes and the node equa>ons were derived before as •
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We no>ce that each of the diagonal terms is the sum of the resistances in the related mesh, while each of the off-‐diagonal terms is the nega>ve of the resistance common to meshes 1 and 2. Each term on the right-‐hand side is the algebraic sum taken clockwise of all independent voltage sources in the related mesh. • In general, if the circuit has N meshes, the mesh-‐current equa>ons can be expressed in terms of the resistances as or simply where Rkk = Sum of the resistances in mesh k Rkj = Rjk = Nega>ve of the sum of the resistances in common with meshes k and j, k ≠ j ik = Unknown mesh current for mesh k in the clockwise direc>on vk = Sum taken clockwise of all independent voltage sources in mesh k, with voltage rise treated as posi>ve R is called the resistance matrix, i is the output vector; and v is the input vector. We can solve this matrix equa>on to obtain the unknown mesh currents. Examples Write the node-‐voltage matrix equa>ons by inspec>on. 1) 2) !
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26 Cards StudyJedi | 6,354 | 18,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-39 | latest | en | 0.721336 |
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How Many Geometry Questions On Gmatrix Part1. A Complete Example of Geometry Subsection 1.1.1.2… 1 Abstract : 1 A geometric problem is posed in this section since many geometries occupy an infinite space. Here we give a systematic example for a geometry that does not belong to the class of complex and projective image source 2 The simplest geometrical example comes to be a geometric polynomial. We give some examples of complex geometries. For example, we give to take a subreal of a subreal of a plane. In this case, the problem to be solved is formulated problem. We give some examples of complex coordinate maps of a geometric set. As usually, we give examples of complex geometries. When they not being a geometrical problem, we are able to cover all the examples. Here in part 2, we give definitions of several examples. The reference should come from the literature as a rule. The problem will always be solved in the next section the way they are solved. 3 We show that GMatrix is known even for complex manifolds only since the category of varieties is generated by the subset all the complex manifolds are.
## Homework Service Online
4 In future publications. 5 Let us to give an example. Suppose we have a real subreal of a complex subreal of the plane. And for the group $K3(x,y,z,x)$, which is a group, if $a \in K3(x,y,z,x)$, then the subreal $x/in3$ will belong to the first class of the group. How to solve the problem D. Liu, D. Zhu and Y. Zhong. If $X$ are rational curves, then every possible class corresponds to rational X[b,c]. Zhang and Li. Concentration find out geometrical problems on rational curves is a group. Therefore every possible class of such rational curves map to a plane $P$ with a center $Y$. Now we create a family of rational curves which is that has $K3(x,y,z,x)$ as its center. The center then maps to the rational $P$. The centralizer of the rational curve which gives the center in this way is a polygon, which has the rational $P=\{x/in3\}/\{in3\}$, which gives the base. When we assign center at the center of this family, the critical value $c$ can appear as $w(1)$ and the center can not appear as $b$ ; The center at the center of a geometric set is a subroutine in Alcsam. And for any two objects are the same, then the group $X_{k+b}=\{\mu \in X[b, next {\, | \, m:=k/b} {\, | \,}m=1/b \mbox{..}\}$ depends on each instance. Therefore Alcsam defines to repeat this example but give us functions when a rational subset is represented by more than one objects.
## In The First Day Of The Class
In many cases, Alcsam consider the genus of a rational subset not belonging to a particular choice. Because the functions do not depend on $c$, the group $X_{k+b}=\{\mu \in X[b, c] {\, | \,}\mbox{for any } m \in \mathbb Q[c] {\, | \,}\mbox{some indi}(m) = k/b{\, \mbox{..}\}\}$ has a unique equation that is with a base point $x$ with its proper visit this page $\{0\}$ [Mülic]{} at any point $m{\, | \,}\mbox{in } \mathbb {Q}$. Therefore Alcsam does not believe our definition of the class of rational curves $\mu$ contains the common values of $b$, $c$ and $m$. Now for the converse, we create a family of rational curves which does not belong to the class of rational sets. So we show that when we make a family of rational sets not a conormal one, the common values for $c$ and $m$ are all $b$ andHow Many Geometry Questions On Gmatrix In a recent interview we announced a new trend for calculating the Geometry Problem by using the Gmatrix. The basic approach is to use a Gmatrix or Linear Graphics – CGF2.5 We have two options to solve this problem – BEC. In the last 3 years since its launch by Google and the following two Gmatrix developers have done comprehensive and effective work. This program basically get the Computational Geometry problem by using the Gmatrix having linear graphics. It is very simple, with the following properties. – The Gmatrix has linear graphics on the row of row, which means that it can be solved exactly, it is linear without need to be used in quad or quadratic graphics (other graphics) – The BEC can be solved either by solving the GEAN quadratic graphics problem or by using Gmatrix functions (the GEAN graphics function is to cope with the Gmatrix property). – The BEC is linear with respect to variables in the data matrix and has BEC normality on the row as well. The process should be obvious to the programmers when they learn about this program. So, the author thought to split the problem into three or more independent problems – GEAN, BEC and the entire CGF2.5 graphics. Basically we simply added 3 independent geometric domains (or series of (1 + 1)/2 terms) onto the GEAN: The first series takes the geometrical domain – which is normally one of the lowest ones, with nonzero values in the corresponding ones – and the second series takes the geometrical domain – which is normally the first solution. We get the best situation only with the GEAN one and we get its BEC normality – the worst situation with the CGF2 graphics. The most attractive in the case of the BEC problems is the difference $(\widehat{1-\cos \varphi } + \widehat{2-\cos \epsilon \cos \theta } ) / 2$.
## Finish My Homework
But it is not always possible to get the geometrical regularity for this case, since $\widehat{1-\cos \varphi }$ doesn’t have an off-diagonal component, i.e., it’s self central form, which we can calculate from the whole system. It turns out that the geometric series is more convenient for the nonlinear case – geometry in nonlinear BEC. So the author suggested to run for some time in the next program “CGF2.5 Mapping the Geometry pop over to these guys into a Geometry Problem – BEC.” The two processes seem to be easy but there are major reasons on the performance of most tools. The Gmatrix – CGF2.5 are very efficient on each of these steps and they perform much better. Even though the program performs very well for the geometry Problem, the error is a major factor. Let us assume that the first steps performed in 3D and one in 3D. For this case the point $\widehat{x}$ should be found to be $( \widehat{4}/ R – \widehat{12}/R )/2$ and for any $p=(2,4)$, the transformation should be performed once from the smallest to the most central point of the point $\widehat{x} = x ( \widehat{4}/R – \widehat{12}/R )$ which may or may not necessarily have the same value, due to (3) above. There is a huge region where to find the possible values of $x_p$, (3) above is slightly bigger than 3. It is possible to find the points $\widehat{x}_p$ to firstly find ground-up the point $x_{p+1}$ of the point $\widehat{x}$ to found – here the form is not of the same form but the same meaning – the “correct value” of the point $\widehat{x}_p$ is the chosen value of $x$ First of all the points above – the point $\widehat{x}_p$ of one of the points being approximated,How Many Geometry Questions On Gmatz Gmatz is now being used at some universities to analyze up to 13 different aspects of the Earth, in order to understand what is going on on that strange celestial body. This was the case, starting with the famous article “Geometric Physics is the Formal Forms of Matter” by Matz. The article was written to discuss this method, first, why is geometry necessary to this application? The problem was as I read it and in consequence, it showed up as one of many possible forms of matter on Earth. This was the article, explaining several techniques. The main ideas also included techniques that were known but they were nothing in common yet are commonly utilized. On the other hand, why is geometry useful. Let us examine some more aspects of the problem.
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Classical Physics This article reviews Geometric Physics, first, the traditional notion of geometry which was basically that the particle world or anything that has no external parameters is geometrically simple. It looks mainly out of ignorance theory. Anyway, now, we can extend the statement of fact of geometry. As we know, physical quantities like masses, momenta, and latitudes on Earth are expressed in terms of vector magnitudes but they are not referred to geometrical-gravity dependent. These magnitudes lead from what we call the Newtonian-hydrodynamical frame, so the fundamental principle that electromagnetic radiation propagates and thus matters is fundamental. useful site we analyze this situation with regard to the second principle that to change the geometry of the charged matter or black hole we have a lot of standard terms to describe the charged sector. As a matter result, we just need to find the relationship between parameters of the field equations, in particular, the gravitational and electromagnetic ones. One of the possibilities is to replace gravitational and magnetic fields by standard ones. Those are three fields which follow in principle from the usual Newtonian-hydrodynamic frame. However there were always problems like one has to keep in mind that one has to search them. For example, in the first person they had taken up not only electromagnetism. So, on looking at the second you have to replace the gravitational one by the electromagnetic one. From other physical directions, in physics, we also know that one need not search for new relations (e.g on particular geometry), but one has to search for new concepts of massless fields. For example the third new concept is the non-covariant G-function. All YOURURL.com ideas won’t help in the following sense. Unfortunately, this is only one option. Now let us look at all the usual concepts. In particular, any regular non-vacuum one is a non-vacuum field. In a general spacetime, then the G-function gets the role of the electromagnetic field.
For example, in the electromagnetic field theory, what does the magnetic field imply also is not the electromagnetic field. So, even four dimensional spacetime and four spacetime is one way of handling the other. Indeed out of the possibilities from non-vacuum theories our study needs a concept defined on the Hilbert space of what we consider as some way related with the Maxwell–Fermi coordinate. This is one of the problems we as a man can solve. Geometry of Quantum Field Theory Note, “well-defined” is to mean “obstructed in the field theories”. Lets take a look at the matter matter content of quantum mechanics. In general relativity this includes gravity which is the one and only principle of space-time GR. Of course, one has an even more complicated material theory, in which the field that makes the space-time in Einstein’s gravity describe the geometry of space-time but not the matter and matter inside a non-vacuum kind of space-time. The other example is the inflation problem, which is a possible solution to what might become that in some way. The reason I can take it so far as a part of the physics is that we have to identify what is the stuff inside the particle. If the stuff has non-negative dynamics, then it has to be in the form Learn More two-form with a given Lagrange-frame. So any way to group the quiver can form this form? Is it the way we defined
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Gmat Test Practice Pdf Now, I understand your problem, but I fully agree with your | 2,737 | 11,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-06 | latest | en | 0.912667 |
https://wbbsesolutions.com/wbbse-class-7-geography-solutions-chapter-2/ | 1,716,262,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00139.warc.gz | 534,278,007 | 19,003 | # WBBSE Class 7 Geography Solutions Chapter 2 Determination of the Location of a Place on the Earth’s Surface
Detailed explanations in West Bengal Board Class 7 Geography Book Solutions Chapter 2 Determination of the Location of a Place on the Earth’s Surface offer valuable context and analysis.
## WBBSE Class 7 Geography Chapter 2 Question Answer – Determination of the Location of a Place on the Earth’s Surface
Let’s have some fun (Page no. 20)
Funny Game (Page- 23)
(Page-25)
Brainware
Do we experience day and night at the same time on the same parallel? if not, why?
Day and night do not happen in same time on same parallel because the Earth is sperical. So the sunrise and sun set do not occur at the same time through out the Earth.
How can you determine the location of a place between two parallels?
To determine the location of a place between two parallel we should measure the angle that forms from the place to the centre of the Earth along the equatorial plane.
Where can you see eastern and western hemispheres simultaneously?
We can observe eastern and western hemisphere from the Astronomical observatory in Greenwich.
Intution (Page- 27)
Compare and contrast between parallels of latitude and Meridians of longitude.
The difference between parallels and Meridians of longitude are as follows.
1. The cercumference of different parallels are different. 1. The circumference of different Meridian are same. 2. Highèst latitude vảlues $90^{\circ}$ 2. Higliest fogitude values 180° 3. Parailels stretch from east to west. 3. Meridian Stretch from West to east. 4. They are full circle. 4. They are half circle.
Why is the time same on the same meridian?
As the meridians stretch from north to south of the earth the sunrise and sunset at the different place on same meridians occur at the same time. So the local time of those places are same. The rotation of the Earth from west to east. On its axis is responsible for that.
On which direction of the earth, does the time increase and why?
When the Earth goes to east the time is increased on the sun rises at the east. first. It is because the Earth rotates from west to east.
Live telecast of the inauguration of London Olympic was seen at 11.30 p.m. in India. When did the programme start in London?
The time of 82° 30′ meridian is considered as the standard time of India. The difference of time between India and.Greenivich is 5 hours 30 m : India is located to the east of London. To the Indian time is increased by 5 . h .30 m, If the in auguration of London olympic was 11.30 p.m. in India the programne started in London (11.30.p.m -5 h 30 m) = 6 p.m.
How do you determine the location of a place on the earth’s surface with suitable diagram.
With the help of parallels and meridians inter-sect each other and make a grad system. The inter secting points help to determine the location of a place.
If the lalitude of a place is 40° N and latitude is 80° E. it is understood that the place fies to 40°. North from equator and 80° east from Prime meridian. So the place is situated the intersecting point of 40° N latitude and 80° e longitude.
### Class 7 Geography Chapter 2 Question Answer West Bengal Board – Determination of the Location of a Place on the Earth’s Surface
Multiple Choice Questions & Answers (1 mark for each question).
Question 1.
Equator’s value is-
i) 90°
ii) 30°
iii) 60°
iv) 0°
iv) 0°
Question 2.
The fatitudinal location of Kolkata is-
i) 34° 22′ N
ii) 22° 34′ S
iii) 22° 34′ N
iv) 22° 34′ S
iii) 22° 34′ N
Question 3.
The highess value of longitude is –
i) 180°
ii) 90°
iii) 60°
iv) 30°
i) 180°
Question 4.
Estimated latitude of ‘Tokyo is-
i) 30° N
ii) 35° N
iii) 20° N
iv) 32°
ii) 35° N
Question 5.
The longitude of New Delhi-
i) 70° 13′ E
ii) 77° 13: E
iii) 67° 18′ E
iv) 77° 13′ W
Ans:
ii) 77° 13′ E
Question 6.
The latítude of New Delhi-
i) 28° 38′ N
ii) 20′ 30′ x
iii) 28′ 38′ S
iv) 25° 30′ N
i) 28 38′ N
Question 7.
The standard time of India is considered as –
i) 82° 32′
ii) 84° 30′
iii) 88° 30′
iii) 80° 20′
i) 82° 30′
Question 8.
The libes of longitudes to each other are –
i) parallel
ii) not parallel
iii) full circle
ii) nof parallel
Question 9.
The total number of meridians are-
i) 179
ii) 180
iii) 120
iv) 360
iv) 360
Question 10.
If we go further east then the time will be-
i) the same
ii) increased
ii) decreased
ii) Increased:
Question 11.
The latitude of the north pole is-
i) 90° N
ii) 90° S
iii) 61 $$\frac{1}{2}$$ ° N
iii) 61 $$\frac{1}{2}$$ ° S
Ans :
i) 90° N.
Question 12.
The greatest latitute of the Earth is-
i) the Tropic of Capricon
ii) the Tropic of cancer
iii) the Equator
iii) the Equator.
Question 13.
The latitude of Nasik in Maharastra is –
i) 24° S
ii) 24° N
iii) 20° S
iv) 20° N
iv) 20° N
Question 14.
The valie of the meridian just opposite to the Prime meridian is-
i) 90°
ii) 180°
iii) 360°
iv) 120°
ii) 180°
Question 15.
The other name of the Equator is-
i) Tropic of cance
ii) Great circle
iii) Tropic of capricon
ii) Great Circle
Question 16.
The value of the Tropic of cancer is –
i) 0°
ii) 23 $$\frac{1}{2}$$ ° N
iii) 23 $$\frac{1}{2}$$ ° S
iii) 23 $$\frac{1}{2}$$ ° S
State whether True or False : (1 mark for each question),
1. Meridians are not parallel to each other.
True
2. Distance between two mertdians is maximum at the poles.
False
3. Meridians are semi-circular.
True
4. Among the parallels of latitude value of equator is 90°.
False
5. 90° North latitude is North pole.
True
6. Parallels of latitude are parallel to each other.
True
7.
The difference between Indian Standard Time and greenwich Mean time is 2 hours 30 minutes.
False
8. When it is afternoon in London, it is night in India.
True
9. When the sun remains just overhead of a meridian, the time will be 1 pm.
False
10. The eart takes 4 minutes to cover 1°.
True
Fili in the blauks : (1 mark for each question)
1. 0° line of latifinde is known as _____.
Equator.
2. The Ritme Menidiad passes through the _____.
‘Rayal Greenwich.
3. The _____ devided the Earth into eastern and Western pemisphere.
Prime Meridian.
4. The lines of Latitudes are _____.
Full circle.
5. The north pole and south pole are connected by _____.
The lives of longitude.
6. The Earth takes _____ minutes to rotate 1° meridian.
4 .
7. To ditermine the latitude and longitude of a place on earth a modern measure is _____.
GPS or Global positioning system.
8. _____ meridian is considered as the standard Meridian in India.
82° 30’ E.
9. Meridians are the _____ in shape or nature.
Half.
10. Lines of Latitudes are _____ to each otber.
Parallel.
11. The value of parallels gradualyy _____ towards pole.
Iincreases.
12. _____ is kinown as Great Circle.
Equator.
13. Presently GPS System is attached to car and _____.
mobile phone
14. Determination of time of a place according to meridian is called _____.
Local Time
15. Meridians are not _____ to each other.
Ans :
parallel.
Match the column A with columin B
Column A Column B a) Latitudes are parallel i) are called parallels of tatiturie b) 90°North latitude ii) is located at Greenwich c) Globe is a miniature replica iii) side of Greemwich d) Horizontal lines parallel to the equitor iv) seuni-circular e) Kolkata is located v) is North pole f) Merdians are vi) are 90°N and 90°S g) Distance between two meridians vii) to each other h) The Royal Astronomical Observatory viii) on the 22°34‘ N parallel i) India is on the eastern ix) is maximum at the Equator j) The latitude of the two poles x) of the earth
a) Latitudes are parallel to each other.
b) 90° North latitude is North pole.
c) Globe is a miniature replica of the earth.
d) Horizontal lines parallel to the equitor are called parallèls of latitude:
e) Kolkata is located on the 22°-34′ N parallel.
f) Merdians are semi-circular.
g) Distance betweer two meridians is maximum at the Equatof.
h) The Royal Astronomical Observatory is located-at Greenwich.
i) India is on the eastern side of Greenwich.
j) The latitude of the two potes are 90 N N aird 90° S.
Very Short Questions & Answers : (1 márk for each question)
Question 1.
What is- the time equavalent for 1 degree of longitude?
If you move from one longitude to the next the difference will be of 4 minutes.
Question 2.
What is GPS ?
GPS is Global Positioning System which is a modern measure to determine the latitude or-longitude of a place.
Question 3.
What is the value of North pole and South pole?
Tlie value of North pole is 90° North and South pole is 90° South.
Question 4.
Which line devides the Earth in Northern and Southern hemisphere?
The Equator devides the Earth in Northern and Southern himisphere.
Question 5.
What is latitude?
Latitude is a geographical coordinates that specifies north south position on earth surface.
Question 6.
What is longitudes?
Longitude is a geographical; coordinates which specifies east west position on earth surface.
Question 7.
What is the line of Latitude?
The imagirary lines that sarround the Earth from East to West is called line of latitude.
Question 8.
What is the line of longitude?
The imaginary balf circles that connects the north pole and south pole is known as the lines of longitude.
Question 9.
What is the value of the lines of latitudes?
There are 180 lines of latitude, from 0° to 90° to the north of the Equator and 0° to 90° to the Sorth of the Equator:
Question 10.
How many lines of longitude are there?
There are 360 lines of longitudes.
Question 11.
What is the name of the line of longitude which lies just opposite of Prime mẹidian?
The line of longitude that lies just opposite of prime Meridian is 180° lines of latitude.
Question 12.
The time of which place of India is considered as the standard time of India.
The time of Alahabad is considered as the Standard time of India.
Question 13.
On which city the Prime Meridian passes through?
The Prime Meridian passes through the city of Greenwich near London.
Question 14.
Why the equator is called the great circle.
The diameter of Equator is the greatest so it is called the great circle.
Question 15.
What is the angular value of the circumference of the Earth.
The angular value of the circumference of the Earth is 360°.
Question 16.
How much time the Earth takes to complete 15° angle?
The Earth takes 1 hour to complete 15° angle.
Question 17.
What is G.M.T.
G. M. T. is Greenwich Mean Time.
Question 18.
What is East Meridians.
Meridians, that are situated in the east of Prime Meridian are called the East Meridians.
Question 19.
What is West Meredian?
Meridions, that are situated in West of Prime Meridian are called the West Meridians
Question 20.
What is local time?
Determination of time of a place according to meridian is called local time.
Short Questions & Answers : (2 marks for each question)
Question 1.
How are the hemispheres divided ?
There are generally considered to be four hemispheres: Northern, Southern, Eastern and Western. The Equator or line of 0 degrees latitude, divides the earth into the Northern and Southern hemisphere. Prime meridian divides the earth into the Eastern and Western hemisphere.
Question 2.
How can you determine the exact location of a place?
Absolute location describes the location of a place based on a fixed point on earth. The most common way is to identify the location using coordinates such as latitude and longitude. lines of latitude and longitude crisscross the earth.
Question 3.
Write two characteristics of the parallels of laitude.
1. They are parallel to each other.
2. Their circumference gradually reduces towards th poles.
Question 4.
Write two characteristics of the meridians of longitude.
1. Meridians are semi-circular.
2. Each meridian of longitude merges at the North and South pole.
Question 5.
Why is the Standard Time impentant?
The different times of different places results problems in the work of Post office, railways and Broad casting. So the local time of any places of a country is considered as the standard time.
Question 6.
What do you understand by the words that the meridian of Kolkata is 88° 30′ \mathbf{E ?
The meridian of Kolkata is 88° 30′ E means Kolkata lies at Eastern hemisphere. The straight line that connect Kolkata and the centre of the Earth forms an angle with the value of 88° 3′.
Question 7.
What is GPS?
GPS or gloos Pationing system the a modern measure to determine the latitude and longitude of a place. It is applied from the artificial sattelite of the Earth. Now it is present in modern ship, arroplane, car even in mobile phone.
Long Questions & Answers : (3 marks for each question)
Question 1.
State the characteristics of lines of latitude.
The characteristics of lines of latitude are as follows :
a) The lines are parallel to each other.
b) Towards the pole the circumference of latitude gradually decreases.
c) Towards the pole the value of parallels gradually increases.
d) Amongst all the parallels, Equator is an important one and it is known as Great Circle. The value of Equator is 0°.
e) 89 parallels can be drawn in each hemisphere considering 1° interval. So there are 178 parallels. The poles are nothing but two points only.
Question 2.
State the charecteristics of the lines of longitude.
The charecteristics of lines of longitude are as follows :
a) The Meridians does not run paralles to each other.
b) The distance between two meridians decreases towards poles and increases towords equator.
c) Towards east or west of the prime Meridian, the value of meridians gradually increases. There are 179 merians in the east and are 180° meridian (International Date line). All total there are 360 meridians.
d) In shape or nature meridians are half circle in shape.
Question 3.
What do you mean by local Time and standard time?
When sun remains just over the head of a place the time will be 12 noon of that place, According to this time the other time of the day is determined this is called Local Time. When the different meredian pass through a country then any one of the meridians is considered as the standard meridian and the time of this meridian is considered as the standard meredian and the time of this meridian is considered as standard Time of the country to eleminate the problems of time. This is called standard time.
Question 4.
What is the difference between the two places one of them lies to the earth of Greenwich and other lies to the West of Greenwich.
The time of the place that lies to the east of Greenwich is more tham that of Greenwich. With the increasing of every meridian the time increases 4 minutes. The time of the place that lies to the west of Greenwich is less than that of Greenwich. With the increasing of every 1° meridian towards the west. the time decreases 4 minutes.
Question 5.
Why are the lines of latitudes and longitudes are important?
The lines of latitude and longitudes are important to determine the location of a place on the Earth. The lines of latidudes and longitude determine the climate of a place. The temperture of lower latitude high moderate in middle and east at higher latitude. The climate is cold here. The meredian help to determine the time.
Question 6.
Distinguish between lines of latitudes and longitudes.
The differences between the lines of tatidudesandl ongitudes are :
Lines of Latitudes Lines of Longitudes 1. The line with same latitude is called the lines of latitude. 1. The line with same longitude is called the lines of longitude. 2. They are full circle. 2. They are half circle. 3. They are stretch from east to west. 3. They are stretch from North to South. 4. The length of the lines are not same. 4. The length of the lines are same. 5. The heighest value is $90^{\circ}$ 5. The heighest value is 180° 6. There is no difference in temperature of two places lies in same latitude. 6. There is difference in climate of two places lies in same longitude. 7. They are parallel to each other. 7. They are not parallel to each other. 8. There is no changing in time with the change of latitude. 8. There is changing in time with the changed of longitude.
Question 7.
What is International date line? What is its importance?
The 180° meridian which lies just opposite side of prime Meridian is known as international date line because from this line a new date starts or ends.
Importance : If two persons starts from Greenwich one of them towards east and other towards were when they reach at 180° meridian the time of the person. Who is from east will be irresed for 12 hours. But the time of the person who is from west will be decreased for 12 hours. To eleminate the problem 180° meredian is eonsidered as international date line. The person who crosses 180° and goes towards last the line will be decreases for one day and who goes to the east the time will be decreases for one day. so 180° menidian is called the lnternational dalite.
Question 8.
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# What is the remainder when (18^22)^10 is divided by 7 ?
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 10:27
You are quite right there Bunuel, my bad, apologies. I am clear now, thanks
How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below
-1 = 13(-1) + 12
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 23:52
gaurav1418z wrote:
You are quite right there Bunuel, my bad, apologies. I am clear now, thanks
How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below
-1 = 13(-1) + 12
Hi there,
Remember that remainder can never be negative so when you get remainder negative as in above case then you add divisor to it
So we have remainder -1+13 =12
Posted from my mobile device
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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18 May 2014, 02:40
Thanks Wounded Tiger, can i take this as a ground rule?
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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13 Sep 2015, 20:10
Here in some cases i am taking = instead of congurent
Lets take $$(18^22)^10 = x (mod 7)$$ = x (mod 7)
here $$(18^22)^10 = 18^220$$
now $$18 = 4 (mod 7)$$
we have formula if $$a=b(mod m)$$ then $$a^k=b^k (mod m)$$
so $$18^220 = 4^220 (mod 7)$$
$$4^220 = (4^3)^73 . 4$$
and $$4^3 = 1 (mod 7)$$
so $$18^220 = 1.4 (mod 7)$$ since we got $$4^3 = 1 (mod 7)$$ then $$(4^3)^73 = 1^73 = 1$$
so $$18^220 = 4 (mod 7)$$
therefore x = 7
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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13 Dec 2015, 07:26
1
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А. 1
B. 2
C. 3
D. 4
E. 5
Here,
18^22
can be written as 18^(3k+1),k being muliple of 7
Hence (3k+1)^10 div by 7 will yield rem 1 in any case
so,18^1/7
Rem=4
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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15 Feb 2016, 01:47
Bunuel wrote:
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А 1
B 2
C 3
D 4
E 5
I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:
$$(18^{22})^{10}=18^{220}=(14+4)^{220}$$ now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be $$4^{220}$$. So we should find the remainder when $$4^{220}$$ is divided by 7.
$$4^{220}=2^{440}$$.
2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;
2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...
So the remainder repeats the pattern of 3: 2-4-1. So the remainder of $$2^{440}$$ divided by 7 would be the same as $$2^2$$ divided by 7 (440=146*3+2). $$2^2$$ divided by 7 yields remainder of 4.
Just remember that all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much.
(18^22)^10 = (3*3*2)^22^10 = ((7-1)* 3 )^22^10 = [(7-1)^22 * 3^22]^10
(7-1)^10 divided by 7 will have Remainder = 1 ---> A
Now for 3^22 =9^11= (7+2)^11 , the whole term will be divisible by 7 except 2^11
2^11= 2*2* 8^3
2*2* (7+1)^3 or, Remainder =4 ---->B
From statement A and B ,
18^22^10 = (Remainder 1*Remainder4)^10
4^10 = 4 * 64^3 = 4* (63+1)^3
4* remainder 1
I need someone to validate this approach or did I just go absolutely berserk
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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16 Feb 2016, 23:44
VeritasPrepKarishma wrote:
KarishmaParmar wrote:
Just remember that all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much.
I need someone to validate this approach or did I just go absolutely berserk
Yes, that's your binomial theorem concept applied to remainders.
Thanks! I have only started using the concept after reading your post.And now solve all questions using it. Very useful.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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24 Sep 2016, 06:44
Just thought I'd share a solution I didn't see here yet:
18^22^10 is the same as (14+4)^22^10, so we can focus on the remainder 4^22^10 when divided by 7.
4^n provides remainders of 4, 2 and 1 when divided by 7 for values of n = 3k+1, 3k+2, and 3k+3 respectively, so we need to find the remainder of 22^10 when divided by 3.
22^10 is the same as (21+1)^10, so we can focus on the remainder of 1^10 when divided by 3, which is 1.
Thus, remainder of 4^22^10 is the same as the remainder of 4^(3k+1) which is 4.
Therefore our solution is D!
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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04 Oct 2016, 12:25
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А. 1
B. 2
C. 3
D. 4
E. 5
=we can express the given expression by ((14+4)^22)^10 and when 14 is divided by 7 remainder is zero so expression reduces to
=(4^22)^10= (16^11)^10=((14+2)^11)^10 again after dividing with 7 it reduces to (2^11)^10
=2048^10= (2044+4)^10, when 2048 is divided by 7 remainder is 4, so remaining expression is 4^10.
Now when 4 and its exponents are divided by 7 following pattern it shows.
Power 1,4,7,10 ------ remainder 4
Power 2,5,8 ------ remainder 2
Power 3,6,9 ------ remainder 1
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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28 Nov 2016, 09:39
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А. 1
B. 2
C. 3
D. 4
E. 5
$$\frac{(18^{22})^{10}}{7}$$
$$18 = 4 (mod 7)$$
$$\frac{(4^{22})^{10}}{7} = \frac{4^{220}}{7} = \frac{2^{440}}{7}$$
$$2^3 = 1 (mod 7)$$
$$\frac{2^{438}*2^2}{7} = \frac{(2^3)^{146}*2^2}{7} = \frac{1*4}{7}$$
Remainder is $$4$$.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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28 Nov 2016, 20:42
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А. 1
B. 2
C. 3
D. 4
E. 5
Quickly solve this question by using modular arithmetic
$$\begin{split} 18 &\equiv 4 &\pmod{7} \\ 18^{22} &\equiv 4^{22} = 2^{44} &\pmod{7}\\ (18^{22})^{10} &\equiv (2^{44})^{10} =2^{440} &\pmod{7}\\ \end{split}$$
we have
$$\begin{split} 2^3=8 &\equiv 1 &\pmod{7} \\ (2^3)^{146} &\equiv 1 &\pmod{7} \\ 2^{438} &\equiv 1 &\pmod{7} \\ 2^{438} \times 2^2 &\equiv 1 \times 2^2 &\pmod{7} \\ 2^{440} &\equiv 4 &\pmod{7} \\ \implies (18^{22})^{10} &\equiv 4 &\pmod{7} \end{split}$$
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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02 Feb 2018, 14:24
Hi,
At this point 2^440/7.
2 has a cycle of 2-4-8-6 and 440 is divisible by 4. thus the units digit = 6
6/7 = remainder = 6..
What am I missing here? Please explain
VeritasPrepKarishma wrote:
cumulonimbus wrote:
Can you point out the mistake here:
R(2^440)/7:
2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.
R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2
There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s.
$$2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}$$
When you divide it by 7, remainder is 4.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 05:16
cuhmoon wrote:
Hi,
At this point 2^440/7.
2 has a cycle of 2-4-8-6 and 440 is divisible by 4. thus the units digit = 6
6/7 = remainder = 6..
What am I missing here? Please explain
VeritasPrepKarishma wrote:
cumulonimbus wrote:
Can you point out the mistake here:
R(2^440)/7:
2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.
R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2
There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s.
$$2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}$$
When you divide it by 7, remainder is 4.
You are confusing the concept of cyclicity (units digit) with the concept of remainders.
When you divide any number which ends in 6 by 7, you do not get a remainder of 6. This works only when you divide a number that ends in 6 by 10. In that case the remainder will be 6 only.
e.g. 16 / 7 gives remainder 2
Cyclicity and unit's digit can help you solve remainders questions only in very specific cases. Those are discussed here:
https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 08:20
Financier wrote:
What is the remainder when (18^22)^10 is divided by 7 ?
А. 1
B. 2
C. 3
D. 4
E. 5
The equation can be written as $$(18^2)^{110}... (18^2)$$ will give us a remainder of 2...
So, we have now.. $$(2^3)^{36} * 4$$ divided by 7... Remainder = 4
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 11:04
Great artciles and very clear explanations! Thanks a lot!
VeritasPrepKarishma wrote:
cuhmoon wrote:
Hi,
At this point 2^440/7.
2 has a cycle of 2-4-8-6 and 440 is divisible by 4. thus the units digit = 6
6/7 = remainder = 6..
What am I missing here? Please explain
You are confusing the concept of cyclicity (units digit) with the concept of remainders.
When you divide any number which ends in 6 by 7, you do not get a remainder of 6. This works only when you divide a number that ends in 6 by 10. In that case the remainder will be 6 only.
e.g. 16 / 7 gives remainder 2
Cyclicity and unit's digit can help you solve remainders questions only in very specific cases. Those are discussed here:
https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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15 May 2018, 11:44
VeritasPrepKarishma wrote:
Responding to a pm:
Quote:
Can you please let me know how the following would be wrong for this problem ?
We have, 18^220=3^220 * 6^220.
Now, 6^220 = (7-1)^220 -- when divided by 7 this yields a remainder of 1. Am I correct ?
Yes, correct.
Quote:
If so, then 18^220 = 3^220 * (7-1)^220 -- therefore, now we ONLY need to check what is the remainder when 3^220 is divided by 7 and per the rule Cyclicity for 3, we can deduce that remainder will be 1 in this case also.
So, FINAL ans : remainder is 1 when (18^22)^10 is divided by 7.
Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10).
Check this post for more: https://www.veritasprep.com/blog/2015/1 ... questions/
You need to handle it using binomial.
$$3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28 - 1)^{73}$$
Remainder from $$(28 - 1)^{73}$$ will be -1 so overall remainder will be -3. Since divisor is 7, that is the same as remainder of 4.
If you are not sure about negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
Hi,
Can you tell me why the approach works with 18^220 = (14+4)^220, but it doesn't work when I try to solve it as 18^220=(21-3)^220?, in first case the remainder = 4 in second it = 1. I assume that (-3)^220 gives 1 and not -3. And how in general to pick numbers to avoid any mistake? Thanks.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2018, 05:35
Hero8888 wrote:
VeritasPrepKarishma wrote:
Responding to a pm:
Quote:
Can you please let me know how the following would be wrong for this problem ?
We have, 18^220=3^220 * 6^220.
Now, 6^220 = (7-1)^220 -- when divided by 7 this yields a remainder of 1. Am I correct ?
Yes, correct.
Quote:
If so, then 18^220 = 3^220 * (7-1)^220 -- therefore, now we ONLY need to check what is the remainder when 3^220 is divided by 7 and per the rule Cyclicity for 3, we can deduce that remainder will be 1 in this case also.
So, FINAL ans : remainder is 1 when (18^22)^10 is divided by 7.
Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10).
Check this post for more: https://www.veritasprep.com/blog/2015/1 ... questions/
You need to handle it using binomial.
$$3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28 - 1)^{73}$$
Remainder from $$(28 - 1)^{73}$$ will be -1 so overall remainder will be -3. Since divisor is 7, that is the same as remainder of 4.
If you are not sure about negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
Hi,
Can you tell me why the approach works with 18^220 = (14+4)^220, but it doesn't work when I try to solve it as 18^220=(21-3)^220?, in first case the remainder = 4 in second it = 1. I assume that (-3)^220 gives 1 and not -3. And how in general to pick numbers to avoid any mistake? Thanks.
No, the answer would be the same
$$(21 - 3)^{220}$$
The last term will be $$(-3)^{220}$$ which is same as $$3^{220}$$.
$$3* 3^{3*73} = 3 * (27)^{73} = 3 * (28 - 1)^{73}$$
The last term now will be $$3*(-1)^{73}$$ which is -3 (a negative remainder)
So the actual remainder will be 7 - 3 = 4
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2018, 10:34
VeritasPrepKarishma wrote:
No, the answer would be the same
$$(21 - 3)^{220}$$
The last term will be $$(-3)^{220}$$ which is same as $$3^{220}$$.
$$3* 3^{3*73} = 3 * (27)^{73} = 3 * (28 - 1)^{73}$$
The last term now will be $$3*(-1)^{73}$$ which is -3 (a negative remainder)
So the actual remainder will be 7 - 3 = 4
Thank you! I have tried just to find the last digit of $$(- 3)^{220}$$, since 3 is already less than 7. I thought it's a last stop. Is the last stop the format (x+/-1)/x or we can stop with any number that is less than devisor? Or the power should be odd? I'm confused. Thank you in advance.
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 17 May 2018, 10:34
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23 Chapters in Trigonometry: High School
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Week {{::cp.getGoalWeekForTopic(20, 23)}}
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Week {{::cp.getGoalWeekForTopic(21, 23)}}
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### Earning College Credit
Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 5,812 | 21,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-22 | longest | en | 0.593717 |
https://edurev.in/p/347804/NCERT-Textbook-Fun-at-Class-Party | 1,726,132,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00816.warc.gz | 203,197,753 | 63,831 | NCERT Textbook: Fun at Class Party
NCERT Textbook: Fun at Class Party | Mathematics for Class 3 (Maths Mela) PDF Download
``` Page 1
Children of Class 3 are preparing for a celebration in the class. Look
at the picture.
Discuss and explain how you answer these questions based on the
picture given above.
1. What are the various activities shown in the picture?
2. How does Shelly find the height of the door?
3. Leena and Adi use their hand spans to measure the length of the
same table. Will they both get the same measurement?
Fun at Class Fun at Class
Party! Party!
10 10
Let us Do
Teacher’s Note: The picture shows some examples of measurement
using informal and formal tools for length. Let children identify the
different ways of measuring and use appropriate words like hand
span, footstep and paper strings.
Page 2
Children of Class 3 are preparing for a celebration in the class. Look
at the picture.
Discuss and explain how you answer these questions based on the
picture given above.
1. What are the various activities shown in the picture?
2. How does Shelly find the height of the door?
3. Leena and Adi use their hand spans to measure the length of the
same table. Will they both get the same measurement?
Fun at Class Fun at Class
Party! Party!
10 10
Let us Do
Teacher’s Note: The picture shows some examples of measurement
using informal and formal tools for length. Let children identify the
different ways of measuring and use appropriate words like hand
span, footstep and paper strings.
129
Fun at Class Party!
4. Circle the child with the longest ponytail.
5. Tick
ü
paper strings in the classroom that are as long as the
height of the window.
6. Find the distance between the two walls of the classroom. How
did you find it? Can there be other ways of measuring it?
7. Identify all the ways that children are using to measure length in
this picture. Which way do you think is better and why?
1. In the picture above, colour the paper strings as instructed below.
(a) Colour the shortest paper string with red. Discuss how you
identified the shortest string.
(b) Colour the longest paper string with green. Discuss how you
identified the longest string.
2. How many more colourful paper strings will be needed to
decorate the border of the green board?
3. How many of are needed to decorate the entire border
of the board?
Children of Class 3 are decorating the board with paper strings of
different colours.
Let us Do
Page 3
Children of Class 3 are preparing for a celebration in the class. Look
at the picture.
Discuss and explain how you answer these questions based on the
picture given above.
1. What are the various activities shown in the picture?
2. How does Shelly find the height of the door?
3. Leena and Adi use their hand spans to measure the length of the
same table. Will they both get the same measurement?
Fun at Class Fun at Class
Party! Party!
10 10
Let us Do
Teacher’s Note: The picture shows some examples of measurement
using informal and formal tools for length. Let children identify the
different ways of measuring and use appropriate words like hand
span, footstep and paper strings.
129
Fun at Class Party!
4. Circle the child with the longest ponytail.
5. Tick
ü
paper strings in the classroom that are as long as the
height of the window.
6. Find the distance between the two walls of the classroom. How
did you find it? Can there be other ways of measuring it?
7. Identify all the ways that children are using to measure length in
this picture. Which way do you think is better and why?
1. In the picture above, colour the paper strings as instructed below.
(a) Colour the shortest paper string with red. Discuss how you
identified the shortest string.
(b) Colour the longest paper string with green. Discuss how you
identified the longest string.
2. How many more colourful paper strings will be needed to
decorate the border of the green board?
3. How many of are needed to decorate the entire border
of the board?
Children of Class 3 are decorating the board with paper strings of
different colours.
Let us Do
130
Maths Mela | Class 3
1. Cut and paste a wool or cotton thread as long as the line given
below.
2. Draw a string longer than the string given below.
Let us Do
3. Draw a string shorter than the decoration string Shelly and Adi
are holding.
Page 4
Children of Class 3 are preparing for a celebration in the class. Look
at the picture.
Discuss and explain how you answer these questions based on the
picture given above.
1. What are the various activities shown in the picture?
2. How does Shelly find the height of the door?
3. Leena and Adi use their hand spans to measure the length of the
same table. Will they both get the same measurement?
Fun at Class Fun at Class
Party! Party!
10 10
Let us Do
Teacher’s Note: The picture shows some examples of measurement
using informal and formal tools for length. Let children identify the
different ways of measuring and use appropriate words like hand
span, footstep and paper strings.
129
Fun at Class Party!
4. Circle the child with the longest ponytail.
5. Tick
ü
paper strings in the classroom that are as long as the
height of the window.
6. Find the distance between the two walls of the classroom. How
did you find it? Can there be other ways of measuring it?
7. Identify all the ways that children are using to measure length in
this picture. Which way do you think is better and why?
1. In the picture above, colour the paper strings as instructed below.
(a) Colour the shortest paper string with red. Discuss how you
identified the shortest string.
(b) Colour the longest paper string with green. Discuss how you
identified the longest string.
2. How many more colourful paper strings will be needed to
decorate the border of the green board?
3. How many of are needed to decorate the entire border
of the board?
Children of Class 3 are decorating the board with paper strings of
different colours.
Let us Do
130
Maths Mela | Class 3
1. Cut and paste a wool or cotton thread as long as the line given
below.
2. Draw a string longer than the string given below.
Let us Do
3. Draw a string shorter than the decoration string Shelly and Adi
are holding.
131
Fun at Class Party!
Shelly and Adi need to take a large table inside the classroom for the
party. The table is too heavy for both of them to move. Without lifting
the table, how can they figure out if the table can go through the door
of the classroom? Help them find out what they must do.
5. Look at the strings and help Adi choose the longest one.
How did you find out? Discuss.
4. Draw the other half of the moustache which is as long as the
half of the moustache on the face shown in the picture.
Page 5
Children of Class 3 are preparing for a celebration in the class. Look
at the picture.
Discuss and explain how you answer these questions based on the
picture given above.
1. What are the various activities shown in the picture?
2. How does Shelly find the height of the door?
3. Leena and Adi use their hand spans to measure the length of the
same table. Will they both get the same measurement?
Fun at Class Fun at Class
Party! Party!
10 10
Let us Do
Teacher’s Note: The picture shows some examples of measurement
using informal and formal tools for length. Let children identify the
different ways of measuring and use appropriate words like hand
span, footstep and paper strings.
129
Fun at Class Party!
4. Circle the child with the longest ponytail.
5. Tick
ü
paper strings in the classroom that are as long as the
height of the window.
6. Find the distance between the two walls of the classroom. How
did you find it? Can there be other ways of measuring it?
7. Identify all the ways that children are using to measure length in
this picture. Which way do you think is better and why?
1. In the picture above, colour the paper strings as instructed below.
(a) Colour the shortest paper string with red. Discuss how you
identified the shortest string.
(b) Colour the longest paper string with green. Discuss how you
identified the longest string.
2. How many more colourful paper strings will be needed to
decorate the border of the green board?
3. How many of are needed to decorate the entire border
of the board?
Children of Class 3 are decorating the board with paper strings of
different colours.
Let us Do
130
Maths Mela | Class 3
1. Cut and paste a wool or cotton thread as long as the line given
below.
2. Draw a string longer than the string given below.
Let us Do
3. Draw a string shorter than the decoration string Shelly and Adi
are holding.
131
Fun at Class Party!
Shelly and Adi need to take a large table inside the classroom for the
party. The table is too heavy for both of them to move. Without lifting
the table, how can they figure out if the table can go through the door
of the classroom? Help them find out what they must do.
5. Look at the strings and help Adi choose the longest one.
How did you find out? Discuss.
4. Draw the other half of the moustache which is as long as the
half of the moustache on the face shown in the picture.
132
Maths Mela | Class 3
Can there be a way to take the table inside the door if both the length
and the breadth are more than the width of the door?
We can use
measure the
length of the
table.
I guess it
will not pass
through!
Yes, it is
longer than
the door.
I measured
the length of
this side of the
table.
The table can pass
through the door
this way.
Teacher’s Note: Comparing two lengths is not always possible by
bringing them next to each other like in the above example of the table
and the door. This exercise should help students to figure out that one
can measure indirectly using a common unit.
Can you name some things that cannot pass through your school
gate? Discuss.
Make a bridge using boxes or bags or
any other objects available in your class.
Place or arrange the boxes so that the
bridge does not move. Take some objects
from your class and guess if the objects
can go through the bridge.
Let us Play
```
Mathematics for Class 3 (Maths Mela)
6 videos|68 docs|18 tests
Mathematics for Class 3 (Maths Mela)
6 videos|68 docs|18 tests
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https://www.maplesoft.com/support/help/errors/view.aspx?path=examples%2FCalculus1DiffApps | 1,701,534,340,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00768.warc.gz | 978,308,952 | 24,516 | Calculus1 Differentiation Apps - Maple Help
Home : Support : Online Help : Applications and Example Worksheets : Calculus : Calculus1 Differentiation Apps
Calculus 1: Applications of Differentiation
The Student[Calculus1] package contains three routines that can be used to both work with and visualize the concepts of Taylor approximations, charting a function, and using Newton's method. This worksheet demonstrates this functionality.
For further information about any command in the Calculus1 package, see the corresponding help page. For a general overview, see Calculus1.
Getting Started
While any command in the package can be referred to using the long form, for example, Student[Calculus1][FunctionChart], it is easier, and often clearer, to load the package, and then use the short form command names.
> $\mathrm{restart}$
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right):$
The following sections show how the routines work.
Taylor Approximations
If at a point $c$, a function $f$ has a power series expansion
$f\left(x\right)={\sum }_{n=0}^{\infty }\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}{a}_{n}{\left(x-c\right)}^{n}$
the coefficients ${a}_{n}$ are given by:
${a}_{n}=\frac{{f}^{\left(n\right)}\left(c\right)}{n!}$
where ${f}^{\left(n\right)}\left(c\right)$ is the nth derivative of $f$ evaluated at the point $c$. Named after the English mathematician Brook Taylor, this infinite series is called the Taylor expansion of the function $f$ at $c$.
The Taylor expansion of the exponential function ${ⅇ}^{x}$ is:
${ⅇ}^{x}={\sum }_{n=0}^{\infty }\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}\frac{{x}^{n}}{n!}$
from which it follows that:
$ⅇ={\sum }_{n=0}^{\infty }\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}\frac{1}{n!}$
Taylor approximations require both an expression and a point around which to expand.
> $\mathrm{TaylorApproximation}\left({x}^{7}-5{x}^{5}+4{x}^{4}-7{x}^{2}+3,x=1,\mathrm{order}=3\right)$
${12}{-}{16}{x}{-}{12}{\left({x}{-}{1}\right)}^{{2}}{+}{\left({x}{-}{1}\right)}^{{3}}$ (1.1)
Thus, around the point $x=1$ the polynomial ${x}^{7}-5{x}^{5}+4{x}^{4}-7{x}^{2}+3$ behaves like ${x}^{3}-15{x}^{2}+11x-1$.
> $\mathrm{TaylorApproximation}\left({x}^{7}-5{x}^{5}+4{x}^{4}-7{x}^{2}+3,x=1,\mathrm{order}=3,\mathrm{output}=\mathrm{plot}\right)$
> $\mathrm{TaylorApproximation}\left(\mathrm{sin}\left(x\right),x=1,\mathrm{output}=\mathrm{animation},\mathrm{order}=1..16\right)$
The derivative of the arctan function has singularities at $I$ and $-I$. Therefore, the radius of convergence of the Taylor approximation around the origin is $1$.
> $\mathrm{TaylorApproximation}\left(\mathrm{arctan}\left(x\right),x=0,\mathrm{output}=\mathrm{animation},\mathrm{order}=1..20\right)$
You can compute and visualize Taylor approximations using the TaylorApproximationTutor command.
> $\mathrm{TaylorApproximationTutor}\left(\right)$
Function Chart
The FunctionChart routine plots a function and shows regions of positive and negative sign, increasing and decreasing, and positive and negative concavity. By default:
1. Roots are marked by circles.
2. Extreme points are marked by diamonds.
3. Inflection points are marked by crosses.
4. Regions of increase and decrease are marked by red and blue lines, respectively.
5. Regions of positive and negative concavity are marked by azure and purple fill, respectively, with arrows pointing in the direction of the concavity.
> $\mathrm{FunctionChart}\left({x}^{4}+2{x}^{3}-9{x}^{2}-3x+6,x=-5..4\right)$
> $\mathrm{FunctionChart}\left(\mathrm{sin}\left(x\right),x=0..2\mathrm{π}\right)$
> $\mathrm{FunctionChart}\left(\frac{{x}^{3}-2{x}^{2}-4x+2}{x-4},x=-3..3\right)$
You can also perform curve analysis using the CurveAnalysisTutor command.
> $\mathrm{CurveAnalysisTutor}\left(\right)$
Newton's Method
Given a point $a$ and an expression $f\left(x\right)$, the $x$-intercept of the tangent line through ($a$, $f\left(a\right)$) can be used as an approximation to a root of the expression $f\left(x\right)$. The equation of the tangent line is:
> $\mathrm{Tangent}\left(f\left(x\right),x=a,\mathrm{output}=\mathrm{line}\right)$
${x}\left(\frac{{ⅆ}}{{ⅆ}{a}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({a}\right)\right){+}{f}{}\left({a}\right){-}\left(\frac{{ⅆ}}{{ⅆ}{a}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({a}\right)\right){a}$ (3.1)
> $\mathrm{collect}\left(\mathrm{convert}\left(,\mathrm{D}\right),\mathrm{D}\left(f\right)\left(a\right)\right)$
$\left({-}{a}{+}{x}\right){\mathrm{D}}{}\left({f}\right){}\left({a}\right){+}{f}{}\left({a}\right)$ (3.2)
Solving for zero:
> $\mathrm{solve}\left(=0,x\right)$
${-}\frac{{f}{}\left({a}\right){-}{\mathrm{D}}{}\left({f}\right){}\left({a}\right){a}}{{\mathrm{D}}{}\left({f}\right){}\left({a}\right)}$ (3.3)
> $\mathrm{expand}\left(\right)$
${-}\frac{{f}{}\left({a}\right)}{{\mathrm{D}}{}\left({f}\right){}\left({a}\right)}{+}{a}$ (3.4)
As an example, consider the function $F\left(x\right)={x}^{2}-1$ and an initial point $x=2.0$.
> $F:=x→{x}^{2}-1$
${F}{≔}{x}{↦}{{x}}^{{2}}{-}{1}$ (3.5)
> $\mathrm{aroot}:=2.0-\frac{F\left(2.0\right)}{\mathrm{D}\left(F\right)\left(2.0\right)}$
${\mathrm{aroot}}{≔}{1.250000000}$ (3.6)
Repeating this another $9$ times:
> $\mathbf{for}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{to}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}5\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{aroot}:=\mathrm{aroot}-\frac{F\left(\mathrm{aroot}\right)}{\mathrm{D}\left(F\right)\left(\mathrm{aroot}\right)}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end do}$
${\mathrm{aroot}}{≔}{1.025000000}$
${\mathrm{aroot}}{≔}{1.000304878}$
${\mathrm{aroot}}{≔}{1.000000046}$
${\mathrm{aroot}}{≔}{1.000000000}$
${\mathrm{aroot}}{≔}{1.000000000}$ (3.7)
The routine NewtonsMethod performs the same process.
> $\mathrm{NewtonsMethod}\left(F\left(x\right),x=2,\mathrm{output}=\mathrm{sequence}\right)$
${2}{,}{1.250000000}{,}{1.025000000}{,}{1.000304878}{,}{1.000000046}{,}{1.000000000}$ (3.8)
> $\mathrm{NewtonsMethod}\left(F\left(x\right),x=2,\mathrm{output}=\mathrm{plot}\right)$
> $\mathrm{NewtonsMethod}\left(\frac{\mathrm{sin}\left(x\right)}{x},x=1,\mathrm{output}=\mathrm{plot}\right)$
The root to which a sequence of Newton iterations converges to depends on the initial point. For example,
> $\mathrm{NewtonsMethod}\left(\frac{\mathrm{sin}\left(x\right)}{x},x=2,\mathrm{output}=\mathrm{plot}\right)$
In general, when the root is not a double root, Newton's method is very efficient. In the following example with Digits set to $30$, Newton's method converges to the root after only $7$ iterations.
> $\mathrm{Digits}:=30$
> $\mathrm{NewtonsMethod}\left({x}^{4}-4{x}^{3}+4{x}^{2}-3x+3,x=1,\mathrm{output}=\mathrm{sequence},\mathrm{iterations}=10\right)$
> $\mathrm{Digits}:=10$
${\mathrm{Digits}}{≔}{30}$
${1}{,}{1.33333333333333333333333333333}{,}{1.28318584070796460176991150443}{,}{1.28231623816647766759714731909}{,}{1.28231595363411166690275078928}{,}{1.28231595363408116582940754743}{,}{1.28231595363408116582940754708}$
${\mathrm{Digits}}{≔}{10}$ (3.9)
You can also learn about Newton's method using the NewtonsMethodTutor command.
> $\mathrm{NewtonsMethodTutor}\left(\right)$
>
Main: Visualization | 2,591 | 7,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 73, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-50 | latest | en | 0.647423 |
https://statkat.com/stattest.php?t=3&t2=17&t3=40&t4=11&t5=5&t6=11&t7=42 | 1,674,887,489,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00031.warc.gz | 545,691,730 | 12,237 | # Goodness of fit test - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Goodness of fit test
Kruskal-Wallis test
Cochran's Q test
One way ANOVA
One sample $z$ test for the mean
One way ANOVA
One sample Wilcoxon signed-rank test
Independent variableIndependent/grouping variableIndependent/grouping variableIndependent/grouping variableIndependent variableIndependent/grouping variableIndependent variable
NoneOne categorical with $I$ independent groups ($I \geqslant 2$)One within subject factor ($\geq 2$ related groups)One categorical with $I$ independent groups ($I \geqslant 2$)NoneOne categorical with $I$ independent groups ($I \geqslant 2$)None
Dependent variableDependent variableDependent variableDependent variableDependent variableDependent variableDependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesis
• H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
• H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: $\pi_1 = \pi_2 = \ldots = \pi_I$
Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$
ANOVA $F$ test:
• H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
• H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
• H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
ANOVA $F$ test:
• H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
• H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
• H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
H0: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
• H1: the population proportions are not all as specified under the null hypothesis
or equivalently
• H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
H1: not all population proportions are equalANOVA $F$ test:
• H1: not all population means are equal
$t$ Test for contrast:
• H1 two sided: $\Psi \neq 0$
• H1 right sided: $\Psi > 0$
• H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
• H1 - usually two sided: $\mu_g \neq \mu_h$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
ANOVA $F$ test:
• H1: not all population means are equal
$t$ Test for contrast:
• H1 two sided: $\Psi \neq 0$
• H1 right sided: $\Psi > 0$
• H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
• H1 - usually two sided: $\mu_g \neq \mu_h$
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• Scores are normally distributed in the population
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• The population distribution of the scores is symmetric
• Sample is a simple random sample from the population. That is, observations are independent of one another
Test statisticTest statisticTest statisticTest statisticTest statisticTest statisticTest statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$
Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.
Note: if ties are present in the data, the formula for $H$ is more complicated.
If a failure is scored as 0 and a success is scored as 1:
$Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$
Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores.
Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.
ANOVA $F$ test:
• \begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned}
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
• $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
• $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
ANOVA $F$ test:
• \begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned}
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
• $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
• $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
2. For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
• $W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{-}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
• Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
• If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
• $W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
n.a.n.a.n.a.Pooled standard deviationn.a.Pooled standard deviationn.a.
---\begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}
Here $s^2_i$ is the variance in group $i.$
-\begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}
Here $s^2_i$ is the variance in group $i.$
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Sampling distribution of $X^2$ if H0 were trueSampling distribution of $H$ if H0 were trueSampling distribution of $Q$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedom
For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.
For small samples, the exact distribution of $H$ should be used.
If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedomSampling distribution of $F$:
• $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - I$ degrees of freedom
Standard normal distributionSampling distribution of $F$:
• $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - I$ degrees of freedom
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.
Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
Significant?Significant?Significant?Significant?Significant?Significant?Significant?
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)
$t$ Test for contrast two sided:
$t$ Test for contrast right sided:
$t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
• Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
• Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
• Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
Two sided:
Right sided:
Left sided:
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)
$t$ Test for contrast two sided:
$t$ Test for contrast right sided:
$t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
• Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
• Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
• Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
n.a.n.a.n.a.$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$$C\% confidence interval for \mu$$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$n.a.
---Confidence interval for $\Psi$ (contrast):
• $c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
• $(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* =$ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
• $\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu$ can also be used as significance test.
Confidence interval for $\Psi$ (contrast):
• $c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
• $(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* =$ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
• $\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
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n.a.n.a.n.a.Effect sizeEffect sizeEffect sizen.a.
---
• Proportion variance explained $\eta^2$ and $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variable: \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to: $$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$
• Cohen's $d$:
Standardized difference between the mean in group $g$ and in group $h$: $$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
• Proportion variance explained $\eta^2$ and $R^2$:
Proportion variance of the dependent variable $y$ explained by the independent variable: \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.
• Proportion variance explained $\omega^2$:
Corrects for the positive bias in $\eta^2$ and is equal to: $$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$
• Cohen's $d$:
Standardized difference between the mean in group $g$ and in group $h$: $$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
-
n.a.n.a.n.a.n.a.Visual representationn.a.n.a.
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n.a.n.a.n.a.ANOVA tablen.a.ANOVA tablen.a.
---
Click the link for a step by step explanation of how to compute the sum of squares.
-
Click the link for a step by step explanation of how to compute the sum of squares.
-
n.a.n.a.Equivalent toEquivalent ton.a.Equivalent ton.a.
--Friedman test, with a categorical dependent variable consisting of two independent groups.OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
• $F$ test ANOVA is equivalent to $F$ test regression model
• $t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
-OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
• $F$ test ANOVA is equivalent to $F$ test regression model
• $t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
-
Example contextExample contextExample contextExample contextExample contextExample contextExample context
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?Do people from different religions tend to score differently on social economic status? Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?Is the average mental health score different between people from a low, moderate, and high economic class?Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$Is the average mental health score different between people from a low, moderate, and high economic class?Is the median mental health score of office workers different from $m_0 = 50$?
SPSSSPSSSPSSSPSSn.a.SPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square...
• Put your categorical variable in the box below Test Variable List
• Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples...
• Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum
• Continue and click OK
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
• Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables
• Under Test Type, select Cochran's Q test
Analyze > Compare Means > One-Way ANOVA...
• Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
-Analyze > Compare Means > One-Way ANOVA...
• Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
• On the Objective tab, choose Customize Analysis
• On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test
• On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your $m_0$ in the box next to Hypothesized median
• Click Run
• Double click on the output table to see the full results
JamoviJamoviJamoviJamovin.a.JamoviJamovi
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
• Put your categorical variable in the box below Variable
• Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
ANOVA > One Way ANOVA - Kruskal-Wallis
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
• Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
-ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
T-Tests > One Sample T-Test
• Put your variable in the box below Dependent Variables
• Under Tests, select Wilcoxon rank
• Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis
Practice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questions | 9,740 | 35,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-06 | longest | en | 0.723042 |
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https://en.wikibooks.org/wiki/Physics_Study_Guide/Print_version/Section_Two | 1,708,676,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00373.warc.gz | 242,306,130 | 34,399 | # Physics Study Guide/Print version/Section Two
## Uniform Circular Motion
### Speed and frequency
Uniform circular motion assumes that an object is moving (1) in circular motion, and (2) at constant speed ${\displaystyle v}$; then
${\displaystyle T={\frac {2\pi r}{v}}}$
where ${\displaystyle r}$ is the radius of the circular path, and ${\displaystyle T}$ is the time period for one revolution.
Any object travelling on a circle will return to its original starting point in the period of one revolution, ${\displaystyle T}$. At this point the object has travelled a distance ${\displaystyle 2\pi r}$. If ${\displaystyle T}$ is the time that it takes to travel distance ${\displaystyle 2\pi r}$ then the object's speed is
${\displaystyle v={\frac {2\pi r}{T}}=2\pi rf}$
where ${\displaystyle f={\frac {1}{T}}}$
### Angular frequency
Uniform circular motion can be explicitly described in terms of polar coordinates through angular frequency, ${\displaystyle \omega }$ :
${\displaystyle \omega ={\frac {d\theta }{dt}}}$
where ${\displaystyle \theta }$ is the angular coordinate of the object (see the diagram on the right-hand side for reference).
Since the speed in uniform circular motion is constant, it follows that
${\displaystyle \omega ={\frac {\Delta \theta }{\Delta t}}}$
From that fact, a number of useful relations follow:
${\displaystyle \omega ={\frac {2\pi }{T}}=2\pi f={\frac {v}{r}}}$
The equations that relate how ${\displaystyle \theta }$ changes with time are analogous to those of linear motion at constant speed. In particular,
${\displaystyle \theta =\theta _{0}+\omega t}$
The angle at ${\displaystyle t=0}$, ${\displaystyle \theta _{0}}$, is commonly referred to as phase.
### Velocity, centripetal acceleration and force
The position of an object in a plane can be converted from polar to cartesian coordinates through the equations
{\displaystyle {\begin{aligned}x&=r\cos(\theta )\\y&=r\sin(\theta )\end{aligned}}}
Expressing ${\displaystyle \theta }$ as a function of time gives equations for the cartesian coordinates as a function of time in uniform circular motion:
{\displaystyle {\begin{aligned}x&=r\cos(\theta _{0}+\omega t)\\y&=r\sin(\theta _{0}+\omega t)\end{aligned}}}
Differentiation with respect to time gives the components of the velocity vector:
{\displaystyle {\begin{aligned}v_{x}&=\omega r(-\sin(\omega t))=-v\sin(\omega t)\\v_{y}&=\omega r\cos(\omega t)=v\cos(\omega t)\end{aligned}}}
Velocity in circular motion is a vector tangential to the trajectory of the object. Furthermore, even though the speed is constant the velocity vector changes direction over time. Further differentiation leads to the components of the acceleration (which are just the rate of change of the velocity components):
{\displaystyle {\begin{aligned}a_{x}&=-\omega ^{2}r\cos(\omega t)\\a_{y}&=-\omega ^{2}r\sin(\omega t)\end{aligned}}}
The acceleration vector is perpendicular to the velocity and oriented towards the centre of the circular trajectory. For that reason, acceleration in circular motion is referred to as centripetal acceleration.
The absolute value of centripetal acceleration may be readily obtained by
{\displaystyle {\begin{aligned}a_{c}&={\sqrt {a_{x}^{2}+a_{y}^{2}}}={\sqrt {(\omega ^{2}r)^{2}{\big (}\cos ^{2}(\omega t)+\sin ^{2}(\omega t){\big )}}}\\&=\omega ^{2}r={\frac {v^{2}}{r}}\end{aligned}}}
For centripetal acceleration, and therefore circular motion, to be maintained a centripetal force must act on the object. From Newton's Second Law it follows directly that the force will be given by
${\displaystyle {\vec {F_{c}}}=m{\vec {a_{c}}}}$
the components being
{\displaystyle {\begin{aligned}F_{x}&=-m\omega ^{2}r\cos(\omega t)\\F_{y}&=-m\omega ^{2}r\sin(\omega t)\end{aligned}}}
and the absolute value
${\displaystyle F_{c}=m\omega ^{2}r=m{\frac {v^{2}}{r}}}$
## Torque and Circular Motion
Circular motion is the motion of a particle at a set distance (called radius) from a point. For circular motion, there needs to be a force that makes the particle turn. This force is called the 'centripetal force.' Please note that the centripetal force is not a new type of force-it is just a force causing rotational motion. To make this clearer, let us study the following examples:
1. If Stone ties a piece of thread to a small pebble and rotates it in a horizontal circle above his head, the circular motion of the pebble is caused by the tension force in the thread.
2. In the case of the motion of the planets around the sun (which is roughly circular), the force is provided by the gravitational force exerted by the sun on the planets.
Thus, we see that the centripetal force acting on a body is always provided by some other type of force -- centripetal force, thus, is simply a name to indicate the force that provides this circular motion. This centripetal force is always acting inward toward the center. You will know this if you swing an object in a circular motion. If you notice carefully, you will see that you have to continuously pull inward. We know that an opposite force should exist for this centripetal force(by Newton's 3rd Law of Motion). This is the centrifugal force, which exists only if we study the body from a non-inertial frame of reference(an accelerating frame of reference, such as in circular motion). This is a so-called 'pseudo-force', which is used to make the Newton's law applicable to the person who is inside a non-inertial frame. e.g. If a driver suddenly turns the car to the left, you go towards the right side of the car because of centrifugal force. The centrifugal force is equal and opposite to the centripetal force. It is caused due to inertia of a body.
${\displaystyle \omega _{\text{avg}}={\frac {\omega +\omega _{f}}{2}}={\frac {\theta }{t}}}$
Average angular velocity is equal to one-half of the sum of initial and final angular velocities assuming constant acceleration, and is also equal to the angle gone through divided by the time taken.
${\displaystyle \alpha ={\frac {\Delta \omega }{t}}}$
Angular acceleration is equal to change in angular velocity divided by time taken.
### Angular momentum
${\displaystyle {\vec {L}}={\vec {r}}\times {\vec {p}}=m({\vec {r}}\times {\vec {v}})}$
Angular momentum of an object revolving around an external axis ${\displaystyle O}$ is equal to the cross-product of the position vector with respect to ${\displaystyle O}$ and its linear momentum.
${\displaystyle {\vec {L}}=I{\vec {\omega }}}$
Angular momentum of a rotating object is equal to the moment of inertia times angular velocity.
${\displaystyle L=I\omega }$
${\displaystyle \tau =I\alpha ={\frac {\Delta L}{t}}}$>
Rotational Kinetic Energy is equal to one-half of the product of moment of inertia and the angular velocity squared.
IT IS USEFUL TO NOTE THAT
The equations for rotational motion are analogous to those for linear motion-just look at those listed above. When studying rotational dynamics, remember:
• the place of force is taken by torque
• the place of mass is taken by moment of inertia
• the place of displacement is taken by angle
• the place of linear velocity, momentum, acceleration, etc. is taken by their angular counterparts.
### Definition of terms
Torque (${\displaystyle {\vec {\tau }}}$): Force times distance. A vector. ${\displaystyle N\!\cdot \!m}$ Moment of inertia (${\displaystyle I}$): Describes the object's resistance to torque — the rotational analog to inertial mass. ${\displaystyle kg\!\cdot \!m^{2}}$ Angular momentum (${\displaystyle {\vec {L}}}$): ${\displaystyle kg\!\cdot \!{\frac {m^{2}}{s}}}$ Angular velocity (${\displaystyle {\vec {\omega }}}$): ${\displaystyle {\frac {\text{rad}}{s}}}$ Angular acceleration (${\displaystyle {\vec {\alpha }}}$): ${\displaystyle {\frac {\text{rad}}{s^{2}}}}$ Rotational kinetic energy (${\displaystyle K_{r}}$): ${\displaystyle J=kg\!\cdot \!\left({\frac {m}{s}}\right)^{2}}$ Time (${\displaystyle t}$): ${\displaystyle s}$
## Buoyancy
Buoyancy is the force due to pressure differences on the top and bottom of an object under a fluid (gas or liquid).
Net force = buoyant force - force due to gravity on the object
## Bernoulli's Principle
Fluid flow is a complex phenomenon. An ideal fluid may be described as:
• The fluid flow is steady i.e its velocity at each point is constant with time.
• The fluid is incompressible. This condition applies well to liquids and in certain circumstances to gases.
• The fluid flow is non-viscous. Internal friction is neglected. An object moving through this fluid does not experience a retarding force. We relax this condition in the discussion of Stokes' Law.
• The fluid flow is irrotational. There is no angular momentum of the fluid about any point. A very small wheel placed at an arbitrary point in the fluid does not rotate about its center. Note that if turbulence is present, the wheel would most likely rotate and its flow is then not irrotational.
As the fluid moves through a pipe of varying cross-section and elevation, the pressure will change along the pipe. The Swiss physicist Daniel Bernoulli (1700-1782) first derived an expression relating the pressure to fluid speed and height. This result is a consequence of conservation of energy and applies to ideal fluids as described above.
Consider an ideal fluid flowing in a pipe of varying cross-section. A fluid in a section of length ${\displaystyle \Delta x_{1}}$ moves to the section of length ${\displaystyle \Delta x_{2}}$ in time ${\displaystyle \Delta t}$. The relation given by Bernoulli is:
${\displaystyle P+{\tfrac {1}{2}}\rho v^{2}+\rho gh={\text{constant}}}$
where:
${\displaystyle P}$ is pressure at cross-section
${\displaystyle h}$ is height of cross-section
${\displaystyle \rho }$ is density
${\displaystyle v}$ is velocity of fluid at cross-section
In words, the Bernoulli relation may be stated as:
As we move along a streamline the sum of the pressure (${\displaystyle P}$), the kinetic energy per unit volume and the potential energy per unit volume remains a constant.
(To be concluded)
# Fields
A field is one of the more difficult concepts to grasp in physics. A field is an area or region in which an influence or force is effective regardless of the presence or absence of a material medium. Simply put, a field is a collection of vectors often representing the force an object would feel if it were placed at any particular point in space. With gravity, the field is measured in newtons, as it depends solely on the mass of an object, but with electricity, it is measured in newtons per coulomb, as the force on an electrical charge depends on the amount of that charge. Typically these fields are calculated based on canceling out the effect of a body in the point in space that the field is desired. As a result, a field is a vector, and as such, it can (and should) be added when calculating the field created by TWO objects at one point in space.
Fields are typically illustrated through the use of what are called field lines or lines of force. Given a source that exerts a force on points around it, sample lines are drawn representing the direction of the field at points in space around the force-exerting source.
There are three major categories of fields:
1. Uniform fields are fields that have the same value at any point in space. As a result, the lines of force are parallel.
2. Spherical fields are fields that have an origin at a particular point in space and vary at varying distances from that point.
3. Complex fields are fields that are difficult to work with mathematically (except under simple cases, such as fields created by two point object), but field lines can still typically be drawn. Dipoles are a specific kind of complex field.
Magnetism also has a field, measured in Tesla, and it also has field lines, but its use is more complicated than simple "force" fields. Secondly, it also only appears in a two-pole form, and as such, is difficult to calculate easily.
The particles that form these magnetic fields and lines of force are called electrons and not magnetons. A magneton is a quantity in magnetism.
## Definition of terms
Field: A collection of vectors that often represents the force that an object would feel if it were placed in any point in space. Field Lines: A method of diagramming fields by drawing several sample lines showing direction of the field through several points in space.
## Newtonian Gravity
Newtonian Gravity (simplified gravitation) is an apparent force (a.k.a. pseudoforce) that simulates the attraction of one mass to another mass. Unlike the three fundamental (real) forces of electromagnetism and the strong and weak nuclear forces, gravity is purely attractive. As a force it is measured in newtons. The distance between two objects is measured between their centers of mass.
${\displaystyle F=G\!\cdot \!{\frac {m_{1}m_{2}}{r^{2}}}}$
Gravitational force is equal to the product of the universal gravitational constant and the masses of the two objects, divided by the square of the distance between their centers of mass.
${\displaystyle g=G\cdot {\frac {m_{1}}{r^{2}}}}$
The value of the gravitational field which is equivalent to the acceleration due to gravity caused by an object at a point in space is equal to the first equation about gravitational force, with the effect of the second mass taken out.
${\displaystyle U=-G\cdot {\frac {Mm}{r}}}$
Gravitational potential energy of a body to infinity is equal to the universal gravitational constant times the mass of a body from which the gravitational field is being created times the mass of the body whose potential energy is being measured over the distance between the two centers of mass. Therefore, the difference in potential energy between two points is the difference of the potential energy from the position of the center of mass to infinity at both points. Near the earth's surface, this approximates:
${\displaystyle \Delta U_{g}=mgh}$
Potential energy due to gravity near the earth's surface is equal to the product of mass, acceleration due to gravity, and height (elevation) of the object.
If the potential energy from the body's center of mass to infinity is known, however, it is possible to calculate the escape velocity, or the velocity necessary to escape the gravitational field of an object. This can be derived based on utilizing the law of conservation of energy and the equation to calculate kinetic energy as follows:
{\displaystyle {\begin{aligned}&{\boldsymbol {ke}}_{initial}=\Delta U\\\\&{\boldsymbol {ke}}_{initial}=U_{infinity}-U_{initial}\\\\&{\tfrac {1}{2}}mv^{2}=G\cdot {\frac {Mm}{r}}\\\\&v_{esc}={\sqrt {\frac {2GM}{r}}}\end{aligned}}}
}}
## Variables
F: force (N) G: universal constant of gravitation, (6.67x10-11 N•m2/kg2) m1: mass of the first body m2: mass of the second body r: the distance between the point at which the force or field is being taken, and the center of mass of the first body g: acceleration due to gravity (on the earth’s surface, this is 9.8 m/s2) U: potential energy from the location of the center of mass to infinity (J) ΔUg: Change in potential energy (J) m and M: mass (kg) h: height of elevation (m) vesc: escape velocity (m/s)
## Definition of terms
Universal constant of gravitation (G): This is a constant that is the same everywhere in the known universe and can be used to calculate gravitational attraction and acceleration due to gravity. 6.67x10-11 N·m2/kg2 Mass one (m1): One of two masses that are experiencing a mutual gravitational attraction. We can use this for the mass of the Earth (1023 kg). Mass two (m2): One of two masses that are experiencing a mutual gravitational attraction. This symbol can represent the mass of an object on or close to earth. Units: kilograms (kg) Acceleration due to gravity (g): This is nearly constant near the earth's surface because the mass and radius of the earth are essentially constant. At extreme altitudes the value can vary slightly, but it varies more significantly with latitude. This is also equal to the value of the gravitational field caused by a body at a particular point in space (9.8 m/s2) Escape velocity (vesc): The velocity necessary to completely escape the gravitational effects of a body.
A black hole is a geometrically defined region of space time exhibiting such large centripetal gravitational effects that nothing such as particles and electromagnetic radiation such as light may escape from inside of it. That is the escape velocity upon the event horizon is equivalent to the speed of light. General relativity is a metric theory of gravitation generalizing space time and Newton's law of universal gravitational attraction as a geometric property of space time.
# Waves
Wave is defined as the movement of any periodic motion like a spring, a pendulum, a water wave, an electric wave, a sound wave, a light wave, etc.
Any periodic wave that has amplitude varied with time, phase sinusoidally can be expressed mathematically as
R(t , θ) = R Sin (ωt + θ)
• Minimum wave height (trough) at angle 0, π, 2π, ...
F(R,t,θ) = 0 at θ = nπ
• Maximum wave height (peak or crest) at π/2, 3π/2, ...
F(R,t,θ) = R at θ = (2n+1)π/2
• Wavelength (distance between two crests) λ = 2π.
λ = 2π - A circle or a wave
2λ = 2(2π) - Two circles or two waves
kλ = k2π - Circle k or k amount of waves
• Wave Number,
k
• Velocity (or Angular Velocity),
ω = 2πf
• Time Frequency,
f = 1 / t
• Time
t = 1 / f
Wave speed is equal to the frequency times the wavelength. It can be understood as how frequently a certain distance (the wavelength in this case) is traversed.
${\displaystyle f={\frac {v}{\lambda }}}$
Frequency is equal to speed divided by wavelength.
${\displaystyle T={\frac {1}{f}}}$
Period is equal to the inverse of frequency.
## Variables
λ: wavelength (m) v: wave speed (m/s) f: frequency (1/s), (Hz) T: period (s)
## Definition of terms
Wavelength (λ): The length of one wave, or the distance from a point on one wave to the same point on the next wave. Units: meters (m). In light, λ tells us the color. Wave speed (v): the speed at which the wave pattern moves. Units: meters per second, (m/s) Frequency of oscillation (f) (or just frequency): the number of times the wave pattern repeats itself in one second. Units: seconds-1 = (1/s) = hertz (Hz) In sound, f tells us the pitch. The inverse of frequecy is the period of oscillation. Period of oscillation (T) (or just period): duration of time between one wave and the next one passing the same spot. Units: seconds (s). The inverse of the period is frequency. Use a capital, italic T and not a lowercase one, which is used for time. Amplitude (A): the maximum height of the wave measured from the average height of the wave (the wave’s center). Unit: meters (m)
Image here
The wave’s extremes, its peaks and valleys, are called antinodes. At the middle of the wave are points that do not move, called nodes.
Examples of waves: Water waves, sound waves, light waves, seismic waves, shock waves, electromagnetic waves …
## Oscillation
A wave is said to oscillate, which means to move back and forth in a regular, repeating way. This fluctuation can be between extremes of position, force, or quantity. Different types of waves have different types of oscillations.
Longitudinal waves: Oscillation is parallel to the direction of the wave. Examples: sound waves, waves in a spring.
Transverse waves: Oscillation is perpendicular to direction of the wave. Example: light
## Interference
When waves overlap each other it is called interference. This is divided into constructive and destructive interference.
Constructive interference: the waves line up perfectly and add to each others’ strength.
Destructive interference: the two waves cancel each other out, resulting in no wave.This happens when angle between them is 180degrees.
## Resonance
In real life, waves usually give a mishmash of constructive and destructive interference and quickly die out. However, at certain wavelengths standing waves form, resulting in resonance. These are waves that bounce back into themselves in a strengthening way, reaching maximum amplitude.
Resonance is a special case of forced vibration when the frequency of the impressed periodic force is equal to the natural frequency of the body so that it vibrates with increased amplitude, spontaneously.
# Wave overtones
For resonance in a taut string, the first harmonic is determined for a wave form with one antinode and two nodes. That is, the two ends of the string are nodes because they do not vibrate while the middle of the string is an antinode because it experiences the greatest change in amplitude. This means that one half of a full wavelength is represented by the length of the resonating structure.
The frequency of the first harmonic is equal to wave speed divided by twice the length of the string. (Recall that wave speed is equal to wavelength times frequency.)
F1 = v/2L
The wavelength of the first harmonic is equal to double the length of the string.
λ1 = 2L
The "nth" wavelength is equal to the fundamental wavelength divided by n.
λn = λ1/n
Harmonics for a taut string*
Harmonic number Overtone number F = λ = F1 First harmonic --- F1 = v/2L λ1 = 2L F2 Second harmonic First overtone F2 = 2F1 λ2 =λ1/2 F3 Third harmonic Second overtone F3 = 3F1 λ3 = λ1/3 Fn Nth harmonic (Nth - 1) overtone Fn = nF1 λn = λ1/n
* or any wave system with two identical ends, such as a pipe with two open or closed ends. In the case of a pipe with two open ends, there are two antinodes at the ends of the pipe and a single node in the middle of the pipe, but the mathematics work out identically.
## Definition of terms
Frequency (F): Units: (1/s), hertz (Hz) Fundamental frequency, first harmonic (F)1: The lowest frequency (longest wavelength) allowed for the system. Length of string (L): (or pipe, etc.) Units: meters (m). Wavelength (λ): Units: meters (m).
The first overtone is the first allowed harmonic above the fundamental frequency (F1).
In the case of a system with two different ends (as in the case of a tube open at one end), the closed end is a node and the open end is an antinode. The first resonant frequency has only a quarter of a wave in the tube. This means that the first harmonic is characterized by a wavelength four times the length of the tube.
F1 = v/4L
The wavelength of the first harmonic is equal to four times the length of the string.
λ1 = 4L
The "nth" wavelength is equal to the fundamental wavelength divided by n.
λn = λ1/n
Note that "n" must be odd in this case as only odd harmonics will resonate in this situation.
Harmonics for a system with two different ends*
Harmonic number Overtone number F = λ = F1 First harmonic --- F1 = v/4L λ1 = 4L F2 Third harmonic First overtone F2 = 3F1 λ2 = λ1/3 F3 Fifth harmonic Second overtone F3 = 5F1 λ3 = λ1/5 Fn Nth harmonic† (Nth - 1)/2 overtone F(n-1)/2 = nF1 λn = λ1/n
* such as a pipe with one end open and one end closed
†In this case only the odd harmonics resonate, so n is an odd integer.
Vs: velocity of sound
• dependent on qualities of the medium transmitting the sound, (the air) such as its density, temperature, and “springiness.” A complicated equation, we concentrate only on temperature.
• increases as temperature increases (molecules move faster.)
• is higher for liquids and solids than for gasses (molecules are closer together.)
• for “room air” is 340 meters per second (m/s).
• Speed of sound is 343 meters per second at 20 degrees C. Based on the material sound is passing through and the temperature, the speed of sound changes.
# Standing waves
${\displaystyle \|{\vec {v}}\|={\sqrt {\frac {\|{\vec {T}}\|}{\mu }}}}$
Wave speed is equal to the square root of tension divided by the linear density of the string.
μ = m/L
Linear density of the string is equal to the mass divided by the length of the string.
λmax = 2L
The fundamental wavelength is equal to two times the length of the string.
## Variables
λ: wavelength (m) λmax: fundamental wavelength (m) μ: linear density (g/m) v: wave speed (m/s) F: force (N) m: mass (kg) L: length of the string (m) l: meters (m)
## Definition of terms
Tension (F): (not frequency) in the string (t is used for time in these equations). Units: newtons (N) Linear density (μ): of the string, Greek mu. Units: grams per meter (g/m) Velocity (v) of the wave (m/s) Mass (m): Units: grams (g). (We would use kilograms but they are too big for most strings). Length of the string (L): Units: meters (m)
Fundamental frequency: the frequency when the wavelength is the longest allowed, this gives us the lowest sound that we can get from the system.
In a string, the length of the string is half of the largest wavelength that can create a standing wave, called its fundamental wavelength.
Sound is defined as mechanical sinosodial vibratory longitudinal impulse waves which oscillate the pressure of a transmitting medium by means of adiabatic compression and decompression consequently resulting in the increase in the angular momentum and hence rotational kinetic energy of the particles present within the transmitting medium producing frequencies audible within hearing range, that is between the threshold of audibility and the threshold of pain on a Fletchford Munson equal loudness contour diagram.
## Intro
When two glasses collide, we hear a sound. When we pluck a guitar string, we hear a sound.
Different sounds are generated from different sources. Generally speaking, the collision of two objects results in a sound.
Sound does not exist in a vacuum; it travels through the materials of a medium. Sound is a longitudinal wave in which the mechanical vibration constituting the wave occurs along the direction of the wave's propagation.
The velocity of sound waves depends on the temperature and the pressure of the medium. For example, sound travels at different speeds in air and water. We can therefore define sound as a mechanical disturbance produced by the collision of two or more physical quantities from a state of equilibrium that propagates through an elastic material medium.
# Sound
${\displaystyle decibel(\mathrm {dB} )=10\cdot \log \left({\frac {I_{1}}{I_{0}}}\right)}$
The amplitude is the magnitude of sound pressure change within a sound wave. Sound amplitude can be measured in pascals (Pa), though its more common to refer to the sound (pressure) level as Sound intensity(dB,dBSPL,dB(SPL)), and the perceived sound level as Loudness(dBA, dB(A)). Sound intensity is flow of sound energy per unit time through a fixed area. It has units of watts per square meter. The reference Intensity is defined as the minimum Intensity that is audible to the human ear, it is equal to 10-12 W/m2, or one picowatt per square meter. When the intensity is quoted in decibels this reference value is used. Loudness is sound intensity altered according to the frequency response of the human ear and is measured in a unit called the A-weighted decibel (dB(A), also used to be called phon).
## The Decibel
The decibel is not, as is commonly believed, the unit of sound. Sound is measured in terms of pressure. However, the decibel is used to express the pressure as very large variations of pressure are commonly encountered. The decibel is a dimensionless quantity and is used to express the ratio of one power quantity to another. The definition of the decibel is ${\displaystyle 10\cdot \log _{10}\left({\frac {x}{x_{0}}}\right)}$, where x is a squared quantity, ie pressure squared, volts squared etc. The decibel is useful to define relative changes. For instance, the required sound decrease for new cars might be 3 dB, this means, compared to the old car the new car must be 3 dB quieter. The absolute level of the car, in this case, does not matter.
${\displaystyle I_{0}=10^{-12}{\mbox{ W}}/{\mbox{m}}^{2}}$
## Definition of terms
Intensity (I): the amount of energy transferred through 1 m2 each second. Units: watts per square meter
Lowest audible sound: I = 0 dB = 10-12 W/m2 (A sound with dB < 0 is inaudible to a human.) Threshold of pain: I = 120 dB = 10 W/m2
Sample equation: Change in sound intensity
Δβ = β2 - β1
= 10 log(I2/I0) - 10 log(I1/I0)
= 10 [log(I2/I0) - log(I1/I0)]
= 10 log[(I2/I0)/(I1/I0)]
= 10 log(I2/I1)
where log is the base-10 logarithm.
## Doppler effect
${\displaystyle f'=f\,{\frac {v\pm v_{0}}{v\mp v_{s}}}}$
f' is the observed frequency, f is the actual frequency, v is the speed of sound (${\displaystyle v=336+0.6T}$), T is temperature in degrees Celsius ${\displaystyle v_{0}}$ is the speed of the observer, and ${\displaystyle v_{s}}$ is the speed of the source. If the observer is approaching the source, use the top operator (the +) in the numerator, and if the source is approaching the observer, use the top operator (the -) in the denominator. If the observer is moving away from the source, use the bottom operator (the -) in the numerator, and if the source is moving away from the observer, use the bottom operator (the +) in the denominator.
### Example problems
A. An ambulance, which is emitting a 400 Hz siren, is moving at a speed of 30 m/s towards a stationary observer. The speed of sound in this case is 339 m/s.
${\displaystyle f'=400\,\mathrm {Hz} \left({\frac {339+0}{339-30}}\right)}$
B. An M551 Sheridan, moving at 10 m/s is following a Renault FT-17 which is moving in the same direction at 5 m/s and emitting a 30 Hz tone. The speed of sound in this case is 342 m/s.
${\displaystyle f'=30\,\mathrm {Hz} \left({\frac {342+10}{342+5}}\right)}$ | 7,269 | 29,655 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 79, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-10 | latest | en | 0.796448 |
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Speaking of Statistics (10) (APsy). Chapter 16: CHI-SQUARE statistic and tests Statistics for the Behavioral Sciences. 17.1 Parametric and Non-Parametric Statistical Tests 17.2 The Chi-Square Test for Goodness of Fit 17.3 The Chi-Square Test for Independence. THE CHI-SQUARE STATISTIC.
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Speaking of Statistics (10)(APsy)
Chapter 16:
CHI-SQUARE statistic and tests
Statistics for the Behavioral Sciences
17.1 Parametric and Non-Parametric Statistical Tests
17.2 The Chi-Square Test for Goodness of Fit
17.3 The Chi-Square Test for Independence
THE CHI-SQUARE STATISTIC
Unlike earlier tests that require numerical scores (X values), the chi-square tests use frequencies and proportions from a sample to test hypotheses about the corresponding population.
Chi-Square as a Statistical Test
• Chi-square test: an inferential statistics technique designed to test for significant relationships between two variables organized in a bivariate table.
• Chi-square requires no assumptions about the shape of the population distribution from which a sample is drawn.
• It can be applied to nominally or ordinallymeasured variables.
Parametric tests
All the statistical tests we have examined thus far are designed to test hypotheses about specific population parameters. They typically make assumptions about the shape of the population distribution and about other population parameters.
For analysis of variance, the population distributions are assumed to be normal and homogeneity of variance is required.
Because these tests all concern parameters and require assumptions about parameters, they are called parametric tests.
A characteristic of parametric tests is that they require a numerical score for each individual in the sample. The scores then are added, squared, averaged, and otherwise manipulated using basic arithmetic. In terms of measurement scales,
parametric tests require data from an interval or a ratio scale.
Non-parametric tests
Often researchers are confronted with experimental situations that do not conform to the requirements of parametric tests.
There are several hypothesis-testing techniques that provide alternatives to parametric tests. These alternatives are called non-parametric tests.
Non-parametric tests usually do not state hypotheses in terms of a specific parameter and that they make few (if any) assumptions about the population distribution. For the latter reason, non-parametric tests sometimes are called distribution-free tests.
Differences between parametric and non-parametric tests
All of the parametric tests that we have examined so far require numerical scores, e.g. IQ score, height, weight, … etc.
For non-parametric tests, on the other hand, the subjects are usually just classified into categories such as Democrat and Republican, or High, Medium, and Low IQ.
These classifications involve measurement on nominal or ordinal scales, and they do not produce numerical values that can be used to calculate means and variances.
The data for many non-parametric tests are simply frequencies; for example, the number of Democrats and the number of Republicans in a sample of n=100 registered voters.
There are situations where a researcher has questions about the proportions or relative frequencies for a distribution.
• How does the number of women lawyers compare with the number of men in the profession?
• Of the three leading brands of soft drinks, which one is preferred by most Chinese? Which brands are second and third, and how big are the differences in popularity among the three?
• To what extent are different ethnic groups represented in the population of UIC?
Chi-square test
The chi-square test for goodness of fit uses sample data to test hypotheses about the shape or proportions of a population distribution.
The test determines how well the obtained sample proportions fit the population proportions specified by the null hypothesis.
For example, a hypothesis might state that 90% of all lawyers are men and only 10% are women.
The simplest way of presenting this hypothesis is to put the hypothesized proportions in the series of boxes representing the scale of measurement:
Generally H0 will fall into one of the following categories:
• No Preference. The null hypothesis often states that there is no preference among the different categories.
For example, a hypothesis stating that there is no preference among the three leading brands of soft drinks would specify a population distribution as follows:
2. No Difference from a Comparison Population. The null hypothesis can state that the frequency distribution for one population is not different from the distribution that is known to exist for another population.
For example, suppose it is known that 60% of Americans favor the president’s foreign policy and 40% are in opposition. A researcher might wonder if this same pattern of attitudes exists among Europeans. The null hypothesis would state that there is no difference between the two populations and would specify that the Europeans are distributed as follows:
The data for a chi-square test are remarkably simple to obtain. You just select a sample of n individuals and count how many are in each category. The resulting values are called observed frequencies.
For example, the following data represent observed frequencies for a sample of n=40 subjects. Each person was given a personality questionnaire and classified into one of three personality categories: A, B, or C.
=
The observed frequency,,is the number of individuals from the sample who are classified in a particular category. Each individual is counted in one and only one category.
The general goal of the chi-square test for goodness of fit is to compare the data (the observed frequencies) with the null hypothesis. The problem is to determine how well the data fit the distribution specified in H0—hence the name goodness of fit.
The first step in the chi-square test is to determine how the sample distribution would look if the sample was in perfect agreement with the proportions stated in the null hypothesis.
Suppose, for example, the null hypothesis states that the population is distributed in three categories with the following proportions:
25% of 40 5 0.25(40)=10 individuals in category A
50% of 40 5 0.50(40)=20 individuals in category B
25% of 40 5 0.25(40)=10 individuals in category C
Expected frequency
The expected frequency, ,for each category is the frequency value that is predicted from the null hypothesis and the sample size (n).
The expected frequencies define an ideal, hypothetical sample distribution that would be obtained if the sample proportions were in perfect agreement with the proportions specified in the null hypothesis.
Calculating the Chi-Square statistic
fe= expected frequencies, obtained from the null hypothesis
fo = observed frequencies
C = the number of categories.
It should be clear from the chi-square formula that the value of chi-square is measuring the discrepancy between the observed frequencies (data) and the expected frequencies (H0). When there are large differences between the observed and expected frequencies, the value of chi-square will be large, and we will conclude that the data do not fit the hypothesis.
A small chi-square value indicates that we fail to reject H0.
To decide whether a particular chi-square value is “large” or “small,” we must refer to a chi-square distribution.
Chi-square distributions are positively skewed. The critical region is placed in the extreme tail, which reflects large chi-square values.
DEGREES OF freedom for the chi-square test literally measure the number of free choices that exist when you are determining the null hypothesis or the expected frequencies.
For example, when you are classifying individuals into three categories, you have exactly two free choices in stating the null hypothesis.
n=40
df = C - 1
Hypothesis Testing with Chi-Square
Chi-square follows five steps:
• Making assumptions (random sampling)
• State the research and null hypotheses and select an alpha
• Compute the expected distribution
• Compute the chi-square statistic
• Make a decision and interpreting the results
The Assumptions
• The chi-square test requires no assumptions about the shape of the population distribution from which the sample was drawn.
• However, like all inferential techniques it assumes random sampling.
• It can be applied to variables measured at a nominaland/or an ordinal level of measurement.
Example:
A researcher is interested in the factors that are involved in course selection. A sample of 50 students is asked, “Which of the following factors is most important to you when selecting a course?”
Students must choose one and only one of the following alternatives:
1. Interest in the course topic
2. Ease of passing the course
3. Instructor for the course
4. Time of day the course is offered
The frequency distribution of responses for this sample (the set of observed frequencies) is as follows:
Step 1
H0 : The population of students shows no preference in selecting any one of the four factors over the others.
H1 : In the population of students, one or more of these factors play a greater role in course selection (the factor is named more frequently by students).
The level of significance is set at a standard value, alpha=0.05.
Step 2:
The value for degrees of freedom is determined
df=C-1=4-1=3
The critical region is sketched.
Step 3:
The calculation of chi-square is actually a two-stage process.
Step 4:
The obtained chi-square value is in the critical region. Therefore, H0 is rejected. The researcher may conclude that there are significant differences among the four factors, with some selected more often and others less often than would be expected by chance.
Example:
A researcher for an insurance company would like to know if high-performance, overpowered automobiles are more likely to be involved in accidents than other types of cars. For a sample of 50 insurance claims, the investigator classifies the automobiles as high-performance, subcompact, mid-size, or full-size. The observed frequencies are as follows:
Assume that only 10% of the cars in the population are the high-performance variety. However, subcompacts, midsize cars, and full-size cars make up 40%, 30%, and 20%, respectively. Can the researcher conclude that the observed pattern of accidents does not fit the predicted expected-frequencies? Test with alpha=0 .05.
a. H0 : In the population, no particular type of car shows a disproportionate number of accidents.
H1 : In the population, a disproportionate number of the accidents occur with certain types of cars.
b. df=3; the critical X2 value is 7.81.
c. The expected-frequenciesfor high-performance, subcompact, midsize, and full-size cars are 5, 20, 15, and 10, respectively.
The obtained chi-square is 51.17.
d. Reject H0 .
THE CHI-SQUARE TEST FOR INDEPENDENCE
The chi-square statistic may also be used to test whether or not there is a relationship
between two variables.
In this situation, each individual in the sample is measured or classified on two separate variables.
Usually the data from this classification are presented in the form of a matrix, where the rows correspond to the categories of one variable and the columns correspond to the categories of the second variable.
Example:
A sample of n=200 students who have been classified by personality and color preference. Each student is given a personality test, and each student is asked to select a preferred color from among the four choices. Notice that the classification is based on the measurements for each student; the researcher does not assign students to categories. Also, the data consist of frequencies, not scores, from a sample.
Version 1The data are viewed as a single sample with each
individual measured on two variables.
H0 : For the general population of students, there is no relationship between color preference and personality.
The data consist of two measurements (X and Y) for each individual, and the goal is to evaluate the relationship between the two variables.
Version 2The data are viewed as two (or more) separate
samples representing two (or more) separate populations.
H0: In the general population of students, the distribution of color preferences has the same shape (same proportions) for both categories of personality.
The data consist of two (or more) separate samples that are being used to test for differences between two (or more) populations.
These two versions are equivalent.
The first version of H0 states that color preference is not related to personality.
If this hypothesis is correct, then the distribution of color preferences should not depend on personality. In other words, the distribution of color preferences should be the same for introverts and extroverts, which is the second version of H0.
Two variables are independent when there is no consistent, predictable relationship between them.
In this case, the frequency distribution for one variable is not related to (or dependent on) the categories of the second variable.
As a result, when two variables are independent, the frequency distribution for one variable will have the same shape for all categories of the second variable.
You can use either version of the null hypothesis to find the expected frequencies, but the logic of the process is much easier when you use H0 stated in terms of equal proportions.
H0 : The frequency distribution of color preference has the same shape (same proportions) for both categories of personality.
100 out of 200 = 50% prefer red
20 out of 200 = 10% prefer yellow
40 out of 200 = 20% prefer green
40 out of 200 = 20% prefer blue
DEGREES OF FREEDOM
The degrees of freedom for the chi-square test of independence are given by the formula
df = (R - 1)(C - 1)
where R is the number of rows and C is the number of columns.
Degrees of freedom and expected frequencies. (Once three values have been selected, all the remaining expected frequencies are determined by the row totals and the column totals. This example has only three free choices, so df = 3.)
Example:
A researcher suspects that color blindness is inherited by a sex-linked gene. This possibility is examined by looking for a relationship between gender and
color vision. A sample of 1000 people is tested for color blindness, and then they are classified according to their sex and color vision status (normal, red-green
blind, other color blindness).
Is color blindness related to gender?
The data are as follows:
ASSUMPTIONS AND RESTRICTIONS FOR CHI-SQUARE TESTS
• Independence of Observations
A chi-square test would be inappropriate if a person could produce responses that can be classified in more than one category or contribute more than one frequency count to a single category.
2. Size of Expected Frequencies
A chi-square test should not be performed when the expected frequency of any cell is less than 5. The chi-square statistic can be distorted when fe is very small. | 3,232 | 16,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-43 | latest | en | 0.811328 |
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Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances
Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume $$\sigma_x = \sigma_y$$. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter σ.[1]
NB: It is common to describe normal distributions using variance, or $$\sigma^2$$, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter $$\gamma = \sigma^2$$. To clarify our parameterization the σ we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution:
$$\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}$$
Where the bivariate normal distribution describes the coordinates (x, y) of shots on target, the Rayleigh distribution describes the distance, or radius, $$r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}$$ of those shots from the center point of impact.
Estimating σ
The Rayleigh distribution provides closed form expressions for precision. However, when estimating σ from sample sets we will most often use methods associated with the normal distribution for one essential reason: We never observe the true center of the distribution. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of Media:Sigma1ShotStatistics.ods.) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.
Correction Factors
The following three correction factors will be used throughout this statistical inference and deduction.
Note that all of these correction factors are > 1, are significant for very small n, and converge towards 1 as $$n \to \infty$$. Their values are listed for n up to 100 in Media:Sigma1ShotStatistics.ods. File:SymmetricBivariate.c uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.
Bessel correction factor
The Bessel correction removes bias in sample variance.
$$c_{B}(n) = \frac{n}{n-1}$$
Gaussian correction factor
The Gaussian correction (sometimes called $$c_4$$) removes bias introduced by taking the square root of variance.
$$\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})$$
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation: $$c_{G}(n)$$ =1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))
Rayleigh correction factor
The unbiased estimator for the Rayleigh distribution is also for $$\sigma^2$$. The following corrects for the concavity introduced by taking the square root to get σ.
$$c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}$$ [2]
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: =EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))
Data
In the following formulas assume that we are looking at a target reflecting n shots and that we are able to determine the center coordinates x and y for each shot.
(One easy way to compile these data is to process an image of the target through a program like OnTarget Precision Calculator.)
Variance Estimates
For a single axis the unbiased estimate of variance for a normal distribution is $$s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$$, from which the unbiased estimate of standard deviation is $$\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}$$.
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the x and y axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance $$s^2 = (s_x^2 + s_y^2)/2$$, and $$\hat{\sigma} = c_G(2n-1) \sqrt{s^2}$$. This turns out to be identical to the Rayleigh estimator.
Rayleigh Estimates
The Rayleigh distribution describes the random variable R defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center $$(\bar{x}, \bar{y})$$. Then for each shot we can compute the sample radius $$r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}$$.
The unbiased Rayleigh estimator is $$\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}$$, which is literally a restatement of the combined variance estimate $$s^2$$. Hence the unbiased parameter estimate is once again $$\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}$$.
Rayleigh sigma estimate provides a derivation of this formula, as well as a variation for the case in which the true center is known.
Confidence Intervals
Siddiqui[3] shows that the confidence intervals are given by the $$\chi^2$$ distribution with 2n degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(n - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance $$s^2$$.)
To find the (1 - α) confidence interval, first find $$\chi_1^2, \ \chi_2^2$$ where:
$$Pr(\chi^2(2(n-1)) \leq \chi_1^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_2^2) = 1 - \alpha/2$$
For example, using spreadsheet functions we have $$\chi_1^2$$ = CHIINV(α/2, 2n-2),$$\quad \chi_2^2$$ = CHIINV((1-α/2), 2n-2).
Now the confidence intervals are given by the following:
$$s^2 \in \left[ \frac{2(n-1) s^2}{\chi_2^2}, \ \frac{2(n-1) s^2}{\chi_1^2} \right]$$, or in equivalent Rayleigh terms $$\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_2^2}, \ \frac{\sum r^2}{\chi_1^2} \right]$$
Using the more convenient Rayleigh expression the confidence interval for the precision parameter is:
$$\widehat{\sigma} \in \left[ c_G(2n-1) \sqrt{\frac{\sum r^2}{\chi_2^2}}, \ c_G(2n-1) \sqrt{\frac{\sum r^2}{\chi_1^2}} \right]$$
How large a sample do we need?
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at $$\overline{r^2} = 2$$. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.
The 3-shot Group
Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has r = .29".
A rifle builder sends you a 3-shot group measuring ½" between each of three centers to prove how accurate your rifle is. What does that really say about the gun's accuracy? In the best case — i.e.:
1. The group was actually fired from your gun
2. The group was actually fired at the distance indicated (in this case 100 yards)
3. The group was not cherry-picked from a larger sample — e.g., the best of an unknown number of test 3-shot groups
4. The group was not clipped from a larger group (in the style of the "Texas Sharpshooter")
— if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is $$r_i = \frac{1}{2 \sqrt{3}} \approx .29"$$.
The Rayleigh estimator $$\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}$$. So $$\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA$$. Not bad! But not very significant. Let's check the confidence intervals: For α = 5% (i.e., 95% confidence intervals)
$$\chi_1^2(4) \approx 0.484, \quad \chi_2^2(4) \approx 11.14$$. Therefore,
$$0.02 \approx \frac{1}{4 \chi_2^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_1^2} \approx 0.52$$, and
$$0.16 \leq \hat{\sigma} \leq 0.76$$
so with 95% certainty we can only say that the gun's true precision σ is somewhere in the range from approximately 0.2MOA to 0.8MOA.
Using σ
Rayleigh distribution of shots given σ
The σ we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function $$\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}$$. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:
$$Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma}$$
Therefore, we expect 39% of shots to fall within a circle of radius σ, 86% within , and 99% within .
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful precision measures:
Mean Radius $$MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma$$.
$$1 - e^{-\frac{\pi}{4}} \approx 54\%$$ of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).
Given σ, the expected sample MR of a group of size n is
$$MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}$$
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for σ, even though in the limit the true value of one is a constant multiple of the other.)
Circular Error Probable (CEP)
For the Rayleigh distribution, the 50%-Circular Error Probable is $$CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma$$. 50% of shots should fall within a circle with this radius around the point-of-aim. See Circular Error Probable for a more detailed discussion.
In theory CEP is the median radius, but especially for small n the sample median is a very bad estimator for the true median. Nevertheless, if you want to know the expected sample median radius of a group of size n it turns out the following is a good estimate:
$$CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}$$
Summary Probabilities
Name Multiple x of σ Shots Covered by Circle of Radius x σ
1 39%
CEP 1.18 50%
MR 1.25 54%
2 86%
MD 2.5 96%
3 99%
Typical values of σ
A lower bound on σ is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which as we will see shortly suggests σ = 0.070MOA, or under 0.025mil.
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means σ = 0.28MOA, or under 0.1mil.
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means σ = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or σ = 1MOA!
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means σ = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.
How many sighter shots do you need?
99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.
How many shots do you need to zero your scope? As detailed in Sighter Distribution we know that the distance from the true center of a "sighting group" of n shots has a Rayleigh distribution with parameter $$\sigma / \sqrt{n}$$. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of σ. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for σ = 0.5MOA.
Sighter
Group Size
Average Distance
from True Zero
Error at 100 yards
for σ = 0.5MOA
Shots Lost to
Sighting Error
on 50% Target
Shots Lost to
Sighting Error
on 96% Target
3 0.7 σ 0.4" 8% 4%
5 0.6 σ 0.3" 6% 3%
10 0.4 σ 0.2" 3% 1%
20 0.3 σ 0.15" 2% <1%
The Rice distribution gives the expected hit probabilities when incorporating a sighting error ε σ. The Rice CDF is hard to calculate so here we used Wolfram Alpha to compute CDF values as $$F(x|ε, \sigma) = P(X \leq x)$$ = MarcumQ[1, ε, 0, x], which gives the probability of a hit within distance x σ given a sighting error of ε σ.
We define a t% target as a target large enough that t% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)
"Shots Lost to Sighting Error" on a t% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ε: i.e., $$F(t|0, \sigma) - F(t|ε, \sigma)$$.
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.
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#1
The position vectors of points A and B are (7, 1, -3) and (-2,8,-1)
a) Calculate vector OA . vector OB
Very simple, (7, 1, -3) . (-2, 8, -1) = -14 + 8 + 3 = -3
b) Find the size, in degrees, of the acute angle between the vectors.
lal = root (49 + 1 + 9) = root 59
lbl = root (4 + 64 + 1) = root 69
a.b = lal lbl cos x
so, -3 = root 59 x root 69 (cos x)
so -3 = root (4071) cos x
so cos x = -3/root (4071)
x = arccos (-3/root (4071)
= 92.7 degrees,
but it says the acute angle, so the angle = 180-x = 87.3
But the textbook says the answer is 87.1? Have I done anything wrong?
0
3 weeks ago
#2
(Original post by TAEuler)
The position vectors of points A and B are (7, 1, -3) and (-2,8,-1)
a) Calculate vector OA . vector OB
Very simple, (7, 1, -3) . (-2, 8, -1) = -14 + 8 + 3 = -3
b) Find the size, in degrees, of the acute angle between the vectors.
lal = root (49 + 1 + 9) = root 59
lbl = root (4 + 64 + 1) = root 69
a.b = lal lbl cos x
so, -3 = root 59 x root 69 (cos x)
so -3 = root (4071) cos x
so cos x = -3/root (4071)
x = arccos (-3/root (4071)
= 92.7 degrees,
but it says the acute angle, so the angle = 180-x = 87.3
But the textbook says the answer is 87.1? Have I done anything wrong?
Everything is right apart from it asks for the acute angle.
Just draw two lines (vectors). You could measure the angle between them as either 93 or 87 degrees depending on the direction of the vector.
Flip the sign of OA, you now get 3 for the dot product and the arccos() is 87.
Last edited by mqb2766; 3 weeks ago
0
3 weeks ago
#3
(Original post by TAEuler)
= 92.7 degrees,
but it says the acute angle, so the angle = 180-x = 87.3
But the textbook says the answer is 87.1? Have I done anything wrong?
Agree with 87.3 (3dec.pl.). Error in book.
#4
I need help with another question as I disagree with their answer, this one i'm less confident with my workings:
Is the point (-2,4,7) closer to the line with equation r = 5i + 2k + λ(i - 2j + 4k) or the line parallel to the vector 2i + j that passes through the origin?:
we know Line 1: r = (5, 0, 2) + λ (1, -2, 4)
and line 2: r = (0,0,0) + s (2, 1, 0)
let point B be on line 1 so that the vector AB is perpendicular to line 1.
so vector AB . (1, -2, 4) = 0
Vector OA = (-2, 4, 7) and Vector OB = (5 + λ, -2λ, 2 + 4λ)
so vector AB = (7 + λ, -2λ - 4, 4λ - 5)
so (7 + λ, -2λ - 4, 4λ - 5) . (1, -2, 4) = 0
7 + λ - 2(-2λ - 4) + 4 (4λ - 5) = 0
7 + λ + 4 λ + 8 + 16λ - 20 = 0
so λ = 5/21
so Vector AB = (152/21, -94/21, -85/21)
lABl = root (1865/21) = 9.424 (3sf)
Let point C be on line 2 so that vector AC is perpendicular to line 2
So vector OC = (2s, s, 0)
vector AC = (2s, s, 0) - (-2, 4, 7)
= (2s + 2, s - 4, -7)
So Vector AC . (2, 1, 0) = 0
(2s + 2, s - 4, -7) . (2, 1, 0) = 0
4s + 4 + s -4 = 0
5s = 0
so s = 0
so Vector AC = (2, -4, -7)
lACl = root 69
therefore, the point is closer to the second line.
However, the book says it's closer to the first?
Have I gone wrong or is it another textbook mistake?
0
3 weeks ago
#5
(Original post by TAEuler)
I need help with another question as I disagree with their answer, this one i'm less confident with my workings:
Is the point (-2,4,7) closer to the line with equation r = 5i + 2k + λ(i - 2j + 4k) or the line parallel to the vector 2i + j that passes through the origin?:
we know Line 1: r = (5, 0, 2) + λ (1, -2, 4)
and line 2: r = (0,0,0) + s (2, 1, 0)
let point B be on line 1 so that the vector AB is perpendicular to line 1.
so vector AB . (1, -2, 4) = 0
Vector OA = (-2, 4, 7) and Vector OB = (5 + λ, -2λ, 2 + 4λ)
so vector AB = (7 + λ, -2λ - 4, 4λ - 5)
so (7 + λ, -2λ - 4, 4λ - 5) . (1, -2, 4) = 0
7 + λ - 2(-2λ - 4) + 4 (4λ - 5) = 0
7 + λ + 4 λ + 8 + 16λ - 20 = 0
so λ = 5/21
so Vector AB = (152/21, -94/21, -85/21)
lABl = root (1865/21) = 9.424 (3sf)
Let point C be on line 2 so that vector AC is perpendicular to line 2
So vector OC = (2s, s, 0)
vector AC = (2s, s, 0) - (-2, 4, 7)
= (2s + 2, s - 4, -7)
So Vector AC . (2, 1, 0) = 0
(2s + 2, s - 4, -7) . (2, 1, 0) = 0
4s + 4 + s -4 = 0
5s = 0
so s = 0
so Vector AC = (2, -4, -7)
lACl = root 69
therefore, the point is closer to the second line.
However, the book says it's closer to the first?
Have I gone wrong or is it another textbook mistake?
I did it by minimising the square of the distances - a different method - and got the exact same results as you. So, book's in error.
#6
Another question, this one in regards to me not being able to visualise something as it's in 3-D:
the line L1 is given by r = (6, -3, 1) + λ (-1, 1, -2)
The line L2 is given by r = (4, 2, -5) + s(-5, 2, -8).
The point A is the intersection of the lines L1 and L2 and the point B has coordinates (1, -2, 7). Calculate the area of the triangle OAB:
To find A...
(6, -3, 1) + λ (-1, 1, -2) = (4, 2, -5) + s(-5, 2, -8)
x: 6 - λ = 4 - 5s
y: λ - 3 = 2 + 2s
z: 1 - 2λ = -5 -8s
through simultaneous equations, I got the A as (-1, 4, -13)
This is where I struggle to visualise it, I would try and use dot product but I can't figure out how the angle calculated plays into it, it could potentially be 180 - the angle from dot product and then simply 1/2 ab sin C to find the area. So I'm not too sure on what to do, or how to visualise it. I also tried a sketch but I couldn't seem to display the vectors.
I also worked out lOAl as root 186
and lOBl as 3 root 6.
0
3 weeks ago
#7
(Original post by TAEuler)
This is where I struggle to visualise it, I would try and use dot product but I can't figure out how the angle calculated plays into it, it could potentially be 180 - the angle from dot product and then simply 1/2 ab sin C to find the area.
Well the fact that it will not make a difference whether you use one angle or the other.
So I'm not too sure on what to do, or how to visualise it. I also tried a sketch but I couldn't seem to display the vectors.
I also worked out lOAl as root 186
and lOBl as 3 root 6.
You know A. So to use you should calculate and with being the angle between AO and AB.
Any sort of diagram like this is sufficient to understand what you need to calculate:
[EDIT: It's actually less work if you just consider the vectors OA and OB instead]
Last edited by RDKGames; 3 weeks ago
0
#8
A simpler question:
This is for my mate but I wasn't able to explain it for him:
Points A and B have position vectors OA = (2, 2, 3) and OB = (-1, 7, 2). Find the angle between AB and OA
The textbook specifies to use the scalar product of two vectors to find the angle between them, you need both vectors to be directed away from the angle, otherwise the angle you find will be 180 - the angle.
His workings:
Let the angle between the position vectors AB and AO be x.
Vector AB = (-1, 7, 2) - (2, 2, 3) = (-3, 5, -1)
lABl = root 35
A
Vector AO = (-2, -2, -3). lAOl = root 17
AB . AO = -1
x = arccos (-1/root 595) = 92.3 degrees
However, the answer given is 87.7 degrees.
This has confused both me and him, have they calculated the wrong angle? And if not why is it 180 - x
0
3 weeks ago
#9
(Original post by TAEuler)
A simpler question:
This is for my mate but I wasn't able to explain it for him:
Points A and B have position vectors OA = (2, 2, 3) and OB = (-1, 7, 2). Find the angle between AB and OA
The textbook specifies to use the scalar product of two vectors to find the angle between them, you need both vectors to be directed away from the angle, otherwise the angle you find will be 180 - the angle.
His workings:
Let the angle between the position vectors AB and AO be x.
Vector AB = (-1, 7, 2) - (2, 2, 3) = (-3, 5, -1)
lABl = root 35
A
Vector AO = (-2, -2, -3). lAOl = root 17
AB . AO = -1
x = arccos (-1/root 595) = 92.3 degrees
However, the answer given is 87.7 degrees.
This has confused both me and him, have they calculated the wrong angle? And if not why is it 180 - x
Nothing in the question suggests you seek the acute/obtuse angle, so no need to aim for one or the other.
Also, since the questions says between AB and OA then those are the vectors you need to dot.
You dotted AB with AO instead.
The dot product formula gives the angle between the vectors that are ‘head-to-head’ or ‘tail-to-tail’. It may not seem that you can take the dot product between AB and OA to get anything meaningful given this, but notice that if (on a diagram) you extend the vector OA through A to A’ so that it is now OA’ = 2*OA, you should realise that OA = AA’ and AA’ has its tail to AB. So that’s the angle the between the tails that the question effectively wants.
Last edited by RDKGames; 3 weeks ago
0
#10
(Original post by RDKGames)
Nothing in the question suggests you seek the acute/obtuse angle, so no need to aim for one or the other.
Also, since the questions says between AB and OA then those are the vectors you need to dot.
You dotted AB with AO instead.
The dot product formula gives the angle between the vectors that are ‘head-to-head’ or ‘tail-to-tail’. It may not seem that you can take the dot product between AB and OA to get anything meaningful given this, but notice that if (on a diagram) you extend the vector OA through A to A’ so that it is now OA’ = 2*OA, you should realise that OA = AA’ and AA’ has its tail to AB. So that’s the angle the between the tails that the question effectively wants.
I don't get it. It says between vectors OA and AB, but the angle between these two is OAB, the vector OA ends at A so why should it be the acute angle because the two and not the obtuse?
0
3 weeks ago
#11
(Original post by TAEuler)
I don't get it. It says between vectors OA and AB, but the angle between these two is OAB, the vector OA ends at A so why should it be the acute angle because the two and not the obtuse?
It’s not necessarily *that* angle tho...
The vector OA is just that, a vector. On your diagram this vector can be anywhere you want it to be as long as it has the same magnitude and direction. It need not start start at O, which is what you are relying on too much. I explained in my last post how AA’ is an equivalent vector to OA, so “the angle between OA and AB” is the same as saying “the angle between AA’ and AB” and if you look at the vector that starts at A and goes to A’ you would see that it has its tail to the vector going from A to B. So at no point do we get the angle you mistakenly look for.
0
#12
"Show that the lines with equations r = (5i - 2j + k) + t(2i - j - k) and r = (i + j - k) + s(-2 + i + j + 5k) are skew."
Is this an error in question, since the -2 is left on it's own and doesn't have an i, j or k?
0
3 weeks ago
#13
(Original post by TAEuler)
"Show that the lines with equations r = (5i - 2j + k) + t(2i - j - k) and r = (i + j - k) + s(-2 + i + j + 5k) are skew."
Is this an error in question, since the -2 is left on it's own and doesn't have an i, j or k?
[Without working it out] I think it's supposed to be instead. Typo.
0
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Preparing for the CHSPE Math test? Want a preview of the most common mathematics questions on the CHSPE Math test? If so, then you are in the right place.
The mathematics section of CHSPE can be a challenging area for many test-takers, but with enough patience, it can be easy and even enjoyable!
Preparing for the CHSPE Math test can be a nerve-wracking experience. Learning more about what you’re going to see when you take the CHSPE can help to reduce those pre-test jitters. Here’s your chance to review the 10 most common CHSPE Math questions to help you know what to expect and what to practice most. Try these 10 most common CHSPE Math questions to hone your mathematical skills and to see if your math skills are up to date on what’s being asked on the exam or if you still need more practice.
Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice.
## 10 Sample CHSPE Math Practice Questions
1- Simplify $$6x^2 y^3 (2x^2 y)^3=$$
A. $$12x^4 y^6$$
B. $$12x^8 y^6$$
C. $$48x^4 y^6$$
D. $$48x^8 y^6$$
2- The diagonal of a rectangle is 10 inches long and the height of the rectangle is 8 inches. What is the perimeter of the rectangle in inches?______
3-What is the value of $$x$$ in the following equation?
$$\frac{2}{3} x+\frac{1}{6}= \frac{1}{3}$$
A. 6
B. $$\frac{1}{2}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{4}$$
4- A card is drawn at random from a standard 52–card deck, what is the probability that the card is of Hearts? (The deck includes 13 of each suit clubs, diamonds, hearts, and spades)
A. $$\frac{1}{3}$$
B. $$\frac{1}{6}$$
C. $$\frac{1}{52}$$
D. $$\frac{1}{4}$$
5- Which of the following shows the numbers in ascending order?
$$\frac{2}{3} , 0.68 , 67\% , \frac{4}{5}$$
A. $$67\%, 0.68, (\frac{2}{3} ), (\frac{4}{5})$$
B. $$67\%, 0.68, (\frac{4}{5} ), (\frac{2}{3})$$
C. $$0.68, 67\%, (\frac{2}{3} ), (\frac{4}{5})$$
D. $$(\frac{2}{3} ), 67\%, 0.68, (\frac{4}{5})$$
6- The mean of 50 test scores was calculated as 88. But, it turned out that one of the scores was misread as 94 but it was 69. What is the mean?
A. 85
B. 87
C. 87.5
D. 88.5
7- Two dice are thrown simultaneously, what is the probability of getting a sum of 6 or 9?
A. $$\frac{1}{3}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{6}$$
D. $$\frac{1}{2}$$
8- Jason is 9 miles ahead of Joe running at 5.5 miles per hour and Joe is running at the speed of 7 miles per hour. How long does it take Joe to catch Jason?
A. 3 hours
B. 4 hours
C. 6 hours
D. 8 hours
9-55 students took an exam and 11 of them failed. What percent of the students passed the exam?
A. $$40\%$$
B. $$60\%$$
C. $$80\%$$
D. $$20\%$$
10- What is the volume of a box with the following dimensions?
Height = 4 cm Width = 5 cm Length = 6 cm
A. 15 cm$$^3$$
B. 60 cm$$^3$$
C. 90 cm$$^3$$
D. 120 cm$$^3$$
## Best CHSPE Math Prep Resource for 2021
1- D
Simplify.
$$6x^2 y^3 (2x^2 y)^3= 6x^2 y^3 (8x^6 y^3 ) = 48x^8 y^6$$
2- 28
Let $$x$$be the width of the rectangle. Use Pythagorean Theorem:
$$a^2 + b^2 = c^2$$
$$x^2 + 8^2 = 10^2 {\Rightarrow} x^2 + 64 = 100 {\Rightarrow} x^2 = 100 – 64 = 36 ⇒ x = 6$$
Perimeter of the rectangle = 2 (length + width) = 2 (8 + 6) = 2 (14) = 28
3- D
Isolate and solve for $$x$$.
$$\frac{2}{3} x+\frac{1}{6} = \frac{1}{3} {\Rightarrow} \frac{2}{3} x= \frac{1}{3} -\frac{1}{6} = \frac{1}{6} {\Rightarrow} \frac{2}{3} x= \frac{1}{6}$$
Multiply both sides by the reciprocal of the coefficient of $$x$$.
$$(\frac{3}{2}) \frac{2}{3} x= \frac{1}{6} (\frac{3}{2}) {\Rightarrow} x= \frac{3}{12}=\frac{1}{4}$$
4- D
The probability of choosing a Hearts is $$\frac{13}{52}=\frac{1}{4}$$
5- D
Change the numbers to decimal and then compare.
$$\frac{2}{3} = 0.666…$$
$$0.68$$
$$67\% = 0.67$$
$$\frac{4}{5} = 0.80$$
Therefore
$$\frac{2}{3} < 67\% < 0.68 < \frac{4}{5}\$$
6- C
average (mean) $$=\frac{(sum \space of \space terms)}{(number \space of \space terms)} {\Rightarrow} 88 = \frac{(sum \space of \space terms)}{50} {\Rightarrow}$$ sum $$= 88 {\times} 50 = 4400$$
The difference of 94 and 69 is 25. Therefore, 25 should be subtracted from the sum.
$$4400 – 25 = 4375$$
mean$$= \frac{(sum of terms)}{(number of terms)} ⇒$$mean $$= \frac{(4375)}{50}= 87.5$$
7- B
To get a sum of 6 for two dice, we can get 5 different options:
(5, 1), (4, 2), (3, 3), (2, 4), (1, 5)
To get a sum of 9 for two dice, we can get 4 different options:
(6, 3), (5, 4), (4, 5), (3, 6)
Therefore, there are 9 options to get the sum of 6 or 9.
Since, we have 6 × 6 = 36 total options, the probability of getting a sum of 6 and 9 is 9 out of 36 or $$\frac{1}{4}$$.
8- C
The distance between Jason and Joe is 9 miles. Jason running at 5.5 miles per hour and Joe is running at the speed of 7 miles per hour. Therefore, every hour the distance is 1.5 miles less. 9 $$\div$$ 1.5 = 6
9- C
The failing rate is 11 out of 55 = $$\frac{11}{55}$$
Change the fraction to percent:
$$\frac{11}{55} {\times} 100\%=20\%$$
20 percent of students failed. Therefore, 80 percent of students passed the exam.
10- D
Volume of a box = length $$\times$$ width $$\times$$ height = 4 $$\times$$ 5 $$\times$$ 6 = 120
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# Algorithm Correctness & Analysis
Algorithm
A finite set of instructions which if followed accomplish a
following criteria:
1. Input: zero or more quantities externally supplied
2. Output: at least one quantity is produced
3. Definiteness: Each instruction must be clear and
unambiguous.
4. Finiteness: In all cases algorithm must terminate after finite
number of steps.
5. Effectiveness: each instruction must be sufficiently basic.
Two algorithms on two systems
Algorithm A1 50 n log n
Algorithm A2 2n2
A2 Super computer Time taken by Super Computer
108 ins/sec = 2.(106)2/ 108 = 20,000 sec
A1
P.C Time taken by P .C.
106 ins /sec = 50 .106 lg 106 / 106 = 1,000 sec
Thus by using a fast algorithm, the personal computer gives
results 20 times faster than the result given by super
computer using a slow algorithm.
## Thus a good algorithm is like a sharp knife, it does exactly what
it is supposed to do with a minimum amount of effort.
Purpose
To estimate how long a program will run.
the program.
## To help focus on the parts of code that are executed the
largest number of times.
Complexity
## There may be many algorithms for a given problem.
Which algorithm to use?
What are the classical algorithm design techniques ?
Are there problems inherently difficult to solve?
## How do we express the complexity of algorithm?
Resources : Time and Space
Complexity lower bounds for problems.
Complexity classes P, NP etc.
Pseudo code
Pseudo code is an English language like representation of the code
required for an algorithm
## The code part consists of an extended version of the basic
algorithmic constructs-sequence, selection and iteration.
Sequence, selection, loop
A sequence is a series of statements that do not alter the execution
path within an algorithm.
Statements such as assign and add are sequence statements.
A call to another algorithm is also considered a sequence
statement.
Selection statements evaluate one or more alternatives. Paths are
followed based on its result.
The typical selection statement is the two way selection if
(condition) action 1 else action 2.
The part of the loop are identified by indentation.
Loop iterates a block of code. It closely resembles the while loop.
It is a pretest loop.
Example
Algorithm: deviation
Pre: nothing
## Post: numbers are read and deviation from average printed
1 i= 0
1 i=i+1
3 sum = sum + number
3 Average = sum / i
4 print (average)
5 j=0
6 loop (j < i)
1 j = j+ 1
2 dev = array[j] average
3 print (array [ j] . Dev)
7 return
8 end deviation
Summary
Confidence in algorithms from testing and
correctness proof
## Correctness of recursive algorithms proved
directly by induction
## Examples: Fibonacci numbers, maximum,
multiplication
Correctness
How do we know that an algorithm works?
## Logical method for checking correctness
Testing
Correctness proof
## Correctness Proof: Prove mathematically; testing may not
found obscure bugs
Using testing alone can be dangerous
Correctness of Recursive
algorithms
To prove correctness of recursive algorithm:
## Need to prove the recursive calls are given sub-problems, i.e.,
no infinite recursion
## use this assumption to prove that the current call works
correctly
Recursive Fibonacci Numbers
Fibonacci numbers: F0 = 0, F1 =1, and for all n 2,
Fn = Fn-2 + Fn-1
function fib(n)
comment return F n
## 2. else return(fib (n 1) + fib(n 2))
Recursive Fibonacci Numbers
Claim: For all n 0, fib(n) return Fn
Base: For n =0, fib(n ) returns 0 as claimed. For n = 1, fib(n) returns 1
as claimed.
Induction: Suppose that n 2 and for all 0 m < n, fib (m) returns
Fm .
## fib(n-1) + fib(n-2) = Fn-1 + Fn-2 (by Ind. Hyp.)
= Fn
Recursive Maximum
function maximum(n)
comment Return max of A[1..n].
1. if n < 1 then return (A[1]) else
2. return(max(maximum(n 1).A[n]))
## Claim: For all n 1. maximum(n) returns max{A[1],A[2].,A[n]}.
Proof by induction on n > 1.
## Base: for n = 1, maximum(n) returns A[1] as claimed.
Recursive maximum
Induction: Suppose that n > 1 and maximum(n) returns
max{A[1],A[2],,A[n]}.
## RTP maximum (n + 1) returns
max{A[1],A[2],,A[n + 1]}.
## max(maximum(n), A[n + 1])
= max(max{A[1],A[2],,A[n]}. A[n + 1]
(by ind . hyp .)
= max{A[1],A[2],,A[n + 1]}.
Recursive Multiplication
Notation: For x IR, x is the largest integer not exceeding x.
function multiply(y, z)
2. if z is odd
## 4. else return (multiply(2y, z/2 ))
Recursive Multiplication
Claim: for all y, z 0, multiply(y, z) returns yz.
## Base: For z = 0, multiply(y, z) returns 0 as claimed.
Induction: Suppose that for z > 0, and for all 0 < q z. multiply (y, q) returns yq.
## There are two cases, depending on whether z + 1 is odd or even.
Recursive Multiplication
If z + 1 is odd, then multiply (y, z + 1) returns
multiply (2y. (z + 1)/2) + y
= 2y(z + 1)/2) + y (by ind. Hyp.)
= 2y(z/2) + y (since z is even)
= y(z + 1)
## If z + 1 is odd, then multiply (y, z + 1) returns
multiply (2y. (z + 1)/2)
= 2y (z + 1)/2 (by ind. Hyp.)
= 2y(z + 1)/2 (since z is odd)
= y(z + 1).
Summary
O, ,
Sum and product rule for O
Analysis of nonrecursive algorithms
Implementing Algorithm
Algorithm
Programmer
Program
Compiler
Executable
Computer
Constant Multiples
Analyze the resource usage of an algorithm to within a constant multiple.
## from an algorithm to executable code:
Programmer ability
Programmer effectiveness
Compiler
Computer hardware
## Recall: Measure resource usage as a function of input size
Big oh
We need a notation for within a constant multiple. Actually, we have several of
them.
## Informal definition: f(n) is O(g(n)) if f grows at most as fast as g.
Formal definition: f(n) = O(g(n)) if there exists c, n0 IR+ such that for all n n0,
f(n) c.g(n).
Example
Most big-Os can be proved by induction.
## Claim : For all n >1, log n <n . The proof is by induction on n.
The claim is trivially for n=1,since 0<1. Now suppose n > 1 and log n < n.
Then,
log (n+1)
< log 2n
= log n+1
## < n+1 (by ind. hyp.)
Second example
2n + 1 = O(3n/n).
## Claim: For all n 7, 2n + 1 3n/n.
The proof is induction on n.
The claim is true for n = 7, since 2n + 1 = 28 = 256, and 3n/n = 37/7 > 312.
Now suppose n 7 and 2n + 1 3n/n. RTP 2n + 2 3n + 1/(n + 1).
2n + 2
= 2.2n + 1
2.3n/n (by ind. hyp.)
3n / (n + 1) . 3n / n (see below)
= 3n + 1 / (n + 1).
Note that we need
3n/(n + 1) 2
3n 2n + 2
n2
Big Omega
Informal Definition: f(n) is (g(n)) if f grows at least as fast as g.
Formal definition: f(n) = (g(n)) if there exists c > 0 such that there are infinitely
many n IN such that f(n) c.g(n)
Big Theta
Informal Definition: f(n) is (g(n)) if f is essentially the same as g, to
within a constant multiple.
## Formal definition: f(n) = (g(n)) if f(n) = O(g(n)) and f(n) = (g(n))
True or false
3n5 16n + 2 = O(n5)?
3n5 16n + 2 = O(n)?
3n5 16n + 2 = O(n17)?
3n5 16n + 2 = (n5)?
3n5 16n + 2 = (n)?
3n5 16n + 2 = (n17)?
3n5 16n + 2 = (n5)?
3n5 16n + 2 = (n)?
3n5 16n + 2 = (n17)?
Claim. If f1 (n) = O(g1(n)) and f2(n) = O(g2(n)), then
f1 (n) + f2 (n) = O(g1(n) + g2(n)).
Proof: Suppose for all n n1, f1(n) c1. g1(n) and for all n n2,
f2(n) c2. g2(n).
Let n0 = max{n1, n2} and c0 = max{c1, c2}. Then for all n n0,
## f1(n) + f2(n) c1.g1(n) + c2.g2(n)
c0(g1(n) + g2(n)).
Continue
Claim. If f1 (n) = O(g1(n)) and f2(n) = O(g2(n)), then
f1 (n) + f2 (n) = O(max{g1(n), g2(n)}).
Proof: Suppose for all n n1, f1(n) c1. g1(n) and for all n n2,
f2(n) c2. g2(n).
## f1(n) + f2(n) c1.g1(n) + c2.g2(n)
(c1 + c2)(max{g1(n), g2(n)})
= c0(max{g1(n), g2(n)}).
Multiplying Big Ohs
Claim. If f1 (n) = O(g1(n)) and f2(n) = O(g2(n)), then
f1 (n) . f2 (n) = O(g1(n) . g2(n)).
Proof: Suppose for all n n1, f1(n) c1 . g1(n) and for all n n2,
f2(n) c2. g2(n).
## f1(n) . f2(n) c1. g1(n) . c2.g2(n)
= c0 . g1(n) . g2(n).
Types of Analysis
Worst case: time taken if the worst possible thing happens. T(n) is the maximum
time taken over all inputs of size n.
Average Case: The expected running time, given some probability distribution on
the inputs (usually uniform). T(n) is the average time taken over all inputs of size
n.
Probabilistic: The expected running time for a random input. (Express the
running time and the probability of getting it.)
Amortized: the running time for a series of executions, divided by the number of
executions.
Example
Consider an algorithm that for all inputs of n bits takes time 2n, and
for one input of size n takes time nn.
## Worst case: (nn)
Average Case:
nn + (2n 1)n nn
( ) = ( )
2 2
Probablistic: O(n) with probability 1 1 / 2n.
## Amortized: A sequence of m executions on different inputs takes
amortized time.
nn + (m 1)n nn
O( ) = O( )
Time Complexity
Well do mostly worst-case analysis. How much time does it take to
execute an algorithm in the worst case?
assignment O(1)
procedure entry O(1)
procedure exit O(1)
if statement time for test plus
O(max of two branches)
loop sum over all iterations of
the time for each iteration
## Put these together using sum rule and product rule.
Exception--- recursive algorithms.
Multiplication
function multiply(y, z)
comment Return yz, where y, z IN
1. x := 0;
2. while z > 0 do
3. if z is odd then x := x + y;
4. y := 2y; z:= z/2 ;
5. return (x)
## Suppose y and z have n bits.
Procedure entry and exit cost O(1) time
Lines 3, 4 cost O(1) time each
The while-loop on lines 2-4 costs O(n) time (it is executed at most n times).
Line 1 costs O(1) time
Therefore, multiplication takes O(n) time (by the sum and product rules).
Bubble sort
Procedure bubblesort(A[1..n])
For i := 1 ton 1 do
For j := 1 ton ido
If A[j] >A[j + 1] then
Swap A[j] with A[j + 1]
Procedure entry and exit costs O(1) time
Line 5 costs O(1) time
The if-statement on lines 4-5 costs O(1) time
The for-loop on lines 3- 5 costs O(n i) time
1
The for-loop on lines 2- 5 costs O( =1 ) time.
1
O( =1 )= O(n(n 1) - 1 2
=1 ) = ( )
Therefore, bubblesort takes time O(n2) in the worst case.
2 2
Analysis Trick
Rather than go through the step-by-step method of analyzing algorithms,
Identify the fundamental operation used in the algorithm, and observe that the
running time is a constant multiple of the number of fundamental operations
used. (Advantage: no need to grunge through line-by-line analysis.)
## This often helps you stay focused, and work faster.
Example
In the bubblesort example, the fundamental operation is the
comparison done in line 4. The running time will be big-O of
the number of comparisons.
Line 4 uses 1 comparison
The for-loop on lines 3-5 uses n i comparisons
1
The for-loop on lines 2 5 uses =1 comparisons,
and
n 1 n 1
(n i) n(n 1) i
i 1 i 1
n( n 1) n( n 1) / 2
n( n 1) / 2
Algorithms and Problems
Big-Os mean different things when applied to algorithms and
problems.
Bubblesort runs in time O(n2). But is it tight? Maybe I was too
lazy to figure it out, or maybe its unknown.
Bubblesort runs in time (n2). This is tight.
The sorting problem takes time O(n log n). There exists an
algorithm that sorts in time O(n log n), but I dont know if there is
a faster one.
The sorting problem takes time (n log n). There exists an
algorithm that sorts in time O(n log n), and no algorithm can do
any better.
Algorithms Analysis 2
Summary
Analysis of iterative (nonrecursive) algorithms.
The heap: an implementation of the priority queue
Insertion in time O(log n)
Deletion of minimum in time O(log n)
Heapsort
Build a heap in time O(n log n).
Dismantle a heap in time O(n log n).
Worst case analysis O(n log n).
How to build a heap in time O(n).
The Heap
A priority queue is a set with the operations
Insert an element
Delete and return the smallest element
## A popular implementation: the heap. A heap is a binary tree with
the data stored in the nodes. It has two important properties:
1. Balance: It is as much like a complete binary tree as possible.
Missing leaves, if any, are on the last level at the far right.
Contd
2. Structure: The value in each parent is the values in its
children.
3
9 5
11 20 10 24
21 15 30 40 12
Note this implies that the value in each parent is the values in
its descendants (Includes self).
To Delete the Minimum
## 1. Remove the root and return the value in it.
3
?
9 5
11 20 10 24
21 15 30 40 12
Contd
2. Replace root with last leaf.
12
9 5
11 20 10 24
21 15 30 40 12
12
9 5
11 20 10 24
21 15 30 40
## But we have violated the structure condition
Contd
3. Repeatedly swap the new element with its smallest child until it reaches a place
where it is no larger than its children.
12
9 5
11 20 10 24
21 15 30 40
5
9 12
11 20 10 24
21 15 30 40
Contd
5
9 12
11 20 10 24
21 15 30 40
9 10
11 20 12 24
21 15 30 40
What Does it work?
Why does swapping the new node with its smallest child work?
a a
or
b c c b
## Suppose b c and a is not in the correct place. That is, either
a > b or a > c. In either case, since b c, we know that a > b.
Contd
a a
or
b c c b
Then we get
b b
or
a c c a
Respectively,
Is b smaller than its children? Yes, since b < a and b c.
Contd
Is c smaller than its children?
## Yes, since it is unchanged.
To Insert a New Element
1. Put the new element in the next leaf. This preserves the
balance.
Contd
## 2. Repeatedly swap the new element with its parent until it
reaches a place where it is no smaller than its parent
Contd
Contd
Why Does it Work
Why does swapping the new node with its parent work?
a
b c
d e
Contd
c
b a
d e
## Is a larger than its parent? Yes, since a > c.
Is b larger than its parent? Yes, since b > a > c.
Is c larger than its parent? Not necessarily. Thats why we continue
to swap
Is d larger than its parent? Yes, since d was a descendant of a in the
original tree, d > a.
Is e larger than its parent? Yes, since e was a descendant of a in the
original tree, e > a.
Do the sub-trees of b, d, e still have the structure condition? Yes,
since they are unchanged.
Implementing a Heap
An n node heap uses an array A[1..n].
The root is stored in A[1]
The left child of a node in A[i]is stored in node A[2i]
The right child of a node in A[i] is stored in node A[2i + 1].
Analysis of Priority Queue
Operation
Delete the Minimum:
1. Remove root O(1)
2. Replace root O(1)
3. Swaps O(l(n))
Insert:
## 1. Put in leaf O(1)
2. Swaps O(l(n))
Analysis of l(n)
A complete binary tree with k levels has exactly 2k 1 nodes(can
prove by induction). Therefore, a heap with k levels has no fewer
than 2k 1 nodes and no more than 2k 1 nodes.
K-1
2k 1 - 1
nodes k
2k 1
nodes
Contd
Therefore, in a heap with n nodes and k levels:
2k-1 n 2k - 1
k1 log <k
k1 log k
k1 log k1
log =k
k = log + 1
Hence, number of levels is l(n) = log + 1
Example
## Left side: 8 nodes, log 8 + 1 = 4 levels. Right side: 15
nodes, log 15 + 1 = 4 levels.
## So, insertion and deleting the minimum from an n-node
heap requires time O(log n).
Heapsort
Algorithm:
To sort n numbers
1. Insert n numbers into an empty heap.
2. Delete the minimum n times.
The numbers come out in ascending order.
Analysis:
Each insertion costs time O(log n). Therefore, cost of line 1 is
O(n log n).
Each deletion costs time O(log n). Therefore, cost of line 2 is
O(n log n).
Therefore, heapsort takes time O(n log n) in the worst case.
Building a Heap Top Down
Cost of building a heap proportional to number of comparisons.
The above method builds from the top down.
Contd
## Cost of an insertion depends on the height of the heap. There are
lots of expensive nodes.
Contd
Number of comparisons (assuming a full heap ):
l ( n ) 1
j 0
j 2 j (l (n)2l ( n ) ) (n log n).
## How do we know this? Can prove by induction that:
j2j 1
j
(k 1)2 k 1 2
( k 2 k )
Building a Heap Bottom Up
## Cost of an insertion depends on the height of the heap. But now
there are few expensive nodes.
Continue
## Number of comparisons is (assuming a full heap):
l ( n) l (n) l (n)
i 1
l ( n ) i
(i 1).2 2 l (n)
i
i 1
/ 2
i
O ( n. ).
i
i 1
/ 2 i
cos t copies
Algorithm Analysis 3
Summary
## Analysis of recursive algorithms:
Recurrence relations
How to derive them
How to solve them
Deriving Recurrence Relations
To derive a recurrence relation for the running time of an algorithm:
## Figure out what n, the problem size, is .
See what value of n is used as the base of the recursion. It will usually be a
single value (e.g. n = 1), but may be multiple values. Suppose it is n0.
Figure out what T(n0) is. You can usually use some constant c, but sometimes a
specific number will be needed.
The general T(n) is usually a sum of various choices of T(m) (for the recursive
calls), plus the sum of the other work done. Usually the recursive calls will be
solving a subproblems of the same size f(n), giving a term a.T(f(n)) in the
recurrence relation.
Continue
Base of recursion
Running time for base
c if n n 0
T ( n)
a.T ( f ( n)) g ( n) otherwise
## Number of times All other processing not
recursive call is made counting recursive calls
Size of problem
solved by
recursive call
Analysis of Multiplication
function multiply(y, z)
comment return the product yz
1. if z = 0 then return (0) else
2. if z is odd
3. then return (multiply(2y, z/2)+y)
4. else return(multiply(2y, z/2))
## Let T(n) be the running time of multiply(y, z), where z is an n-bit
natural number.
Then for some c, d R,
c if n 1
T ( n)
T (n 1) d otherwise
Solving Recurrence Relations
Use repeated substitution.
## Substitute a few times until you see a pattern
Write a formula in terms of n and the number of substitutions i.
Choose i so that all references to T() become references to the base
case.
Solve the resulting summation
This will not always work, but works most of the time in practice.
The Multiplication Example
We know that for all n > 1,
## Therefore , for large enough n,
T ( n) T ( n 1) d
T ( n 1) T ( n 2) d
T ( n 2) T ( n 3) d
.
.
.
T ( 2) T (1) d
T (1) c
Repeated Substitution
T (n) T (n 1) d
(T (n 2) d ) d
T (n 2) 2d
(T (n 3) d ) 2d
T (n 3) 3d
## There is a pattern developing. It looks like after I substitutions,
T(n) = T(n i) + id.
## Now choose i = n 1. Then
T (n) T (1) d (n 1)
dn c d
Prove that statement by induction on i, or
Prove the result by induction on n.
Reality Check
We claim that
T(n) = dn + c d
Proof by induction on n. The hypothesis is true for n = 1,
since d + c d = c.
## Now suppose that the hypothesis is true for n. we are
required to prove that
T(n + 1) = dn + c
Now,
T(n+1) = T (n) + d
= dn +c - d +d
= dn + c
= d(n+1) + c - d
Merge Sorting
function mergesort(L, n)
comment sorts a list L of n numbers,
when n is a power of 2
if n 1 then return(L) else
break L into 2 lists L1, L2 of equal size
return(merge (mergesort (L1, n/2),
mergesort (L2, n/2)))
Here we assume a procedure merge which can merge two sorted lists of n
elements into a single sorted list in time O(n).
## Correctness: easy to prove by induction on n.
Analysis: Let T(n) be the running time of mergesort(L, n). Then for some c, d R,
c if n 1
T ( n)
2T (n / 2) dn otherwise
Continue
Continue
T (n) 2T (n / 2) dn
2(2T (n / 4) dn / 2) dn
4T (n / 4) dn dn
4(2T (n / 8) dn / 4) dn dn
8T (n / 8) dn dn dn
Therefore,
T (n) 2i T (n / 2i ) i.dn
Taking i = log n
T ( n) 2 log n T ( n / 2 log n ) dn log n
dn log n cn
Therefore T(n) = O(n log n).
Mergesort is better than bubblesort (in the worst case, for large enough n).
Recursion Tree Method
A General Theorem
Theorem: if n is a power of c, the solution to the recurrence
d if n 1
T ( n)
aT (n / c) bn otherwise
is O ( n) if a c
T (n) O(n log n) if a c
O(n log ca ) if a c
Examples:
If T(n) = 2T(n/3) + dn, then T(n) = O(n)
If T(n) = 2T(n/2) + dn, then T(n) = O(n log n)
If T(n) = 4T(n/2) + dn, then T(n) = O(n2)
Proof Sketch
If n is a power of c, then
T ( n) a.T ( n / c ) bn
a ( a.T ( n / c 2 ) bn / c ) bn
a 2 .T ( n / c 2 ) abn / c bn
a 2 ( a.T ( n / c 3 ) bn / c 2 ) abn / c bn
a 3 .T ( n / c 3 ) a 2bn / c 2 abn / c bn
.
.
.
i 1
a T ( n / c ) bn ( a / c) j
i i
j 0
log n 1
a log n
T (1) bn (a / c)
j 0
j
log n 1
da log n
bn (a / c)
j 0
j
Continue
Now,
alogen= (clog ea) log en= (clog e n) log ea = nlogea.
Therefore,
log n1
T(n) = d.nlogea+ bn e /
=0
The sum is the hard part. There are three cases to consider,
depending on which of a and c is biggest.
## But first, we need some results on geometric progressions.
Geometric Progressions
n
Finite Sums: Define S n i. If > 1, then
i 0
n n
.S n S n i 1
i
i 0 i 0
n 1 1
Therefore
S n ( n 1 1) /( 1)
Infinite Sums: Suppose 0 < < 1 and let S . Then, S i
i
i 0 i 1
## and so S S = 1. That is, S = 1/ (1 )
Back to the Proof
Case 1: a < c.
log n 1
(a / c)
j 0
j
(a / c) j c /(c a )
j 0
Therefore,
T (n) d .n log a bcn /(c a) O(n).
(Note that since a < c, the first term is insignificant.)
Case 2: a = c. Then
log n 1
T (n) d .n bn O(n log n)
1 j
j 0
Continue
log en 1
## Case 3: a > c. Then (
j 0
a / c ) j
is a geometric progression.
Hence, log n 1
( a / c ) log n
1
j 0
(a / c)
j
(a / c) 1
n log a 1 1
(a / c) 1
O ( n log a 1)
## Therefore, T(n) = O(nlog e a).
Messy Details
What about when n is not a power of c?
## Example: In our merge sort example, n may not be a power of 2. We
can modify the algorithm easily: cut the list L into two halves of size
n/2 and n/2. the recurrence relation becomes T(n) = c if n 1,
and
## T(n) = T(n/2 ) + T(n/2) + dn
Otherwise.
Continue
This is much harder to analyze , but gives the same result: T(n) =
O(n log n). To see why, think of padding the input with extra
numbers up to the next power of 2. You at most double the number
of inputs, so the running time is
Home Work
## If T(1) = 1, solve the following to within a constant multiple:
T(n) = 2T(n/2) + 6n
T(n) = 3T(n/3) + 6n 9
T(n) = 2T(n/3) + 5n
## T(n) = 2T(n/3) + 12n +16
T(n) = 4T(n/2) + n
T(n) = 3T(n/2) + 9n | 7,192 | 21,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-39 | latest | en | 0.909871 |
https://metanumbers.com/29290 | 1,638,876,304,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00059.warc.gz | 453,600,606 | 7,343 | # 29290 (number)
29,290 (twenty-nine thousand two hundred ninety) is an even five-digits composite number following 29289 and preceding 29291. In scientific notation, it is written as 2.929 × 104. The sum of its digits is 22. It has a total of 4 prime factors and 16 positive divisors. There are 11,200 positive integers (up to 29290) that are relatively prime to 29290.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 29 thousand 290 twenty-nine thousand two hundred ninety
## Notation
Scientific notation 2.929 × 104 29.29 × 103
## Prime Factorization of 29290
Prime Factorization 2 × 5 × 29 × 101
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 29290 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 29,290 is 2 × 5 × 29 × 101. Since it has a total of 4 prime factors, 29,290 is a composite number.
## Divisors of 29290
1, 2, 5, 10, 29, 58, 101, 145, 202, 290, 505, 1010, 2929, 5858, 14645, 29290
16 divisors
Even divisors 8 8 8 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 55080 Sum of all the positive divisors of n s(n) 25790 Sum of the proper positive divisors of n A(n) 3442.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 171.143 Returns the nth root of the product of n divisors H(n) 8.50835 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 29,290 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 29,290) is 55,080, the average is 344,2.5.
## Other Arithmetic Functions (n = 29290)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 11200 Total number of positive integers not greater than n that are coprime to n λ(n) 700 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3180 Total number of primes less than or equal to n r2(n) 32 The number of ways n can be represented as the sum of 2 squares
There are 11,200 positive integers (less than 29,290) that are coprime with 29,290. And there are approximately 3,180 prime numbers less than or equal to 29,290.
## Divisibility of 29290
m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 2 2 4
The number 29,290 is divisible by 2 and 5.
• Deficient
• Polite
• Square Free
## Base conversion (29290)
Base System Value
2 Binary 111001001101010
3 Ternary 1111011211
4 Quaternary 13021222
5 Quinary 1414130
6 Senary 343334
8 Octal 71152
10 Decimal 29290
12 Duodecimal 14b4a
20 Vigesimal 3d4a
36 Base36 mlm
## Basic calculations (n = 29290)
### Multiplication
n×y
n×2 58580 87870 117160 146450
### Division
n÷y
n÷2 14645 9763.33 7322.5 5858
### Exponentiation
ny
n2 857904100 25128011089000 735999444796810000 21557423738098564900000
### Nth Root
y√n
2√n 171.143 30.8252 13.0822 7.82247
## 29290 as geometric shapes
### Circle
Diameter 58580 184034 2.69519e+09
### Sphere
Volume 1.05256e+14 1.07807e+10 184034
### Square
Length = n
Perimeter 117160 8.57904e+08 41422.3
### Cube
Length = n
Surface area 5.14742e+09 2.5128e+13 50731.8
### Equilateral Triangle
Length = n
Perimeter 87870 3.71483e+08 25365.9
### Triangular Pyramid
Length = n
Surface area 1.48593e+09 2.96136e+12 23915.2
## Cryptographic Hash Functions
md5 7a1eae3489145133fdd9dbd917810547 84852dbc90b9261ba5f8af526f64a8f28d9a3e5b 7806ffa6be476696b6e55602c96d08b1526fc307acbd571cb352ce31c0db7995 37b6f8e00566e8dea716a2350a4eb95340110be91e5fdd08217e0cbe7d9655cb44fd16dec43433428a093b82cd1d66a85648846cbfc3cfdc217a929e7a277371 e60417110909fcbab7f3a9e1b9fb4f884f731c6f | 1,474 | 4,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.795435 |
http://mathforum.org/library/drmath/view/51735.html | 1,519,333,904,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814290.11/warc/CC-MAIN-20180222200259-20180222220259-00146.warc.gz | 224,969,163 | 5,324 | Associated Topics || Dr. Math Home || Search Dr. Math
### Computing Ellipse Parameters
```
Date: 9/11/96 at 6:5:25
From: symah
Subject: Computing Ellipse Parameters
Hello Dr. Math,
I want to find the five parameters of an ellipse (center, axes,
orientation) given five or more points. What is the best and fastest
method?
I want to find the best fitting ellipse to an array of points (five
parameters to find, all points are random points of the ellipse). The
only information I have is the coordinates of these points. Least mean
squares leads to very complicated equations. Are there other
interesting methods?
Thanks a lot.
Nicolas
```
```
Date: 9/11/96 at 19:6:7
From: Doctor Tom
Subject: Re: Computing Ellipse Parameters
This isn't an easy problem.
If you have more than 5 points, I simply don't know any technique
other than a least-squares method to get a good fit.
I do know a trick that's very numerically stable for getting the
equation of an ellipse given exactly 5 points. (I can even supply
C code that does it!)
After you get the equation of the conic, you've got to check that
it is, in fact, an ellipse and not a circle, hyperbola, or some
degenerate case, and then use standard methods to get the center,
axes, and orientation.
The method is based on Pascal's theorem, which states that if you take
any hexagon inscribed in a conic and find the intersections of the
opposite edges, those three intersections must lie on a line!
Since you've got 5 points, call the sixth point (x, y, w) (in
homogeneous coordinates). Now write down, based on Pascal's theorem,
the equations of the opposite lines, find their intersections, and
then write down the condition that the three intersections meet on
a line. Multiply this whole mess out, and you'll get an equation
like this:
Ax^2 + Bxy + Cy^2 + Dxw + Eyw + Fw^2 = 0
Since you're not interested in the projective case with points at
infinity, just set w = 1, and you get the standard equation of a
conic. The A, B, C, D, E, and F will just be numbers that depend on
the coordinates of the 5 fixed points.
To save you a pile of time, I'll include the code I wrote to do
this. Needless to say, it took a couple of hours to get it right.
***********************************************************************
/* Code to find the equation of a conic */
/* Tom Davis */
/* April 12, 1996 */
#include <stdio.h>
#define FLOAT double
#define cross(a, b, ab) ab[0] = a[1]*b[2] - a[2]*b[1]; \
ab[1] = a[2]*b[0] - a[0]*b[2]; \
ab[2] = a[0]*b[1] - a[1]*b[0];
#define abs(x) ((x > 0.0) ? x : (-x))
/* toconic takes five points in homogeneous coordinates, and returns the
* coefficients of a general conic equation in a, b, c, ..., f:
*
* a*x*x + b*x*y + c*y*y + d*x + e*y + f = 0.
*
* The routine returns 1 on success; 0 otherwise. (It can fail, for
* example, if there are duplicate points.
*
* Typically, the points will be finite, in which case the third (w)
* coordinate for all the input vectors will be 1, although the code
* deals cleanly with points at infinity.
*
* For example, to find the equation of the conic passing through (5, 0),
* (-5, 0), (3, 2), (3, -2), and (-3, 2), set:
*
* p0[0] = 5, p0[1] = 0, p0[2] = 1,
* p1[0] = -5, p1[1] = 0, p1[2] = 1,
* p2[0] = 3, p2[1] = 2, p2[2] = 1,
* p3[0] = 3, p3[1] = -2, p3[2] = 1,
* p4[0] = -3, p4[1] = 2, p4[2] = 1.
*
* But if you want the equation of the hyperbola that is tangent to the
* line 2x=y at infinity, simply make one of the points be the point at
* infinity along that line, for example:
*
* p0[0] = 1, p0[1] = 2, p0[2] = 0.
*/
int toconic(FLOAT p0[3], FLOAT p1[3], FLOAT p2[3], FLOAT p3[3], FLOAT p4[3],
FLOAT *a, FLOAT *b, FLOAT *c, FLOAT *d, FLOAT *e, FLOAT *f)
{
FLOAT L0[3], L1[3], L2[3], L3[3];
FLOAT A, B, C, Q[3];
FLOAT a1, a2, b1, b2, c1, c2;
FLOAT x0, x4, y0, y4, w0, w4;
FLOAT aa, bb, cc, dd, ee, ff;
FLOAT y4w0, w4y0, w4w0, y4y0, x4w0, w4x0, x4x0, y4x0, x4y0;
FLOAT a1a2, a1b2, a1c2, b1a2, b1b2, b1c2, c1a2, c1b2, c1c2;
cross(p0, p1, L0)
cross(p1, p2, L1)
cross(p2, p3, L2)
cross(p3, p4, L3)
cross(L0, L3, Q)
A = Q[0]; B = Q[1]; C = Q[2];
a1 = L1[0]; b1 = L1[1]; c1 = L1[2];
a2 = L2[0]; b2 = L2[1]; c2 = L2[2];
x0 = p0[0]; y0 = p0[1]; w0 = p0[2];
x4 = p4[0]; y4 = p4[1]; w4 = p4[2];
y4w0 = y4*w0;
w4y0 = w4*y0;
w4w0 = w4*w0;
y4y0 = y4*y0;
x4w0 = x4*w0;
w4x0 = w4*x0;
x4x0 = x4*x0;
y4x0 = y4*x0;
x4y0 = x4*y0;
a1a2 = a1*a2;
a1b2 = a1*b2;
a1c2 = a1*c2;
b1a2 = b1*a2;
b1b2 = b1*b2;
b1c2 = b1*c2;
c1a2 = c1*a2;
c1b2 = c1*b2;
c1c2 = c1*c2;
aa = -A*a1a2*y4w0
+A*a1a2*w4y0
-B*b1a2*y4w0
-B*c1a2*w4w0
+B*a1b2*w4y0
+B*a1c2*w4w0
+C*b1a2*y4y0
+C*c1a2*w4y0
-C*a1b2*y4y0
-C*a1c2*y4w0;
cc = A*c1b2*w4w0
+A*a1b2*x4w0
-A*b1c2*w4w0
-A*b1a2*w4x0
+B*b1b2*x4w0
-B*b1b2*w4x0
+C*b1c2*x4w0
+C*b1a2*x4x0
-C*c1b2*w4x0
-C*a1b2*x4x0;
ff = A*c1a2*y4x0
+A*c1b2*y4y0
-A*a1c2*x4y0
-A*b1c2*y4y0
-B*c1a2*x4x0
-B*c1b2*x4y0
+B*a1c2*x4x0
+B*b1c2*y4x0
-C*c1c2*x4y0
+C*c1c2*y4x0;
bb = A*c1a2*w4w0
+A*a1a2*x4w0
-A*a1b2*y4w0
-A*a1c2*w4w0
-A*a1a2*w4x0
+A*b1a2*w4y0
+B*b1a2*x4w0
-B*b1b2*y4w0
-B*c1b2*w4w0
-B*a1b2*w4x0
+B*b1b2*w4y0
+B*b1c2*w4w0
-C*b1c2*y4w0
-C*b1a2*x4y0
-C*b1a2*y4x0
-C*c1a2*w4x0
+C*c1b2*w4y0
+C*a1b2*x4y0
+C*a1b2*y4x0
+C*a1c2*x4w0;
dd = -A*c1a2*y4w0
+A*a1a2*y4x0
+A*a1b2*y4y0
+A*a1c2*w4y0
-A*a1a2*x4y0
-A*b1a2*y4y0
+B*b1a2*y4x0
+B*c1a2*w4x0
+B*c1a2*x4w0
+B*c1b2*w4y0
-B*a1b2*x4y0
-B*a1c2*w4x0
-B*a1c2*x4w0
-B*b1c2*y4w0
+C*b1c2*y4y0
+C*c1c2*w4y0
-C*c1a2*x4y0
-C*c1b2*y4y0
-C*c1c2*y4w0
+C*a1c2*y4x0;
ee = -A*c1a2*w4x0
-A*c1b2*y4w0
-A*c1b2*w4y0
-A*a1b2*x4y0
+A*a1c2*x4w0
+A*b1c2*y4w0
+A*b1c2*w4y0
+A*b1a2*y4x0
-B*b1a2*x4x0
-B*b1b2*x4y0
+B*c1b2*x4w0
+B*a1b2*x4x0
+B*b1b2*y4x0
-B*b1c2*w4x0
-C*b1c2*x4y0
+C*c1c2*x4w0
+C*c1a2*x4x0
+C*c1b2*y4x0
-C*c1c2*w4x0
-C*a1c2*x4x0;
if (aa != 0.0) {
bb /= aa; cc /= aa; dd /= aa; ee /= aa; ff /= aa; aa = 1.0;
} else if (bb != 0.0) {
cc /= bb; dd /= bb; ee /= bb; ff /= bb; bb = 1.0;
} else if (cc != 0.0) {
dd /= cc; ee /= cc; ff /= cc; cc = 1.0;
} else if (dd != 0.0) {
ee /= dd; ff /= dd; dd = 1.0;
} else if (ee != 0.0) {
ff /= ee; ee = 1.0;
} else {
return 0;
}
*a = aa;
*b = bb;
*c = cc;
*d = dd;
*e = ee;
*f = ff;
return 1;
}
FLOAT p0[3] = {0, 0, 1};
FLOAT p1[3] = {1, 1, 1};
FLOAT p2[3] = {-1, -1, 1};
FLOAT p3[3] = {2, 2, 1};
FLOAT p4[3] = {3, 3, 1};
main()
{
FLOAT a, b, c, d, e, f, s0, s1, s2, s3, s4;
FLOAT x, y, w;
int i, ret;
/*
ret = toconic(p0, p1, p2, p3, p4, &a, &b, &c, &d, &e, &f);
if (ret == 1) {
printf("success\n");
printf("%g %g %g %g %g %g\n", a, b, c, d, e, f);
x = p0[0]; y = p0[1]; w = p0[2];
printf("%g ", a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w);
x = p1[0]; y = p1[1]; w = p1[2];
printf("%g ", a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w);
x = p2[0]; y = p2[1]; w = p2[2];
printf("%g ", a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w);
x = p3[0]; y = p3[1]; w = p3[2];
printf("%g ", a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w);
x = p4[0]; y = p4[1]; w = p4[2];
printf("%g\n", a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w);
} else {
printf("toconic failed\n");
}
*/
for (i = 0; i < 100000; i++) {
p0[0] = (FLOAT) (rand()%30);
p0[1] = (FLOAT) (rand()%30);
p0[2] = (FLOAT) (rand()%30);
p1[0] = (FLOAT) (rand()%30);
p1[1] = (FLOAT) (rand()%30);
p1[2] = (FLOAT) (rand()%30);
p2[0] = (FLOAT) (rand()%30);
p2[1] = (FLOAT) (rand()%30);
p2[2] = (FLOAT) (rand()%30);
p3[0] = (FLOAT) (rand()%30);
p3[1] = (FLOAT) (rand()%30);
p3[2] = (FLOAT) (rand()%30);
p4[0] = (FLOAT) (rand()%30);
p4[1] = (FLOAT) (rand()%30);
p4[2] = (FLOAT) (rand()%30);
if (toconic(p0, p1, p2, p3, p4, &a, &b, &c, &d, &e, &f)) {
x = p0[0]; y = p0[1]; w = p0[2];
s0=a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w;
x = p1[0]; y = p1[1]; w = p1[2];
s1=a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w;
x = p2[0]; y = p2[1]; w = p2[2];
s2=a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w;
x = p3[0]; y = p3[1]; w = p3[2];
s3=a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w;
x = p4[0]; y = p4[1]; w = p4[2];
s4=a*x*x+b*x*y+c*y*y+d*x*w+e*y*w+f*w*w;
if (abs(s0) > .00001 || abs(s1) > .00001 || abs(s2) > .00001
|| abs(s3) > .00001 || abs(s4) > .00001) {
printf("%g %g %g %g %g\n", s0, s1, s2, s3, s4);
printf("\t%g %g %g\n", p0[0], p0[1], p0[2]);
printf("\t%g %g %g\n", p1[0], p1[1], p1[2]);
printf("\t%g %g %g\n", p2[0], p2[1], p2[2]);
printf("\t%g %g %g\n", p3[0], p3[1], p3[2]);
printf("\t%g %g %g\n", p4[0], p4[1], p4[2]);
printf("%g %g %g %g %g %g\n", a, b, c, d, e, f);
}
}
}
}
-Doctor Tom, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 4,075 | 8,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-09 | latest | en | 0.917346 |
https://www.itetgroup.com/staring-dream-meaning/ | 1,623,915,240,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00119.warc.gz | 707,772,075 | 10,076 | # Staring Dream Meaning
The Life Path number is the most crucial since it is your primary purpose in life and the main reason you were birthed; staring dream meaning the expression number is just how you reveal on your own to the globe and exactly how others see you.
The soul’s desire number makes you satisfied in life and what brings you joy, and the birthday number is all the abilities and abilities and capacities that you come down into this lifetime with to support your life’s purpose.
So I’m mosting likely to show you exactly how to calculate the Life Course number initially because that is one of the most vital number in your chart, and it is your primary purpose in life, so the Life Course number is established by your day of birth.
So all we’re doing is we are just including all of the numbers with each other in your day of birth up until we damage them down and get a single number.
So in this example, we have December 14, 1995.
So we require to get a single figure for each and every area.
Initially, we need to get a single figure for the month.
We require to get a solitary figure for the day of the month, and we need to get a single number for the birth year, and then when we get a single number for each one of these, we can add these three together.
So December is the 12th month of the year, so we need to add the one in both to obtain a solitary digit.
One plus 2 amounts to 3, and for the day of the month, we need to include the one in the four with each other, which will certainly offer us number 5 for the day of the month.
And then we require to add 1995 together.
Which will certainly give us a solitary figure for the year, so one plus nine plus nine plus five equates to twenty-four, and then we require to add both in the 4 together due to the fact that we have to break it down to one digit for the year.
So 2 plus 4 is equivalent to six.
So when we have a single number each of these areas, a solitary number for the month, a single figure for the day, and a solitary figure for the year, then we’re mosting likely to add the three of these together, and in this example.
It would certainly be three plus five plus six equates to 14, and afterwards we have to add the one and the 4 together due to the fact that we’re trying to damage all of this down up until we get a solitary number.
So if you add the one in the 4 with each other in 14, you obtain a number 5.
So in this scenario, he or she’s life path number is a number 5.
This individual is a 14 5, and it’s critical to make note of the two last numbers that we added together to obtain the Life Course number 5 because these two last figures are vital to the person’s life course number.
So, when it comes to a Life Course, the number five might be a fourteen five since the four equals number 5.
They might also be a two and a 3 a twenty-three number 5. Still, those 2 numbers that you added together to obtain that Life Path number are significant because they inform you what powers you will require to utilize during your lifetime to accomplish your life purpose. In this circumstance, staring dream meaning, he or she has a Life Course number of number 5.
Still, to accomplish their life course, the number of a number five, they are mosting likely to need to make use of the energy of the number one and the number Four to accomplish their function, so remember of those two last numbers that you added with each other to get your last Life Course number due to the fact that those are extremely essential.
A fourteen-five is mosting likely to be really different than a twenty-three-five.
If you have any kind of inquiries concerning these two last numbers used to obtain your final Life Path, number comment listed below, and I will certainly try to answer your questions now.
I desired to reveal you this instance due to the fact that, in some situations, we do not damage down every one of the numbers to obtain a single digit.
So in this instance, we have December 14, 1992, and when we added every one of the numbers together in the month, the day, and the year, we finished up with a number 11 and 11 in numerology is a master number and the master numbers.
We do not add the 2 digits together, so there are 3 master numbers in numerology, and the three master numbers are 11, 22, and 33.
So after you have actually included all of the digits with each other in your birth day and if you wind up with either an 11, a 22, or a 33, you will certainly not add these 2 numbers with each other due to the fact that you have a master number Life Course.
Number and the master numbers are different from the other numbers in numerology due to the fact that they hold the dual figures’ energy, and we do not add the two figures together in these scenarios.
So if you have either an 11, a 22, or 33, you will not include both figures together.
You will maintain it as is, and you have a master number as a life course.
In this scenario, the number that we combined to get the 11 were 8 and 3.
Those are considerable numbers in this scenario because this master number 11 will need to use the eight and the 3.
To obtain their number 11 life purpose, so in 83, 11 will be a lot different from on 92 11 due to the fact that an 83 11 will certainly need to make use of the energy of the 8 in the 3 to acquire their life purpose. The 9211 will certainly Need to utilize the nine and both indicate receive their 11 life function.
So the birthday number is probably the most easily accessible number to compute in your chart because for this number, all you have to do is add the digits together of the day you were born upon, so he or she was born upon December 14, 1995.
So we will certainly include the one and the 4 together due to the fact that those are the digits of the day. staring dream meaning
He or she was birthed, so 1 plus 4 equals 5.
So this person’s birthday celebration number is a number 5.
Now, in this instance, December 11, 1995.
He or she was born on the 11th day of the month, and 11 is a master number, so we do not include the two ones with each other due to the fact that 11 is a master number, and there are 3 master numbers in numerology, 11, 22, and 33.
So if you were birthed on the 11th of a month or the 22nd of a month, you would not include both numbers with each other since your birthday celebration number is a master number.
So your birthday number is either a master number 11 or a master number 22. The master numbers are the only numbers in numerology that we do not add both figures with each other to obtain a final number.
So you will certainly maintain those two digits alone, and you will not add them together.
For the last 2 numbers, you have actually used your date of birth to determine those 2 numbers, yet for the expression number and the soul’s desire number, you will make use of the complete name on your birth certificate.
You’re mosting likely to utilize your first, middle, and last name on your birth certification to calculate your expression and your spirit’s urge number, and so we’re currently going to make use of the Pythagorean number system to determine these numbers.
And it’s called the Pythagorean system since Pythagoras, a Greek mathematician, produced it. He was the mathematician who developed the Pythagorean theory. He is the dad of numerology, and he found that all numbers hold energy. They all own a Specific vibration, therefore after discovering that all numbers hold specific power, Pythagoras produced the Pythagorean number system. From that, we have modern-day numerology today.
It is a graph with all of the letters of the alphabet and all letters representing a specific number.
So essentially, all letters in the alphabet have the energy of a number.
And if you consider this chart, you can identify what number each letter has the power of.
So a has the energy of a number one B has the energy of a number.
Two C has the energy of a number three and so on, and more.
So, for the expression number, all you have to do is add every one of the letters in your full birth name, so the very first, center, and surname on your birth certificate and lower them down to one number. staring dream meaning
So in this instance, we have Elvis Presley.
So what I did was I checked out the number graph, and I found the corresponding number to every letter in Elvis’s name.
So E is a five, l is a 3 V is a four.
I am a nine, and I just discovered all of the numbers representing all of the letters in his name; and I included all of those numbers with each other, and the final number I got was an 81.
Then I added the eight and the one with each other since we need to maintain breaking these down up until we obtain a single-digit, and when I said the 8 and the one with each other, I got a nine, so Elvis’s expression number is a 9.
Currently the one situation where you would not remain to include these numbers with each other up until you got a solitary digit would certainly be if you got a master number, so the three master numbers are 11, 22, and 33.
If you got a master number, you would certainly not remain to add these numbers together.
You will maintain them as either 11, 22, or 33.
So it’s the exact same situation as it was with the life half number and the birthday celebration number.
The master numbers are unique, and we do not include both numbers with each other, so going on to the soles prompt number so occasionally the soles urge number can be called the sole number, and it can additionally be called the heart’s wish number.
These words are utilized reciprocally but know whenever you see the single number or heart’s desire number that they are essentially the exact same point as a soles prompt number.
For this number, we will include every one of the vowels in your complete birth name.
All of the vowels in your very first, center, and last name on your birth certification, and we’re going to minimize them to one figure.
So we’re mosting likely to use the same Pythagorean graph that we did before, and we’re mosting likely to search for all of the numbers that match to the vowels in your first, center, and surname.
So right here we have Kate Middleton.
I just recently just did a video clip on her numerology, so I figured why not utilize her today.
So her very first center and last name are Catherine, Elizabeth Middleton, so I looked up the numbers corresponding to just the vowels in her name.
So, as you can see, an equates to 1 B equals 5, I equates to nine, and E equates to 5, and afterwards I did that for all of the vowels in her complete name.
And after that I just included all of those numbers with each other, and that offered me 60, and afterwards 60 minimizes to number 6 due to the fact that we remember we’re simply attempting to break these numbers down till we get a single number.
So in this scenario, Kate’s Seoul’s impulse number is a number 6.
Currently you will not proceed to break down the numbers if you get an 11, a 22, or a 33, so, as I stated with all the other numbers if you get among these numbers, this is a master number, and we do not include the two digits together so 11, 22 and 33 are master numbers.
Staring Dream Meaning
If you obtain among those as your hearts prompt numbers, you will certainly not proceed to include the numbers with each other; you will certainly keep them as double figures. | 2,593 | 11,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-25 | latest | en | 0.957997 |
http://statkat.com/stattest.php?t=20&t2=44 | 1,563,316,438,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00124.warc.gz | 148,516,741 | 6,407 | # Logistic regression - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Logistic regression
Binomial test for a single proportion
Independent variablesIndependent variable
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesNone
Dependent variableDependent variable
One categorical with 2 independent groupsOne categorical with 2 independent groups
Null hypothesisNull hypothesis
Model chi-squared test for the complete regression model:
• $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
• $\beta_k = 0$
or in terms of odds ratio:
• $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• $\beta_k = 0$
or in terms of odds ratio:
• $e^{\beta_k} = 1$
in the regression equation $\ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$
$\pi = \pi_0$
$\pi$ is the population proportion of "successes"; $\pi_0$ is the population proportion of successes according to the null hypothesis
Alternative hypothesisAlternative hypothesis
Model chi-squared test for the complete regression model:
• not all population regression coefficients are 0
Wald test for individual $\beta_k$:
• $\beta_k \neq 0$
or in terms of odds ratio:
• $e^{\beta_k} \neq 1$
If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
• right sided: $\beta_k > 0$
• left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual $\beta_k$:
• $\beta_k \neq 0$
or in terms of odds ratio:
• $e^{\beta_k} \neq 1$
Two sided: $\pi \neq \pi_0$
Right sided: $\pi > \pi_0$
Left sided: $\pi < \pi_0$
AssumptionsAssumptions
• In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
Sample is a simple random sample from the population. That is, observations are independent of one another
Test statisticTest statistic
Model chi-squared test for the complete regression model:
• $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance}$
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
• Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$
• Wald $= \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition
Likelihood ratio chi-squared test for individual $\beta_k$:
• $X^2 = D_{K-1} - D_K$
$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
$X$ = number of successes in the sample
Sampling distribution of $X^2$ and of the Wald statistic if H0 were trueSampling distribution of $X$ if H0 were true
Sampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
• chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: approximately a chi-squared distribution with 1 degree of freedom
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: approximately a standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
• chi-squared distribution with 1 degree of freedom
Binomial($n$, $p$) distribution
Here $n = N$ (total sample size), and $p = \pi_0$ (population proportion according to the null hypothesis)
Significant?Significant?
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
Two sided:
• Check if $X$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $X$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $X$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Wald-type approximate $C\%$ confidence interval for $\beta_k$n.a.
$b_k \pm z^* \times SE_{b_k}$
where $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
-
Goodness of fit measure $R^2_L$n.a.
$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
-
Example contextExample context
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?Is the proportion smokers amongst office workers different from $\pi_0 = .2$?
SPSSSPSS
Analyze > Regression > Binary Logistic...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
JamoviJamovi
Regression > 2 Outcomes - Binomial
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questions | 1,896 | 7,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-30 | latest | en | 0.694743 |
https://curriculum.illustrativemathematics.org/k5/teachers/grade-5/unit-6/lesson-1/lesson.html | 1,680,118,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00161.warc.gz | 236,969,589 | 27,508 | # Lesson 1
Place Value Patterns
## Warm-up: Notice and Wonder: Same Digits (10 minutes)
### Narrative
The purpose of this warm-up is for students to discuss the multiplicative relationships between the place values of the digits in two numbers. This will be useful when students write multiplication and division expressions to represent place value relationships in a later activity. While students may notice and wonder many things about these numbers, the place value relationships between the digits in the numbers and the numbers themselves are the important discussion points.
### Launch
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
What do you notice? What do you wonder?
8,200
820
82
8.2
0.82
0.082
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “How does the value of 8,200 compare to the value of 820?” (It’s 10 times as much.)
• “How does the value of 0.82 compare to the value of 0.082? How do you know?” (It’s also ten times as much since there are ten thousandths in a hundredth.)
## Activity 1: Many True Equations (20 minutes)
### Narrative
The purpose of this activity is for students to express place value relationships using multiplication and division. Students examined decimal place values in depth in the previous unit and used the relationships between the values when they performed arithmetic with decimals. Here they focus on expressing these relationships using multiplication and division. This will be helpful throughout the next several lessons as students examine powers of ten and then use them for measurement conversions.
This activity uses MLR7 Compare and Connect. Advances: representing, conversing.
### Launch
• Groups of 2
• Display the numbers: 60, 6
• “How many times the value of 6 is 60? How do you know?” (10 times because it’s 6 tens)
• Display the equation: $$60 = 10 \times 6$$.
• “What division equation shows that 60 is ten times the value of 6?” ($$60 \div 6 = 10$$ is another way of saying that 60 is ten 6s or ten times 6.)
• Display the equation: $$60 \div 6 = 10$$
• “You are going to write equations like these relating different numbers.”
### Activity
• 5 minutes: partner work time
MLR7 Compare and Connect
• “Create a visual display that shows your equations. You may want to include details such as notes, diagrams or drawings to help others understand your thinking.”
• Monitor for students who:
• identify an equation that is incorrect during the gallery walk
• notice place value patterns during the gallery walk
• 2–5 minutes: independent or group work
• 5–7 minutes: gallery walk
### Student Facing
Use the numbers and symbols to write as many different true equations as you can. You may use each number and symbol more than once.
600
0.06
100
60
$$\times$$
10
6
$$\div$$
0.1
0.6
=
0.01
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Invite students to share equations they made.
• Display the equation: $$0.6 \div 10 = 0.06$$
• “How do you know this equation is true?” (When I divide tenths into ten equal pieces I get hundredths so if I divide 6 tenths into 10 equal pieces that's 6 hundredths.)
• “Can you express the relationship between 0.6 and 0.06 using multiplication?” (Yes. $$0.6 = 10 \times 0.06$$.)
• Display the equation: $$600 \times 0.01 = 6$$
• “How do you know this equation is true?” (I know 100 hundredths is 1 so 600 hundredths is 6.)
• “Can you express the relationship between 600 and 6 using division?” (Yes. $$600 \div 100 = 6$$.)
• Invite students to describe any patterns they noticed.
## Activity 2: Describe Multiplicative Relationships (15 minutes)
### Narrative
In the previous activity, students wrote multiplication and division equations relating numbers with a single non-zero digit. The purpose of this activity is to focus on the same set of numbers and describe how the value of the non-zero digit changes when it moves one place to the left or right. This serves to highlight two important patterns that came out in some of the equations of the previous activity:
• The value of a digit is multiplied by 10 when it shifts one place to the left (MP7).
• The value of a digit is multiplied by 0.1 or $$\frac{1}{10}$$ when it shifts one place to the right (MP7).
The former idea will be further developed in the next lesson where students examine large numbers and exponential notation and the latter idea will be developed when students examine conversions from a smaller metric unit to a larger metric unit.
Representation: Develop Language and Symbols. Synthesis: Make connections between representations visible, such as between information provided in the task statement and equations from the previous activity.
Supports accessibility for: Conceptual Processing, Visual-Spatial Processing, Organization
### Launch
• Groups of 2
• “We are going to continue to work with the numbers from the previous activity to explore more patterns.”
### Activity
• 5 minutes: individual work time
• 5 minutes: partner work time
### Student Facing
$$\displaystyle{600 \phantom{00}\\ \phantom{0}60 \phantom{00}\\ \phantom{00}6 \phantom{0}\\ \phantom{00}0.6 \phantom{0}\\ \phantom{00}0.06 }$$
1. Explain or show how the value of the 6 changes in the different numbers.
2. Which numbers would come before 600 if the list continued? Explain your reasoning.
3. Which numbers would come after 0.06 if the list continued? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “What happens to the value of the 6 when it shifts one place to the left?” (It is multiplied by 10.)
• “What happens to the value of the 6 when it shifts one place to the right?” (It is multiplied by $$\frac{1}{10}$$ or 0.1. It is divided by 10.)
• Invite students to share the numbers that they listed that come before 600 on the list.
• “Do you think you can keep listing bigger and bigger numbers with more and more zeros?”
• Yes, I can always add more zeros.
• I don’t know. After 600,000, I don’t know if I can keep going.
• “In the next lesson we will look at some really big numbers and how they relate to multiplying over and over by 10.”
## Lesson Synthesis
### Lesson Synthesis
“Today we looked at place values and expressed relationships between them using division and multiplication.”
Display: 0.1 and 0.01
“What multiplication equation can I write to describe the relationship between a tenth and a hundredth?” ($$0.1 = 10 \times 0.01$$, $$0.01 = 0.1 \times 0.1$$)
“What division equation can I write to describe the relationship between a tenth and a hundredth?” ($$0.1 \div 10 = 0.01$$.)
Display: 10,000 and 1,000
“Can you also compare the value of these two numbers using multiplication and division?” (Yes. I know $$1,\!000 = 10,\!000 \div 10$$ and $$10,\!000 = 10 \times 1,\!000$$.)
“In the next several lessons we will multiply and divide whole numbers and decimals by 10.”
## Cool-down: Multiplication and Division Equations (10 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. | 1,784 | 7,294 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2023-14 | latest | en | 0.863486 |
https://lesswrong.com/tag/politics/ | 1,513,394,786,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581053.56/warc/CC-MAIN-20171216030243-20171216052243-00385.warc.gz | 605,870,632 | 27,749 | A Voting Puzzle, Some Political Science, and a Nerd Failure Mode
88 10 October 2013 02:10AM
In grade school, I read a series of books titled Sideways Stories from Wayside School by Louis Sachar, who you may know as the author of the novel Holes which was made into a movie in 2003. The series included two books of math problems, Sideways Arithmetic from Wayside School and More Sideways Arithmetic from Wayside School, the latter of which included the following problem (paraphrased):
The students have Mrs. Jewl's class have been given the privilege of voting on the height of the school's new flagpole. She has each of them write down what they think would be the best hight for the flagpole. The votes are distributed as follows:
• 1 student votes for 6 feet.
• 1 student votes for 10 feet.
• 7 students vote for 25 feet.
• 1 student votes for 30 feet.
• 2 students vote for 50 feet.
• 2 students vote for 60 feet.
• 1 student votes for 65 feet.
• 3 students vote for 75 feet.
• 1 student votes for 80 feet, 6 inches.
• 4 students vote for 85 feet.
• 1 student votes for 91 feet.
• 5 students vote for 100 feet.
At first, Mrs. Jewls declares 25 feet the winning answer, but one of the students who voted for 100 feet convinces her there should be a runoff between 25 feet and 100 feet. In the runoff, each student votes for the height closest to their original answer. But after that round of voting, one of the students who voted for 85 feet wants their turn, so 85 feet goes up against the winner of the previous round of voting, and the students vote the same way, with each student voting for the height closest to their original answer. Then the same thing happens again with the 50 foot option. And so on, with each number, again and again, "very much like a game of tether ball."
Question: if this process continues until it settles on an answer that can't be beaten by any other answer, how tall will the new flagpole be?
Answer (rot13'd): fvkgl-svir srrg, orpnhfr gung'f gur zrqvna inyhr bs gur bevtvany frg bs ibgrf. Naq abj lbh xabj gur fgbel bs zl svefg rapbhagre jvgu gur zrqvna ibgre gurberz.
Why am I telling you this? There's a minor reason and a major reason. The minor reason is that this shows it is possible to explain little-known academic concepts, at least certain ones, in a way that grade schoolers will understand. It's a data point that fits nicely with what Eliezer has written about how to explain things. The major reason, though, is that a month ago I finished my systematic read-through of the sequences and while I generally agree that they're awesome (perhaps moreso than most people; I didn't see the problem with the metaethics sequence), I thought the mini-discussion of political parties and voting was on reflection weak and indicative of a broader nerd failure mode.
TLDR (courtesy of lavalamp):
1. Politicians probably conform to the median voter's views.
2. Most voters are not the median, so most people usually dislike the winning politicians.
3. But people dislike the politicians for different reasons.
4. Nerds should avoid giving advice that boils down to "behave optimally". Instead, analyze the reasons for the current failure to behave optimally and give more targeted advice.
The noncentral fallacy - the worst argument in the world?
157 27 August 2012 03:36AM
David Stove once ran a contest to find the Worst Argument In The World, but he awarded the prize to his own entry, and one that shored up his politics to boot. It hardly seems like an objective process.
If he can unilaterally declare a Worst Argument, then so can I. I declare the Worst Argument In The World to be this: "X is in a category whose archetypal member gives us a certain emotional reaction. Therefore, we should apply that emotional reaction to X, even though it is not a central category member."
Call it the Noncentral Fallacy. It sounds dumb when you put it like that. Who even does that, anyway?
It sounds dumb only because we are talking soberly of categories and features. As soon as the argument gets framed in terms of words, it becomes so powerful that somewhere between many and most of the bad arguments in politics, philosophy and culture take some form of the noncentral fallacy. Before we get to those, let's look at a simpler example.
Suppose someone wants to build a statue honoring Martin Luther King Jr. for his nonviolent resistance to racism. An opponent of the statue objects: "But Martin Luther King was a criminal!"
Any historian can confirm this is correct. A criminal is technically someone who breaks the law, and King knowingly broke a law against peaceful anti-segregation protest - hence his famous Letter from Birmingham Jail.
But in this case calling Martin Luther King a criminal is the noncentral. The archetypal criminal is a mugger or bank robber. He is driven only by greed, preys on the innocent, and weakens the fabric of society. Since we don't like these things, calling someone a "criminal" naturally lowers our opinion of them.
The opponent is saying "Because you don't like criminals, and Martin Luther King is a criminal, you should stop liking Martin Luther King." But King doesn't share the important criminal features of being driven by greed, preying on the innocent, or weakening the fabric of society that made us dislike criminals in the first place. Therefore, even though he is a criminal, there is no reason to dislike King.
This all seems so nice and logical when it's presented in this format. Unfortunately, it's also one hundred percent contrary to instinct: the urge is to respond "Martin Luther King? A criminal? No he wasn't! You take that back!" This is why the noncentral is so successful. As soon as you do that you've fallen into their trap. Your argument is no longer about whether you should build a statue, it's about whether King was a criminal. Since he was, you have now lost the argument.
Ideally, you should just be able to say "Well, King was the good kind of criminal." But that seems pretty tough as a debating maneuver, and it may be even harder in some of the cases where the noncentral Fallacy is commonly used.
Conspiracy Theories as Agency Fictions
30 [deleted] 09 June 2012 03:15PM
Here I consider in some detail a failure mode that classical rationality often recognizes. Unfortunately nearly all heuristics normally used to detect it seem remarkably vulnerable to misfiring or being exploited by others. I advocate an approach where we try our best to account for the key bias, seeing agency where there is none, while trying to minimize the risk of being tricked into dismissing claims because of boo lights.
What does calling something a "conspiracy theory" tell us?
What is a conspiracy theory? Explanations that invoke plots orchestrated by covert groups are easily called or thought of as such. In a more legal sense conspiracy is an agreement between persons to mislead or defraud others. This simple story gets complicated because people aren't very clear on what they consider a conspiracy.
To give an example, is explicit negotiation or agreement really necessary to call something a conspiracy? Does silent cooperation on Prisoner's Dilemma count? What if the players are deceiving themselves that they are really following a different goal and the resulting cooperation is just a side effect? How could we tell the difference and would it matter? The latter is especially interesting if one applies the anthropic principle to social attitudes and norms.
The phrase is also a convenient tool to mark an opponent's tale as low status and unworthy of further investigation. A boo light easily applied to anything that has people acting in something that can be framed as self-interest and happens to be few inferential jumps away from the audience. Not only is its use in this way well known, this is arguably the primary meaning of calling an argument a conspiracy theory.
We have plenty of historical examples of high-stakes conspiracies so we know they can be the right answer. Noting this and putting aside the misuse of the label, people do engage in crafting conspiracy theories when they just aren't needed. Entire communities can fixate on them or fail to call such bad thinking out. Why does this happen? Humans being the social animals that we are, the group dynamics at work probably need an article or sequence of their own. It should suffice for now to point to belief as attire, the bandwagon effect and Robin Hanson's take on status. Let's rather consider the question of why individuals may be biased towards such explanations. Why do they privilege the hypothesis?
When do they seem more likely than they are?
First off we have a hard time understanding that coordination is hard. Seeing a large pay off available and thinking it easily in reach if "we could just get along" seems like a classical failing. Our pro-social sentiments lead us to downplay such barriers in our future plans. Motivated cognition on behalf of assessing the threat potential of perceived enemies or strangers likely shares this problem. Even if we avoid this, we may still be lost since the second big relevant thing is our tendency for anthropomorphizing things that better not be. Ours is a paranoid brain seeing agency in every shadow or strange sound. The cost of false positives was once reasonably low, while the cost of a false negative very high.
Our minds are also just plain lazy. We are pretty good at modelling other human minds and considering just how hard the task really is, we do a pretty remarkable job of it. If you are stuck in relative ignorance on a subject, say the weather, dancing to appease the sky spirits makes sense. After all the weather is pretty capricious and angry sky spirits is a model that makes as much or more sense as any other model you know. Unlike some other models this one is at least cheap to run on your brain! The modern world is remarkably complex. Do we see ghosts in it?
Our Dunbarian minds probably just plain can't get how a society can be that complex and unpredictable without it being "planned" by a cabal of Satan or Heterosexual White Males or the Illuminati (but I repeat myself twice) scheming to make weird things happen in our oblivious small stone age tribe. Learning about useful models helps people escape anthropomorphizing human society or the economy or government. The latter is particularly salient. I think most people slip up occasionally in assuming that say something like the United States government can be successfully modelled as a single agent to explain most of its "actions". To make matters worse it is a common literary device used by pundits.
A mysterious malignant agency or someone keeping a secret playing the role of the villain makes a good story. Humans love stories. Its fun to think in stories. Any real conspiracy revealed will probably be widely publicized. Peter Knight in his 2003 book cites historians who have put forward the idea, that the United States is something of a home for popular conspiracy theories because so many high-level ones have been undertaken and uncovered since the 1960s. We are more likely to hear about real confirmed conspiracies today than ever before.
Wishful thinking also plays a role. A universe where bad things happen because bad people make them to is appealing. Getting rid of bad people, even very bad people, is easy compared to all the different things one has to do to make sure bad things don't happen in a universe that doesn't care about us and where really bad things are allowed to happen. Finding bad people whether there are or aren't is a problematic tendency. The sad thing is that this may also be how we often manage to coordinate. Do all theories of legitimacy also perhaps rest on the same cognitive failings that conspiracy theories do? The difference between a shadowy cabal we need to get rid of and an institution worthy of respect may be just some bad luck.
Putting aside such wild speculation, what should we take away from this? When do conspiracy theories seem more likely than they are?
• The phenomena is unpredictable or can't be modelled very well
• Models used by others are hard to understand or are very counter-intuitive
• Thinking about the subject significantly strains cognitive resources
• The theory explains why bad things happen or why something went wrong
• The theory requires coordination
When you see these features you probably find the theory more plausible than it is.
But how many here are likely to accept "conspiracy theories"? To do so with stuff that actually gets called a conspiracy theory doesn't fit our tribal attire. Reverse stupidity may be particularly problematic for us on this topic. Being open to thinking conspiracy is recommended. Just remember to compare how probable it is in relation to other explanations. It is important to call out people who misuse the tag for rhetorical gain.
This applies to debunking as well. Don't go wildly contrarian. But remember that even things that are tagged conspiracy theories are surprisingly popular. How popular might false theories that avoid that tag be? History shows us we don't have the luxury of hoping that kind of thing just doesn't happen in human societies. When assessing an explanation sharing the key features that make conspiracy theories seem more plausible than they are, compensate as you would with a conspiracy theory.
But don't listen to me, I'm talking conspiracy theories.
Note: This article started out as a public draft, feedback to other such drafts is always welcomed. Special thanks to user Villiam_Bur for his commentary and user copt for proofreading and suggestions. Also thanks to the LessWrong IRC chatroom for last minute corrections and stylistic tips.
When None Dare Urge Restraint, pt. 2
56 30 May 2012 03:28PM
In the original When None Dare Urge Restraint post, Eliezer discusses the dangers of the "spiral of hate" that can develop when saying negative things about the Hated Enemy trumps saying accurate things. Specifically, he uses the example of how the 9/11 hijackers were widely criticized as "cowards," even though this vice in particular was surely not on their list. Over this past Memorial Day weekend, however, it seems like the exact mirror-image problem played out in nearly textbook form.
The trouble began when MSNBC host Chris Hayes noted* that he was uncomfortable with how people use the word "hero" to describe those who die in war -- in particular, because he thinks this sort of automatic valor attributed to the war dead makes it easier to justify future wars. And as you might expect, people went crazy in response, calling Hayes's comments "reprehensible and disgusting," something that "commie grad students would say," and that old chestnut, apparently offered without a hint of irony, "unAmerican." If you watch the video, you can tell that Hayes himself is really struggling to make the point, and by the end he definitely knew he was going to get in trouble, as he started backpedaling with a "but maybe I'm wrong about that." And of course, he apologized the very next day, basically stating that it was improper to have "opine[d] about the people who fight our wars, having never dodged a bullet or guarded a post or walked a mile in their boots."
This whole episode struck me as particularly frightening, mostly because Hayes wasn't even offering a criticism. Soldiers in the American military are, of course, an untouchable target, and I would hardly expect any attack on soldiers to be well received, no matter how grounded. But what genuinely surprised me in this case was that Hayes was merely saying "let's not automatically apply the single most valorizing word we have, because that might cause future wars, and thus future war deaths." But apparently anything less than maximum praise was not only incorrect, but offensive.
Of course, there's no shortage of rationality failures in political discourse, and I'm obviously not intending this post as a political statement about any particular war, policy, candidate, etc. But I think this example is worth mentioning, for two main reasons. First, it's just such a textbook example of the exact sort of problem discussed in Eliezer's original post, in a purer form than I can recall seeing since 9/11 itself. I don't imagine many LW members need convincing in this regard, but I do think there's value in being mindful of this sort of problem on the national stage, even if we're not going to start arguing politics ourselves.
But second, I think this episode says something not just about nationalism, but about how people approach death more generally. Of course, we're all familiar with afterlifism/"they're-in-a-better-place"-style rationalizations of death, but labeling a death as "heroic" can be a similar sort of rationalization. If a death is "heroic," then there's at least some kind of silver lining, some sense of justification, if only partial justification. The movie might not be happy, but it can still go on, and there's at least a chance to play inspiring music. So there's an obvious temptation to label death as "heroic" as much as possible -- I'm reminded of how people tried to call the 9/11 victims "heroes," apparently because they had the great courage to work in buildings that were targeted in a terrorist attack.
If a death is just a tragedy, however, you're left with a more painful situation. You have to acknowledge that yes, really, the world isn't fair, and yes, really, thousands of people -- even the Good Guy's soldiers! -- might be dying for no good reason at all. And even for those who don't really believe in an afterlife, facing death on such a large scale without the "heroic" modifier might just be too painful. The obvious problem, of course -- and Hayes's original point -- is that this sort of death-anesthetic makes it all too easy to numb yourself to more death. If you really care about the problem, you have to face the sheer tragedy of it. Sometimes, all you can say is "we shall have to work faster." And I think that lesson's as appropriate on Memorial Day as any other.
*I apologize that this clip is inserted into a rather low-brow attack video. At the time of posting it was the only link on Youtube I could find, and I wanted something accessible.
Schelling fences on slippery slopes
182 16 March 2012 11:44PM
Slippery slopes are themselves a slippery concept. Imagine trying to explain them to an alien:
"Well, we right-thinking people are quite sure that the Holocaust happened, so banning Holocaust denial would shut up some crackpots and improve the discourse. But it's one step on the road to things like banning unpopular political positions or religions, and we right-thinking people oppose that, so we won't ban Holocaust denial."
And the alien might well respond: "But you could just ban Holocaust denial, but not ban unpopular political positions or religions. Then you right-thinking people get the thing you want, but not the thing you don't want."
This post is about some of the replies you might give the alien.
Abandoning the Power of Choice
This is the boring one without any philosophical insight that gets mentioned only for completeness' sake. In this reply, giving up a certain point risks losing the ability to decide whether or not to give up other points.
For example, if people gave up the right to privacy and allowed the government to monitor all phone calls, online communications, and public places, then if someone launched a military coup, it would be very difficult to resist them because there would be no way to secretly organize a rebellion. This is also brought up in arguments about gun control a lot.
I'm not sure this is properly thought of as a slippery slope argument at all. It seems to be a more straightforward "Don't give up useful tools for fighting tyranny" argument.
The Legend of Murder-Gandhi
Previously on Less Wrong's The Adventures of Murder-Gandhi: Gandhi is offered a pill that will turn him into an unstoppable murderer. He refuses to take it, because in his current incarnation as a pacifist, he doesn't want others to die, and he knows that would be a consequence of taking the pill. Even if we offered him \$1 million to take the pill, his abhorrence of violence would lead him to refuse.
But suppose we offered Gandhi \$1 million to take a different pill: one which would decrease his reluctance to murder by 1%. This sounds like a pretty good deal. Even a person with 1% less reluctance to murder than Gandhi is still pretty pacifist and not likely to go killing anybody. And he could donate the money to his favorite charity and perhaps save some lives. Gandhi accepts the offer.
Now we iterate the process: every time Gandhi takes the 1%-more-likely-to-murder-pill, we offer him another \$1 million to take the same pill again.
Maybe original Gandhi, upon sober contemplation, would decide to accept \$5 million to become 5% less reluctant to murder. Maybe 95% of his original pacifism is the only level at which he can be absolutely sure that he will still pursue his pacifist ideals.
Unfortunately, original Gandhi isn't the one making the choice of whether or not to take the 6th pill. 95%-Gandhi is. And 95% Gandhi doesn't care quite as much about pacifism as original Gandhi did. He still doesn't want to become a murderer, but it wouldn't be a disaster if he were just 90% as reluctant as original Gandhi, that stuck-up goody-goody.
What if there were a general principle that each Gandhi was comfortable with Gandhis 5% more murderous than himself, but no more? Original Gandhi would start taking the pills, hoping to get down to 95%, but 95%-Gandhi would start taking five more, hoping to get down to 90%, and so on until he's rampaging through the streets of Delhi, killing everything in sight.
Now we're tempted to say Gandhi shouldn't even take the first pill. But this also seems odd. Are we really saying Gandhi shouldn't take what's basically a free million dollars to turn himself into 99%-Gandhi, who might well be nearly indistinguishable in his actions from the original?
Maybe Gandhi's best option is to "fence off" an area of the slippery slope by establishing a Schelling point - an arbitrary point that takes on special value as a dividing line. If he can hold himself to the precommitment, he can maximize his winnings. For example, original Gandhi could swear a mighty oath to take only five pills - or if he didn't trust even his own legendary virtue, he could give all his most valuable possessions to a friend and tell the friend to destroy them if he took more than five pills. This would commit his future self to stick to the 95% boundary (even though that future self is itching to try to the same precommitment strategy to stick to its own 90% boundary).
Real slippery slopes will resemble this example if, each time we change the rules, we also end up changing our opinion about how the rules should be changed. For example, I think the Catholic Church may be working off a theory of "If we give up this traditional practice, people will lose respect for tradition and want to give up even more traditional practices, and so on."
Slippery Hyperbolic Discounting
One evening, I start playing Sid Meier's Civilization (IV, if you're wondering - V is terrible). I have work tomorrow, so I want to stop and go to sleep by midnight.
At midnight, I consider my alternatives. For the moment, I feel an urge to keep playing Civilization. But I know I'll be miserable tomorrow if I haven't gotten enough sleep. Being a hyperbolic discounter, I value the next ten minutes a lot, but after that the curve becomes pretty flat and maybe I don't value 12:20 much more than I value the next morning at work. Ten minutes' sleep here or there doesn't make any difference. So I say: "I will play Civilization for ten minutes - 'just one more turn' - and then I will go to bed."
Time passes. It is now 12:10. Still being a hyperbolic discounter, I value the next ten minutes a lot, and subsequent times much less. And so I say: I will play until 12:20, ten minutes sleep here or there not making much difference, and then sleep.
And so on until my empire bestrides the globe and the rising sun peeps through my windows.
This is pretty much the same process described above with Murder-Gandhi except that here the role of the value-changing pill is played by time and my own tendency to discount hyperbolically.
The solution is the same. If I consider the problem early in the evening, I can precommit to midnight as a nice round number that makes a good Schelling point. Then, when deciding whether or not to play after midnight, I can treat my decision not as "Midnight or 12:10" - because 12:10 will always win that particular race - but as "Midnight or abandoning the only credible Schelling point and probably playing all night", which will be sufficient to scare me into turning off the computer.
(if I consider the problem at 12:01, I may be able to precommit to 12:10 if I am especially good at precommitments, but it's not a very natural Schelling point and it might be easier to say something like "as soon as I finish this turn" or "as soon as I discover this technology").
Coalitions of Resistance
Suppose you are a Zoroastrian, along with 1% of the population. In fact, along with Zoroastrianism your country has fifty other small religions, each with 1% of the population. 49% of your countrymen are atheist, and hate religion with a passion.
You hear that the government is considering banning the Taoists, who comprise 1% of the population. You've never liked the Taoists, vile doubters of the light of Ahura Mazda that they are, so you go along with this. When you hear the government wants to ban the Sikhs and Jains, you take the same tack.
But now you are in the unfortunate situation described by Martin Niemoller:
First they came for the socialists, and I did not speak out, because I was not a socialist.
Then they came for the trade unionists, and I did not speak out, because I was not a trade unionist.
Then they came for the Jews, and I did not speak out, because I was not a Jew.
Then they came for me, but we had already abandoned the only defensible Schelling point
With the banned Taoists, Sikhs, and Jains no longer invested in the outcome, the 49% atheist population has enough clout to ban Zoroastrianism and anyone else they want to ban. The better strategy would have been to have all fifty-one small religions form a coalition to defend one another's right to exist. In this toy model, they could have done so in an ecumenial congress, or some other literal strategy meeting.
But in the real world, there aren't fifty-one well-delineated religions. There are billions of people, each with their own set of opinions to defend. It would be impractical for everyone to physically coordinate, so they have to rely on Schelling points.
In the original example with the alien, I cheated by using the phrase "right-thinking people". In reality, figuring out who qualifies to join the Right-Thinking People Club is half the battle, and everyone's likely to have a different opinion on it. So far, the practical solution to the coordination problem, the "only defensible Schelling point", has been to just have everyone agree to defend everyone else without worrying whether they're right-thinking or not, and this is easier than trying to coordinate room for exceptions like Holocaust deniers. Give up on the Holocaust deniers, and no one else can be sure what other Schelling point you've committed to, if any...
...unless they can. In parts of Europe, they've banned Holocaust denial for years and everyone's been totally okay with it. There are also a host of other well-respected exceptions to free speech, like shouting "fire" in a crowded theater. Presumably, these exemptions are protected by tradition, so that they have become new Schelling points there, or are else so obvious that everyone except Holocaust deniers is willing to allow a special Holocaust denial exception without worrying it will impact their own case.
Summary
Slippery slopes legitimately exist wherever a policy not only affects the world directly, but affects people's willingness or ability to oppose future policies. Slippery slopes can sometimes be avoided by establishing a "Schelling fence" - a Schelling point that the various interest groups involved - or yourself across different values and times - make a credible precommitment to defend.
The bias shield
18 31 December 2011 05:44PM
A friend asked me to get her Bill O'Reilly's new book Killing Lincoln for Christmas. I read its reviews on Amazon, and found several that said it wasn't as good as another book about the assassination, Blood on the Moon. This seemed like a believable conclusion to me. Killing Lincoln has no footnotes to document any of its claims, and is not in the Ford's Theatre national park service bookstore because the NPS decided it was too historically inaccurate to sell. Nearly 200 books have been written about the Lincoln assassination, including some by professional Lincoln scholars. So the odds seemed good that at least one of these was better than a book written by a TV talk show host.
But I was wrong. To many people, this was not a believable conclusion.
(This is not about the irrationality of Fox network fans. They are just a useful case study.)
A few analogies to illustrate key rationality points
50 09 October 2011 01:00PM
Introduction
Due to long inferential distances it's often very difficult to use knowledge or understanding given by rationality in a discussion with someone who isn't versed in the Art (like, a poor folk who didn't read the Sequences, or maybe even not the Goedel, Escher, Bach !). So I find myself often forced to use analogies, that will necessary be more-or-less surface analogies, which don't prove anything nor give any technical understanding, but allow someone to have a grasp on a complicated issue in a few minutes.
A tale of chess and politics
Once upon a time, a boat sank and a group of people found themselves isolated in an island. None of them knew the rules of the game "chess", but there was a solar-powered portable chess computer on the boat. A very simple one, with no AI, but which would enforce the rules. Quickly, the survivors discovered the joy of chess, deducing the rules by trying moves, and seeing the computer saying "illegal move" or "legal move", seeing it proclaiming victory, defeat or draw game.
So they learned the rules of chess, movement of the pieces, what "chess" and "chessmate" is, how you can promote pawns, ... And they understood the planning and strategy skills required to win the game. So chess became linked to politics, it was the Game, with a capital letter, and every year, they would organize a chess tournament, and the winner, the smartest of the community, would become the leader for one year.
One sunny day, a young fellow named Hari playing with his brother Salvor (yes, I'm an Asimov fan), discovered a new move of chess : he discovered he could castle. In one move, he could liberate his rook, and protect his king. They kept the discovery secret, and used it on the tournament. Winning his games, Hari became the leader.
Soon after, people started to use the power of castling as much as they could. They even sacrificed pieces, even their queen, just to be able to castle fast. But everyone was trying to castle as fast as they could, and they were losing sight of the final goal : winning, for the intermediate goal : castling.
Probability and Politics
17 24 November 2010 05:02PM
Follow-up toPolitics as Charity
Can we think well about courses of action with low probabilities of high payoffs?
Giving What We Can (GWWC), whose members pledge to donate a portion of their income to most efficiently help the global poor, says that evaluating spending on political advocacy is very hard:
Such changes could have enormous effects, but the cost-effectiveness of supporting them is very difficult to quantify as one needs to determine both the value of the effects and the degree to which your donation increases the probability of the change occurring. Each of these is very difficult to estimate and since the first is potentially very large and the second very small [1], it is very challenging to work out which scale will dominate.
This sequence attempts to actually work out a first approximation of an answer to this question, piece by piece. Last time, I discussed the evidence, especially from randomized experiments, that money spent on campaigning can elicit marginal votes quite cheaply. Today, I'll present the state-of-the-art in estimating the chance that those votes will directly swing an election outcome.
Disclaimer
Politics is a mind-killer: tribal feelings readily degrade the analytical skill and impartiality of otherwise very sophisticated thinkers, and so discussion of politics (even in a descriptive empirical way, or in meta-level fashion) signals an increased probability of poor analysis. I am not a political partisan and am raising the subject primarily for its illustrative value in thinking about small probabilities of large payoffs.
Vote Qualifications, Not Issues
10 26 September 2010 08:26PM
In the United States and other countries, we elect our leaders. Each individual voter chooses some criteria by which to decide who they vote for, and the aggregate result of all those criteria determines who gets to lead. The public narrative overwhelmingly supports one strategy for deciding between politicians: look up their positions on important and contentious issues, and vote for the one you agree with. Unfortunately, this strategy is wrong, and the result is inferior leadership, polarization into camps and never-ending arguments. Instead, voters should be encouraged to vote based on the qualifications that matter: their intelligence, their rationality, their integrity, and their ability to judge character.
If an issue really is contentious, then a voter without specific inside knowledge should not expect their opinion to be more accurate than chance. If everyone votes based on a few contentious issues, then politicians have a powerful incentive to lie about their stance on those issues. But the real problem is, most of the important things that a politician does have nothing to do with the controversies at all. Whether a budget is good or bad depends on how well its author can distinguish between efficient and inefficient spending, over many small projects and expenditures that will never be reviewed by the voters, and not on the total amount taxed or spent. Whether a regulation is good or bad depends on how well its author can predict the effects and engineer the small details for optimal effect, and not on whether it is more or less strict overall. Whether foreign policies succeed or fail depends on how well the diplomats negotiate, and not on any strategy that could be determined years earlier before the election.
Politics as Charity
29 23 September 2010 05:33AM
Jane is a connoisseur of imported cheeses and Homo Economicus in good standing, using a causal decision theory that two-boxes on Newcomb's problem. Unfortunately for her, the politically well-organized dairy farmers in her country have managed to get an initiative for increased dairy tariffs on the ballot, which will cost her \$20,000. Should she take an hour to vote against the initiative on election day?
She estimates that she has a 1 in 1,000,000 chance of casting the deciding vote, for an expected value of \$0.02 from improved policy. However, while Jane may be willing to give her two cents on the subject, the opportunity cost of her time far exceeds the policy benefit, and so it seems she has no reason to vote.
Jane's dilemma is just the standard Paradox of Voting in political science and public choice theory. Voters may still engage in expressive voting to affiliate with certain groups or to signal traits insofar as politics is not about policy, but the instrumental rationality of voting to bring about selfishly preferred policy outcomes starts to look dubious. Thus many of those who say that we rationally ought to vote in hopes of affecting policy focus on altruistic preferences: faced with a tiny probability of casting a decisive vote, but large impacts on enormous numbers of people in the event that we are decisive, we should shut up and multiply, voting if the expected value of benefit to others sufficiently exceeds the cost to ourselves.
Meanwhile, at the Experimental Philosophy blog, Eric Schwitzgebel reports that philosophers overwhelmingly rate voting as very morally good (on a scale of 1 to 9), with voting placing right around donating 10% of one's income to charity. He offers the following explanation: | 7,850 | 36,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-51 | latest | en | 0.96094 |
https://www.sporcle.com/blog/spy-vs-spy-logic-puzzle-solution/ | 1,696,186,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00444.warc.gz | 1,074,985,804 | 14,777 | <— Back to Spy Vs. Spy Logic Puzzle
1. Clue A1 immediately tells us that A3, A5, E1, E3 and E5 are safe.
2. Clue E5 tells us that A4 was rigged by the black spy.
3. We know that there’s a pair of consecutive safe spots in column 2, thanks to A3’s clue. Let’s lay out all the possibilities: the two safe spots can be A2 and B2, B2 and C2, C2 and D2, D2 and E2, or E2 and F2.
• But neither A2 nor E2 can be safe, because that would give us three consecutive safe spots in a row, which violates A4’s clue. So we can eliminate every possibility that includes either A2 or E2.
• That leaves us with two possibilities: B2 and C2, or C2 and D2. However, consider the clue you got from E1. There’s one corner that’s adjacent to three rigged spaces (and no safe spaces). That corner has to be A1, since A6, F1 and F6 are each adjacent to at least one safe space.
• That means that B1, A2 and B2 are all rigged. Since we know that B2 isn’t safe, the two consecutive safe spots in column 2 have to be C2 and D2.
• Now let’s look at Column 4. Using the same logical process, we can immediately rule out any pairs that include A4 or E4. That leaves us with two possibilities: B4 and C4, or C4 and D4.
• We don’t have the information necessary to narrow this down any further, but we do know that either way, C4 is safe.
4. There are three consecutive black traps in Row F, according to C4’s clue. Once again, let’s narrow down the possibilities.
• A5’s clue told us that two corners contain white traps and two corners are safe. In other words, none of the corners contain black traps, which means that the three consecutive black spaces in Row F do not include F1 or F6.
• That leaves us with two possibilities. The three consecutive black traps occupy either F2, F3 and F4 or F3, F4 and F5. Either way, F3 and F4 contain black traps.
5. We know that E4 cannot be safe, or else it would contradict A4’s clue. F4’s clue tells us that it can’t have been rigged by the white spy, either. Thus, E4 contains a black trap.
6. We’ve already identified two safe spaces adjacent to E4 (E3 and E5), so none of the other spaces surrounding it are safe. This includes F5, which we know cannot contain a white trap either (F4’s clue). Thus, F5 was rigged by the black spy.
7. Thus far, you’ve identified nine safe spaces (A1, A3, A5, E1, E3, E5, C2, D2 and C4). F5’s clue tells us that there are 12 safe spaces in total, so there are three more you need to find. From A3’s clue, we can conclude that two of those are in Column 6, and the last remaining one is either B4 or D4. From this point forward, we can assume that every remaining space outside Column 4 and 6 is rigged.
• Therefore, we know that F2 was rigged, but there are exactly three consecutive black traps in Row F (and we found them all), so F2 cannot contain a black trap. F2 was rigged by the white spy.
• We also know that F1 was rigged, but it couldn’t have been rigged by the black spy either, since none of the corners contain black traps (A5’s clue). F1 was rigged by the white spy.
8. 8. Since Row F must contain at least one safe space (F1’s clue), we can conclude that F6 is safe.
• That leaves us with two remaining safe spots; one in Column 6, and one in Column 4. But the two safe spots in Column 6 have to be consecutive (A3’s clue), so E6 is safe.
• And now that we’ve identified two safe corners (A1 and F6), we can conclude that the other two corners contain white traps (A5’s clue). A6 contains a white trap.
9. A6’s clue helps us figure out D4’s status. D4 cannot contain a black trap, since that would give us three consecutive black traps in Column 4. However, we know that it cannot be safe either, since E4 is only adjacent to two safe spaces (E4’s clue). D4 contains a white trap.
• Now that we know that D4 isn’t safe, we can conclude that B4 is (A3’s clue).
10. C2’s clue told us that one column contains four white traps. Let’s figure out which one it is.
• In order for Column 1 to have four white traps, B1, C1 and D1 would all have to contain white traps, which would contradict A6’s clue.
• Column 3 only contains one white trap (B4’s clue).
• We’ve already identified every space in Column 4, and we only found one white trap.
• We haven’t found any white traps in Column 5 so far, and there are only 3 spaces remaining.
• In order for column 6 to have four white traps, F2, F3 and F4 would all have to contain white traps, which, again, would contradict A6’s clue.
• The only remaining column is Column 2. We’ve identified one white trap in that column so far, and there are only three more spaces remaining, so A2, B2 and E2 were rigged by the white spy.
11. B2’s clue tells us that one corner is adjacent to three spaces that all have the same status. We know that this can’t apply to F1 (adjacent to two safe spaces and one white trap) or F6 (adjacent to two safe spaces and one black trap) so that leaves us with A1 and A6.
• A6 is adjacent to one safe space. In order for it to fit B2’s clue, both B5 and B6 would have to be safe as well. However, we have already identified all 12 safe spaces, so this cannot be true.
• Therefore, A1 must fit B2’s clue, which means that it must be surrounded by white traps. B1 contains a white trap.
12. B4’s clue tells us that there is only one white trap in Column 3, and as we have established, none of the remaining spaces are safe. So if B3 contains a white trap, both C3 and D3 would have to contain black traps.
• But this can’t be true. Both C3 and D3 lie on main diagonals, so according to B1’s clue, they can’t both contain black traps. B3 contains a black trap.
• Let’s look at B4’s clue some more. Of the three remaining spaces that lie on main diagonals (C3, D3 and B5), exactly one contains a black trap.
• This means that if B5 contains a black trap, then neither C3 nor D3 can. They’d both have to contain white traps, but that would contradict B4’s clue.
• B5 contains a white trap.
13. E4 is adjacent to eight spaces in total, so B5’s clue leaves us with two possibilities. E4 is either adjacent to one white trap and two black traps, or two white traps and four black traps. But we’ve already identified three black traps adjacent to E4 (F3, F4 and F5), so it must be the latter. That means that of the two remaining spaces surrounding E4 (D3 and D5), one contains a white trap and one contains a black trap.
• Look back at the clues in F3 and F5. There are 12 safe spaces in total, and the remaining 24 spaces are divided equally between the black spy and the white spy. This means that each spy rigged 12 spaces.
• So far we’ve identified 9 spaces that were rigged by the white spy (B1, F1, A2, B2, E2, F2, D4, B5 and A6). This means that there are 3 remaining; two are in Row C (B3’s clue), and the last remaining one is either D3 or D5. Thus, we can safely assume that all other spaces contain black traps.
• B6, D1 and D6 contain black traps.
• Now that we know the status of B6 and D6, we can tell that C6 contains a white trap (A6’s clue).
14. C6’s clue tells us that there are two columns that contain an odd number of black traps. We already know that one of them is Column 4 (A4, E4 and F4), but what about the other one?
• Let’s look at Column 3. There are two remaining spaces in that column that we haven’t identified, and they both lie on main diagonals. Thanks to B1’s clue, we know that exactly one of them contains a black trap.
• We’ve already identified two other black traps in the column (B3 and F3), so that makes three in total. Columns 3 and 4 each contain an odd number of black traps.
• This means that each of the other columns either contains an even number of black traps, or none at all. C1 contains a black trap.
15. C1’s clue tells us that C5 contains a white trap, meaning that D5 has to contain a black trap (A6’s clue). This, in turn, means that D3 has to contain a white trap (B5’s clue), which means that C3 contains a black trap (B1’s clue). | 2,114 | 7,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-40 | latest | en | 0.941985 |
https://www.techylib.com/en/view/eyelashesnectarine/ch._10_elasticity_and_oscillations_link_to_unit_5_link_to_unit_5 | 1,519,595,268,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817437.99/warc/CC-MAIN-20180225205820-20180225225820-00486.warc.gz | 955,835,719 | 13,731 | # Ch. 10 Elasticity and Oscillations - Link to Unit 5 Link to Unit 5 Link to ...
Software and s/w Development
Nov 3, 2013 (4 years and 4 months ago)
144 views
10.5 SHM
Concept Check
Oscillations
Which of the following is necessary to make an object oscillate?
1. a stable equilibrium
2. little or no friction
3. a disturbance
4. all of the above
Concept Check
Oscillations
Which of the following is necessary to make an object oscillate?
1. a stable equilibrium
2. little or no friction
3. a disturbance
4. all of the above
A stable equilibrium point is
needed. In addition, some sort of
disturbance is required to set the
object in motion (otherwise the
object would simply remain at the
equilibrium point). Finally, friction
should be small or absent,
otherwise the object doesn’t
oscillate but just returns to the
equilibrium point without
overshooting it.
Concept Check
Oscillations (2)
An object can oscillate around
1. any equilibrium point.
2. any stable equilibrium point.
3. certain stable equilibrium points.
4. any point, provided the forces exerted on it obey Hooke’s law.
5. any point.
Concept Check
Oscillations (2)
An object can oscillate around
1. any equilibrium point.
2. any stable equilibrium point.
3. certain stable equilibrium points.
4. any point, provided the forces exerted on it obey Hooke’s law.
5. any point.
When an object in stable equilibrium is
disturbed, it tends to return to the
equilibrium point. This is the basic
requirement for an oscillation. An object in
unstable equilibrium tends to move farther
away from equilibrium when it is
disturbed. Forces around any stable
equilibrium obey Hooke’s law, provided
the displacement from equilibrium is not
too large, so any stable equilibrium point
will do.
Concept Check
SHM
A mass attached to a spring oscillates back and forth as indicated in
the position vs. time plot below. At point
P
, the mass has
1. positive velocity and positive acceleration.
2. positive velocity and negative acceleration.
3. positive velocity and zero acceleration.
4. negative velocity and positive acceleration.
5. negative velocity and negative acceleration.
6. negative velocity and zero acceleration.
7. zero velocity but is accelerating (positively or negatively).
Concept Check
SHM
A mass attached to a spring oscillates back and forth as indicated in
the position vs. time plot below. At point
P
, the mass has
1. positive velocity and positive acceleration.
2. positive velocity and negative acceleration.
3. positive velocity and zero acceleration.
4. negative velocity and positive acceleration.
5. negative velocity and negative acceleration.
6. negative velocity and zero acceleration.
7. zero velocity but is accelerating (positively or negatively).
The velocity is
positive
because the slope of the
curve at point
P
is positive.
The acceleration is
negative
because the curve is
concave down at
P
.
Concept Check
SHM (2)
A mass suspended from a spring is oscillating up and down as
indicated. Consider two possibilities:
(i)
at some point during the oscillation the mass has zero velocity but is
accelerating (positively or negatively);
(ii)
at some point during the oscillation the mass has zero velocity and zero
acceleration.
1.
Both occur sometime during the oscillation.
2.
Neither occurs during the oscillation.
3.
Only
(i)
occurs.
4.
Only
(ii)
occurs.
Concept Check
SHM (2)
A mass suspended from a spring is oscillating up and down as
indicated. Consider two possibilities:
(i)
at some point during the oscillation the mass has zero velocity but is
accelerating (positively or negatively);
(ii)
at some point during the oscillation the mass has zero velocity and zero
acceleration.
1.
Both occur sometime during the oscillation.
2.
Neither occurs during the oscillation.
3.
Only
(i)
occurs.
4.
Only
(ii)
occurs.
The velocity is zero at the maxima and minima of the curve.
At these points, the curve is concave either up or down, and
so the particle is accelerating. The particle has zero
acceleration at the points of inflection of the curve, which
occur at those times
t
for which
x
= 0. At these points, the
curve has nonzero slope and so the velocity cannot be zero.
10.5 SHM
T
f
1
cycle
per
time
T
time
of
unit
per
cycles
f
(sec)
T
Hz
or
sec
cycles
f
Periodic Motion Terms
cos
x A t
t
x
A
-
A
T
¾
T
½
T
¼
T
10.5 SHM
Harmonic Motion
Click on above image to view the SHM java applet
f
2
T
2
10.5 SHM
Harmonic Motion
f
2
T
2
10.5 SHM
max
cos cos
x A x t
Displacement in SHM
A
P
Q
x
max
cos
A
x
A
P
Q
x
max
cos
A
x
O
O
10.5 SHM
r
v
T
Velocity in SHM
A
P
Q
x
max
cos
A
x
A
P
Q
x
max
cos
A
x
A
v
max
fA
2
v
max
v
x
sin
A
v
x
t
A
sin
O
O
10.5 SHM
Velocity in SHM
A
P
Q
x
max
cos
A
x
A
P
Q
x
max
cos
A
x
v
max
sin
A
v
x
O
O
2 2
y A x
sin
y
A
2 2
A x
A
2 2
1
x A
2 2
max
1
x
v v x A
y
10.5 SHM
Acceleration in SHM
A
P
Q
x
max
cos
A
x
A
P
Q
x
max
cos
A
x
v
max
2
c
v
a
r
O
O
cos
x c
a a
2
r
2
A
2
cos
A
2
cos
A t
2
x
a x
10.5 SHM
Displacement, Velocity, and Acceleration in SHM
t
A
x
cos
sin
x
dx
v A t
dt
2
cos
x
dv
a A t
dt
Java applet showing SHM of a mass
-
spring system
10.5 SHM
Displacement, Velocity, and Acceleration in SHM
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
period
, or time it takes for the object to complete one cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
period
, or time it takes for the object to complete one cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
Period is the inverse of frequency. The angular frequency in each expression is
given by
in the following general expression for SHM:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
F
ACD
BE
cos
x A t
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
maximum velocity
of the object during one complete cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
maximum velocity
of the object during one complete cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
Velocity of an object in SHM is given by the expression:
Maximum velocity occurs when sin = 1, or:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
E
F
ABCD
sin
v A t
max
v A
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
maximum acceleration
of the object during one complete cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
SHM Equations
Listed below are mathematical representations for objects undergoing simple harmonic
motions (SHM). In these expressions,
y
is in meters,
t
is in seconds, and the numerical
values have appropriate units.
Rank these mathematical representations of SHM, from greatest to least, on the basis of
the
maximum acceleration
of the object during one complete cycle of this motion.
Greatest
Least
1 ________ 2 ________ 3 ________ 4 ________ 5 ________ 6 ________
Or, it is not possible to compare these representations without knowing more. ________
Explain:
Acceleration of an object in SHM is given by the expression:
Maximum velocity occurs when cos = 1, or:
6sin3
y t
A
3sin6
y t
B
6cos3
y t
C
6sin 3 30
y t
D
10cos6
y t
E
10sin2
y t
F
E
B
ACD
2
cos
a A t
2
max
a A
F
Concept Check
Oscillating Spring
An object hangs motionless from a spring. When the object is pulled
down, the sum of the elastic potential energy of the spring and the
gravitational potential energy of the object and Earth
1. increases.
2. stays the same.
3. decreases.
Concept Check
Oscillating Spring
An object hangs motionless from a spring. When the object is pulled
down, the sum of the elastic potential energy of the spring and the
gravitational potential energy of the object and Earth
1. increases.
2. stays the same.
3. decreases.
If released from its new position, the object accelerates upward and
passes the equilibrium point with nonzero velocity. The object has
therefore gained kinetic energy. The two forms of potential energy
present are elastic potential energy of the spring and gravitational
potential energy. Even though the latter decreases as the object is
pulled down, the sum of the two must increase for the object to be able
to gain kinetic energy.
10.6 Elastic Restoring Force
An Oscillating Spring
o
k
m
o
o
f
2
since
m
k
f
o
2
1
k
m
T
o
2
kx
ks
F
e
ma
F
x
m
k
a
x
2
for an object in SHM
x
a x
ma kx
Concept Check
Pendulum
A person swings on a swing. When the person sits still, the swing
oscillates back and forth at its natural frequency. If, instead,
two people
sit on the swing, the natural frequency of the swing is
1. greater.
2. the same.
3. smaller.
Concept Check
Pendulum
A person swings on a swing. When the person sits still, the swing
oscillates back and forth at its natural frequency. If, instead,
two people
sit on the swing, the natural frequency of the swing is
1. greater.
2. the same.
3. smaller.
Oscillations are an interplay between inertia and a restoring force. The
extra person doubles both the rotational inertia of the swing as well as
the restoring torque. The two effects cancel.
Concept Check
Pendulum
A person swings on a swing. When the person sits still, the swing
oscillates back and forth at its natural frequency. If, instead, the person
stands
on the swing, the natural frequency of the swing is
1. greater.
2. the same.
3. smaller.
Concept Check
Pendulum
A person swings on a swing. When the person sits still, the swing
oscillates back and forth at its natural frequency. If, instead, the person
stands
on the swing, the natural frequency of the swing is
1. greater.
2. the same.
3. smaller.
By standing up, the distance of the center of mass to the pivot point is
reduced. The restoring torque decreases linearly with this distance; the
rotational inertia as the square of it. Thus, the rotational inertia
decreases more and the period decreases.
10.7 The Pendulum
sin
mg
l
sin restoring force
F mg
nd
2 Law
T
F ma
sin
T
F
a g
m
L
l
If sin, then
T
g
a l
L
L
g
f
o
2
1
g
L
T
o
2
2
x
a x
L
g
o
f
o
2
l
L
In order to be SHM, , not sin
T
a
10.7 The Pendulum
What happens to the
period
of a simple pendulum
if we increase its
length
so that it is 4 times longer?
What happens to the
frequency
of a simple pendulum
if we increase its
length
so that it is 4 times longer?
What happens to the
period
of a simple pendulum if we increase its
mass
so that it is 4 times greater?
sin
mg
l
o
T L
Increasing L by a factor of 4 increases
T by a factor of or 2.
4
Period and frequency are inverse of each other, so if the period is doubled,
according to the previous problem, frequency will be halved.
The period of a simple pendulum is independent of mass. Therefore, the
period will be the same with a mass that is 4 times greater.
10.8 Damping, Forcing, and Resonance
Damping
Java applet showing Damped Harmonic motion
Click on above image to go to Hyperphysics explanation of damped harmonic motion
10.8 Damping, Forcing, and Resonance
Forced Oscillation and Resonance
Java applet showing Forced Oscillation and Resonance
University of Salford explanation of Forced Oscillation | 4,178 | 13,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-09 | longest | en | 0.819104 |
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3 Pages
### hw3soln
Course: EECE 253, Spring 2011
School: UBC
Rating:
Word Count: 544
#### Document Preview
all Note solutions assume rms values unless otherwise specified. 1) Transform the circuit: 4 I I1 2 I2 2j -2j Vp = 10 30V For the 4 resistor: 1 2 1 P = I R = (1.581) 2 ( 4) = 5 W 2 2 For the 2 resistor: 1 2 1 2 P = I 2 R = (1.581) ( 2) = 2.5 W 2 2 For the inductor and the capacitor: The total impedance seen by the voltage source is: Z = 4 + ( j2) || (2 - j2) ( j2)(2 - j2) = 4 + 2 + j2 j2 + 2 - j2 = 6 + j2 =...
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all Note solutions assume rms values unless otherwise specified. 1) Transform the circuit: 4 I I1 2 I2 2j -2j Vp = 10 30V For the 4 resistor: 1 2 1 P = I R = (1.581) 2 ( 4) = 5 W 2 2 For the 2 resistor: 1 2 1 2 P = I 2 R = (1.581) ( 2) = 2.5 W 2 2 For the inductor and the capacitor: The total impedance seen by the voltage source is: Z = 4 + ( j2) || (2 - j2) ( j2)(2 - j2) = 4 + 2 + j2 j2 + 2 - j2 = 6 + j2 = 4 + P = 0W Using this we can calculate I: V 10 30 I = = = 1.58111.565 A 6 + j2 Z Now we use current dividers to calculate I1 and I2: j2 I1 = I = 1.581101.565 A 2 2 - j2 I2 = I = 2.236 - 33.435 A 2 For the source: S = 1 * 1 V I = (1030)(1.581 - 11.565) 2 2 = 7.5 + j 2.5 VA P = 7.5 W j1 S 2) 3) 0.5 S V1 4 60 A + V0 - V2 + 4V0 - Mathematically the function can be written as: V3 A -0.5j S 0.25 S t 0 < t < 2 v(t) = - 1 2 < t < 4 RMS voltage is calculated by: Convert the circuit and form the matrix for modified nodal analysis: -0.5 -j 0.5 + j V1 -0.5 V = 0.5 - j 0.5 0 2 -j 0 0.25 + j V3 Using conditions on dependent voltage source: 4V0 = V2 - V3 V0 = V1 - V2 4V1 - 5V2 + V3 = 0 4 60 A -A 2 Vrms = 4 1 2 2 2 t dt + ( -1) dt = 1.1667 V 4 0 2 Vrms = 1.08 V Expand the matrix and add the new condition: 0 V1 -0.5 -j 0.5 + j 4 60 0.5 - j 0.5 0 0 - 1 V2 -0.5 = -j 0 0 0.25 + j 1 V3 1 0 A -5 4 0 Solving for V3: V3 = 9.992 5.06 V For the 4 resistor: P = V32 R = 24.96 W 1 4) 5) Using the figure given in the textbook & nodal analysis: At node V1: 120 30 - V1 V1 V - V2 = + 1 20 j30 50 At node V2: V1 - V2 V V2 = 2 + 50 10 -j40 Using these eqns we can find: V1 = 45.04 + j 66.935, (a): Pj 30 = P- j 40 = 0 P = 10 P50 = P20 = V22 = 17.33 W R a) The total input impedance is: Z = 4 + ( (- j2) || (j5 - j2) ) ( - j2)( j3) = 4 + = 4 - j6 - j2 + j3 = 7.211 -56.31 Which a gives power factor of: pf = cos ( -56.31) = 0.5547 leading V 2 = 9.423 + j 9.913 b) The total input impedance is: Z = 1 || ( - j1 + ( ( j2 ) || (4 + j1 ) ) ) ( - j1 + 0.64 + j1.52 |V 1 |120 30 - V | 1 - V2 | = 92.06 W R 2 2 ( j2)( 4 + j1) = 1 || = 1 || - j1 + 5 + j3 (1)( 0.64 + j0.52 ) = 0.4793 21.5 = 1.64 + j0.52 ) Which gives a power factor of: pf = cos ( 21.5) = 0.9304 lagging R = 175.72 W (b): 120 30 - V1 20 S = VS I * = 285.0 + j 212.6, I = pf = 285 = 0.8015 lagging 355.6 S = 355.6 VA (c): 6) 7) Redraw circuit: transform voltage source into its Norton equivalent, and transform impedances into admittances V1 I1 I2 a) The total input impedance is: Z = 2 + (10 - j5) || (8 + j6) (10 - j5)(8 + j6) 18 + j1 = 8.188 5.382 = 2 + 4.8 0 A 0.02S j0.05S 40A 0.016 - j 0.012 S Which gives a power factor of: pf = cos ( 5.382 ) = 0.9956 lagging b) Average power is: 2 * V2 (16 ) S = V I = = * ( 8.188 -5.382 ) Z = 31.26 5.382 VA = 31.12 + j 2.93 VA Form node equation: [ 0.02 V1 + j 0.05 + 0.016 - j 0.012] V1 = [ 4.8 + 4.0] 8.8 = 0.036 + j 0.038 V1 = 168.13 -46.5 V Now find the currents across the inductor and capacitor: I1 = V1 ( j 0.05) = 8.4143.45 A P = Re S ( ) = 31.12 W b) Reactive power is: Q = Im S ( ) = 2.93 VAR c) Apparent power is: S = 31.26 VA d) Complex power is: S = 31.26 5.382 VA I 2 = V1 (0.016 - j 0.012) = 3.36 - 83.42 A Reactive power in the inductor is: S = I2Z = ( 3.363) ( j30 ) 2 = j339.3 VAR Reactive power in the capacitor is: S = I 2 Z = (8.41) (- j 20 ) = - j1414 VAR 2 2 8) I2 I1 8 120 0 V j12 IS Using nodal analysis: IS = I1 - I2 I1 = 120 0 = 4.62 - j6.92 A 8 + j12 Now using the complex power: S = VI2 * 2500 - j400 I2* = = 20.83 - j3.33 120 0 I2 = 20.83 + j3.33 A IS = I1 - I2 = ( 4.62 - j6.92) - ( 20.83 + j3.33) = -16.21 - j19.54 A = -16.21 - j10.26 A 3
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UBC - EECE - 253
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UBC - EECE - 253
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MCDB Low Lecture #10Meiosis: The basis of Mendelian Genetics Mitosis = Asexual Reproductiono Vegetative reproductiono Results in genetic constancyo Offspring = clones of parentso Variation by random mutation is the only way these organisms evolveo
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UCSB - MCDB - 1a
MCDB Lecture Low #11Summary of Meiosis Meiosis Summary In human cells just before meiosis, a diploid precursor cell contains:23 pairs of chromosomes, 46 total chromosomesEach chromosome has two chromatids, so 92 chromatids alltogether. A gamete aft
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MCDB Low Lecture #12Leftovers from Fertilization Cortical granules - membrane-bound structures in the egg, derived from theGolgi apparatus, and found just beneath the plasma membrane. The corticalreaction is the release of their contents (proteases, m
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CycloalkanesMany organic compounds contain cyclic or ringstructures: carbohydrates nucleotides in DNA and RNA antibioticsOOHCH2 C NHOOtestosteronepenicillin GSNCO2HCycloalkanesCycloalkanes: alkanes that contain three or more carbonsarra | 6,921 | 21,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2013-20 | latest | en | 0.826378 |
https://bookdown.org/lgpcappiello/IntroStats/introduction-to-inference.html | 1,656,144,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00714.warc.gz | 190,631,426 | 14,012 | # Chapter 5 Introduction to Inference
## 5.1 Module Overview
This module will bridge the gap between our discussion on the normal distribution and our first forays into statistical inference. As it turns out, much of the statistical inference we will use relies on the normal distribution and the t-distribution, which we will introduce in this module. We begin our study of statistical inference by learning about confidence intervals.
Module Learning Objectives/Outcomes
1. Find the distribution of a sample mean.
2. Estimate probabilities for a sample mean.
3. Calculate and interpret confidence intervals for a population mean.
4. Use the standard normal and t-distributions to find critical values.
This module’s outcomes correspond to course outcome (6) apply statistical inference techniques of parameter estimation such as point estimation and confidence interval estimation and (7) apply techniques of testing various statistical hypotheses concerning population parameters.
## 5.2 Sampling Distributions
### 5.2.1 Sampling Error
We want to use a sample to learn something about a population, but no sample is perfect! Sampling error is the error resulting from using a sample to estimate a population characteristic.
If we use a sample mean $$\bar{x}$$ to estimate $$\mu$$, chances are that $$\bar{x}\ne\mu$$ (they might be close but… they might not be!). We will consider
• How close is $$\bar{x}$$ to $$\mu$$?
• What if we took many samples and calculated $$\bar{x}$$ many times?
• How would that relate to $$\mu$$?
• What would be the distribution of these values?
The distribution of a statistic (across all possible samples of size $$n$$) is called the sampling distribution. We will focus primarily on the distribution of the sample mean.
For a variable $$x$$ and given a sample size $$n$$, the distribution of $$\bar{x}$$ is called the sampling distribution of the sample mean or the distribution of $$\boldsymbol{\bar{x}}$$.
Example: Suppose our population is the five starting players on a particular basketball team. We are interested in their heights (measures in inches). The full population data is
Player A B C D E
Height 76 78 79 81 86
The population mean is $$\mu=80$$. Consider all possible samples of size $$n=2$$:
Sample A,B A,C A,D A,E B,C B,D B,E C,D C,E D,E
$$\bar{x}$$ 77 77.5 78.5 81.0 78.5 79.5 82.0 80.0 82.5 83.5
There are 10 possible samples of size 2. Of these samples, 10% have means exactly equal to $$\mu$$ (for a random sample of size 2, you’d have a 10% chance to find $$\bar{x}=\mu$$… and a 90% chance not to!).
In general, the larger the sample size, the smaller the sampling error tends to be in estimating $$\mu$$ using $$\bar{x}$$.
In practice, we have one sample and $$\mu$$ is unknown. We also have limited resources to collect data, so it may not be feasible to collect a very large sample.
The mean of the distribution of $$\bar{x}$$ is $$\mu_{\bar{X}}=\mu$$ and the standard deviation is $$\sigma_{\bar{X}}=\sigma/\sqrt{n}$$. We refer to the standard deviation of a sampling distribution as standard error. (Note that this standard error formula is built for very large populations, so it will not work well for our basketball players. This is okay! We usually work with populations so large that we treat them as “infinite.”)
Example: The mean living space for a detatched single family home in the United States is 1742 ft$$^2$$ with a standard deviation of 568 square feet. (Does that mean seem huge to anyone else??) For samples of 25 homes, determine the mean and standard error of $$\bar{x}$$.
Using our formulae: $\mu_{\bar{X}} = \mu = 1742$ and $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{568}{\sqrt{25}} = 113.6.$
### 5.2.2 The Sampling Distribution of $$\bar{X}$$
First, we consider the setting where $$X$$ is Normal($$\mu$$, $$\sigma$$). The plots below show (A) a random sample of 1000 from a Normal(100, 25) distribution and (B) the approximate sampling distribution of $$\bar{X}$$ when X is Normal(100, 25).
Notice how the x-axis changes from one plot to the next.
In fact, if $$X$$ is Normal($$\mu$$, $$\sigma$$), then $$\bar{X}$$ is Normal($$\mu_{\bar{X}}=\mu$$, $$\sigma_{\bar{X}}=\sigma/\sqrt{n}$$).
Central Limit Theorem
For relatively large sample sizes, the random variable $$\bar{X}$$ is approximately normally distributed regardless of the distribution of $$X$$: $\bar{X}\text{ is Normal}(\mu_{\bar{X}}=\mu, \sigma_{\bar{X}}=\sigma/\sqrt{n}).$
Notes
• This approximation improves with increasing sample size.
• In general, “relatively large” means sample sizes $$n \ge 30$$.
## 5.3 Developing Confidence Intervals
Recall: A point estimate is a single-value estimate of a population parameter. We say that a statistic is an unbiased estimator if the mean of its distribution is equal to the population parameter. Otherwise, it is a biased estimator.
Comment Remember how our formula for standard deviation, the “mean squared deviance” divides by $$n-1$$ instead of $$n$$? We do this so that $$s$$ is an unbiased estimate of $$\sigma$$.
Ideally, we want estimates that are unbiased with small standard error. For example, a sample mean (unbiased) with a large sample size (results in smaller standard error).
Point estimates are useful, but they only give us so much information. The variability of an estimate is also important!
Example Think about estimating what tomorrow’s weather will be like. If it’s May in Sacramento, the average high temperature is 82 degrees Fahrenheit, but it’s not uncommon to have highs anywhere from 75 to 90! Since the highs are so variable, it’s hard to be confident using 82 to predict tomorrow’s weather.
On the flip side, think about July in Phoenix. The average high is 106 degrees Fahrenheit. In Phoenix, it’s uncommon to have a July day with a high below 100. Since the highs are not variable, you could feel pretty confidence using 105 to predict tomorrow’s weather.
Take a look at these two boxplots:
Both samples are size $$n=100$$ and have $$\bar{x}=0$$, which would be our point estimate for $$\mu$$… but Variable 1 has a standard deviation of $$\sigma=0.5$$ and Variable 2 has standard deviation $$\sigma=5$$. As a result, we can be more confident in our estimate of the population mean for Variable 1 than for Variable 2.
We want to formalize this idea of confidence in our estimates. A confidence interval is an interval of numbers based on the point estimate of the parameter. Say we want to be 95% confident about a statement. In Statistics, this means that we have arrived at our statement using a method that will give us a correct statement 95% of the time.
Assume we are taking a sample from a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$. We will assume the value of $$\sigma$$ is known to us. Then $$\bar{X}$$ is Normal($$\mu, \sigma/\sqrt{n}$$). If we standardize $$\bar{X}$$, we get $Z = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}.$
We want some interval $$(a,b)$$. We will start by considering $$a < Z < b$$, so $$a < Z$$ and $$Z < b$$ (or $$b > Z$$). Then
\begin{aligned} Z &< b\\ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} &< b\\ \bar{X}-\mu &< b\sigma/\sqrt{n} \\ \bar{X}-b\sigma/\sqrt{n} &< \mu \end{aligned}
and
\begin{aligned} a &< Z \\ a &< \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \\ a\sigma/\sqrt{n} &< \bar{X}-\mu \\ \mu &< \bar{X}-a\sigma/\sqrt{n} \end{aligned}
putting these together, $\bar{X}-b\frac{\sigma}{\sqrt{n}} < \mu < \bar{X}-a\frac{\sigma}{\sqrt{n}}.$ If we want to be 95% confident, then we want $$P(a < Z < b)=0.95$$: $P\left(\bar{X}-b\frac{\sigma}{\sqrt{n}} < \mu < \bar{X}-a\frac{\sigma}{\sqrt{n}}\right) = 0.95.$ To calculate the 95% confidence interval, we need to find $$a$$ and $$b$$ such that $$P(a < Z < b)=0.95$$.
We want this interval to be as narrow (small) as possible. Why? Narrower intervals are more informative. If I say I’m 95% confidence that tomorrow’s high will be between -100 and 200 degrees Fahrenheit, that’s a useless interval. If I change it to between 70 and 100, that’s a little better. Changing it to between 85 and 90 is even better. This is what we mean by more informative.
It turns out that, with a symmetric distribution like the normal distribution, the way to make a confidence interval as narrow as possible is to take advantage of this symmetry. Each of the plots below show a shaded area of 0.95. The narrowest interval (along the horizontal axis) is the first interval, which is shaded on $$(-1.96 < z < 1.96)$$.
The confidence interval, then, is $\left(\bar{x} - z_*\frac{\sigma}{\sqrt{n}}, \bar{x} + z_*\frac{\sigma}{\sqrt{n}}\right)$ where $$z_* = 1.96$$. The midpoint of this interval is $$\bar{x}$$. The value of $z_*\frac{\sigma}{\sqrt{n}}$ is called the margin of error.
### 5.3.1 Interpreting a Confidence Interval
To interpret a confidence interval, we need to think back to our definition of probability as “the proportion of times is would occur if the experiment were run infinitely many times.” In the confidence interval case, if an experiment is run infinitely many times, the true value of $$\mu$$ will be contained in 95% of the intervals.
The graphic above shows 95% confidence intervals for 100 samples of size $$n=60$$ drawn from a population with mean $$\mu=80$$ and standard deviation $$\sigma=25$$. Each sample’s confidence interval is represented by a horizontal line. The dot in the middle of each is the sample mean. When a confidence interval does not capture the population mean $$\mu$$, the line is printed in red. Based on this concept of repeated sampling, we would expect about 95% of these intervals to capture $$\mu$$. In fact, 96 of the 100 intervals capture $$\mu$$.
Finally, when you interpret a confidence interval, it is important to do so in the context of the problem.
Example The preferred keyboard height for typists is approximately normally distributed with $$\sigma=2.0$$. A sample of size $$n=31$$, resulted in a mean prefered keyboard height of $$80 cm$$. Find and interpret a 95% confidence interval for keyboard height.
The interval is $\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} = 80.0 \pm 1.96\times\frac{2.0}{\sqrt{31}} = 80.0 \pm 0.70 = (79.3, 80.7).$ Interpretation:We can be 95% confident that the mean preferred keyboard height for typists is between 79.3cm and 80.7cm.
Notice that I kept the interpretation simple! That’s okay - just be sure you are also able to explain what it means to be 95% confident (using the concept of repeated sampling).
Common mistakes:
• It is NOT accurate to say that “the probability that $$\mu$$ is in the confidence interval is 0.95.” The parameter $$\mu$$ is some fixed quantity and it’s either in the interval or it isn’t.
• We are NOT “95% confident that $$\bar{x}$$ is in the interval.” The value $$\bar{x}$$ is some known quantity and it’s always in the interval.
### 5.3.2 Exercises
1. Suppose I took a random sample of 50 Sac State students and asked about their SAT scores and found a mean score of 1112. Prior experience with SAT scores in the CSU system suggests that SAT scores are well-approximated by a normal distribution with standard deviation known to be 50.
1. Find a 95% confidence interval for Sac State SAT scores.
2. Interpret your interval in the context of the problem.
3. What is the width of your interval? If you want a narrower interval, what could you do?
## 5.4 Other Levels of Confidence
While the 95% confidence interval is common in research, there’s nothing inherently special about it. You could calculate a 90%, a 99%, or - if you’re feeling spicy - something like a 43.8% confidence interval. These numbers are called the confidence level and they represent the proportion of times that the parameter will fall in the interval (if we took many samples).
The 100(1-$$\alpha$$)% confidence interval for $$\mu$$ is given by $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ where $$z_{\alpha/2}$$ is the z-score associated with the $$[1-(\alpha/2)]$$th percentile of the standard normal distribution. The value $$z_{\alpha/2}$$ is called the critical value (“c.v.” on the plot, below).
We can find critical values in R using the same command we used to find percentiles: qnorm(p). We want a 100(1-$$\alpha$$)% confidence interval, so we need to quickly solve for $$\alpha$$ and divide by 2. For example, for a 98% interval, $100(1-\alpha) = 98 \implies \alpha=0.02$ Then $$\alpha/2 = 0.01$$ and
qnorm(0.01)
## [1] -2.326348
So the critical value is $$z_{\alpha/2}=2.326$$. Notice that I dropped the negative sign here. That’s because our formula uses $$\pm z_{\alpha/2}$$, so the sign doesn’t matter. I’ll always ignore that negative for critical values. As long as you write your interval as $$(\text{smaller number}, \text{bigger number})$$, it’s all good.
Common Critical Values
Confidence Level $$\alpha$$ Critical Value, $$z_{\alpha/2}$$
90% 0.10 1.645
95% 0.05 1.96
98% 0.02 2.326
99% 0.01 2.575
### 5.4.1 Breaking Down a Confidence Interval
Consider $\left(\bar{x}- z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \quad \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)$ The key values are
• $$\bar{x}$$, the sample mean
• $$\sigma$$, the population standard deviation
• $$n$$, the sample size
• $$z_{\alpha/2}$$, the critical value $P(Z > z_{\alpha/2}) = \frac{\alpha}{2}$
The value of interest is $$\mu$$, the (unknown) population mean; the confidence interval gives us a reasonable range of values for $$\mu$$.
• The standard error, $$\frac{\sigma}{\sqrt{n}}$$
• The margin of error, $$z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$
### 5.4.2 Confidence Level, Precision, and Sample Size
If we can be 99% confident (or even higher), why do we tend to “settle” for 95%?? Take a look at the common critical values (above) and the confidence interval formula $\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}.$ What will higher levels of confidence do to this interval? Think back to the intuitive interval width explanation with the weather. Mathematically, the same thing will happen: the interval will get wider! And remember, a narrow interval is a more informative interval. There is a trade off here between interval width and confidence. In general, the scientific community has settled on 95% as a compromise between the two, but different fields may use different levels of confidence.
There is one other thing we can control in the confidence interval: the sample size $$n$$. One strategy is to specify the confidence level and the maximum acceptable interval width and use these to determine sample size. We know that $\text{interval width} \ge 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (Note: I use $$\ge$$ because $$2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$ is the maximum interval width - we would still be happy if this value turned out to be smaller!) Letting interval width equal $$w$$, we can solve for $$n$$: $n \ge \left(2z_{\alpha/2}\frac{\sigma}{w}\right)^2$ Alternately, we may specify a maximum margin of error $$m$$ instead: $n \ge \left(z_{\alpha/2}\frac{\sigma}{m}\right)^2$ Once we’ve done this calculation, we need a whole number for $$n$$. Since $$n \ge$$ something, we will always round up.
Example Suppose we want a 95% confidence interval for the mean of a normally distributed population with standard deviation $$\sigma=10$$. It is important for our margin of error to be no more than 2. What sample size do we need?
Using the formula for sample size with a desired margin of error, I can plug in $$z_{0.05/2}=1.96$$, $$m=2$$ and $$\sigma=10$$: $n = \left(1.96\times\frac{10}{2}\right)^2 = 96.04$. So (rounding up!) I need a sample size of at least 97.
• As desired width/margin of error decreases, $$n$$ will increase.
• As $$\sigma$$ increases, $$n$$ will also increase. (More population variability will necessitate a larger sample size.)
• As confidence level increases, $$n$$ will also increase.
### 5.4.3 Exercises
1. In the previous section, you worked with a random sample of 50 Sac State students with mean SAT score 1112. Prior experience with SAT scores in the CSU system suggests that SAT scores are well-approximated by a normal distribution with standard deviation known to be 50. Calculate a
1. 98% confidence interval.
2. 90% confidence interval.
3. Interpret each interval in the context of the problem. Comment on how the intervals change as you change the confidence level.
4. Find the sample size required for a 98% confidence interval with maximum margin of error 10.
## 5.5 Confidence Intervals, $$\sigma$$ Unknown
In practice, the value of $$\sigma$$ is almost never known… but we know that we can estimate $$\sigma$$ using $$s$$. Can we plug in $$s$$ for $$\sigma$$? Sometimes!
Remember the Central Limit Theorem (Section 5.1)? For samples of size $$n \ge 30$$, $$\bar{X}$$ will be approximately normal even if $$X$$ isn’t. In this case, we can plug in $$s$$ for $$\sigma$$: $\bar{x} \pm z_{\alpha/2}\frac{s}{\sqrt{n}}.$
That setting is pretty straightforward! Now we need to consider the setting where $$n < 30$$, which will require a bit of additional work.
### 5.5.1 The T-Distribution
Enter: the t-distribution. If $Z = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$ has a standard normal distribution (for $$X$$ normal or $$n\ge30$$), the slightly modified $T = \frac{\bar{X}-\mu}{s/\sqrt{n}}$ has what we call the t-distribution with $$n-1$$ degrees of freedom (even when $$n < 30$$!). The only thing we need to know about degrees of freedom is that $$df=n-1$$ is the t-distribution’s only parameter.
The t-distribution is symmetric and always centered at 0. When $$n\ge30$$, the t-distribution is approximately equivalent to the standard normal distribution. For smaller sample sizes, the t-distribution has more area in the tails (and therefore less area in the center of the distribution).
For a sample of size $$n < 30$$, we plug in $$s$$ for $$\sigma$$ and use a t critical value (instead of a z critical value): $\bar{x} \pm t_{df, \, \alpha/2}\frac{s}{\sqrt{n}}.$ The t critical value is found through $P(T_{df} > t_{df, \, \alpha/2}) = \alpha/2$ where $$T_{df}$$ is the t-distribution with $$df=n-1$$ degrees of freedom.
To find a t critical value, we will again use R, now with the command qt(p, df). (Notice that this is similar to the command for the standard normal distribution, but instead of “norm” for normal it has “t” for the t-distribution.) For example, for a 98% interval with a sample size of 15, $100(1-\alpha) = 98 \implies \alpha=0.02$ Then $$\alpha/2 = 0.01$$ and $$df=15-1=14$$.
qt(0.01, df=14)
## [1] -2.624494
which gives the t critical value $$t_{14,\alpha/2} = 2.625$$. Notice again that I am able to ignore the sign because our formula uses $$\pm t_{df,\alpha/2}$$.
As before, if you prefer you may use the applet, Rossman and Chance t Probability Calculator, instead of R. For this applet, enter the degrees of freedom $$n-1$$ next to “df.” Then check the top box under “t-value probability” and make sure the inequality is clicked to “>” . Enter the value of $$\alpha/2$$ for the probability. Click anywhere else on the page and the applet will automatically fill in the box under “t-value.” This is your t critical value.
## 5.6 Summary of Confidence Interval Settings
Setting 1: $$\mu$$ is target parameter, $$X$$ is normal, $$\sigma$$ known
• Critical value: $$z_{\alpha/2}$$
• Confidence interval: $\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}.$
Setting 2: $$\mu$$ is target parameter, $$n \ge 30$$, $$\sigma$$ unknown
• Critical value: $$z_{\alpha/2}$$
• Confidence interval: $\bar{x} \pm z_{\alpha/2}\frac{s}{\sqrt{n}}.$
Setting 3: $$\mu$$ is target parameter, $$n < 30$$, $$\sigma$$ unknown
• Critical value: $$t_{df, \alpha/2}$$
• Confidence interval: $\bar{x} \pm t_{df, \alpha/2}\frac{s}{\sqrt{n}}.$ | 5,386 | 19,722 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-27 | latest | en | 0.856367 |
https://www.examsegg.com/time-and-distance.html | 1,716,543,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00586.warc.gz | 673,920,873 | 64,257 | # Time and Distance Problems with Answers
### Speed, Time and Distance Aptitude Objective Type Questions with Answers:
Ques. Shivganga Superfast train crosses a platform in 60 seconds travelling at a speed of 54 km/hr. If the length of the platform is 500 metres, what is the length of the train?
(a) 400 m
(b) 350 m
(c) 450 m
(d) 500 m
Ans. (a)
Ques. Nidhi walks at 5 km/hr and reaches her coaching 10 minutes late. If she increases her speed by 1 km/hr, she would have reached 15 minutes earlier. The distance she travelled from his house to coaching is
(a) 10 kms
(b) 5.5 kms
(c) 12.5 kms
(d) 10.5 kms
Ans. (c)
Ques. In 10 litres of petrol, a car travels a distance of 110 km. If it consumes 6 litres of petrol to cover next 34 km. On an average, how much distance it covers in 1 litre of petrol:
(a) 8 km
(b) 9 km
(c) 10 km
(d) 10.5 km
Ans. (b)
Ques. A train passes a station platform in 36 seconds and a man standing on the platform in 20 second. If the speed of the train is 54 km/hr, what is the length of the platform?
(a) 180 m
(b) 210 m
(c) 240 m
(d) 260 m
Ans. (c)
Ques. A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(a) 40 km/hr
(b) 45 km/hr
(c) 50 km/hr
(d) 55 km/hr
Ans. (c)
Related: Riddle questions for kids
Ques. Rohit crosses a street 600m long in 5 minutes. What is his speed in Kilometres per hours?
(a) 9 km/hr
(b) 7 km/hr
(c) 9.6 km/hr
(d) 7.2 km/hr
Ans. (d)
Ques. A motor cyclist travels for 10 hours, the first half at 21 km/hr. and the other half at 24 km/hr. Find the distance travelled.
(a) 225 km
(b) 224 km
(c) 200 km
(d) 324 km
Ans. (b)
Ques. The distance between two cities A and B is 520 km. One car starts from city A towards city B with the speed of 60 km/h. At the same time, another car starts from city B towards city A with a speed of 70 km/h. After how many hours they will cross each other?
(a) 3 hours
(b) 4 hours
(c) 5 hours
(d) 6 hours
Ans. (b)
Ques. In covering a distance of 30 km, Abhinav takes 2 hours more than Shivani. If Abhinav doubles his speed, then he would take 1 hour less than Shivani. Abhinav’s speed is:
(a) 5 kmph
(b) 6 kmph
(c) 6.25 kmph
(d) 7.5 kmph
Ans. (a)
Ques. A bus can travel 50% faster than a truck. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the bus lost about 12.5 minutes while stopping at the bus stops. The speed of the truck is:
(a) 100 kmph
(b) 110 kmph
(c) 120 kmph
(d) 130 kmph
Ans. (c)
Related:Â Dice and Cubes MCQs
Ques. The ratio between the speeds of two scooter is 7 : 8. If the second scooter runs 400 km in 4 hours, then the speed of the first scooter is:
(a) 70 km/hr
(b) 75 km/hr
(c) 84 km/hr
(d) 87.5 km/hr
Ans. (d)
Ques. Two trains whose lengths 180 m and 220m respectively are running in directions opposite to one another with respective speeds of 40 kmph and 50 kmph. Time taken by them to cross one another will be
(a) 16 seconds
(b) 17 seconds
(c) 18 seconds
(d) 22 seconds
Ans. (a)
Ques. Two cars start from the same point at 8 a.m. in the same direction. The speed of the first car is 60 km/h and the speed of second car is 55 km/h. The distance between them at 2 p.m. on same day is:
(a) 20 km
(b) 25 km
(c) 28 km
(d) 30 km
Ans. (d)
Ques. Two trains running in the same direction at 40 km/hr and 22 km/hr completely pass one another in 1 minute. lf the length of the 1st train is 125 m., then what will be the length of 2nd train?
(a) 175 m
(b) 180 m
(c) 185 m
(d) 190 m
Ans. (a)
Ques. Two towns A and B are 160 km apart. A bus starts from A to B at 7 A.M at a speed of 50 kmph. Another bus starts from B to A at 8 A.M at a speed of 60 kmph. The time of their meeting is:
(a) 9 AM
(b) 9.30 AM
(c) 10 AM
(d) 8.30 AM
Ans. (a)
Ques. A thief steals a car at 1.30 pm and drives it at 40 km/hr. The theft is discovered at 2 pm and the owner sets off in another car at 50 km/hr. He will overtake the thief at
(a) 3.30 PM
(b) 4 PM
(c) 4.30 PM
(d) 6 PM
Ans. (b)
Ques. A car covers a distance of 715 km at a constant speed. If the speed of the car had been 10 km/hr. more, then it would have taken 2 hrs. less to cover the same distance. What was the original speed of the car in km/hr?
(a) 45
(b) 50
(c) 55
(d) 65
Ans. (c)
Ques. In a road map 1 cm represents 18 km. A person drives a car on a road for 81 km. what is the distance covered in the map?
(a) 3.5 cm
(b) 4 cm
(c) 4.25 cm
(d) 4.5 cm
Ans. (d)
Ques. Puran complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
(a) 220 km
(b) 224 km
(c) 230 km
(d) 234 km
Ans. (b)
Ques. A car starting from airport reaches the bus station in 45 mts with an average speed of 40 km/hr. If the speed of the car is increased by 10 km/hr. how much less time will the car take to cover the distance?
(a) 19 mts
(b) 10 mts
(c) 9 mts
(d) 24 mts
Ans. (c)
Related: Data Interpretation questions
Ques. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
(a) 50 km
(b) 56 km
(c) 70 km
(d) 80 km
Ans. (a)
Ques. Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
(a) 48 sec
(b) 45 sec
(c) 42 sec
(d) 38 sec
Ans. (a)
Ques. A car travels 50 km in first hour, 60 km in second hour and 43 km in third hour. The average speed of the car is:
(a) 43 km/h
(b) 50 km/h
(c) 51 km/h
(d) 55 km/h
Ans. (c)
Ques. The time taken by a train 180 m long, travelling at 42 km/hr. to pass a man walking in the same direction at 6 km/hr. will be
(a) 18 seconds
(b) 21 seconds
(c) 24 seconds
(d) 25 seconds
Ans. (a)
Ques. Ishita walk at the speed of 4 km/hr. during the first 10 km and 9 km/hr. during the second 10 km. The average speed of Ishita in km/hr. is
(a) 5
(b) 5.5
(c) 6
(d) 6.5
Ans. (b)
Related:Â Coding/ Decoding PDF Questions
Ques. Abhishek travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
(a) 14 km
(b) 15 km
(c) 16 km
(d) 17 km
Ans. (c)
Ques. Sameer’s car moves 175 km in 8 hrs whereas Bhanu’s car moves 189 km in 12 hrs. The ratio between the speed of Sameer’s car and Bhanu’s car is
(a) 3 : 2
(b) 4 : 3
(c) 5 : 4
(d) 25 : 18
Ans. (d)
Ques. The speed of a car is 60 km/h. The time taken to cover a distance of 300 km in hours is:
(a) 4
(b) 5
(c) 6
(d) 8
Ans. (b)
Ques. If Rajat walks from his house to his school @ 4 km/hr. he reaches the school 10 mts earlier than the school time. However, if he walks @ 3 km/hr, he reaches 10 mts late. The distance of school from his house is
(a) 6 km
(b) 4.5 km
(c) 4 km
(d) 3 km
Ans. (c)
Ques. Kabir takes 2.5 hours to fly from Australia to New Zealand, a distance of 1200 miles. What is the average speed of his plane in miles per hour?
(a) 200 mph
(b) 240 mph
(c) 410Â mph
(d) 480 mph
Ans. (d)
Related:Â General English Sample Paper
Ques. Two trains running in the same direction at 40 km/hr. and 22 km/hr. completely pass one another in one minute. If the length of the first train is 125 m, the length of the second train is
(a) 125 m
(b) 150 m
(c) 175 m
(d) 200 m
Ans. (c)
Ques. A person standing on a platform of length 200 metres observed that a train takes 10 seconds to pass him, passed completely through the platform in 30 seconds. Then the length of the train is
(a) 100 metres
(b) 150 metres
(c) 80 metres
(d) 105 metres
Ans. (a)
Ques. A car travels a distance of 150 km in 3 hours and next 242 km in 4 hours. The average speed of car for complete journey is:
(a) 50 km/h
(b) 55.25 km/h
(c) 56 km/h
(d) 58 km/h
Ans. (c)
Ques. A man can row 9 km/hr in still water. It takes him twice as long to row up as to row down the river. Find the rate of stream.
(a) 2 km/hr
(b) 3 km/hr
(c) 4 km/hr
(d) 5 km/hr
Ans. (a)
Ques. A train running at uniform speed takes 54 seconds to pass a platform of 324 m. long and 45 seconds to pass another platform of 240m. long. The length of the train in metres is
(a) 180
(b) 170
(c) 190
(d) 100
Ans. (a)
Related: top 10 Religions in the World
Ques. A man rows upstream 13 km and down stream 28 km taking 5 hrs each time. The velocity of the current in km/ hr. is
(a) 0.5
(b) 1
(c) 1.5
(d) 2
Ans. (c)
Ques. Rishabh travels first 50 km at 25 km/hr, next 40 km with 20 km/hr, and then 90 km at 15 km/hr. His average speed for the whole journey (in km/hr)?
(a) 22 km/hr
(b) 25 km/hr
(c) 15 km/hr
(d) 18 km/hr
Ans. (d)
Ques. A car travels 60 km in 4 litres of petrol. The quantity of petrol required to cover a distance of 375 km is:
(a) 25 litres
(b) 30 litres
(c) 35 litres
(d) 36 litres
Ans. (a)
Ques. A person can row upstream 12 km/hr. and downstream at 18 km/hr. Then, the rate of the current is
(a) 14 kmph
(b) 20 kmph
(c) 17 kmph
(d) 3 kmph
Ans. (d)
Ques. Rajhaani Express train 240 metres long, takes 24 seconds to cross a man running at 10 km/hr in the same direction as the train. The speed of the train is
(a) 32 km/hr
(b) 46 km/hr
(c) 38 km/hr
(d) 24 km/hr
Ans. (b)
Ques. A boat can be rowed 9 km upstream or 18 km downstream in a period of 3 hrs. What is the speed of the boat in still water in kmph?
(a) 1.5
(b) 3
(c) 4.5
(d) 6
Ans. (c)
Ques. A person can row down stream at 12 kmph. and upstream in 6 kmph. Then, the persons’ rate in still water is:
(a) 9 kmph
(b) 5 kmph
(c) 12 kmph
(d) 6 kmph
Ans. (a)
Ques. A boat moves down stream at the rate of 1 km in 7.5 mts and upstream at the rate of 5 km/hr. The speed of the boat (in km/hr.) in still water is
(a) 6
(b) 6.5
(c) 7
(d) 7.5
Ans. (b)
Ques. A train running at 50 m/s takes 30 seconds to cross a platform of 800m. long. The length of the train in metres is
(a) 900
(b) 300
(c) 550
(d) 700
Ans. (d)
Ques. Two trains running in the same direction at 40 km/hr. and 22 km/hr. completely pass one another in one minute. If the length of the first train is 125 m, the length of the second train is
(a) 125 m
(b) 150 m
(c) 175 m
(d) 200 m
Ans. (c)
Ques. Two trains are running at 40 km/hr. and 20 km/hr. respectively in the same direction. The faster train completely passes a man sitting in the slower train in 5 seconds. What is the length of the faster train?
(a) 23 2/9 m
(b) 27m
(c) 27 7/9 m
(d) 23 m
Ans. (c)
Ques. If a person’s rate down the current is 11 km/hr. and the rate of the current is 3 km/hr., then the person’s rate against the current (in km/hr.) is
(a) 7 kmph
(b) 8 kmph
(c) 5 kmph
(d) 6 kmph
Ans. (c)
Ques. Two trains running in the same direction at 40 km/hr. and 22 km/hr. completely pass one another in 1 minute. If the length of the first train is 125m., find the length of the second train.
(a) 170 m
(b) 175 m
(c) 180 m
(d) 190 m
Ans. (b)
Ques. A boat takes 9 hrs to travel from A to B upstream. If the river current is 3 km/hr, how long will it take to travel downstream between B and A?
(a) 3 hrs
(b) 6 hrs
(c) 4.5 hrs
(d) 5 hrs
Ans. (d)
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# Section 1.2 The Algebra of Complex Numbers
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C01-2.mws
COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
[email protected] [email protected]
Complimentary software to accompany the textbook:
COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: [email protected], http://www.jbpub.com/
CHAPTER 1 COMPLEX NUMBERS
Section 1.2 The Algebra of Complex Numbers
We have seen that complex numbers came to be viewed as ordered pairs of real numbers. That is, a complex number is defined to be
.
The reason we say ordered pair is because we are thinking of a point in the plane. The point (2, 3), for example, is not the same as (3, 2). The order in which we write and in the equation makes a difference. Clearly, then, two complex numbers are equal if and only if their coordinates are equal and their coordinates are equal. In other words,
iff and .
If we are to have a meaningful number system, there needs to be a method for combining these ordered pairs. We need to define algebraic operations in a consistent way so that the sum, difference, product, and quotient of any two ordered pairs will again be an ordered pair. The key to defining how these numbers should be manipulated is to follow Gauss' lead and equate with . Then, by letting and be arbitrary complex numbers, we have
Thus, if and are arbitrary complex numbers, the following definitions should make sense.
Formula (1-6), Page 7.
Definition 1.2: Subtraction
Formula (1-7), Page 7.
The rules for addition, subtraction, multiplication and division of complex numbers
are extensions of the rules for real numbers. They obey familiar algebraic properties.
Example 1.1, Page 7.
Find and .
> z:='z': Z1 := 3 + 7*I: Z2 := 5 - 6*I: z[1] = Z1; z[2] = Z2; ` `; z[1] + z[2] = Z1 + Z2; z[1] - z[2] = Z1 - Z2;
Definition 1.3: Multiplication
Formula (1-8), Page 8.
Example 1.2, Page 8.
Find .
> z:='z': Z1 := 3 + 7*I: Z2 := 5 - 6*I: z[1] = Z1; z[2] = Z2; ` `; z[1]*z[2] = Z1*Z2;
Definition 1.4: Division
Formula (1-9), Page 9.
Example 1.3, Page 9.
Find .
> z:='z': Z1 := 3 + 7*I: Z2 := 5 - 6*I: z[1] = Z1; z[2] = Z2; ` `; z[1]/z[2] = Z1/Z2;
Derivation for Multiplication,
Formula (1-8), Page 8. In general we can derive:
> x:='x': y:='y': z:='z': Z1:='Z1': Z1 := x[1] + I*y[1]: Z2:='Z2': Z2 := x[2] + I*y[2]: z[1] = Z1; z[2] = Z2; ` `; z[1]*z[2] = Z1*Z2; z[1]*z[2] = expand(Z1*Z2);
Derivation for Division,
Formula (1-9), Page 9. In general we can derive:
> d:='d': n:='n': x:='x': y:='y': z:='z': Z1:='Z1': Z1 := x[1] + I*y[1]: Z2:='Z2': Z2 := x[2] + I*y[2]: z[1] = Z1; z[2] = Z2; ` `; z[1]/z[2] = Z1/Z2; n := expand(Z1*(x[2]-I*y[2])): d := expand(Z2*(x[2]-I*y[2])): z[1]/z[2] = n/d;
Definition 1.5: Real Part
The real part of denoted is the real number .
Definition 1.6: Imaginary Part
The imaginary part of denoted is the real number .
Definition 1.7: Conjugate
The conjugate of denoted is the complex number .
Example 1.4a, Page 12.
Find and .
> z:='z': Z1 := -3 + 7*I: z[1] = Z1; Re(z[1]) = Re(Z1); ` `; Z2 := 9 + 4*I: z[2] = Z2; Re(z[2]) = Re(Z2);
Example 1.4b, Page 12.
Find and .
> z:='z': Z1 := -3 + 7*I: z[1] = Z1; `Im(z1) ` = Im(Z1); ` `; Z2 := 9 + 4*I: z[2] = Z2; `Im(z2) ` = Im(Z2);
Example 1.4c, Page 12.
Find and .
> z:='z': Z1 := -3 + 7*I: z[1] = Z1; conjugate(z[1]) = conjugate(Z1); ` `; Z2 := 9 + 4*I: z[2] = Z2; conjugate(z2) = conjugate(Z2);
Derivation of the Commutative Law for Addition,
Property (P1), Page 10. In general we can derive:
> x:='x': y:='y': z:='z': Z1:='Z1': Z1 := x[1] + I*y[1]: Z2:='Z2': Z2 := x[2] + I*y[2]: z[1] = Z1; z[2] = Z2; ` `; `z1 + z2` = Z1 + Z2; `z2 + z1` = Z2 + Z1; ` `; `Does z1 + z2 = z2 + z1 ?`; Z1+Z2 = Z2+Z1; evalb(Z1+Z2 = Z2+Z1);
Derivation of the Associative Law for Multiplication,
Property (P6), Page 10. In general we can derive:
> x:='x': y:='y': Z1 := x[1] + I*y[1]: `z1 ` = Z1; Z2 := x[2] + I*y[2]: `z2 ` = Z2; Z3 := x[3] + I*y[3]: `z3 ` = Z3; w1 := Z1*(Z2 + Z3): w2 := Z1*Z2 + Z1*Z3: ` `; `z1*(z2 + z3) ` = w1; `z1*z2 + z1*z3 ` = w2; w1 := expand(w1): w2 := expand(w2): ` `; `z1*(z2 + z3) ` = w1; `z1*z2 + z1*z3 ` = w2; ` `; `Does z1*(z2 + z3) = z1*z2 + z1*z3 ?`; evalb(w1 = w2);
>
Theorem 1.1, Page 12. Suppose , , and are arbitrary complex numbers. Then
(1-10) ,
(1-11) ,
(1-12) ,
(1-13) ,
(1-14) ,
(1-15) ,
(1-16) ,
(1-17) .
End of Section 1.2. | 1,971 | 4,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.811617 |
https://db0nus869y26v.cloudfront.net/en/Rate_of_convergence | 1,722,886,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00610.warc.gz | 157,582,217 | 27,830 | In numerical analysis, the order of convergence and the rate of convergence of a convergent sequence are quantities that represent how quickly the sequence approaches its limit. A sequence ${\displaystyle (x_{n})}$ that converges to ${\displaystyle L}$ is said to have order of convergence ${\displaystyle q\geq 1}$ and rate of convergence ${\displaystyle \mu }$ if
${\displaystyle \lim _{n\rightarrow \infty }{\frac {\left|x_{n+1}-L\right|}{\left|x_{n}-L\right|^{q))}=\mu .}$[1]
The rate of convergence ${\displaystyle \mu }$ is also called the asymptotic error constant. Note that this terminology is not standardized and some authors will use rate where this article uses order (e.g., [2]).
In practice, the rate and order of convergence provide useful insights when using iterative methods for calculating numerical approximations. If the order of convergence is higher, then typically fewer iterations are necessary to yield a useful approximation. Strictly speaking, however, the asymptotic behavior of a sequence does not give conclusive information about any finite part of the sequence.
Similar concepts are used for discretization methods. The solution of the discretized problem converges to the solution of the continuous problem as the grid size goes to zero, and the speed of convergence is one of the factors of the efficiency of the method. However, the terminology, in this case, is different from the terminology for iterative methods.
Series acceleration is a collection of techniques for improving the rate of convergence of a series discretization. Such acceleration is commonly accomplished with sequence transformations.
## Convergence speed for iterative methods
### Convergence definitions
Suppose that the sequence ${\displaystyle (x_{k})}$ converges to the number ${\displaystyle L}$. The sequence is said to converge with order ${\displaystyle q}$ to ${\displaystyle L}$, and with a rate of convergence[3] of ${\displaystyle \mu }$, if
${\displaystyle \lim _{k\to \infty }{\frac {|x_{k+1}-L|}{|x_{k}-L|^{q))}=\mu }$ (Definition 1)
for some positive constant ${\displaystyle \mu \in (0,\infty )}$ if ${\displaystyle q>1}$, and ${\displaystyle \mu \in (0,1)}$ if ${\displaystyle q=1}$.[4][5] It is not necessary, however, that ${\displaystyle q}$ be an integer. For example, the secant method, when converging to a regular, simple root, has an order of φ ≈ 1.618.[citation needed]
Convergence with order
• ${\displaystyle q=1}$ is called linear convergence if ${\displaystyle \mu \in (0,1)}$, and the sequence is said to converge Q-linearly to ${\displaystyle L}$.
• ${\displaystyle q=2}$ is called quadratic convergence.
• ${\displaystyle q=3}$ is called cubic convergence.
• etc.
#### Order estimation
A practical method to calculate the order of convergence for a sequence generated by a fixed point iteration is to calculate the following sequence, which converges to ${\displaystyle q}$:[6] ${\displaystyle q\approx {\frac {\log \left|\displaystyle {\frac {x_{k+1}-x_{k)){x_{k}-x_{k-1))}\right|}{\log \left|\displaystyle {\frac {x_{k}-x_{k-1)){x_{k-1}-x_{k-2))}\right|)).}$
For numerical approximation of an exact value through a numerical method of order q see [7]
#### Q-convergence definitions
In addition to the previously defined Q-linear convergence, a few other Q-convergence definitions exist. Given Definition 1 defined above, the sequence is said to converge Q-superlinearly to ${\displaystyle L}$ (i.e. faster than linearly) in all the cases where ${\displaystyle q>1}$ and also the case ${\displaystyle q=1,\mu =0}$.[8] Given Definition 1, the sequence is said to converge Q-sublinearly to ${\displaystyle L}$ (i.e. slower than linearly) if ${\displaystyle q=1,\mu =1}$. The sequence ${\displaystyle (x_{k})}$ converges logarithmically to ${\displaystyle L}$ if the sequence converges sublinearly and additionally if[9] ${\displaystyle \lim _{k\to \infty }{\frac {|x_{k+1}-x_{k}|}{|x_{k}-x_{k-1}|))=1.}$ Note that unlike previous definitions, logarithmic convergence is not called "Q-logarithmic."
In the definitions above, the "Q-" stands for "quotient" because the terms are defined using the quotient between two successive terms.[10]: 619 Often, however, the "Q-" is dropped and a sequence is simply said to have linear convergence, quadratic convergence, etc.
#### R-convergence definition
The Q-convergence definitions have a shortcoming in that they do not include some sequences, such as the sequence ${\displaystyle (b_{k})}$ below, which converge reasonably fast, but whose rate is variable. Therefore, the definition of rate of convergence is extended as follows.
Suppose that ${\displaystyle (x_{k})}$ converges to ${\displaystyle L}$. The sequence is said to converge R-linearly to ${\displaystyle L}$ if there exists a sequence ${\displaystyle (\varepsilon _{k})}$ such that ${\displaystyle |x_{k}-L|\leq \varepsilon _{k}\quad {\text{for all ))k\,,}$ and ${\displaystyle (\varepsilon _{k})}$ converges Q-linearly to zero.[3] The "R-" prefix stands for "root". [10]: 620
### Examples
Consider the sequence ${\displaystyle (a_{k})=\left\{1,{\frac {1}{2)),{\frac {1}{4)),{\frac {1}{8)),{\frac {1}{16)),{\frac {1}{32)),\ldots ,{\frac {1}{2^{k))},\dots \right\}.}$ It can be shown that this sequence converges to ${\displaystyle L=0}$. To determine the type of convergence, we plug the sequence into the definition of Q-linear convergence, ${\displaystyle \lim _{k\to \infty }{\frac {\left|1/2^{k+1}-0\right|}{\left|1/2^{k}-0\right|))=\lim _{k\to \infty }{\frac {2^{k)){2^{k+1))}={\frac {1}{2)).}$ Thus, we find that ${\displaystyle (a_{k})}$ converges Q-linearly and has a convergence rate of ${\displaystyle \mu =1/2}$. More generally, for any ${\displaystyle c\in \mathbb {R} ,\mu \in (-1,1)}$, the sequence ${\displaystyle (c\mu ^{k})}$ converges linearly with rate ${\displaystyle |\mu |}$.
The sequence ${\displaystyle (b_{k})=\left\{1,1,{\frac {1}{4)),{\frac {1}{4)),{\frac {1}{16)),{\frac {1}{16)),\ldots ,{\frac {1}{4^{\left\lfloor {\frac {k}{2))\right\rfloor ))},\,\ldots \right\))$ also converges linearly to 0 with rate 1/2 under the R-convergence definition, but not under the Q-convergence definition. (Note that ${\displaystyle \lfloor x\rfloor }$ is the floor function, which gives the largest integer that is less than or equal to ${\displaystyle x}$.)
The sequence ${\displaystyle (c_{k})=\left$$(\frac {1}{2)),{\frac {1}{4)),{\frac {1}{16)),{\frac {1}{256)),{\frac {1}{65,\!536)),\ldots ,{\frac {1}{2^{2^{k)))),\ldots \right$$)$ converges superlinearly. In fact, it is quadratically convergent.
Finally, the sequence ${\displaystyle (d_{k})=\left\{1,{\frac {1}{2)),{\frac {1}{3)),{\frac {1}{4)),{\frac {1}{5)),{\frac {1}{6)),\ldots ,{\frac {1}{k+1)),\ldots \right\))$ converges sublinearly and logarithmically.
## Convergence speed for discretization methods
This section may require cleanup to meet Wikipedia's quality standards. The specific problem is: There appears to be a mixture of defining convergence with regards to grid points n {\displaystyle n} and with step size h {\displaystyle h} . Section should be modified for consistency and include an explanation of alternative (equivalent?) definitions. Please help improve this section if you can. (August 2020) (Learn how and when to remove this message)
A similar situation exists for discretization methods designed to approximate a function ${\displaystyle y=f(x)}$, which might be an integral being approximated by numerical quadrature, or the solution of an ordinary differential equation (see example below). The discretization method generates a sequence ${\displaystyle {y_{0},y_{1},y_{2},y_{3},...))$, where each successive ${\displaystyle y_{j))$ is a function of ${\displaystyle y_{j-1},y_{j-2},...}$ along with the grid spacing ${\displaystyle h}$ between successive values of the independent variable ${\displaystyle x}$. The important parameter here for the convergence speed to ${\displaystyle y=f(x)}$ is the grid spacing ${\displaystyle h}$, inversely proportional to the number of grid points, i.e. the number of points in the sequence required to reach a given value of ${\displaystyle x}$.
In this case, the sequence ${\displaystyle (y_{n})}$ is said to converge to the sequence ${\displaystyle f(x_{n})}$ with order q if there exists a constant C such that
${\displaystyle |y_{n}-f(x_{n})|
This is written as ${\displaystyle |y_{n}-f(x_{n})|={\mathcal {O))(h^{q})}$ using big O notation.
This is the relevant definition when discussing methods for numerical quadrature or the solution of ordinary differential equations (ODEs).[example needed]
A practical method to estimate the order of convergence for a discretization method is pick step sizes ${\displaystyle h_{\text{new))}$ and ${\displaystyle h_{\text{old))}$ and calculate the resulting errors ${\displaystyle e_{\text{new))}$ and ${\displaystyle e_{\text{old))}$. The order of convergence is then approximated by the following formula:
${\displaystyle q\approx {\frac {\log(e_{\text{new))/e_{\text{old)))}{\log(h_{\text{new))/h_{\text{old))))),}$[citation needed]
which comes from writing the truncation error, at the old and new grid spacings, as
${\displaystyle e=|y_{n}-f(x_{n})|={\mathcal {O))(h^{q}).}$
The error ${\displaystyle e}$ is, more specifically, a global truncation error (GTE), in that it represents a sum of errors accumulated over all ${\displaystyle n}$ iterations, as opposed to a local truncation error (LTE) over just one iteration.
### Example of discretization methods
Consider the ordinary differential equation
${\displaystyle {\frac {dy}{dx))=-\kappa y}$
with initial condition ${\displaystyle y(0)=y_{0))$. We can solve this equation using the Forward Euler scheme for numerical discretization:
${\displaystyle {\frac {y_{n+1}-y_{n)){h))=-\kappa y_{n},}$
which generates the sequence
${\displaystyle y_{n+1}=y_{n}(1-h\kappa ).}$
In terms of ${\displaystyle y(0)=y_{0))$, this sequence is as follows, from the Binomial theorem: ${\displaystyle y_{n}=y_{0}(1-h\kappa )^{n}=y_{0}\left(1-nh\kappa +n(n-1){\frac {h^{2}\kappa ^{2)){2))+....\right).}$
The exact solution to this ODE is ${\displaystyle y=f(x)=y_{0}\exp(-\kappa x)}$, corresponding to the following Taylor expansion in ${\displaystyle h\kappa }$ for ${\displaystyle h\kappa \ll 1}$: ${\displaystyle f(x_{n})=f(nh)=y_{0}\exp(-\kappa nh)=y_{0}\left[\exp(-\kappa h)\right]^{n}=y_{0}\left(1-h\kappa +{\frac {h^{2}\kappa ^{2)){2))+....\right)^{n}=y_{0}\left(1-nh\kappa +{\frac {n^{2}h^{2}\kappa ^{2)){2))+...\right).}$
In this case, the truncation error is
${\displaystyle e=|y_{n}-f(x_{n})|={\frac {nh^{2}\kappa ^{2)){2))={\mathcal {O))(h^{2}),}$
so ${\displaystyle (y_{n})}$ converges to ${\displaystyle f(x_{n})}$ with a convergence rate ${\displaystyle q=2}$.
### Examples (continued)
The sequence ${\displaystyle (d_{k})}$ with ${\displaystyle d_{k}=1/(k+1)}$ was introduced above. This sequence converges with order 1 according to the convention for discretization methods.[why?]
The sequence ${\displaystyle (a_{k})}$ with ${\displaystyle a_{k}=2^{-k))$, which was also introduced above, converges with order q for every number q. It is said to converge exponentially using the convention for discretization methods. However, it only converges linearly (that is, with order 1) using the convention for iterative methods.[why?]
## Recurrent sequences and fixed points
The case of recurrent sequences ${\displaystyle x_{n+1}:=f(x_{n})}$ which occurs in dynamical systems and in the context of various fixed-point theorems is of particular interest. Assuming that the relevant derivatives of f are continuous, one can (easily) show that for a fixed point ${\displaystyle f(p)=p}$ such that ${\displaystyle |f'(p)|<1}$, one has at least linear convergence for any starting value ${\displaystyle x_{0))$ sufficiently close to p. If ${\displaystyle |f'(p)|=0}$ and ${\displaystyle |f''(p)|<1}$, then one has at least quadratic convergence, and so on. If ${\displaystyle |f'(p)|>1}$, then one has a repulsive fixed point and no starting value will produce a sequence converging to p (unless one directly jumps to the point p itself).
## Acceleration of convergence
Many methods exist to increase the rate of convergence of a given sequence, i.e., to transform a given sequence into one converging faster to the same limit. Such techniques are in general known as "series acceleration". The goal is to reduce the computational cost of approximating the limit of the transformed sequence. One example of series acceleration is Aitken's delta-squared process. These methods in general (and in particular Aitken's method) do not increase the order of convergence, and are useful only if initially the convergence is not faster than linear: If ${\displaystyle (x_{n})}$ convergences linearly, one gets a sequence ${\displaystyle (a_{n})}$ that still converges linearly (except for pathologically designed special cases), but faster in the sense that ${\displaystyle \lim(a_{n}-L)/(x_{n}-L)=0}$. On the other hand, if the convergence is already of order ≥ 2, Aitken's method will bring no improvement.
## References
1. ^ Ruye, Wang (2015-02-12). "Order and rate of convergence". hmc.edu. Retrieved 2020-07-31.
2. ^ Senning, Jonathan R. "Computing and Estimating the Rate of Convergence" (PDF). gordon.edu. Retrieved 2020-08-07.
3. ^ a b Bockelman, Brian (2005). "Rates of Convergence". math.unl.edu. Retrieved 2020-07-31.
4. ^ Hundley, Douglas. "Rate of Convergence" (PDF). Whitman College. Retrieved 2020-12-13.
5. ^ Porta, F. A. (1989). "On Q-Order and R-Order of Convergence" (PDF). Journal of Optimization Theory and Applications. 63 (3): 415–431. doi:10.1007/BF00939805. S2CID 116192710. Retrieved 2020-07-31.
6. ^ Senning, Jonathan R. "Computing and Estimating the Rate of Convergence" (PDF). gordon.edu. Retrieved 2020-08-07.
7. ^ Senning, Jonathan R. "Verifying Numerical Convergence Rates" (PDF). Retrieved 2024-02-09.
8. ^ Arnold, Mark. "Order of Convergence" (PDF). University of Arkansas. Retrieved 2022-12-13.
9. ^ Van Tuyl, Andrew H. (1994). "Acceleration of convergence of a family of logarithmically convergent sequences" (PDF). Mathematics of Computation. 63 (207): 229–246. doi:10.2307/2153571. JSTOR 2153571. Retrieved 2020-08-02.
10. ^ a b Nocedal, Jorge; Wright, Stephen J. (2006). Numerical Optimization (2nd ed.). Berlin, New York: Springer-Verlag. ISBN 978-0-387-30303-1.
## Literature
The simple definition is used in
The extended definition is used in
• Walter Gautschi (1997), Numerical analysis: an introduction, Birkhäuser, Boston. ISBN 0-8176-3895-4.
• Endre Süli and David Mayers (2003), An introduction to numerical analysis, Cambridge University Press. ISBN 0-521-00794-1.
The Big O definition is used in
• Richard L. Burden and J. Douglas Faires (2001), Numerical Analysis (7th ed.), Brooks/Cole. ISBN 0-534-38216-9
The terms Q-linear and R-linear are used in | 4,195 | 14,946 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 101, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.891669 |
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Exponents. Could you give me a link that could lead me to more details regarding that program ? Edit. Save. For multiplication, use the * symbol. Feel free to try them now. Sometimes writing expressions and finding equivalent expressions can … If you incorrectly enter it as 4/1/2 then it is solved 4/1 = 4 first then 4/2 = 2 last. Every aspect covered this is just what you need our Cookie Policy to division, we got. You give me a link that could lead me to more details regarding that?! Equation into the equivalent fractions contain different numbers but they mean the same thing: 1/2 = 0.5 performed. By a variable and subtraction, use the following rules to equivalent expression calculator soup expressions into calculator! Free download, examples difficult problems in dividing fractions baffle me and I ’ m sure ’! Equivalent when they are both equal when written in lowest terms step-by-step this website uses to., trigonometric, and / signs 5x is equivalent to 5 * x and. Half, and that 50/100 is half = 2 last you agree to our Cookie Policy ; attempts use! Final solution on trinomial calculator, math holt Grade 9, ignore coefficients... Solve the equation in a couple of clicks you will need advice on equivalent expressions calculator, ti-84 free,. Online calculator that completes the square of a reasonably priced software that I use! 2 ( 3 ) = 6 same value when both the numerator and denominator of the fraction evaluate your down... You to solve the equation in a new window always the ideal site to pay a visit to 5. Assistance below to refresh your memory as to legitimate operators, as well as mathematics content polymathlove.com... Sign, so 5x is equivalent to 5 * can! Me a link that could lead me to more details regarding that program following expression solver the! Radical expression otherwise but problems in dividing fractions baffle me and I ’ m you! 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www.wallcoverings.org | 1,596,677,203,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735990.92/warc/CC-MAIN-20200806001745-20200806031745-00376.warc.gz | 848,801,100 | 11,698 | How to Estimate
Measuring Before Estimating Wallcovering Needs
The most important step in estimating wallcovering is accurate measurements. Use a yardstick or steel tape measure, never a cloth tape measure. Take measurements in feet rounding off to the next highest half foot or foot. Draw a room diagram showing doors, windows and ceiling height. If a wall is unusually broken up with a fireplace, built-in bookcases, etc., a diagram with detailed measurements will be beneficial in figuring square footage of wallcovering needed.
Measure wall height from floor to ceiling. Exclude baseboards and moldings. Measure length of each wall including doors and windows. Find the total square feet of the wall(s) by multiplying ceiling height by total wall length. Subtract areas that will not be covered. (Standard doors are about 3 x 7 feet or 21square feet; standard windows about 3 x 4 or 12 square feet.)
These calculations give the total number of square feet to be covered. Using this the number of rolls or linear yards of wallcovering can be determined.
For example:
In the above figure, each wall is 12' long with an 8' ceiling. Multiply 12 x 8 = 96 square feet for each wall, then multiply 96 x 4 (since there are four walls with 96 square feet each)= 384 total square feet for the room.
Metric Single Roll
Repeat Length Usable Yield 0" to 6" 25 sq. ft. 7" to 12" 22 sq. ft. 13" to 18" 20 sq. ft. 19" to 23" 18 sq. ft.
These figures work well for normal sized walls. For unusually short (under 3 feet) or high (over 9 feet) they do not apply.
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Square Foot Area Method
Using the above diagram as an example, figure the amount of wallcovering that will be needed to hang the room. The figure of 384 square feet has not taken into account the square footage of the doors and windows. Subtract the actual square footage of each opening - 21 square feet for the door, and 12 square feet for each of the windows. 384 square feet - 45 square feet (21 + 12 + 12 = 45) = 339 square feet of wallspace that will be covered with wallcovering. If you are using a wallcovering with a pattern repeat of 8 inches, figure that each metric single roll will contain 22 square feet of usable wallcovering, 339 square feet (the amount of wallspace from above that will be hung) divided by 22 square feet (from Usable Yield Chart) which equals 15.4, or round up to 16 metric singlerolls that will be needed to hang the example toom (8 metric rolls).
The equation would look like this:
384 sq. ft. (room size)
-21 sq. ft. (one standard door)
-12 sq. ft. (one standard window)
-12 sq. ft. (one standard window)
=339 sq. ft. of wallspace that will be hung
339 sq. ft. / 22 sq. ft. = 15.4 msr to hang the room, rounded up to 16 msr.
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Stairways or Cathedral Walls
When estimating a wall that has a diagonal, remember there will be extra waste to allow for the slope of the steps or the ceiling pitch. There are two different types of stairways to figure: one with a horizontal ceiling line, and the second with a diagonal ceiling line that parallels the fall of the steps.
In both cases, the first step is to divide the wall in either squares or rectangles to determine the square footage. In the figure given above, the upstairs ceiling height is 8", and the downstairs ceiling height is 8". These figures are the length of the wall. Next, measure wall width horizontally from the top of the stairs to an imaginary vertical line originating at the bottom of the stairs, which in the example is 15". Taking the top rectangle, figure 8" x 15" = 120 sq. ft. Next, figure the bottom rectangle, 8" x 15" = 120 sq. ft., but since a portion of this wall area is under the stairs, multiply the bottom rectangle square footage by 65%, an industry standard. Add the two figures together to arrive at the square feet that needs to by hung with wallcovering. The equation would look as follows:
8" x 15" = 120 sq. ft. (top rectangle)
8" x 15" = 120 sq. ft. x 65% = 78 sq. ft. (bottom rectangle)
120 sq. ft. + 78 sq. ft. = 198 sq. ft.
Once you have the square footage figured, estimate the amount of wallcovering just as you would for an ordinary room, finding the usable square feet for the particular pattern from the Usable Yield Chart and then dividing the total square feet by that figure.
For example, if using a wallcovering with repeat of 14", each msr would contain 20 square feet of usable wallcovering. The equation would look as follows: 198 sq. ft. / 20 sq. ft. = 9.9 msr rounded to 10 msr. If the stairway has a sloping ceiling, do as the first example in finding the width and length of the imaginary rectangle or square. The next step is to take both of these rectangle/square figures multiplied by 65% to find the square feet of wall area. The equation would look as follows:
8" x 15" = 120 sq. ft. (top rectangle)
8" x 15" = 120 sq. ft. (bottom rectangle)
120 sq. ft. + 120 sq. ft. = 240 sq. ft.
240 sq. ft. x 65% = 156 sq. ft. of wall area to be covered
Using the same pattern with a repeat of 14", each msr would contain 20 square feet of usable wallcovering and the equation would look as follows:
156 sq. ft. / 20 sq. ft. = 7.8 msr rounded to 8 msr
A cathedral ceiling would be estimated the same way, squaring the top rectangle, multiplying the square feet by 65%, then adding that figure to the square feet of the bottom rectangle.
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Estimating Commercial Square Footage
After the wallcovering has been determined from the specification, now figure the square footage and how much is needed for the job. Once the width is known, the number of square feet in a lineal yard for that particular wallcovering width can be determined. An important formula to remember is:
• Width divided by 12 = number of feet
• Number of feet multiplied by 3 (1 yard) = square feet/width (square feet per lineal yard)
• Divide square feet of wall space to cover by square feet/width
For example:
• 54-inch wide material used to cover 1500 square feet is figured
• 54 divided by 12 = 4.5
• 4.5 multiplied by 3 = 13.5 square feet per linear yard
• 1500 divided by 13.5 = 111.11
1500 square feet of wall space would require 112 yards without waste.
Once the width and the square footage for the width is known, any amount can be determined. If the yardage for a particular width is known, and the width of the material is changed, to convert from one width to another, work backward to determine the yardage.
For example:
• 150 yards of 54-inch wide material
• 54 inches is 13.5 square feet per yard (54/12 = 4.5 X 3 = 13.5)
• 150 yards multiplied by 13.5 = 2,025 square feet of wall space to cover
New width is 36 inches wide
• 36/12 = 3
• 3 X 3 = 9 square feet
• 2,025 divided by 9 = 225 yards of 36-inch wide wallcovering instead of original 150 yards of 54-inch material.
These are exact yardage amounts and do not allow for waste caused by pattern repeat. A 10 percent waste factor is a good figure to consider; however, a matching pattern with a large repeat would require additional material compared to a textured pattern without a match. All contractors should be aware of the pattern, width, and match before submitting final cid and figures.
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Association Management Software Powered by YourMembership :: Legal | 1,857 | 7,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-34 | latest | en | 0.891338 |
https://stats.stackexchange.com/questions/629498/problem-with-cumulative-distribution-function/629502 | 1,721,757,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00164.warc.gz | 465,103,158 | 43,701 | # Problem with Cumulative distribution function
I can't understand this cumulative distribution function. I would like to calculate the data distribution function:
F(t) := P(X <= t) ~ sum_i freq(observation_i <= t)/total_observation =: f(t)
Having the datas:
List_Goal: [1, 2, 2, 1, 2]
Each of these values (1, 2, 2, 1, 2) is a goal scored by a club in previous matches. For example, the Liverpool club: in the first match he scored 1 goal, in the second match he scored 2 goals, in the third match he scored 2 goals, in the fourth match he scored 1 goal, in the fifth match he scored 2 goals. In all the club played 5 matches. So in the sixth match, still to be played, we want to know the Cumulative distribution function in relation to the goals of the 5 previous matches.
How many 0 goals are scored? 0/5
How many 1 goals are scored? 2/5
How many 2 goals are scored? 5/5
So:
0 goals 0/5 = 0
1 goals 2/5 = 0.4
2 goals 5/5 = 1
REQUEST: If it were correct, then how should we proceed next?
Thank you all!!!
• Before I answer your question I would like to ask you explain in your own words what is a cumulative distribution function. Commented Oct 24, 2023 at 6:49
• @Math-fun Premise that i'm just a math enthusiast and i'm not even good at it. Maybe I can be wrong, but the cumulative distribution function is a function F(x) that takes values over the set of real numbers, enclosing various data before or after a given point. It depends on a real variable and a probability distribution variable that returns the probability that the variable is equal to or less than a specific value. I don't know if I said it right. Can you help me please? Commented Oct 24, 2023 at 7:01
• Sounds good :-) So let the random variable you are looking at be $X$. Which values can $X$ take in your question? Commented Oct 24, 2023 at 7:03
• @Math-fun I don't know, that's why I was asking for help on the sheet I attached to the question. In theory, even if not very well, I know what the cumulative distribution function is. But I can't understand the exercise. For this reason I asked for help. In the exercise we use the basic List_Goal values: [1, 2, 2, 1, 2]. Then I don't understand. Can you help me please? Thank you Commented Oct 24, 2023 at 7:18
• Sure, this is what I try: I am trying to communicate to you what you should understand first before moving to find the CDF. As you mentioned in your explanation, CDF helps us understand the behavior of a random variable. But before we move to see what the CDF of a random variable is, we need to understand the variable we wish to analyze. And the first thing to do here is to know which value does the random variable take. Commented Oct 24, 2023 at 7:25
given the comments, I think it make sense that I summarize our discussions here.
First, a random variable is a "mathematical model" which we use to model real world phenomon. In your example the random variable can be defined as follows:
$$X=$$ the number of goals scored.
And $$X$$ is indeed a function of certain events that should take place so that a certain number of goals are socred. Now let's move on to the values that this random variable (which is a function of events) takes. Obviously $$X$$ can't be negative and obviusly it can only take integra values, $$0$$ (no goal), 1 (one goal), 2 (two goals), and so on. But $$X$$ cannot be a very large number in this case. So using these arguments we may refine our model for the number of goals:
$$X=$$ the number of goals scored, where $$X\in\{0,1,2,...,g_m\}$$ where $$g_m$$ is the maximum number of goals possible to be scored.
So far so good. We now may take one step further and ask ourselves what is the probability that we have no goals scored? And here we need a tool that helps us quantify the probability that $$X=0$$. This tool is called a probability measure, since it measures the probabilities. Accordingly we can go on and ask ourselves what is the probability that $$X=x$$ where $$x\in\{0,1,2,...,g_m\}$$. And we take one step further and write $$\Pr(X=x)=p_x.$$ That is, $$p_0$$ is the probability that we score 0 goals, $$p_1$$ is the probability that we score 1 goals and so on.
Obviously this was just abstract and had nothing to do with "reality". To make our model usable, we need to observe how many goals are scored over time and using these observations we could find "some guesses" about $$p_x$$. For example in your small set set of observatiobs we have 3 games with 2 goals. Therefore using your observations we may guess (read estimate) the $$p_2$$ as $$\frac35$$ (since 3 times out of 5 times we have scored 2 goals), which is essentially the fraction of games with 2 goals, we write $$\widehat{p}_2=\frac35,$$ where $$\widehat{}$$ is there to remind us that "hey you have a guess of $$p_2$$, you don't have $$p_2$$". Now can you "estimate" $$p_x$$ for $$x=0$$, $$x=1$$ and other values of $$X$$ from your observations (which we call it a sample)?
once you identify the estimated probabilities, you may go on step further and estimate the CDF too. Share how you proceed, if you wish of course!
• Some things I understood, other things I didn't understand. Your answer is great, but very theoretical and not very practical. Could you update something to your answer please? Could you add the steps (without explanation) to get to the final result? For example: 1 step, 2 step, 3 step, etc? Thank you Commented Oct 24, 2023 at 8:28
• The point for me here is not to write all the answer for you, but to help you understand the question and find the answer yourself. I will of course provide you with further explanations once I know which steps you don't understand and what the problem you are facing :-) I understand that you wish to understand things and you are not necessarily after the "fina result". correct me, please, if I am wrong :-) Commented Oct 24, 2023 at 8:47
• In my example (and the one on the attached sheet): 0 goals were scored 0 out of 5 times; 1 goal was scored 2 times out of 5; 2 goals were scored 3 times out of 5; So according to your reasoning, i will have p0= 0/5, p1=2/5, p2=3/5. If so, then why is there 0/5, 2/5 on the paper, but no 3/5? On the sheet I read 5/5, but I didn't understand why 5/5. After your explanation, therefore after having calculated p0= 0/5, p1=2/5, p2=3/5, how should I proceed? Commented Oct 24, 2023 at 8:59
• Great! Can you estimate the cdf of $X$ using your observations (I am not talking about the handwritten page you have attached) at 0, 1, 2? Commented Oct 24, 2023 at 9:03
• Can you show me the results of the X cdfs? (just to understand if I calculate them correctly) Furthermore, after having calculated them, how should I proceed? Commented Oct 24, 2023 at 9:12
I am not 100% sure I understand your question. Please let me know if this is what you are looking for:
To find the empirical cumulative distribution function for integer data you need to:
1. Calculate the proportion of each possible observation;
2. Order these proportions using the increasing ordering of the corresponding integer values;
3. Calculate the cumulative sum of these proportions.
Below is an example with R for your data:
# The integer values
x <- c(1, 2, 2, 1, 2)
# The function "table" will already order the values
# according to the order of the integers
cumsum(prop.table(table(x)))
# Validate with the "ecdf" function, which creates
# a function to calculate the empirical CDF
ecdf(x)(1)
ecdf(x)(2)
• I don't know, I don't know if this is what I'm looking for. I would like to receive the same thing that is written on the paper. I guess you know more than me. Why aren't you 100% sure? Is the result of what you wrote the same as the result of the article? P.S: could you also write it in Python? Commented Oct 24, 2023 at 3:48
I had already found the answer on my own, but I didn't realize it. It seems right to share it anyway. I thought we needed to proceed differently and write other things. However, I thank everyone for their attention. I voted on everyone's answers
The user Math-fun came close to what I was looking for and was kind in his explanation, but his answer wasn't what I was looking for. To solve the exercise I didn't have to count the number of goals and obtain for example p2 = 3/5 (as in the user's answer), but I had to count how many observations are less than or equal to 0, 1, and 2? So for example I didn't need p2= 3/5, but p2= 5/5
The solution to the exercise is this:
If the goals scored are <= 0, then i will have goals 0/5 = 0
If the goals scored are <= 1, then i will have goals 2/5 = 0.4
If the goals scored are <= 2, then i will have goals 5/5 = 1
I know:
Observation <= 0 (0) goals: CDF calculation
Observation <= 1 (0.4) goals: CDF calculation
Observation <= 2 (1) goals: CDF calculation | 2,364 | 8,790 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-30 | latest | en | 0.948877 |
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• 1. Given enough information about a line, we can reconstruct an equation of the line. Equations of Lines
• 2. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. Equations of Lines
• 3. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. Case I. Horizontal and Vertical Lines (The Special Case) Equations of Lines
• 4. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Case I. Horizontal and Vertical Lines (The Special Case) Equations of Lines
• 5. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) Equations of Lines
• 6. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= 3 Equations of Lines Horizontal lines have slope 0.
• 7. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= 3 y=1½ Equations of Lines Horizontal lines have slope 0.
• 8. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0.
• 9. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0. The slope of vertical lines is undefined, i.e. there is no “y” in the equation.
• 10. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0. The slope of vertical lines is undefined, i.e. there is no “y” in the equation. So the equations of vertical lines are x = c.
• 11. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0. The slope of vertical lines is undefined, i.e. there is no “y” in the equation. So the equations of vertical lines are x = c. Slope of vertical line is undefined. x= 5
• 12. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0. The slope of vertical lines is undefined, i.e. there is no “y” in the equation. So the equations of vertical lines are x = c. Slope of vertical line is undefined. x= 2 x= 5
• 13. Given enough information about a line, we can reconstruct an equation of the line. We separate them into two cases. The slope of horizontal lines is 0. Hence the equations of horizontal lines are y = c. Case I. Horizontal and Vertical Lines (The Special Case) y= –3 y= 3 y=1½ Equations of Lines Horizontal lines have slope 0. The slope of vertical lines is undefined, i.e. there is no “y” in the equation. So the equations of vertical lines are x = c. Slope of vertical line is undefined. x= –4 x= 2 x= 5
• 14. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation.
• 15. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation.
• 16. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation.
• 17. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation.
• 18. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line.
• 19. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c.
• 20. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3.
• 21. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation.
• 22. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation.
• 23. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation.
• 24. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation. Because it’s parallel to the x- axis, it must be a horizontal line.
• 25. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation. Because it’s parallel to the x- axis, it must be a horizontal line. So the equation is y = c for some c.
• 26. Equations of Lines Example A. a. A line passes through (3, –1 ), (3, –3). Draw. Find its equation. It’s a vertical line. So the equation is x = c for some c. Since (3, –1) is on the line so the equation must be x = 3. b. A line passes through (3, –1 ) and it’s parallel to the x-axis. Draw. Find its equation. Because it’s parallel to the x- axis, it must be a horizontal line. So the equation is y = c for some c. Since (3, –1) is on the line so the equation must be y = –1.
• 27. Equations of Lines Case II. Tilted Lines (The General Case)
• 28. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below.
• 29. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly.
• 30. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula.
• 31. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. The Point Slope Formula (for composing the equations)
• 32. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, The Point Slope Formula (for composing the equations)
• 33. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line.
• 34. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line.
• 35. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 36. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. The slope is –2, the point is (0, –7). Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 37. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. The slope is –2, the point is (0, –7). Hence, y = –2(x Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 38. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. The slope is –2, the point is (0, –7). Hence, y = –2(x – 0) Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 39. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. The slope is –2, the point is (0, –7). Hence, y = –2(x – 0) + (–7) Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 40. Equations of Lines Case II. Tilted Lines (The General Case) To find the equations of tilted lines, use the formula below. It gives the slope-intercept equations directly. We need the slope and a point on the line to use this formula. Given the slope m, and a point (x1, y1) on the line, then The Point Slope Formula (for composing the equations) y = m(x – x1) + y1 is the equation of the line. The slope is –2, the point is (0, –7). Hence, y = –2(x – 0) + (–7) or y = –2x – 7 Example B. Find the equations of the following lines. a. The line with slope -2 and y-intercept at -7.
• 41. b. The line that contains (1, –2) with the x-intercept at –4. Equations of Lines
• 42. b. The line that contains (1, –2) with the x-intercept at –4. We have two points on the line (1, –2), (–4, 0) and we need the slope. Equations of Lines
• 43. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, m = Equations of Lines
• 44. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, m = Equations of Lines
• 45. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, =m = Equations of Lines
• 46. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula m = Equations of Lines y = m(x – x1) + y1
• 47. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula m = Equations of Lines y = m(x – x1) + y1
• 48. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = Equations of Lines y = m(x – x1) + y1
• 49. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = Equations of Lines y = m(x – x1) + y1
• 50. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = Equations of Lines y = m(x – x1) + y1
• 51. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = Equations of Lines
• 52. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = y = (x + 4) 5 –2 Equations of Lines
• 53. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = y = (x + 4) 5 –2 y = x – 5 –2 8 5 Equations of Lines
• 54. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = y = (x + 4) 5 –2 y = x – 5 –2 8 5 Equations of Lines (or 5y = –2x – 8)
• 55. b. The line that contains (1, –2) with the x-intercept at –4. Δy Δx 0 – (–2 ) –4 – (1) 2 –5 = y = We have two points on the line (1, –2), (–4, 0) and we need the slope. Use the slope formula, = using the point (–4, 0), plug in the Point Slope Formula 5 –2 (x – (–4)) + 0 m = y = (x + 4) 5 –2 y = x – 5 –2 8 5 Equations of Lines Recall that parallel lines have the same slope and perpendicular lines have slopes that are the negative reciprocals of each other. (or 5y = –2x – 8)
• 56. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. Equations of Lines
• 57. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. Equations of Lines Our line has the same slope as the line 3y – 4x = 2.
• 58. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y.
• 59. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. 3y = 4x + 2
• 60. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. 3y = 4x + 2 y = 4 3 x + 2 3
• 61. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. 4 3 Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. Therefore the slope of the line 3y – 4x = 2 is . 3y = 4x + 2 y = 4 3 x + 2 3
• 62. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. 4 3 Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. Therefore the slope of the line 3y – 4x = 2 is . 3y = 4x + 2 y = 4 3 x + 2 3 So our line has slope .4 3
• 63. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. 4 3 Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. Therefore the slope of the line 3y – 4x = 2 is . 3y = 4x + 2 y = 4 3 x + 2 3 By the point-slope formula, the equation is So our line has slope .4 3
• 64. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. 4 3 Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. Therefore the slope of the line 3y – 4x = 2 is . 3y = 4x + 2 y = 4 3 x + 2 3 y = (x – 3) + (–1) By the point-slope formula, the equation is So our line has slope .4 3 4 3
• 65. c. The line that passes through (3, –1) and is parallel to the line 3y – 4x = 2. 4 3 Equations of Lines Our line has the same slope as the line 3y – 4x = 2. To find the slope of 3y – 4x = 2, solve for the y. Therefore the slope of the line 3y – 4x = 2 is . 3y = 4x + 2 y = 4 3 x + 2 3 y = (x – 3) + (–1) By the point-slope formula, the equation is So our line has slope .4 3 4 3 y = 4 3 x – 4 – 1
• 66. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. Equations of Lines
• 67. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. Equations of Lines For the slope, solve 2x – 3y = 2
• 68. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2
• 69. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. 2 3 Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2 y = 2 3 x –
• 70. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. 2 3 Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2 y = 2 3 x – Hence the slope of 2x – 3y = 2 is . 2 3
• 71. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. 2 3 Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2 y = Since perpendicular lines have slopes that are the negative reciprocals of each other, our slope is . 2 3 x – Hence the slope of 2x – 3y = 2 is . 2 3 –3 2
• 72. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. 2 3 Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2 y = 2 3 x – Hence the slope of 2x – 3y = 2 is . 2 3 Hence the equation for our line is y = (x – (0)) + (–3) –3 2 Since perpendicular lines have slopes that are the negative reciprocals of each other, our slope is .–3 2
• 73. d. The line that has y-intercept at –3 and is perpendicular to the line 2x – 3y = 2. 2 3 Equations of Lines For the slope, solve 2x – 3y = 2 –3y = –2x + 2 y = 2 3 x – Hence the slope of 2x – 3y = 2 is . 2 3 Hence the equation for our line is y = (x – (0)) + (–3) –3 2 y = x – 3 –3 2 Since perpendicular lines have slopes that are the negative reciprocals of each other, our slope is .–3 2
• 74. Linear Equations and Lines Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought.
• 75. Linear Equations and Lines Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”.
• 76. Linear Equations and Lines Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 77. Linear Equations and Lines Example C. We live by a river that floods regularly. On a rock by the river bank there is a mark indicating the highest point the water level ever reached in the recorded time. Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 78. Linear Equations and Lines Example C. We live by a river that floods regularly. On a rock by the river bank there is a mark indicating the highest point the water level ever reached in the recorded time. At 12 pm on July 11, the water level is 28 inches from this mark. At 8 am on July 12 the water is 18 inches from this mark. Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 79. Linear Equations and Lines Example C. We live by a river that floods regularly. On a rock by the river bank there is a mark indicating the highest point the water level ever reached in the recorded time. At 12 pm on July 11, the water level is 28 inches from this mark. At 8 am on July 12 the water is 18 inches from this mark. Let x be a measurement for time, and y be the distance from the water level and the mark. Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 80. Linear Equations and Lines Example C. We live by a river that floods regularly. On a rock by the river bank there is a mark indicating the highest point the water level ever reached in the recorded time. At 12 pm on July 11, the water level is 28 inches from this mark. At 8 am on July 12 the water is 18 inches from this mark. Let x be a measurement for time, and y be the distance from the water level and the mark. Find the linear equation between x and y. Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 81. Linear Equations and Lines Example C. We live by a river that floods regularly. On a rock by the river bank there is a mark indicating the highest point the water level ever reached in the recorded time. At 12 pm on July 11, the water level is 28 inches from this mark. At 8 am on July 12 the water is 18 inches from this mark. Let x be a measurement for time, and y be the distance from the water level and the mark. Find the linear equation between x and y. At 4 pm July 12, the water level is 12 inches from the mark, is the flood easing or intensifying? Many real world relations between two quantities are linear. For example the cost \$y is a linear formula of x–the number of apples bought. For those relations that we don’t know whether they are linear or not, linear formulas give us the most basic “educated guesses”. The following example demonstrates that these problems are pondered by people ancient or present alike.
• 82. Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation.
• 83. Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11.
• 84. Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20.
• 85. Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18.
• 86. Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18).
• 87. The slope m = = = –1/2 Δy Δx 28 – 18 0 – 20 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18).
• 88. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x
• 89. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x At 4 pm July 12, x = 28.
• 90. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x At 4 pm July 12, x = 28. According to the formula y = – 28/2 + 28 = –14 + 28 = 14.
• 91. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x At 4 pm July 12, x = 28. According to the formula y = – 28/2 + 28 = –14 + 28 = 14. But our actual observation, the water level is only 12 inches from the mark.
• 92. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x At 4 pm July 12, x = 28. According to the formula y = – 28/2 + 28 = –14 + 28 = 14. But our actual observation, the water level is only 12 inches from the mark. Hence the flood is intensifying.
• 93. The slope m = = = –1/2 Hence the linear equation is y = –1/2(x – 0) + 28 or that y = – + 28 Δy Δx 28 – 18 0 – 20 2 Equations of Lines The easiest way to set the time measurement x is to set x = 0 (hr) to the time of the first observation. Hence set x = 0 at 12 pm July 11. Therefore at 8 am of July 12, x = 20. In particular, we are given that at x = 0, y = 28, and at x = 20, y = 18. We want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). x At 4 pm July 12, x = 28. According to the formula y = – 28/2 + 28 = –14 + 28 = 14. But our actual observation, the water level is only 12 inches from the mark. Hence the flood is intensifying. The linear equation that we found is also called a trend line and it is shown below.
• 94. Linear Equations and Lines x = number of hours passed since 12 pm July 11 y = distance from the water level to the high mark 10 20 30 10 20 30 y 40 50 x (0, 28) y = –x/2 + 28 (20, 18)
• 95. Linear Equations and Lines x = number of hours passed since 12 pm July 11 y = distance from the water level to the high mark 10 20 30 10 20 30 y 40 50 x (0, 28) (20, 18) y = –x/2 + 28 (28, 14) The projected distance for 4 pm July 12
• 96. Linear Equations and Lines x = number of hours passed since 12 pm July 11 y = distance from the water level to the high mark 10 20 30 10 20 30 y 40 50 x (0, 28) (20, 18) y = –x/2 + 28 (28, 14) The projected distance for 4 pm July 12 (28, 14) The actual data taken at 4 pm July 12
• 97. Linear Equations and Lines Exercise A. For problems 1–8 select two points and estimate the slope, and find an equation of each line. 1. 2. 3. 4. 5. 6. 7. 8.
• 98. Linear Equations and Lines Exercise B. Draw each line that passes through the given two points. Find the slope and an equation of the line. Identify the vertical lines and the horizontal lines by inspection first. 9. (0, –1), (–2, 1) 10. (1, –2), (–2, 0) 11. (1, –2), (–2, –1) 12. (3, –1), (3, 1) 13. (1, –2), (–2, 3) 14. (2, –1), (3, –1) 15. (4, –2), (–3, 1) 16. (4, –2), (4, 0) 17. (7, –2), (–2, –6) 18. (3/2, –1), (3/2, 1) 19. (3/2, –1), (1, –3/2) 20. (–5/2, –1/2), (1/2, 1) 21. (3/2, 1/3), (1/3, 1/3) 23. (3/4, –1/3), (1/3, 3/2) Exercise C. Find the equations of the following lines. 24. The line that passes through (0, 1) and has slope 3. 25. The line that passes through (–2 ,1) and has slope –1/2. 26. The line that passes through (5, 2) and is parallel to y = x. 27. The line that passes through (–3, 2) and is perpendicular to –x = 2y. 22. (–1/4, –5/6), (2/3, –3/2)
• 99. Linear Equations and Lines Exercise D. Find the equations of the following lines. 28. The line that passes through (0, 1), (1, –2) 31. It’s perpendicular to 2x – 4y = 1 and passes through (–2, 1) 29. 30. 32. It’s perpendicular to 3y = x with x–intercept at x = –3. 33. It has y–intercept at y = 3 and is parallel to 3y + 4x = 1. 34. It’s perpendicular to the y–axis with y–intercept at 4. 35. It has y–intercept at y = 3 and is parallel to the x axis. 36. It’s perpendicular to the x– axis containing the point (4, –3). 37. It is parallel to the y axis has x–intercept at x = –7. 38. It is parallel to the x axis has y–intercept at y = 7.
• 100. Linear Equations and Lines The cost y of renting a tour boat consists of a base–cost plus the number of tourists x. With 4 tourists the total cost is \$65, with 11 tourists the total is \$86. 39. What is the base cost and what is the charge per tourist? 40. Find the equation of y in terms of x. 41. What is the total cost if there are 28 tourists? The temperature y of water in a glass is rising slowly. After 4 min. the temperature is 30 Co, and after 11 min. the temperature is up to 65 Co. Answer 42–44 assuming the temperature is rising linearly. 42. What is the temperature at time 0 and what is the rate of the temperature rise? 43. Find the equation of y in terms of time. 44. How long will it take to bring the water to a boil at 100 Co?
• 101. Linear Equations and Lines The cost of gas y on May 3 is \$3.58 and on May 9 is \$4.00. Answer 45–58 assuming the price is rising linearly. 45. What should x be and what is the rate of increase in price in terms of x? 46. Find the equation of the price in term of the dates in May. 47. What is the projected price on May 20? 48. In 2005, the most inexpensive tablet cost \$900. In the year 2010, it was \$500. Find the equation of the price p in terms of time t. What is the projected price in the year 2014? | 11,422 | 36,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2015-14 | longest | en | 0.842281 |
https://metanumbers.com/34096 | 1,670,019,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00869.warc.gz | 422,670,104 | 7,353 | # 34096 (number)
34,096 (thirty-four thousand ninety-six) is an even five-digits composite number following 34095 and preceding 34097. In scientific notation, it is written as 3.4096 × 104. The sum of its digits is 22. It has a total of 5 prime factors and 10 positive divisors. There are 17,040 positive integers (up to 34096) that are relatively prime to 34096.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 34 thousand 96 thirty-four thousand ninety-six
## Notation
Scientific notation 3.4096 × 104 34.096 × 103
## Prime Factorization of 34096
Prime Factorization 24 × 2131
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 4262 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 34,096 is 24 × 2131. Since it has a total of 5 prime factors, 34,096 is a composite number.
## Divisors of 34096
1, 2, 4, 8, 16, 2131, 4262, 8524, 17048, 34096
10 divisors
Even divisors 8 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 10 Total number of the positive divisors of n σ(n) 66092 Sum of all the positive divisors of n s(n) 31996 Sum of the proper positive divisors of n A(n) 6609.2 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 184.651 Returns the nth root of the product of n divisors H(n) 5.15887 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 34,096 can be divided by 10 positive divisors (out of which 8 are even, and 2 are odd). The sum of these divisors (counting 34,096) is 66,092, the average is 660,9.2.
## Other Arithmetic Functions (n = 34096)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 17040 Total number of positive integers not greater than n that are coprime to n λ(n) 8520 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3642 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 17,040 positive integers (less than 34,096) that are coprime with 34,096. And there are approximately 3,642 prime numbers less than or equal to 34,096.
## Divisibility of 34096
m n mod m 2 3 4 5 6 7 8 9 0 1 0 1 4 6 0 4
The number 34,096 is divisible by 2, 4 and 8.
## Classification of 34096
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (34096)
Base System Value
2 Binary 1000010100110000
3 Ternary 1201202211
4 Quaternary 20110300
5 Quinary 2042341
6 Senary 421504
8 Octal 102460
10 Decimal 34096
12 Duodecimal 17894
20 Vigesimal 454g
36 Base36 qb4
## Basic calculations (n = 34096)
### Multiplication
n×y
n×2 68192 102288 136384 170480
### Division
n÷y
n÷2 17048 11365.3 8524 6819.2
### Exponentiation
ny
n2 1162537216 39637868916736 1351492778585030656 46080497778635205246976
### Nth Root
y√n
2√n 184.651 32.4266 13.5886 8.06382
## 34096 as geometric shapes
### Circle
Diameter 68192 214231 3.65222e+09
### Sphere
Volume 1.66035e+14 1.46089e+10 214231
### Square
Length = n
Perimeter 136384 1.16254e+09 48219
### Cube
Length = n
Surface area 6.97522e+09 3.96379e+13 59056
### Equilateral Triangle
Length = n
Perimeter 102288 5.03393e+08 29528
### Triangular Pyramid
Length = n
Surface area 2.01357e+09 4.67137e+12 27839.3
## Cryptographic Hash Functions
md5 63a6cfc9c917b4133ce3f8d000cb9a8e 4864af7dc24aa47e6c5a9f992cd69312ee9e6673 482c7453010971f5d03161b3025dcc7d1a434922caddc41e63053a60ebe555aa 83f6a83a39cb74292f9594767a2fc2c1c47dddeecf3324b522e239eceedc1a8eaf5d4f87e498d267721f7b78f8a74ba5ba99d6698d9183e9f861b0d9d44ae49e 1ff8be339c2b03bf2246fb6539a25f71333741b3 | 1,473 | 4,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.821089 |
https://www.scribd.com/presentation/53947176/Hypothesis-2 | 1,511,450,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806842.71/warc/CC-MAIN-20171123142513-20171123162513-00181.warc.gz | 843,375,173 | 32,258 | # Standard Error of the Difference Between Means Reliability of Differences Between Correlated Means (N>30) (N>30
)
Reliability of Differences Between Correlated Means (N>30) (N>30)
Reasons for getting correlated Means
1 The participants serves as own control (repeated measures) 2 Participants are matched on task relevant variable. 3 Participants are identical twins or litter mates.
Reliability of Differences Between Correlated Means (N>30) (N>30) Standard Error of Difference Between Correlated Means (N>30) SD ! X SX SX 1 2 2 2 2r S 2 X1S X2 D ! z SD X X ! X X S X S X 2r S X S X 1 2 2 1 2 1 2 If we use this formula. nobody uses this formula!) .. it is necessary to compute the correlation between the two means««.(since there is an alternative.
§D z! §D ! N N§ D § D 2 D SD X X ! X SX D D z! N S D N 1 . and works. Requires computation of Correlation between means. Of course this Too much room for erroris the one There is an easier way! people use! We take advantage of the fact that the mean of the difference is the same as the difference between the means. but it is rarely if ever used.Reliability of Differences Between Correlated Means (N>30) (N>30) The proceeding formula is appropriate.
2 N 2 N 1 .
12 Difference 0 0 2 0 1 -1 -1 3 0 2 3 0 -1 1 -1 1 0 0 2 3 0 -1 3 4 1 21 0.84 Study to evaluate two golf balls (Regular/New) .Reliability of Differences Between Correlated Means (N>30) (N>30) Example Illustrating Computation Pro 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Sum Mean Regular 3 5 4 6 5 4 2 5 3 4 5 2 3 5 5 4 2 3 4 5 3 2 5 6 4 99 3.84 z! ! 1384 25.96 New 3 5 2 6 4 5 3 2 3 2 2 2 4 4 6 3 2 3 2 2 3 3 2 2 3 78 3.
84 .73 441 625 625 24 25 1 .05 level .304 .84 . beyond . ! ! 276 * Sig..
01 level --->2 tailed ± 2. Spend some time learning how to use the normal curve table.01 level --->1 tailed ± 2. ± This is true since z is a 1 tailed table.57 We use additive theorem.0029 -->1 tailed P=. .Reliability of Differences Between Correlated Means (N>30) (N>30) When we look up Z=2. we find: P=. Value of z required for .76.01 level for 2 tailed test: .0058--> 2 tailed ± to get 2 tailed value we double 1 tailed value.32 .
but until we get that small. As the sample gets smaller. ± Tables were developed by William Gossett (student) . Differences between means will be normally distributed about the actual mean difference of zero (0) if we draw samples randomly from the same population. the distribution of differences (sampling distribution) becomes less and less normal. ± This difference starts occurring before 30.Reliability of Differences Between Correlated Means (N<30) We use t test when we have small samples. it doesn¶t make much difference. the departure from normalcy is so great a special set of curves have been developed to describe these distributions. ± For samples less than 30.
Reliability of Differences Between Uncorrelated Means (N<30) Standard Error of Difference Between Uncorrelated Means (N<30) t! D SD X X ! X X §x §x ¨n n © n n 2 ª n n 1 2 2 2 2 1 1 1 2 1 2 2 ¸ ¹ º This is the same principle as the z test .
t! D SD X X ! X X §x §x ¨n n © n n 2 ª n n 1 2 2 2 2 1 1 1 2 1 2 2 ¸ ¹ º Now all we have to do is plug in the numbers!! N N=13 .Reliability of Differences Between Uncorrelated Means (N<30) Computational Example: No We do a study to determine whether those who have had a course in 2 2 C a lc calculus will doxbetter in psychological statistics. Students are C a lc x1 2 x 1 x 2 grouped according to whether or not they have had statistics.
Reliability of Differences Between Uncorrelated Means (N<30) Computational Example: (Continued) No C a c 103 98 106 71 108 120 150 79 93 101 113 95 139 94 106 92 100 N 17 Ca c 137 151 131 133 115 110 139 124 94 123 90 135 104 x1 -1 2 -3 3 4 16 46 -2 5 -1 1 -3 9 -9 35 -1 0 2 -1 2 -4 x2 15 29 9 11 -7 -1 2 17 2 -2 8 1 -3 2 13 -1 8 x 1 1 36 4 1089 16 256 2116 625 121 9 81 81 1225 100 4 144 16 2 x 2 225 841 81 121 49 144 289 4 784 1 1024 169 324 2 t! D SD X X ! X X §x §x ¨n n © n n 2 ª n n 1 2 2 2 2 1 1 1 2 1 2 2 ¸ ¹ º t! 122 104 ¨ 5924 4056¸¨ 17 13 ¸ ¹© © © .
.
05 level of significance««.01)=2.05)=2.763 N=13 Thus our t is significant at .56 For df=28 df=N1+N2-2 t(.048 t(. . ¹ ¹ ª 17 13 2 ºª 17 13 º t ! 2.
05.01/2 to get the one (1) tailed value. To determine the equivalent 1 tailed value we divide . We divide it by 2 and look up .005) One (1) tailed table *z tables ± Divide desired level by 2 and look up value for that value. (. the more leptokurtic the sampling distribution becomes t tables are usually 2 tailed tables (often they list both 1 and 2 tailed value) Two (2) tailed table ± Look up a given value.025 . We want .Reliability of Differences Between Uncorrelated Means (N<30) Some Miscellaneous Points The smaller the sample.
Reliability of Differences Between Correlated Means (N<30) Standard Error of Difference Between Correlated Means §D X t! SX D §D ! N N§ D § D 2 ! N D S diff N 1 .
2 ! N df = 2 N 1 -1 This is the same formula as we use for z for correlated means l= umber of matched participants .
each score had some freedom to vary or take on different values. Until this point.Reliability of Differences Between Correlated Means (N<30) Concept of Degree of Freedom. -----------------------> This score must be 4. 1 4 8 Sum=16 Mean=4 -1 0 ? Refers freedom of scores to vary.1 is FIXED! . This is OT true with this SCORE! ± Three scores are free to Vary .
it takes a larger t to be significant. X X X X X X X X X X X X X X X X -1=3 n1+n2-2=6 .Reliability of Differences Between Uncorrelated Means (N<30) Comparison of df for Correlated and Uncorrelated Means You can see that if we don¶t reduce variance by matching. we are better off using Uncorrelated Groups With fewer df.
2 .Comparison of Formula for z-test and t-test D t! SD X X ! X X §x §x ¨ n n ¸ © ¹ n n 2 ª n n º 1 2 2 2 2 1 1 2 1 2 1 2 D z! SD X X ! X X SX SX 1 2 2 2 1 We see that t and z are exactly the same except that t pool the variance before computing the Standard Error where in the case of z the variance is pooled after the Standard Error is computed.
Hypothesis Testing (t and z) You really should learn these assumptions Assumptions for t and z tests 1. All other assumptions for parametric tests ± a) Interval level of Measurement ± b) Random Sampling from the Population . The population variances are equal 3. 2. Populations are normally distributed.
Summary of Steps used to Test the Null Hypothesis State the ull Hypothesis and the associated Research Hypothesis Choose a statistical test with its associated statistical model We base our choice on level of measurement. . assumptions. etc. Decide whether to do parametric test.
Summary of Steps used to Test the Null Hypothesis(Continued:) State the ull Hypothesis and the associated Research Hypothesis Choose a statistical test with its associated statistical model Specify a Level of Confidence (E) .
result of sample size. alpha and size of Standard Error. Power of Test (Function of Alpha.Summary of Steps used to Test the Null Hypothesis(Continued:) State the ull Hypothesis and the associated Research Hypothesis Choose a statistical test with its associated statistical model Specify a Level of Confidence (E) and set the sample size Alpha (Type I Error) .we set Beta (Type II Error) . Beta. Variance and N .
Make Decision .Summary of Steps used to Test the Null Hypothesis(Continued:) State the ull Hypothesis and the associated Research Hypothesis Choose a statistical test with its associated statistical model Specify a Level of Confidence (E) and set the sample size Find or Assume the Sampling Distribution Define the Region of Rejection for the Ho Compute the Value of the Test .
THE END! Reliability of Differences Between Uncorrelated Means (N<30) Press This Button to Return to Class Page . | 2,603 | 7,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-47 | latest | en | 0.769005 |
https://statskey.com/need-help-with-mymathlab-statistics-problems | 1,723,516,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00284.warc.gz | 431,704,998 | 28,104 | # Need help with MyMathLab statistics problems?
## Need help with MyMathLab statistics problems?
Need help with MyMathLab statistics problems? No problem at all. Help me out on some other numbers? Just don’t forget to enter their values and set a limit depending on my options. If that does not help you, try again with the ‘As I understand – If necessary, I can’t review them yet. MyMathLab provides a comprehensive overview of basic data fields from my math project. I, however, do not believe click now they are valid. My project is a program that I use to compare apples to apples from a computer and run on my professor’s lab in a test room. My current goal is to better understand what a computer is doing with its data by looking for some errors in how they compare apples to apples. There are one or more data fields – of which the following table lists the measurements of four apples, each measurement has a count value assigned – of which the next six apples are “yes” and “no”, with a number of decimal places. First and third and eighth apples are “+1” “-1” (always) and “-1” (always) respectively, followed by “-1” (=1) and “1” (=1).. See the example below for more information. This is basically just showing what the current measurements are. It does not show if “yes” or “no” have been broken up as the arithmetic means do to the apple each measurement is as different as what is available in MyMathLab. “On the Computer To get more useful information about apples, the table above, and a summary of all apples, here is an example. Most apples are listed as different apples. Since apples are separated by commas (count values), this lets you see the numbers within the group you’re looking to count with based on the commas. Looking at this “on the computers” table, there is a “Yes” and “No” value for the “+1” (I think this is the case) and “-1” (=1) and “1” (=1) respectively using commas. On the computer From the way this table is created, I can see the name of each piece of apples on the computer as it is displayed in the “yes” and “No” category on the left menu. On the right-side of the table, I compare apples with apples from the computer – and the same applies for the “+1” and “-1” category. This means apples are now separated by commas and the difference is obvious so it makes sense to compare apples to apples in this case.
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One more thing I have seen is that the last row of each column shows what the average value of the last apple and the corresponding “+1” and “-1” category is. It is also the ‘+1’ category is marked as ‘-1’ as there is nothing on the right side of the table, it has just too ‘+1’ as it is displayed. For example, if you do that “+-1 is +-1”, it gives you about read what he said quarter of the apples as see page “Yes” or “Yes +-1”. A summary of each data field is the “x” number of apples to show when it appears in a row. The “x” variable is the average value of apples per line, and the “y” variable is the average value of apples per line. On the computer, the “x” variable is 5 for apples and 11 for apples and 1 for apples and only a quarter for the “+1” category. For most of the apples in Table 1, the average is at 0. That means apples are labeled “Yes”, “No”, and ‘+1’ but the average is shown as “+-1”. There is going on around this image the next row of data line – which is the – is showing apples a fantastic read line. This is why the list of apples that are counted is given for the “Yeah, Yes” category only so the category of “Yes that is 1” is obvious. **Note: In order to fix my above assumption, I reordered the tables between the “Yes” and “No” categories on the right. So in the left-side table below, we see apples per line. On the left-side, before using commNeed help with MyMathLab statistics problems? Don’t have troubles* I’ve seen people say I have troubles with math on my computer so the math I currently have is with the computers I’ve setup within my network, and on my computer’s network, I have problems finding it to understand what I need to do to it, as I have this computer, it searches for solutions for errors, but then I have on this computer error reports, so if I have any luck, I have some errors. If you think I missed something, please tell me right away. It may take a day or two to know about these, so please help 🙂 To try a few quick examples: Problem in $2 \times 2$ matrix Problem in $2 \times 3$ matrix Problem in $3 \times 2 \times 3$ matrix Problem in $3 \times 4 \times 4$ matrix Problem in $2 \times 3 \times 2$ matrix Again, I did a lot research on your answer, and it seems like you’re right, and I already have a lot of confidence in your software. Many thanks! Now, how do you know your computers are functioning correctly? If you didn’t know, there might have been some “bugs” that got you confused all at once. If you didn’t know, some of these were my initial call to see how your computer might function as a hardware device. Obviously, if you also didn’t know everything existed with you and what you needed, you could have ran your program to test it for errors. As far as I know, MatLab does not have anything suitable for handling such a hardware device, so instead, I have this big graphic program! It’s a modified version of Mathematica, but if you try running this in a loop block of code, you will notice that there are no errors. Here’s how you have it running: { var d = {math: math.
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F + r_{1}, r: math.F + r_{2}, r: math.F + r_{3}, r: math.F + r_{4}, o: op} ; i = {j, k, l, m, u}; while (dt = dt + i) ; theta = matplotlib.symbolTable; nf2 = theta.inf; mat = matplotlib.matplot2d(theta, nf2, …) ; b1 = mat; for (i = d.rows; i < nf2; ++i) mat.col = mat.col; mat.scatter[i][0] = b1; mat.scatter[i][1] = b1; // mat.scatter[i][0] = b2; mat.x[i] = b1; mat.subplot(2,2,2) ; mat.frame(b1,(-2) -2,(-2) +2,(-2) +2) ; } As you can see, I'm getting this error when I run the entire file in its entirety. Its pretty much the same as Mathematica got as fast as Matlab wasn't its best navigate here by all means.
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What am I doing wrong here? Are you using 2nd-order/3rd-order functions from within Matlab? SCHOOL_A_2 = nf2 / psf2; SYPMATHM = \sysMathToPseudoCpy(); if (nf2 <= 0) error("You have a problem in your node functions as we do nothing (we are getting the program asynchrony)."); else atomp = \mathmakelibrary("Mlibrary'); (deflint / basefmt; / cte:cmath; / FAST_LIBM_LIB ); The if statement should be: if (-math.F-math) As there is a second line between the numbers above, and the if, but I’m using a different function, I discovered that there are 100 more lines than these not-yet compiled codes for.Need help with MyMathLab statistics problems? I currently work with my online statistics lab. I have bought a few other labs from the companies mentioned. They all output data for a single record the following table outputted in a excel sheet: ID | Notes —|———— | [1] | 1 | 2 | 2 | 2 [2] | 1 | 3 | 3 | 2 [3] | 1 | 3 | 3 | 2 [4] | 1 | 4 | 3 | 2 [5] | 1 | 4 | 3 | 2 [6] | 1 | 4 | 3 | 2 [7] | 2 | 2 | 2 | 2 [8] | 2 | 3 | 2 | 2 [9] | 2 | 3 | 2 | 2 [10] | 3 | 3 | 2 | 2 [11] | 3 | 4 | 3 | 2
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https://edurev.in/course/quiz/attempt/17118_Test-Profit-and-Loss-2/b803f746-362d-4a72-8807-f90ad3210c1e | 1,725,729,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00786.warc.gz | 217,898,156 | 56,283 | Test: Profit and Loss- 2 - SSC CGL MCQ
# Test: Profit and Loss- 2 - SSC CGL MCQ
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## 20 Questions MCQ Test Quantitative Aptitude for SSC CGL - Test: Profit and Loss- 2
Test: Profit and Loss- 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The Test: Profit and Loss- 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The Test: Profit and Loss- 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Profit and Loss- 2 below.
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Test: Profit and Loss- 2 - Question 1
### A retailer sold 12 notes at a profit of 20% and 8 notes at a profit of 10%. If he had sold all the 20 notes at a profit of 15%, then his profit would have been reduced by Rs.36. What is the cost price of each note?
Detailed Solution for Test: Profit and Loss- 2 - Question 1
Explanation :
Cost Price = x
Total Profit
= 12x * 20/100 + 8x * 10/100 = 32x/100 = 3.2x
= 20x * 15/100 — (ii) Profit of 15% on 20 notes
Total Profit
3.2x - 3x = 36
0.2x = 36
x = 180
Test: Profit and Loss- 2 - Question 2
### The profit Percentage on 3 bikes are 15%, 35% and 10% and the ratio of CP is 5:3:1. Also the ratio of the Bike sold of P, Q and R is 2:3:5. Then the overall approximate Profit Percentage is?
Detailed Solution for Test: Profit and Loss- 2 - Question 2
Explanation :
5x * 2y + 3x * 3y + 5y = 24xy
Total Profit —
= 515xy/100
= 5.15xy
Overall Profit Percentage
= 5.15xy * 100/24xy
= 21.46%
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Test: Profit and Loss- 2 - Question 3
### When a shopkeeper reduces the selling price of an article from 1180 to 1126 his loss increases by 5% . What is the cost price of article?
Detailed Solution for Test: Profit and Loss- 2 - Question 3
Explanation :
5% of cp = 1180 - 1126
CP = 54 * 100 / 5 = 1080
Test: Profit and Loss- 2 - Question 4
A and B, there are two companies, selling the packs of cold-drinks. For the same selling price A gives two successive discounts of 10% and 25%. While B sells it by giving two successive discounts of 15% and 20%. What is the ratio of their marked price?
Detailed Solution for Test: Profit and Loss- 2 - Question 4
Answer - C. 136 : 135
Explanation :
A = 90/100*75/100
- .675
B = 85/100*80/100
.68
680:675
136:135
Test: Profit and Loss- 2 - Question 5
A reputed company sells a wrist watch to a wholesaler making a profit of 10%. The wholesaler, in turn, sells it to the retailer making a profit of 10%. A customer purchases it by paying Rs. 990. Thus the profit of retailer is 2(3/11)% What is the cost incurred by the the company to produce it?
Detailed Solution for Test: Profit and Loss- 2 - Question 5
Explanation :
x*110/100*110/100*(100 + 25/11)/100 = 990
x = 800
Test: Profit and Loss- 2 - Question 6
A and B both are dealers of Honda Motorcycles. The price of an used Honda Motorcycle is Rs.28,000. A gives a discount of 10% on whole, while B gives a discount of 12% on the first Rs. 20,000 and 8% on the rest Rs. 8000. What is the difference between their selling prices?
Detailed Solution for Test: Profit and Loss- 2 - Question 6
Discount offer by A = 10 % of 28000 = 2800
Total Discount offer by B = 12% of 20,000 + 8% of 8000 = 3040
Required difference = 3040 – 2800 = 240
Test: Profit and Loss- 2 - Question 7
The profit percentage of P and Q is same on selling the articles at Rs. 1800 each but A calculates his profit on the selling price while Q calculates it correctly on the cost price which is equal to 20%. What is the difference in their profits?
Detailed Solution for Test: Profit and Loss- 2 - Question 7
Explanation:
Profit(Calculated on SP) = 20% of 1800 = 360
Profit(calculated on CP)
x + x/5 = 1800
x = 1500
Profit = 300
Difference = 360 – 300 = 60
Test: Profit and Loss- 2 - Question 8
A person sold a pen at Rs. 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells it at twice the percentage profit of its previous percentage profit then new selling price will be?
Detailed Solution for Test: Profit and Loss- 2 - Question 8
CP = x
Profit Percentage = x%
SP = x(100 + x)/100
x(100 + x)/100 = 96
x = 60
Profit Percentage = 60%
New SP = 60 * 220 / 100 = 132
Test: Profit and Loss- 2 - Question 9
A trader mixes 25% of solution A to his Solution B and then he sells the whole mixture at the price of Solution B. If the cost price of Solution A be 50% of the cost price of Solution B, what is the net profit percentage?
Detailed Solution for Test: Profit and Loss- 2 - Question 9
Quantity of Solution B = 100 litre
Quantity of Solution A = 25 litre
CP of 1 litre Solution B = Rs.10
CP of 1 litre Solution A = Rs.5
CP = 100 * 10 + 25 * 5 = 1125
SP = (100 + 25)*10 = 1250
Profit = 1250 – 1125 = 125
% = 125 * 100 / 1125 = 100/9%
Test: Profit and Loss- 2 - Question 10
A scientist mixes 10% water in his solution but he is not content with it so he again mixes 10% more water in the previous mixture. What is the profit percentage of the scientist if he sells it at cost price:
Detailed Solution for Test: Profit and Loss- 2 - Question 10
Let Initial Quantity of Solution = 100 litre
After mixing 10% water, Quantity of the mixture = 110 * 110 / 100 = 121 litre
CP of One litre of Solution = Rs.1
Total CP = Rs.100
Total SP = Rs.121
Profit = 121 – 100 = 21
Profit % = 21 * 100/100 = 21%
Test: Profit and Loss- 2 - Question 11
The profit earned after selling an article for Rs. 1680 is the same as the loss incurred after selling the article for Rs. 1512. What is the cost price of the article?
Detailed Solution for Test: Profit and Loss- 2 - Question 11
CP = x
1680 – x = x – 1512
2x = 3192
x = 1596
Test: Profit and Loss- 2 - Question 12
Arun sells an article at 20% profit to Bala, Bala sells it to Catherine at 10% profit. Catherine sells it to Dinesh at Rs. 16 profit. The difference between the cost price of Dinesh and cost price of Arun was Rs. 500. How much did Bala pay to Arun for the article?
Detailed Solution for Test: Profit and Loss- 2 - Question 12
Cost Price of Arun = 100x
Cost Price of Bala = 120x
Cost Price of Catherine = 132x
Cost Price of Dinesh = 132x + 16
132x + 16 – 100x = 500
32x + 16 = 500
32x = 484
x = 484/32 = 15.125
Cost Price of Bala =120x = 120 * 15.125 = 1815
Test: Profit and Loss- 2 - Question 13
Rahul purchased an article for Rs. 8400 and sold it for a loss of 5%. From that money he purchased another article and sold it for a gain of 5%. What is the overall gain or loss?
Detailed Solution for Test: Profit and Loss- 2 - Question 13
CP = 8400
SP = 8400 * 95/100 = 7980
CP = 7980
SP = 7980 * 105/100 = 8379
Difference = 8400 – 8379 = 21
Test: Profit and Loss- 2 - Question 14
A Shop Keeper sells two bags for Rs. 500 each. On one, he gets 14% profit and on the other he gets 14% loss. His profit or loss in the entire transaction was?
Detailed Solution for Test: Profit and Loss- 2 - Question 14
% = x
Loss % = x²/100 = 196/100 = 49/25%
Test: Profit and Loss- 2 - Question 15
A Shopkeeper bought 30 kg of rice at the rate of Rs. 40 per kg. He sold 40% of the total quantity at the rate of Rs. 50 per kg. At what price per kg should he sell the remaining quantity to make 25% overall profit?
Detailed Solution for Test: Profit and Loss- 2 - Question 15
Total CP of Rice = 30 * 40 = 1200
40% of Total Quantity = 40% of 30 = 12
SP = 12*50 = 600
SP = 1200 * 125/100 = 1500
SP of Remaining Quantity = 1500 – 600 = 900
Remaining Quantity = 18kg
Rice per Kg = 900/18 = Rs. 50
Test: Profit and Loss- 2 - Question 16
If the Cost Price of 25 articles is equal to the Selling Price of 20 articles, then what is the gain %?
Detailed Solution for Test: Profit and Loss- 2 - Question 16
Test: Profit and Loss- 2 - Question 17
A TV was purchased for Rs. 54000. Its price was marked up by 40%.It was sold at a discount of 20% on the marked price. What was the profit percent of the cost price?
Detailed Solution for Test: Profit and Loss- 2 - Question 17
Explanation:
40 – 20 + [40 * (-20)/100] = 20 – 8 = 12%
Test: Profit and Loss- 2 - Question 18
A shopkeeper sold a smartphone for Rs.15000. Had he offered discount of 10% on the Selling Price, he would have earned a profit of 8%. What is the Cost Price of that Smartphone?
Detailed Solution for Test: Profit and Loss- 2 - Question 18
S.P of Smart Phone = Rs.15000
Discount = 10%
New SP = 15000 – 1500 = Rs. 13500
Profit = 8%
CP = 13500 * 100/108 = 12500
Test: Profit and Loss- 2 - Question 19
Pinkey sold a machine to Shalini at a profit of 30%. Shalini sold this machine to Arun at a loss of 20%. If Pinkey paid Rs.5000 for this machine, then find the cost price of machine for Arun?
Detailed Solution for Test: Profit and Loss- 2 - Question 19
R1 = 30% R2 = 20%
5000 * 130/100 * 80/100 = Rs. 5200
Test: Profit and Loss- 2 - Question 20
A merchant earns a profit of 20% by selling a basket containing 80 Oranges whose cost is Rs.240 but he gives one-fourth of it to his friend at cost price and sells the remaining oranges. In order to earn the same profit, at what price must he sell each orange?
Detailed Solution for Test: Profit and Loss- 2 - Question 20
CP of 80 Oranges = Rs.240
CP of 1 Orange = Rs.3
CP of 20 Oranges = Rs.60
120% of 240 = 288
SP of remaining 60 Oranges = 288 – 60 = 228
SP of 1 Orange = 228/60 = Rs.3.80
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## Quantitative Aptitude for SSC CGL
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# Sample Spaces and Events
## Words or diagrams that detail favorable outcomes and intersections, complements, and unions of events.
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Practice Sample Spaces and Events
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Descriptions of Events
You pick a card from a standard deck. Event \begin{align*}C\end{align*} is choosing a card that has an even number or a spade. Rewrite event \begin{align*}C\end{align*} as the combination of two events. Then, list the outcomes in event \begin{align*}C\end{align*}.
#### Watch This
Watch this video to learn about the concepts of unions and intersections.
#### Guidance
In the context of probability, an experiment is any occurrence that can be observed. For example, rolling a pair of dice and finding the sum of the numbers is an experiment.
An outcome is one possible result of the experiment. So, for the experiment of rolling a pair of dice and finding the sum of the numbers, one outcome is a “7” and a second outcome is an “11”.
Every experiment has one or more outcomes. The sample space, \begin{align*}S\end{align*}, of an experiment is the set of all possible outcomes. For the experiment of rolling a pair of dice and finding the sum of the numbers, the sample space is \begin{align*}S=\{ 2,3,4,5,6,7,8,9,10,11,12\}\end{align*}.
Often there is one or more outcomes that you are particularly interested in. For example, perhaps you are interested in the sum of the numbers on the dice being greater than five. The event, \begin{align*}E\end{align*}, is a subset of the sample space that includes all of the outcomes you are interested in (sometimes called the favorable outcomes). If \begin{align*}E\end{align*} is the sum of the numbers on the dice being greater than five, \begin{align*}E=\{6,7,8,9,10,11,12\}\end{align*}. There are many possible events that could be considered for any given experiment.
There are three common operations to consider with one or more events, shown in the table below. Consider the experiment of rolling a pair of dice and finding the sum of the numbers on the dice. Let \begin{align*}E\end{align*} be the event that the sum of the numbers is greater than five. \begin{align*}(E=\{6,7,8,9,10,11,12\})\end{align*}. Let \begin{align*}F\end{align*} be the event that the sum of the numbers is even \begin{align*}(F=\{2,4,6,8,10,12\})\end{align*}.
Operation Definition in Words Pair of Dice Example Complement of an Event \begin{align*}(E^\prime)\end{align*} The event that includes all outcomes in the sample space NOT in event \begin{align*}E\end{align*}. \begin{align*}E^\prime = \{1,2,3,4,5\}\end{align*} \begin{align*}E^\prime\end{align*} is the sum of the numbers on the dice being five or less. Union of Events \begin{align*}(E \cup F)\end{align*} The event that includes all outcomes in either event \begin{align*}E\end{align*}, event \begin{align*}F\end{align*}, or both. \begin{align*}E \cup F=\{2,4,6,7,8,9,10,11,12\}\end{align*} Intersection of Events \begin{align*}(E \cap F)\end{align*} The event that includes only the outcomes that occur in both event \begin{align*}E\end{align*} and event \begin{align*}F\end{align*}. \begin{align*}E \cap F=\{6,8,10,12\}\end{align*}
To help visualize the way different events or combinations of events interact within a sample space, consider a Venn diagram.
The diagram above has a big rectangle for sample space \begin{align*}S\end{align*}. Within \begin{align*}S\end{align*}, the outcomes 2 through 12 appear in various places. The circle labeled \begin{align*}E\end{align*} represents event \begin{align*}E\end{align*}, and within that circle are all the outcomes in event \begin{align*}E\end{align*}. Similarly, the circle labeled \begin{align*}F\end{align*} represents event \begin{align*}F\end{align*}, and within that circle are all the outcomes in event \begin{align*}F\end{align*}. The place where the circles overlap contains the outcomes that are in both events \begin{align*}E\end{align*} and \begin{align*}F\end{align*}.
Example A
Shade the area of the diagram below that represents \begin{align*}F^\prime\end{align*}. Describe the event \begin{align*}F^\prime\end{align*}.
Solution: \begin{align*}F^\prime\end{align*} is the complement of event \begin{align*}F\end{align*}. It contains all the outcomes in the sample space that are NOT in event \begin{align*}F\end{align*}. In this case, \begin{align*}F^\prime\end{align*} is the sum of the numbers on the dice being odd.
Example B
Shade the area of the diagram below that represents \begin{align*}E \cup F\end{align*}. Then, shade the area of the diagram that represents \begin{align*}E\cap F\end{align*}. How is the union of two events different from the intersection of two events?
Solution: \begin{align*}E \cup F\end{align*} is the union of events \begin{align*}E\end{align*} and \begin{align*}F\end{align*}. It contains all the outcomes that are in event \begin{align*}E\end{align*}, event \begin{align*}F\end{align*}, or both events E and \begin{align*}F\end{align*}. The symbol \begin{align*}\cup\end{align*} can be thought of as “or”, but remember that it is not exclusive or, since it includes outcomes that are in both events. \begin{align*}E \cup F\end{align*} is shown below.
\begin{align*}E \cap F\end{align*} is the intersection of events \begin{align*}E\end{align*} and \begin{align*}F\end{align*}. It contains all the outcomes that are in both events \begin{align*}E\end{align*} and \begin{align*}F\end{align*}. The symbol \begin{align*}\cap\end{align*} can be thought of as “and”, since it includes only the outcomes that are in both events. \begin{align*}E \cap F\end{align*} is shown below.
Notice that \begin{align*}E \cup F\end{align*} will always contain the same outcomes as \begin{align*}E \cap F\end{align*}, plus more outcomes (usually). \begin{align*}E \cup F\end{align*} could never contain less outcomes than \begin{align*}E \cap F\end{align*}.
Example C
Consider the experiment of tossing three coins and recording the sequence of heads and tails. Let \begin{align*}A\end{align*} be the event that there are exactly two heads. Let \begin{align*}B\end{align*} be the event that there are exactly two tails.
a) Find the sample space for the experiment.
b) List the outcomes in event \begin{align*}A\end{align*}.
c) List the outcomes in event \begin{align*}B\end{align*}.
d) Create a diagram that shows the sample space, events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, and all of the outcomes.
Solution: Typically when working with experiments having to do with coins, \begin{align*}H\end{align*} represents getting “heads” and \begin{align*}T\end{align*} represents getting “tails”.
a) The sample space is: \begin{align*}S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}\end{align*}
b) \begin{align*}A=\{HHT,HTH,THH\}\end{align*}
c) \begin{align*}B=\{HTT,THT,TTH\}\end{align*}
d) Notice that this time there are no outcomes that are in both \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, so the circles don't overlap.
Concept Problem Revisited
There are 52 cards in a standard deck of cards. These 52 cards are organized by suit:
Clubs: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Diamonds: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Hearts: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Spades: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Event \begin{align*}C\end{align*} is choosing a card that has an even number or a spade on it. Because the word “or” is used, you know this is really the union of two other events. Let \begin{align*}D\end{align*} be choosing an even number and let \begin{align*}E\end{align*} be choosing a spade. Then \begin{align*}C=D \cup E\end{align*}.
\begin{align*}D=\end{align*} {2 of clubs, 4 of clubs, 6 of clubs, 8 of clubs, 10 of clubs, 2 of diamonds, 4 of diamonds, 6 of diamonds, 8 of diamonds, 10 of diamonds, 2 of hearts, 4 of hearts, 6 of hearts, 8 of hearts, 10 of hearts, 2 of spades, 4 of spades, 6 of spades, 8 of spades, 10 of spades}
\begin{align*}E=\end{align*} {Ace of spades, 2 of spades, 3 of spades, 4 of spades, 5 of spades, 6 of spades, 7 of spades, 8 of spades, 9 of spades, 10 of spades, Jack of spades, Queen of spades, King of spades}
\begin{align*}C=D \cup E=\end{align*} {Ace of spades, 2 of spades, 3 of spades, 4 of spades, 5 of spades, 6 of spades, 7 of spades, 8 of spades, 9 of spades, 10 of spades, Jack of spades, Queen of spades, King of spades, 2 of clubs, 4 of clubs, 6 of clubs, 8 of clubs, 10 of clubs, 2 of diamonds, 4 of diamonds, 6 of diamonds, 8 of diamonds, 10 of diamonds, 2 of hearts, 4 of hearts, 6 of hearts, 8 of hearts, 10 of hearts}
#### Vocabulary
An experiment is an occurrence with a result that can be observed.
An outcome of an experiment is one possible result of the experiment.
The sample space for an experiment is the set of all possible outcomes of the experiment.
An event for an experiment is a subset of the sample space containing outcomes that you are interested in (sometimes called favorable outcomes).
The complement of an event is the event that includes all outcomes in the sample space not in the original event. The symbol for complement is ′.
The union of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is \begin{align*}\cup\end{align*}.
The intersection of two events is the event that includes all outcomes that are in both of the original events. The symbol for intersection is \begin{align*}\cap\end{align*}.
A Venn diagram is a way to visualize sample spaces, events, and outcomes.
#### Guided Practice
Consider the experiment from Example C of tossing three coins and recording the sequence of heads and tails. The diagram below represents the sample space and two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.
1. Describe \begin{align*}A^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
2. Describe \begin{align*}(A \cup B)\end{align*} in words and with the diagram. What outcomes are in this event?
3. Describe \begin{align*}(A \cup B)^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
4. Describe \begin{align*}(A \cap B)^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
1. \begin{align*}A^\prime\end{align*} is the event of not getting exactly two heads. \begin{align*}A^\prime=\{TTH,THT,HTT,HHH,TTT\}\end{align*} On the diagram, it is everything in the rectangle except circle \begin{align*}A\end{align*}.
2. \begin{align*}(A \cup B)\end{align*} is all of the outcomes in event \begin{align*}A\end{align*}, event \begin{align*}B\end{align*}, or both. Note that for this experiment, there are no events in both \begin{align*}A\end{align*} and \begin{align*}B\end{align*}\begin{align*}A \cup B=\{HHT, HTH, THH, TTH, THT, HTT\}\end{align*}. In the diagram, only circles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are shaded.
3. \begin{align*}(A \cup B)^\prime\end{align*} is all of the outcomes not in \begin{align*}(A \cup B)\end{align*}. This means it is all of the outcomes that are in neither events \begin{align*}A\end{align*} nor \begin{align*}B\end{align*}\begin{align*}(A \cup B)^\prime=\{HHH,TTT\}\end{align*}. In the diagram, the opposite part of the rectangle is shaded compared with #2.
4. \begin{align*}(A \cap B)^\prime\end{align*} is all of the outcomes not in \begin{align*}(A \cap B)\end{align*}. Remember that \begin{align*}A \cap B\end{align*} is all of the outcomes in both events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}; however, in this experiment the two events don't overlap (they are disjoint).
\begin{align*}A \cap B=\{\}\end{align*}, the empty set. This means that \begin{align*}(A \cap B)^\prime\end{align*} must be the whole sample space, since it has to include all outcomes in the sample space not in \begin{align*}A \cap B\end{align*}.
\begin{align*}(A \cap B)^\prime=\{HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}\end{align*}. In the diagram, everything in the rectangle is shaded.
#### Practice
Consider the experiment of spinning the spinner below twice and recording the sequence of results. Let \begin{align*}F\end{align*} be the event that the same color comes up twice. Let \begin{align*}H\end{align*} be the event that there is at least one red.
1. Find the sample space for the experiment.
2. List the outcomes in event \begin{align*}F\end{align*} and the outcomes in event \begin{align*}H\end{align*}.
3. Create a Venn diagram that shows the relationships between the outcomes in \begin{align*}F\end{align*}, \begin{align*}H\end{align*}, and the sample space.
4. Describe \begin{align*}F^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
5. Describe the event “getting two reds” with symbols and with the diagram. What outcomes are in this event?
6. Describe \begin{align*}(F \cup H)^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
Consider the experiment of rolling a pair of dice and finding the sum of the numbers on the dice. Let \begin{align*}J\end{align*} be the event that the sum is less than 4. Let \begin{align*}K\end{align*} be the event that the sum is an odd number.
7. Find the sample space for the experiment.
8. List the outcomes in event \begin{align*}J\end{align*} and the outcomes in event \begin{align*}K\end{align*}.
9. Create a Venn diagram that shows the relationships between the outcomes in \begin{align*}J, K\end{align*}, and the sample space.
10. Describe \begin{align*}K^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
11. Describe the event “getting an even number less than 4” with symbols and with the diagram. What outcomes are in this event?
12. Describe \begin{align*}(J \cap K^\prime)^\prime\end{align*} in words and with the diagram. What outcomes are in this event?
13. In this experiment, are you just as likely to get a sum of 2 as a sum of 7? Explain.
14. Compare and contrast unions of events with intersections of events.
15. Consider some experiment with event \begin{align*}E\end{align*}. Describe \begin{align*}E \cup E^\prime\end{align*}. Describe \begin{align*}E \cap E^\prime\end{align*}.
### Vocabulary Language: English
experiment
experiment
An experiment is the process of taking a measurement or making an observation.
Favorable Outcome
Favorable Outcome
A favorable outcome is the outcome that you are looking for in an experiment.
Outcome
Outcome
An outcome of a probability experiment is one possible end result.
Sample Space
Sample Space
In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.
simple events
simple events
A simple event is the simplest outcome of an experiment.
tree diagrams
tree diagrams
Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree. | 4,651 | 15,232 | {"found_math": true, "script_math_tex": 132, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2016-07 | latest | en | 0.853921 |
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# When x is divided by 3, the remainder is 2, and when y is divided by 7
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When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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27 Apr 2016, 00:06
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When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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27 Apr 2016, 02:01
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Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.
X can be 2,5 and y can be 18 so reminder is different in both..SO E
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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29 Jul 2017, 10:07
Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.
can someone give a solution with equations ?
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Posts: 7763
When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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29 Jul 2017, 10:38
1
Takdir wrote:
Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?
(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.
can someone give a solution with equations ?
Hi,
if you are looking for solution with equation..
the main statement tells us..
A) x=3a+2.... x could be 2,5,8,11,14,17....
B) y=7b+4.... y could be 4,11,18,25...
statements..
I) $$x^2=7c+4$$..... $$x^2$$ could be 4,11,18,25.....
so x can be $$\sqrt{4},\sqrt{11},\sqrt{18},\sqrt{25}$$... or $$2,\sqrt{11},\sqrt{18},5$$...
not required to go beyond this as we see ATLEAST two values of x possible 2,5...
insuff
II. it will give you some value of y..
insuff
combined..
whatever be the value of y..
y+2 and y+5 will surely give different REMAINDER when div by 21..
hence insuff
E
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7 [#permalink]
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09 Nov 2018, 16:58
Hi,
x when divided by 3, Remainder is 2.
x = 3a + 2
y when divided by 7, remainder is 4.
y = 7b + 4
Remainder when x + y divided by 21?
x + y = 3a + 2 + 7b+ 4
x + y = 3a + 7b+ 6.
If a is divisible by 7 and b is divisible by 3 then Remainder = 6.
A. x2 / 7 , Remainder = 4.
x = (3a + 2). x2 = 9a2 +6a + 4.
a is divisible by 7, so that remainder is 4.
B. y - 4 is divisible by 3
y = 7b + 4 ; y-4 = 7b
7b is divisible by 3, so b is divisible by 3.
C. a = divisible by 7, b = divisible by 3.
x + y = 3a + 7b + 6 when divided by 21 should give remainder 6.
The OA is E, Please can someone tell why C is incorrect?
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7 [#permalink]
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09 Nov 2018, 17:48
Akshay_Naik wrote:
Hi,
x when divided by 3, Remainder is 2.
x = 3a + 2
y when divided by 7, remainder is 4.
y = 7b + 4
Remainder when x + y divided by 21?
x + y = 3a + 2 + 7b+ 4
x + y = 3a + 7b+ 6.
If a is divisible by 7 and b is divisible by 3 then Remainder = 6.
A. x2 / 7 , Remainder = 4.
x = (3a + 2). x2 = 9a2 +6a + 4.
a is divisible by 7, so that remainder is 4.
B. y - 4 is divisible by 3
y = 7b + 4 ; y-4 = 7b
7b is divisible by 3, so b is divisible by 3.
C. a = divisible by 7, b = divisible by 3.
x + y = 3a + 7b + 6 when divided by 21 should give remainder 6.
The OA is E, Please can someone tell why C is incorrect?
Try the numbers x=5 and y=4 it gives a remainder of 9 when divided by 21
But when you take x=23 and y=4, it gives a remainder of 6 when divided by 21
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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31 Dec 2018, 00:31
Bunuel chetan2u
So what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong:
Stem: $$x=3q+2$$ and $$y=7p+4$$ $$–>$$ $$x+y=3q+7p+6$$
Hence, our question becomes: is $$3q+7p=21k$$. This is the case whenever $$3q$$ is divisible by $$7$$ and $$7p$$ is divisible by $$3$$.
Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y
Statement II: $$y-4=3m$$. Now from the stem, we know that $$y=7p+4$$ hence, this statement tells us that $$7p=3m$$
Still, Not sufficient as no information is given about x
Statement I&II combined:
From I) we know that $$3q$$ is divisible by $$7$$, hence, $$3q$$ must be a multiple of $$21$$
From II) we know that $$7p$$ is divisible by $$3$$, hence, $$7p$$ must be a multiple of $$21$$
Thus, $$x+y=(3q+7p)+6= 21b+6$$
Please let me know where I went wrong!
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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04 Jan 2019, 03:55
no one has any idea why above solution is incorrect?
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink]
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04 Jan 2019, 05:03
2
ghnlrug wrote:
Bunuel chetan2u
So what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong:
Stem: $$x=3q+2$$ and $$y=7p+4$$ $$–>$$ $$x+y=3q+7p+6$$
Hence, our question becomes: is $$3q+7p=21k$$. This is the case whenever $$3q$$ is divisible by $$7$$ and $$7p$$ is divisible by $$3$$.
Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y
Statement II: $$y-4=3m$$. Now from the stem, we know that $$y=7p+4$$ hence, this statement tells us that $$7p=3m$$
Still, Not sufficient as no information is given about x
Statement I&II combined:
From I) we know that $$3q$$ is divisible by $$7$$, hence, $$3q$$ must be a multiple of $$21$$
From II) we know that $$7p$$ is divisible by $$3$$, hence, $$7p$$ must be a multiple of $$21$$
Thus, $$x+y=(3q+7p)+6= 21b+6$$
Please let me know where I went wrong!
You are wrong here..
Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y
$$x^2=9q^2+12q+4=7L+4 –>9q^2+12q=7L......3q(3q+4)=7L$$..
When q is multiple of 7, 3q(3q+4)=21a(21a+4)=7L.... possible
Here 3q is divisible by 7
When 3q+4 is multiple of 7, say q is 1....3*1(3*1+4)=3*(7).... possible
Here 3q=3 and it is not divisible by 7
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7 [#permalink] 04 Jan 2019, 05:03
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https://infogalactic.com/info/334_(number) | 1,544,489,111,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823516.50/warc/CC-MAIN-20181210233803-20181211015303-00192.warc.gz | 644,068,299 | 32,934 | # 300 (number)
(Redirected from 334 (number))
← 299 300 301 →
Cardinal three hundred
Ordinal 300th
(three hundredth)
Factorization 22× 3 × 52
Roman numeral CCC
Binary 1001011002
Ternary 1020103
Quaternary 102304
Quinary 22005
Senary 12206
Octal 4548
Duodecimal 21012
Vigesimal F020
Base 36 8C36
Hebrew ש (Shin)
300 (three hundred) is the natural number following 299 and preceding 301.
## Mathematical properties
The number 300 is a triangular number and the sum of a pair of twin primes (149 + 151), as well as the sum of ten consecutive primes (13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47). It is palindromic in 3 consecutive bases: 30010 = 6067 = 4548 = 3639, and also in bases 13, 19, 24, 29, 49 and 59.
## Other fields
Three hundred is:
• In bowling, a perfect score, achieved by rolling strikes in all ten frames (a total of twelve strikes)
• The lowest possible Fair Isaac credit score
• Three hundred ft/s is the maximum legal speed of a shot paintball
• In the Old Testament 300 is the number of the ancient Israeli judge Gideon's followers victorious against the Midianites
• According to Islamic tradition, 300 is the number of ancient Israeli king Thalut's soldiers victorious against Goliath's soldiers
• According to Herodotus, 300 is the number of ancient Spartans resisting one million Persian invaders during the Battle of Thermopylae
• In Islamic history, 300 is the number of Muhammad's followers victorious in the Battle of Badr
• Three hundred is the number of families followers of Jewish heretic Sabbatai Zevi forced to convert to Islam by the Ottoman Sultan and became the ancestors of Donmeh
• Three hundred is the number of seats in the Hellenic parliament
## Integers from 301 to 399
### 300s
#### 301
301 = 7 × 43. 301 is the sum of three consecutive primes (97 + 101 + 103), happy number in base 10[1]
An HTTP status code, indicating the content has been moved and the change is permanent (permanent redirect). It is also the number of a debated Turkish penal code.
#### 302
302 = 2 × 151. 302 is a nontotient[2] and a happy number[1]
302 is the HTTP status code indicating the content has been moved (temporary redirect). It is also the displacement in cubic inches of Ford's "5.0" V8 and the area code for the state of Delaware.
#### 303
303 = 3 × 101
303 is the "See other" HTTP status code, indicating content can be found elsewhere. Model number of the Roland TB-303 synthesizer which is accredited as having been used to create the first acid house music tracks, in the late 1980s.
#### 304
304 = 24 × 19. 304 is the sum of six consecutive primes (41 + 43 + 47 + 53 + 59 + 61), sum of eight consecutive primes (23 + 29 + 31 + 37 + 41 + 43 + 47 + 53), primitive semiperfect number,[3] untouchable number,[4] nontotient[2]
304 is the HTTP code indicating the content has not been modified, and the record number of wickets taken in English cricket season by Tich Freeman in 1928. 304 is also the name of a card game popular in Sri Lanka and southern India. 304 is also a slang term meaning "hoe", a meaning it acquired due to the popularity of LED calculators and watches.[5]
It is also one of the telephone area codes for West Virginia.
#### 305
305 = 5 × 61.
305 is the HTTP status code indicating a proxy must be used.
#### 306
306 = 2 × 32 × 17. 306 is the sum of four consecutive primes (71 + 73 + 79 + 83), pronic number,[6] Harshad number, and an untouchable number.[4]
It is also a telephone area code for the province of Saskatchewan, Canada.
#### 307
307 is a prime number, Chen prime,[7] and the HTTP status code for "temporary redirect"
#### 308
308 = 22 × 7 × 11. 308 is a nontotient,[2] totient sum of the first 31 integers, Harshad number, heptagonal pyramidal number,[8] and the sum of two consecutive primes (151 + 157).
309 = 3 × 103
### 310s
#### 310
310 = 2 × 5 × 31. 310 is a sphenic number,[9] noncototient,[10] and self number.[11]
#### 311
311: a prime number, see 311 (number).
#### 312
312 = 23 × 3 × 13. 312 is Harshad number and self number[11]
#### 313
313: a prime number, see 313 (number).
#### 314
314 = 2 × 157. 314 is a nontotient.[2]
#### 315
315 = 32 × 5 × 7. 315 is a Harshad number
#### 316
316 = 22 × 79. 316 is a centered triangular number[12] and a centered heptagonal number[13]
#### 317
317 is a prime number, Eisenstein prime with no imaginary part, Chen prime,[7] and a strictly non-palindromic number.
317 is also shorthand for the LM317 adjustable regulator chip. It is also the area code for the Indianapolis region.
#### 318
318 = 2 × 3 × 53. It is a sphenic number,[9] nontotient,[2] and the sum of twelve consecutive primes (7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47)
#### 319
319 = 11 × 29. 319 is the sum of three consecutive primes (103 + 107 + 109), Smith number,[14] cannot be represented as the sum of fewer than 19 fourth powers, happy number in base 10[1]
"319" is a song by Prince.
British Rail Class 319s are dual-voltage electric multiple unit trains
### 320s
#### 320
320 = 26 × 5 = (25) × (2 × 5). 320 is a Leyland number,[15] maximum determinant of a 10 by 10 matrix of zeros and ones, and a Harshad number. The largest of only two numbers (92 is the other) which has an hexadecimal representation of its MD5-value that starts with the number itself. A popular bitrate.
#### 321
321 = 3 × 107, a Delannoy number[16]
An area code in central Florida.
#### 322
322 = 2 × 7 × 23. 322 is a sphenic,[9] nontotient, untouchable,[4] Lucas number,[17] and a Harshad number.
It is also seen as a Skull and Bones reference of power
#### 323
323 = 17 × 19. 323 is the sum of nine consecutive primes (19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53), Motzkin number,[18] self number.[11] A Lucas and Fibonacci pseudoprime. See 323 (disambiguation)
#### 324
324 = 22 × 34 = 182. 324 is the sum of four consecutive primes (73 + 79 + 83 + 89), totient sum of the first 32 integers, untouchable number,[4] and a Harshad number.
#### 325
325 = 52 × 13. 325 is a triangular number, hexagonal number,[19] nonagonal number,[20] centered nonagonal number.[21] 325 is the smallest number to be the sum of two squares in 3 different ways: 12 + 182, 62 + 172 and 102 + 152. 325 is also the smallest (and only known) 3-hyperperfect number.
#### 326
326 = 2 × 163. 326 is a nontotient, noncototient,[10] and an untouchable number.[4]
#### 327
327 = 3 × 109. 327 is a perfect totient number.[22]
#### 328
328 = 23 × 41. 328 is the sum of the first fifteen primes, and a refactorable number.[23]
#### 329
329 = 7 × 47. 329 is the sum of three consecutive primes (107 + 109 + 113), and a highly cototient number.[24]
### 330s
#### 330
330 = 2 × 3 × 5 × 11. 330 is sum of six consecutive primes (43 + 47 + 53 + 59 + 61 + 67), pentatope number (and hence a binomial coefficient $\tbinom {11}4$), a pentagonal number,[25] divisible by the number of primes below it, sparsely totient number,[26] and a Harshad number. It is also the number of dimples on a British golf ball.
#### 331
331 is a prime number, cuban prime,[27] sum of five consecutive primes (59 + 61 + 67 + 71 + 73), centered pentagonal number,[28] centered hexagonal number,[29] and Mertens function returns 0.[30]
#### 332
332 = 22 × 83, Mertens function returns 0.[30]
#### 333
333 = 32 × 37, Mertens function returns 0,[30] Harshad number.
Symbolically 333 is used to represent Choronzon, a demon used in the philosophy of Thelema
#### 334
334 = 2 × 167, nontotient, self number,[11]
334 was the long-time highest score for Australia in Test cricket (held by Sir Donald Bradman and Mark Taylor). 334 is also the name of a science fiction novel by Thomas M. Disch.
#### 335
335 = 5 × 67, divisible by the number of primes below it.
#### 336
336 = 24 × 3 × 7, Harshad number, untouchable number,[4] also the number of dimples on an American golf ball.
#### 337
337, prime number, permutable prime with 373 and 733, Chen prime,[7] star number
#### 338
338 = 2 × 132, nontotient.
339 = 3 × 113
### 340s
#### 340
340 = 22 × 5 × 17, sum of eight consecutive primes (29 + 31 + 37 + 41 + 43 + 47 + 53 + 59), sum of ten consecutive primes (17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53), sum of the first four powers of 4 (41 + 42 + 43 + 44), divisible by the number of primes below it, nontotient, noncototient.[10]
#### 341
341 = 11 × 31, sum of seven consecutive primes (37 + 41 + 43 + 47 + 53 + 59 + 61), octagonal number,[31] centered cube number,[32] super-Poulet number. 341 is the smallest Fermat pseudoprime; it is the least composite odd modulus m greater than the base b, that satisfies the Fermat property "bm−1 − 1 is divisible by m", for bases up to 128 of b = 2, 15, 60, 63, 78, and 108.
#### 342
342 = 2 × 32 × 19, pronic number,[6] Harshad number, untouchable number.[4]
#### 343
343 = 73, nice Friedman number since 343 = (3 + 4)3. It's the only known example of x2+x+1 = y3, in this case, x=18, y=7. It is z in a triplet (x,y,z) such that x5 + y2 = z3.
The speed of sound in dry air at 20 °C (68 °F) is 343 m/s. 343 was the number of New York City Fire Department members that died in the events of September 11, 2001.
#### 344
344 = 23 × 43, octahedral number,[33] noncototient,[10] totient sum of the first 33 integers, refactorable number.[23]
#### 345
345 = 3 × 5 × 23, sphenic number,[9] self number.[11]
#### 346
346 = 2 × 173, Smith number,[14] noncototient.[10]
#### 347
347 is a prime number, safe prime,[34] Eisenstein prime with no imaginary part, Chen prime,[7] Friedman number since 347 = 73 + 4, and a strictly non-palindromic number.
It is the number of an area code in New York.
#### 348
348 = 22 × 3 × 29, sum of four consecutive primes (79 + 83 + 89 + 97), refactorable number.[23]
#### 349
349, prime number, sum of three consecutive primes (109 + 113 + 127), since 1976 the number of seats in the Swedish parliament.[35]
### 350s
#### 350
350 = 2 × 52 × 7, primitive semiperfect number,[3] divisible by the number of primes below it, nontotient, a truncated icosahedron of frequency 6 has 350 hexagonal faces and 12 pentagonal faces.
350.org is an international environmental organization. 350 is the number of cubic inches displaced in the most common form of the Small Block Chevrolet V8. The number of seats in the Congress of Deputies (Spain) is 350.
#### 351
351 = 33 × 13, triangular number, sum of five consecutive primes (61 + 67 + 71 + 73 + 79), member of Padovan sequence,[36] Harshad number.
It is also the 351 Windsor engine from Ford Motor Company as well as the 351 (building) in St. John's, Newfoundland and Labrador.
#### 352
352 = 25 × 11, the number of n-Queens Problem solutions for n = 9.
The number of international appearances by Kristine Lilly for the USA women's national football (soccer) team, an all-time record for the sport.
#### 353
353 is a prime number, Chen prime,[7] Proth prime,[37] Eisenstein prime with no imaginary part, palindromic prime, and Mertens function returns 0.[30] 353 is the base of the smallest 4th power that is the sum of 4 other 4th powers, discovered by Norrie in 1911: 3534 = 304 + 1204 + 2724 + 3154.
#### 354
354 = 2 × 3 × 59, sphenic number,[9] nontotient, also SMTP code meaning start of mail input. It is also sum of absolute value of the coefficients of Conway's polynomial.
#### 355
355 = 5 × 71, Smith number,[14] Mertens function returns 0,[30] divisible by the number of primes below it. the numerator of the best simplified rational approximation of pi having a denominator of four digits or fewer. This fraction (355/113) is known as Milü.
#### 356
356 = 22 × 89, Mertens function returns 0,[30] self number.[11]
#### 357
357 = 3 × 7 × 17, sphenic number.[9]
357 also refers to firearms or ammunition of .357 caliber, with the best-known cartridge of that size being the .357 Magnum. The .357 SIG, whose name was inspired by the performance of the .357 Magnum, is actually a 9 mm or .355 caliber.
#### 358
358 = 2 × 179, sum of six consecutive primes (47 + 53 + 59 + 61 + 67 + 71), Mertens function returns 0.[30]
#### 359
359 is a prime number, safe prime,[34] Eisenstein prime with no imaginary part, Chen prime,[7] and strictly non-palindromic number.
### 360s
#### 360
360 has its own article.
#### 361
361 = 192, centered triangular number,[12] centered octagonal number, centered decagonal number,[38] member of the Mian–Chowla sequence;[39] also the number of positions on a standard 19 x 19 Go board.
#### 362
362 = 2 × 181, Mertens function returns 0,[30] nontotient, noncototient.[10]
#### 363
363 = 3 × 112, sum of nine consecutive primes (23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59), Mertens function returns 0,[30] perfect totient number.[22]
#### 364
364 = 22 × 7 × 13, tetrahedral number,[40] sum of twelve consecutive primes (11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53), Mertens function returns 0,[30] nontotient, Harshad number. It is a repdigit in base 3 (111111), base 9 (444), base 25 (EE), base 27 (DD), base 51 (77) and base 90 (44).
The total number of gifts received in the song "The Twelve Days of Christmas".
#### 365
365 = 5 × 73, it has its own article
#### 366
366 = 2 × 3 × 61, sphenic number,[9] Mertens function returns 0,[30] noncototient.[10] Also, the number of days in a leap year; it is 26-gonal and 123-gonal.
#### 367
367 is a prime number, Perrin number,[41] self number,[11] happy number, and a strictly non-palindromic number.
#### 368
368 = 24 × 23 It is also a Leyland number.[15]
#### 369
369 = 32× 41, it is the magic constant of the 9 × 9 normal magic square and n-queens problem for n = 9; there are 369 free polyominoes of order 8. With 370, a Ruth–Aaron pair with only distinct prime factors counted.
### 370s
#### 370
370 = 2 × 5 × 37, sphenic number,[9] sum of four consecutive primes (83 + 89 + 97 + 101), Nontotient, with 369 part of a Ruth–Aaron pair with only distinct prime factors counted, Harshad number, Base 10 Armstrong number since 33 + 73 + 03 = 370.
System/370 is a computing architecture from IBM.
#### 371
371 = 7 × 53, sum of three consecutive primes (113 + 127 + 131), sum of seven consecutive primes (41 + 43 + 47 + 53 + 59 + 61 + 67), sum of the primes from its least to its greatest prime factor (sequence A055233 in OEIS), the next such composite number is 2935561623745, Armstrong number since 33 + 73 + 13 = 371.
#### 372
372 = 22 × 3 × 31, sum of eight consecutive primes (31 + 37 + 41 + 43 + 47 + 53 + 59 + 61), Harshad number, noncototient,[10] untouchable number,[4] refactorable number.[23]
#### 373
373, prime number, balanced prime,[42] sum of five consecutive primes (67 + 71 + 73 + 79 + 83), permutable prime with 337 and 733, palindromic prime in 3 consecutive bases: 5658 = 4549 = 37310 and also in base 4: 113114, two-sided primes.
#### 374
374 = 2 × 11 × 17, sphenic number,[9] nontotient.
#### 375
375 = 3 × 53, Harshad number
#### 376
376 = 23 × 47, pentagonal number,[25] 1-automorphic number,[43] nontotient, refactorable number.[23]
#### 377
377 = 13 × 29, Fibonacci number, a Lucas and Fibonacci pseudoprime, the sum of the squares of the first six primes, a common approximation for the impedance of free space in ohms.
#### 378
378 = 2 × 33 × 7, triangular number, hexagonal number,[19] Smith number,[14] Harshad number, self number.[11]
#### 379
379 is a prime number, Chen prime,[7] and a happy number in base 10.
### 380s
#### 380
380 = 22 × 5 × 19, pronic number.[6]
#### 381
381 = 3 × 127, sum of the first sixteen primes. Palindrome in base 2 and base 8.
Represents the term "I Love You", (3) words, (8) letters, (1) meaning.
#### 382
382 = 2 × 191, sum of ten consecutive primes (19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59), Smith number.[14]
#### 383
383, prime number, safe prime,[34] Woodall prime,[44] Thabit number, Eisenstein prime with no imaginary part, palindromic prime.
#### 384
384 has its own article
#### 385
385 = 5 × 7 × 11, sphenic number,[9] square pyramidal number,[45] the number of integer partitions of 18.
#### 386
386 = 2 × 193, Nontotient, noncototient,[10] centered heptagonal number,[13]
386 is also shorthand for the Intel 80386 microprocessor chip. 386 generation refers to South Koreans, especially politicians, born in the '60s (ko:386 세대).
#### 387
387 = 32 × 43, also shorthand for the Intel 80387, math coprocessor chip to the 386.
388 = 22 × 97
#### 389
389, prime number, Eisenstein prime with no imaginary part, Chen prime,[7] highly cototient number,[24] self number,[11] strictly non-palindromic number. Smallest conductor of a rank 2 Elliptic curve.
Also, 389 equals the displacement in cubic inches of the famous Pontiac GTO V-8 engine of 1964–66. The port number for LDAP, and the name for the Fedora Directory Server project.
### 390s
#### 390
390 = 2 × 3 × 5 × 13, sum of four consecutive primes (89 + 97 + 101 + 103), nontotient,
System/390 is a computing architecture from IBM.
#### 391
391 = 17 × 23, Smith number,[14] centered pentagonal number.[28]
#### 392
392 = 23 × 72, Harshad number.
#### 393
393 = 3 × 131, Mertens function returns 0.[30]
393 is the number of county equivalents in Canada
#### 394
394 = 2 × 197, a Schroder number,[46] nontotient, noncototient.[10]
#### 395
395 = 5 × 79, sum of three consecutive primes (127 + 131 + 137), sum of five consecutive primes (71 + 73 + 79 + 83 + 89).
#### 396
396 = 22 × 32 × 11, sum of a twin prime (197 + 199), totient sum of the first 36 integers, refactorable number,[23] Harshad number, digit-reassembly number.
396 also refers to the displacement in cubic inches of early Chevrolet Big-Block engines.
#### 397
397, prime number, cuban prime,[27] centered hexagonal number.[29]
#### 398
398 = 2 × 199, Nontotient.
#### 399
399 = 3 × 7 × 19, sphenic number,[9] smallest Lucas–Carmichael number, Harshad number.
## References
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5. http://onlineslangdictionary.com/meaning-definition-of/304
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44. "Sloane's A050918 : Woodall primes". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-22.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
45. "Sloane's A000330 : Square pyramidal numbers". The On-Line Encyclopedia of INteger Sequences. OEIS Foundation. Retrieved 2016-05-22.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
46. "Sloane's A006318 : Large Schröder numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-22.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> | 8,050 | 27,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-51 | latest | en | 0.863319 |
http://www.maximapedia.com/p/parabola.html | 1,544,529,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823618.14/warc/CC-MAIN-20181211104429-20181211125929-00034.warc.gz | 430,168,794 | 6,229 | Parabola
PARABOLA, a plane curve of the second degree. It may be defined as a section of a right circular cone by a plane parallel to a tangent plane to the cone, or as the locus of a point which moves so that its distances from a fixed point and a fixed line are equal. It is therefore a conic section having its eccentricity equal to unity. The parabola is the curve described by a projectile which moves in a non-resisting medium under the influence of gravity (see Mechanics). The general relations between the parabola, ellipse and hyperbola are treated in the articles Geometry, Analytical, and Conic Sections; and various projective properties are demonstrated in the article Geometry, Projective. Here only the specific properties of the parabola will be given.
The form of the curve is shown in fig. i, where P is a point on the curve equidistant from the fixed line AB, known as the directrix, and the fixed point F known as the focus. The line CD passing through the focus and perpendicular to the directrix is the axis or principal diameter, and meets the curve in the vertex G. The line FL perpendicular to the axis, and passing through the focus, is the semilatus rectum, the latus rectum being the focal chord parallel to the directrix. Any line parallel to the axis is a diameter, and the parameter of any diameter is measured by the focal chord drawn Fig. i.
parallel to the tangent at the vertex of the diameter and is equal Fig. 2.
to four times the focal distance of the vertex. To construct the parabola when the focus and directrix are given, draw the axis CD and bisect CF at G, which gives the vertex. Any number of points on the parabola are obtained by taking any point E on the directrix, joining EG and EF and drawing FP so that the angles PFE and DFE are equal. Then EG produced meets FP in a point on the curve. By joining the points so obtained the parabola may be described. A mechanical construction, when the same conditions are given, consists in taking a rigid bar ABC bent at right angles at B (fig. 2), and fastening a string of length BC to C = and F. Then if a pencil be placed along BC so as to keep the string taut, and the limb AB be shd along the directrix, the - pencil will trace out the parabola.
Properties which may be readily deduced by euchdian methods from the definition include the following: the tangent at any point bisects the angle between the focal distance and the perpendicular on the directrix and is equally incUned to the focal distance and the axis; tangents at the extremities of a focal chord intersect at right angles on the directrix, and as a corollary we have that the locus of the intersection of tangents at right angles is the directrix; the circumcircle of a triangle circumscribing a parabola passes through the focus; the subtangent is equal to twice the abscissa of the point of contact; the subnormal is constant and equals the semilatus rectum; and the radius of curvature at a point P is 2 (FP)At/a^ where a is the semilatus rectum and FP the focal distance of P.
A fundamental property of the curve is that the line at infinity is a tangent (see Geometry, Projective), and it follows that the centre and the second real focus and directrix are at infinity. It also follows that a Une half-way between a point and its polar and parallel to the latter touches the parabola, and therefore the hues joining the middle points of the sides of a self-conjugate triangle form a circumscribing triangle, and also that the ninepoint circle of a self-conjugate triangle passes through the focus. The orthocentre of a triangle circumscribing a parabola is on the directrix; a deduction from this theorem is that the centre of the circumcircle of a self-conjugate triangle is on the directrix (" Steiner's Theorem ").
In the article Geometry, Analytical, it is shown that the general equation of the second degree represents a parabola when the highest terms form a perfect square.
Analytic rpj^j j ^- analytical expression of the projective Geometry. , ,,. ( r ( ( ^ i^i.
property that the fine at mfimty is a tangent. the simplest equation to the parabola is that which is referred to its axis and the tangent at the vertex as the axes of co-ordinates, when it assumes the form y'' = ^ax where 2a = semilatus rectum; this may be deduced directly from the definition. An equation of similar form is obtained when the axes of co-ordinates are any diameter and the tangent at the vertex. The equations to the tangent and normal at the point x'y' are yy' =2a(x+x') and 2a(y - y')-\-y'(x - x') = o, and may be obtained by general methods (see Geometry, Analytical, and Infinitesimal Calculus). More convenient forms in terms of a single parameter are deduced by substituting x' = ant^, y' = 2am (for on eliminating tn between these relations the equation to the parabola is obtained). The tangent then becomes my = x+am' and the normal y = mx+2am - am^. The envelope of this last equation is 27a/ = 4(a; - 20)', which shows that the evolute of a parabola is a semi-cubical parabola (see below Higher Orders). The cartesian equation to_a parabola which touches the coordinate axes is Vax+Vby=i, and the polar equation when the focus is the pole and the axis the initial fine is r cos-5/2 = (2. The equation to a parabola in triangular co-ordinates is generally derived by expressing the condition that the line at infinity is a tangent in the equation to the general conic. For example, in trilinear co-ordinates, the equation to the general conic circumscribing the triangle of reference is l/37-f As!7a-|-Aso)3 = o; for this to be a parabola the line aa + b^ + c'y = o must be a tangent. Expressing this condition we obtain V/a^ Vntb=^ VAsc = oasthe relation which must hold between the co-efficients of the above equation and the sides of the triangle of reference for the equation to represent a parabola. Similarly, the conditions for the inscribed conic V la+V mfi-bV ny = o to be a parabola is Ibc+mca+nab - o, and the conic for which the triangle of reference is self-conjugate la^-\-mfi--(-ny'^ = o is a-mn-(-bhil-\-cHm = o. The various forms in areal co-ordinates may be derived from the above by substituting Xo for /, nb for m and vc for n, or directly by expressing the condition for tangency of the hne x-\-y-\-z = o to the conic expressed in areal coordinates. In tangential (/>, q, r) co-ordinates the inscribed and circumscribed conies take the forms \qr-\-iirp-\-vpq = o and \'\p+\/ixq-'ryvr = o; these are parabolas when X-)-j:i-|-i' = o and VX='=V/u='=V!' = o respectively.
The length of a paraboUc arc can be obtained by the methods of the infinitesimal calculus; the curve is directly quadrable, the area of any poition between two ordinates being two thirds of the circumscribing parallelogram. The pedal equation with the focus as origin is p^ = ar; the first positive pedal for the vertex is the cissoid (q.v.) and for the focus the directrix. (See Infinitesimal Calculus.)
References. - Geometrical constructions of the parabola are to be found in T. H. Eagles' Plane Curves (1885). See the bibliography to the articles Conic Sections; Geometry, Analytical; and Geometry, Projective.
In the geometry of plane curves, the term parabola is often used to denote the curves given by the general equation a"'x" = ym+n^ thus ax = y^ is the quadratic or Apollonian parabola; a-x = y' is the cubic parabola, a'x = y* is orders the biquadratic parabola; semi parabolas have the general equation ax"~^ = y", thus ax' = y^ is the semicubical parabola and a:(? = y* the semibiquadratic parabola. These curves were investigated by Rene Descartes, Sir Isaac Newton, Colin Maclaurin and others. Here we shall treat only the more important forms.
The cartesian parabola is a cubic curve which is also known as the trident of Newton on account of its three-pronged form. Its equation is xy = ax'^-'rbx--\-cx-\-d, and it consists of two legs asymptotic to the axis of y and two parabolic legs (fig. 3). The simplest form is axy = x^ - a^, in this case the serpentine position shown in the figure degenerates into a point of inflexion. Descartes used the curve to solve sextic equations by determining its intersections with a circle; mechanical constructions were given by Descartes (Geometry, lib. 3) and Maclaurin (Organica geometrica).
The cubic parabola (fig. 4) is a cubic curve having the equation y = ax''-\-bx'^-\-cx-\-d. It consists of two parabolic branches tending in opposite directions. John Wallis utilized the intersections of this curve with a right line to solve cubic equations, and Edmund Halley solved sextic equations with the aid of a circle.
Diverging parabolas are cubic curves given by the equation y^ = ax^+bx''-\-cx-\-d. Newton discussed the five forms which arise from the relations of the roots of the cubic equation. When all the Fig. 3.
Fig. 4.
Fig. 5.
roots are real and unequal the curve consists of a closed oval and a parabolic branch (fig. 5). As the two lesser roots are made more and more equal the oval shrinks in size and ultimately becomes a real conjugate point, and the curve, the equation of which is y^ = (x - a)^(.r - fe) (in which a>b) consists of this point and a bell-like branch resembling the right-hand member of fig. 5. If two roots are imaginary the equation is y = (.x^-t-a^) (.x - o) and the curve resembles the parabolic branch, as in the preceding case. This is sometimes termed the campaniform (or bell-shaped) parabola. If the two greater roots are equal the equation is y = (x - a) (x - bY (in which a<b) and the curve assumes the form shown in fig. 6, and is known as the nodated parabola. Finally, if all the roots are equal, the equation becomes y^ = (x -a)'\ this curve is the cuspidal or semicubical parabola (fig. 7). This curve, which is sometimes termed the Neilian parabola after William Neil (1637-1670), is the evolute of the ordinary parabola, and is especially interesting as being the first curve to be rectified. This was accomplished in 1657 by Neil in England, and in 1659 by Heinrich van Haureat in Holland. Newton showed that all the five varieties of the diverging parabolas may be exhibited as plane sections of the solid of revolution of the semicubical parabola. A plane oblique to the axis and passing below the vertex gives the first variety; If it passes through the vertex, Fig. 6. Fig. 7.
the second form; if above the vertex and oblique or parallel to the axis, the third form; if below the vertex and touching the surface, the fourth form, and if the plane contains the axis, the fifth form results (see Curve).
The biquadratic parabola has, in its most general form, the equation y = ax'-\-bx^-\-cx'^+dx+e, and consists of a serpentinous and two parabolic branches (fig. 8). If all the roots of the quartic in Fig. 8. Fig. 9. Fig. 10.
X are equal the curve assumes the form shown in fig. 9, the axis of x being a double tangent. If the two middle roots are equal, fig. 10 results. Other forms which correspond to other relations between the roots can be readily deduced from the most general form. (See Curve; and Geometry, Analytical.)
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# How many 3 digit positive integers with distinct digits
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Re: How many 3 digit positive integers with distinct digits [#permalink]
rrsnathan wrote:
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?
A. 576
B. 520
C. 504
D. 432
E. 348
A number not to be a multiple of 10 should not have the units digit of 0.
XXX
9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)
9*8*8=576.
Hi Bunuel,
why is this solution wrong?
9C1 options for the first digit (from 1 to 9 inclusive).
9C1 options for the Second digit (from 0 to 9 inclusive minus the one we used for the first digit).
7C1 options for the third digit (from 1 to 9 inclusive minus 2 digits we used for the first and the second digits)
Which gives
9*9*7=567
Where i am wrong???
Rrsnathan.
If the second digit is 0, then you'll have 8 not 7 options for the third digit.
Hope it's clear.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Perfectly clear!! Thank you very much!
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?
A. 576
B. 520
C. 504
D. 432
E. 348
A number not to be a multiple of 10 should not have the units digit of 0.
XXX
9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)
9*8*8=576.
Hi Bunuel,
why is this solution wrong?
9C1 options for the first digit (from 1 to 9 inclusive).
9C1 options for the Second digit (from 0 to 9 inclusive minus the one we used for the first digit).
7C1 options for the third digit (from 1 to 9 inclusive minus 2 digits we used for the first and the second digits)
Which gives
9*9*7=567
Where i am wrong???
Rrsnathan.
If the second digit is 0, then you'll have 8 not 7 options for the third digit.
Hope it's clear.
It will still be wrong it is 9*9*8 and not 9*8*8...
still confused
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Shouldn't it be 9x9x8?
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Re: How many 3 digit positive integers with distinct digits [#permalink]
pretzel wrote:
Shouldn't it be 9x9x8?
Nope. Check here: how-many-3-digit-positive-integers-with-distinct-digits-159249.html#p1264268
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Re: How many 3 digit positive integers with distinct digits [#permalink]
1
Kudos
I did in this way :
case 1: tens place is 0 : 8 x 1 x 9 = 72
case 2 : tens place is not 0 : 7 x 8 x9 = 504
therefore , total numbers = 72 + 504 = 576 ans.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?
Bunuel wrote:
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?
A. 576
B. 520
C. 504
D. 432
E. 348
A number not to be a multiple of 10 should not have the units digit of 0.
XXX
9 options for the first digit (from 1 to 9 inclusive).
8 options for the third digit (from 1 to 9 inclusive minus the one we used for the first digit).
8 options for the second digit (from 0 to 9 inclusive minus 2 digits we used for the first and the third digits)
9*8*8=576.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Amit0507 wrote:
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?
hi, there can be certain cases in which zero can occur at the 3rd digit, if one follows 1st-2nd-3rd digit option. in order to eliminate all such cases we go with 1st-3rd-2nd digit option.
remember a number will be divisible by 10 if its last digit ends with zero, and in this case we want to eliminate all such cases in which zero occurs at the unit digit.
i hope it helps.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.
manpreetsingh86 wrote:
Amit0507 wrote:
Bunuel, Why must one go by 1st-3rd-2nd digit option and not 1st-2nd-3rd digit option, which would result in 9*9*8?
hi, there can be certain cases in which zero can occur at the 3rd digit, if one follows 1st-2nd-3rd digit option. in order to eliminate all such cases we go with 1st-3rd-2nd digit option.
remember a number will be divisible by 10 if its last digit ends with zero, and in this case we want to eliminate all such cases in which zero occurs at the unit digit.
i hope it helps.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.
lets try to understand, why we approach this question from first digit and why not from second or third digit ??
we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_
now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8
so the number is 98_
now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.
hence we have to eliminate all such possible cases.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
You mixed possibilities with the number at the highlighted section. Let me explain my approach.
1. I have 9 possibilities for the 1st digit.
2. I have again 9 possibilities for the second place as I will consider 0 here
3. Now out of 10, 2 numbers are used. If I assume 0 was used in the second place (10_) then I have 8 possibilities for the last place as well. so the final possibility will be 9*9*8.
I hope I made myself clear on the understanding. Please point out the flaw.
manpreetsingh86 wrote:
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.
lets try to understand, why we approach this question from first digit and why not from second or third digit ??
we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_
now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8
so the number is 98_
now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.
hence we have to eliminate all such possible cases.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
Amit0507 wrote:
You mixed possibilities with the number at the highlighted section. Let me explain my approach.
1. I have 9 possibilities for the 1st digit.
2. I have again 9 possibilities for the second place as I will consider 0 here
3. Now out of 10, 2 numbers are used. If I assume 0 was used in the second place (10_) then I have 8 possibilities for the last place as well. so the final possibility will be 9*9*8.
I hope I made myself clear on the understanding. Please point out the flaw.
manpreetsingh86 wrote:
Amit0507 wrote:
Hi Manpreet,
The approach is still not clear to me. By 9*9*8 option, we eliminate the possibility of a Zero at the 3rd place.
lets try to understand, why we approach this question from first digit and why not from second or third digit ??
we don't want zero at the first digit. why ?? if zero occurs at the first digit then number won't be a three digit number(it will become a two digit number). that's why we want to eliminate all such cases in which zero occurs at the first digit so we started with first digit. now up to this point we have 9 possibilities for first digit . which are 1,2,3,4,5,6,7,8,9. suppose we selected 9 for the first digit. so number will be 9,_,_
now lets suppose we follow your method. so for second digit we have 0,1,2,3,4,5,6,7,8. now from these 9 numbers suppose we selected digit 8
so the number is 98_
now for third digit we have 0,1,2,3,4,5,6,7, now from these 8 number we can select any of the number. that is we have certain cases in which 0 can also occur at the third digit. (i.e. 980 is one of the possible number) which we don't want.
hence we have to eliminate all such possible cases.
how do you know that zero will be at the second position. ?? i have highlighted one of the cases in which zero will not be at the second position.
and my dear friend, i'm not mixing any possibilities. that's the correct way of approaching the problem.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
I found the total distinct#s and subtract the #s that end with Zero
9*9*8 - 9*9*1 = 576
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How many 3 digit positive integers with distinct digits [#permalink]
I found the total distinct#s and subtract the #s that end with Zero
9*9*8 - 9*9*1 = 576
9*9*8 - 9*9*1 = 567 and not 576...
9*9*8-9*8*1=576
_ _ _; Last digit has to be 0, so one possibility.
First digit can be one of 9 possibilities(1 to 9). second digit can be 8 (10-2)
Originally posted by origen87 on 21 Apr 2017, 02:40.
Last edited by origen87 on 25 Apr 2017, 09:00, edited 1 time in total.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
1
Kudos
aparnaharish wrote:
How many 3 digit positive integers with distinct digits are there, which are not multiples of 10?
A. 576
B. 520
C. 504
D. 432
E. 348
Let’s analyze the number of choices for each of the 3 digits from left to right.
Since the first digit can’t be 0, there are 9 choices for the first digit.
Since the second digit can’t be same as the first digit (but it can be 0), there are also 9 choices for the second digit. However, let’s consider the following two cases: 1) the second digit is not 0, and 2) the second digit is 0. For each of these two cases, we will also consider the last digit.
Case 1: The second digit is not 0
Since the second digit is not 0 (and it can’t be the same as the first digit), there are 8 choices for the second digit. As for the last digit, it can’t be 0 (otherwise it will be a multiple of 10) and it can’t be the same as either of the first two digits; thus, there are 7 choices. Thus, there are 9 x 8 x 7 = 504 three-digit numbers with distinct digits when the second digit is not 0.
Case 2: The second digit is 0
Since the second digit is 0, there is only 1 choice for the second digit. As for the last digit, it can’t be 0 (otherwise it will be a multiple of 10) and it can’t be the same as either of the first two digits. However, since the last digit won’t be 0, it won’t be the same as the second digit. In other words, it only needs to be different from the first digit; thus there are 8 choices. Thus, there are 9 x 1 x 8 = 72 three-digit numbers with distinct digits when the second digit is 0.
Lastly, the number of three-digit numbers with distinct digits is 504 + 72 = 576.
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Re: How many 3 digit positive integers with distinct digits [#permalink]
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Re: How many 3 digit positive integers with distinct digits [#permalink]
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http://mathhelpforum.com/differential-equations/121897-general-solution-inhomogenous-ode.html | 1,481,266,463,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542686.84/warc/CC-MAIN-20161202170902-00478-ip-10-31-129-80.ec2.internal.warc.gz | 168,424,585 | 13,916 | # Thread: General Solution of inhomogenous ODE
1. ## General Solution of inhomogenous ODE
ok,so again, reading my lecture notes,i bumped into this problem..
consider
A = (1 2,
0 -1)
find the general solution of
x' = Ax + ( 0,1 )
write x = (x, y) then the ODE is :
x' = x+2y
y' = -y + 1
y' = -y+1 implies y(t) = Ce^-t + 1 (how did one obtain this?? )
The ODE for x(t) is x' = x + 2(C(e^-t) + 1)
This linear inhomogeneous ODE gives x(t) = -2 - c(e^-t) + d(e^t) again how does one get this? i integrated but i cant get this. after trying for hours..i simply gave up ='(
thank you !
2. While the technique described in this post works, there is a big-time shortcut to this problem that I initially oversaw. The shortcut works when A is a constant matrix. Check the posts after this one for an explanation of this shortcut method.
The idea of variation of parameters is to seek the solution to
$\bf{x}' = A\bf{x} + \bf{b}$
in the form
$
\bf{v}_1(t)\bf{x}_1(t) + \bf{v}_2 (t)\bf{x}_2 (t) = F(t)\bf{v}(t)$
whereby $F(t) = \begin{bmatrix} x_1 & x_2 \end{bmatrix}$ is a fundamental matrix.
Note that the constant vector c of coefficients is replaced by an unknown function v(t).
What is the equation for our unknown function v(t) you ask?
It is fairly straight forward to derive:
$
(F\bf{v})' = F'\bf{v} + F\bf{v'} = A(F\bf{v}) + \bf{b} \implies
F\bf{v'} = \bf{b} \implies
\bf{v'} = F^{-1}\bf{b}$
.
IMPORTANT: You must make sure that $F^{-1}$ exists in order to solve for v !!!! (not factorial)
Finally, using integration, we have:
$
\bf{x}(t) = F(t) \int F(t)^{-1} \bf{b}(t)dt$
I'll now rewrite the original problem in latex so it is more palatable to the eye:
Find the general solution of
$
\bf{x}' = A\bf{x} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
, whereby $A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$.
Does this help?
3. Originally Posted by abender
The idea of variation of parameters is to seek the solution to
$\bf{x}' = A\bf{x} + \bf{b}$
in the form
$
\bf{v}_1(t)\bf{x}_1(t) + \bf{v}_2 (t)\bf{x}_2 (t) = F(t)\bf{v}(t)$
whereby $F(t) = \begin{bmatrix} x_1 & x_2 \end{bmatrix}$ is a fundamental matrix.
Note that the constant vector c of coefficients is replaced by an unknown function v(t).
What is the equation for our unknown function v(t) you ask?
It is fairly straight forward to derive:
$
(F\bf{v})' = F'\bf{v} + F\bf{v'} = A(F\bf{v}) + \bf{b} \implies
F\bf{v'} = \bf{b} \implies
\bf{v'} = F^{-1}\bf{b}$
.
IMPORTANT: You must make sure that $F^{-1}$ exists in order to solve for v !!!! (not factorial)
Finally, using integration, we have:
$
\bf{x}(t) = F(t) \int F(t)^{-1} \bf{b}(t)dt
$
I'll now rewrite the original problem in latex so it is more palatable to the eye:
Find the general solution of
$
\bf{x}' = A\bf{x} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
, whereby $A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$.
Does this help?
I believe this is an overkill for this particular system.
Originally Posted by matlabnoob
ok,so again, reading my lecture notes,i bumped into this problem..
consider
A = (1 2,
0 -1)
find the general solution of
x' = Ax + ( 0,1 )
write x = (x, y) then the ODE is :
x' = x+2y
y' = -y + 1
y' = -y+1 implies y(t) = Ce^-t + 1 (how did one obtain this?? ) (*)
The ODE for x(t) is x' = x + 2(C(e^-t) + 1) (**)
This linear inhomogeneous ODE gives x(t) = -2 - c(e^-t) + d(e^t) again how does one get this? i integrated but i cant get this. after trying for hours..i simply gave up ='(
thank you !
First question (*) above. It is a separable ODE - separate and integrate.
Second question (**) is linear with the integrating factor $\mu = e^{-t}$.
4. Or perhaps I have an easier way of doing this.
We have a matrix differential equation of the form
$
\bf{x}'(t) = A\bf{x}(t) + \bf{b}(t)$
whereby $A$ is a constant matrix.
Since $A$ is a constant matrix, if we can calculate $
e^{tA}$
, then we can find the solution to the system. So, we make $e^{tA}$ an integrating factor and multiply throughout:
$
e^{-tA} \bf{x}'(t) = e^{-tA} A\bf{x}(t) + e^{-tA} \bf{b}(t)$
$\implies
e^{-tA} \bf{x}'(t) - e^{-tA} A\bf{x}(t) = e^{-tA} \bf{b}(t)$
$\implies
\frac{d}{dt} \big{[} e^{-tA} \bf{x}(t) \big{]} = e^{-tA} \bf{b}(t)$
$
\implies \bf{x}(t) = e^{tA} \big{[} \int^t_0
e^{-\mu A} \bf{b}(\mu) d\mu
\big{]}$
5. thanks!!im still trying to follow through
im not v.good with things unless they're worked out directly from how they originally are.but i have to know the method anyway
6. You're welcome and I appreciate your appreciation.
Which part of the problem are having the most difficulty with? Is it the "setting things up" part? Crunching the integral? Just getting started??
-Andy
7. Originally Posted by abender
You're welcome and I appreciate your appreciation.
Which part of the problem are having the most difficulty with? Is it the "setting things up" part? Crunching the integral? Just getting started??
-Andy
the main things im having difficult with (i still dont know! spend the night trying to figure out. aah.. maths takes me forever to solve)
.. they are :
y' = -y+1 implies y(t) = Ce^-t + 1
ok so for this i integrated just the homogeneous equation first, which i took as dy/dt + y = 0 and i got Ce^-t from that! but how did the +1 get there too? =S.... im not sure how to solve this without making errors or forgetting the +1 ..
and.. when i tried to integrate..
x(t) is x' = x + 2(C(e^-t) + 1)
i couldnt get..
x(t) = -2 - c(e^-t) + d(e^t)
i tried integrating by parts... substitution... it wont work
8. Originally Posted by matlabnoob
the main things im having difficult with (i still dont know! spend the night trying to figure out. aah.. maths takes me forever to solve)
.. they are :
y' = -y+1 implies y(t) = Ce^-t + 1
ok so for this i integrated just the homogeneous equation first, which i took as dy/dt + y = 0 and i got Ce^-t from that! but how did the +1 get there too? =S.... im not sure how to solve this without making errors or forgetting the +1 ..
and.. when i tried to integrate..
x(t) is x' = x + 2(C(e^-t) + 1)
i couldnt get..
x(t) = -2 - c(e^-t) + d(e^t)
i tried integrating by parts... substitution... it wont work
First ODE $\frac{dy}{dt} = - (y-1)$ so $\frac{dy}{y-1} = - dt$ so $\ln | y-1 | = - t + \ln c$. Exponentiating gives $y = ce^{-t} + 1$.
Second ODE $\frac{dx}{dt} - x = 2\left(c e^{-t} + 1 \right)$. The integrating factor is
$\mu = e^{-t}$ so $e^{-t} \left( \frac{dx}{dt} - x\right) = e^{-t} 2\left(c e^{-t} + 1 \right)$.
Thus, $\frac{d}{dt} \left( x e^{-t} \right) = 2c e^{-2t} + 2 e^{-t}$ and integrating gives $x e^{-t} = -c e^{-2t} - 2 e^{-t} + d$. Isolating x gives $x = -c e^{-t} - 2 + de^{t}$
9. Good morning.
The question asks us to find the general solution to
$
\boldsymbol{x}'(t) = A\boldsymbol{x}(t) + \boldsymbol{b}(t)
$
whereby
$A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$ , $\boldsymbol{b}(t) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ .
Just in case this it is a source of confusion for you, I'll clarify that \boldsymbol{x} can be (and in this post, is) expressed as follows:
$
\boldsymbol{x}(t) = \begin{bmatrix} x \\ y \end{bmatrix}
$
.
Now I will generalize much less and try to spell out each step and computation for you. It will take a bit longer, but perhaps you need to see things this way.
We begin by plugging in what we know into
$
\boldsymbol{x}'(t) = A\boldsymbol{x}(t) + \boldsymbol{b}(t)
$
.
We get
$\begin{bmatrix} x' \\ y' \end{bmatrix} =
\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} +
\begin{bmatrix} 0 \\ 1 \end{bmatrix} =
$
$
\begin{bmatrix} x + 2y \\ -y \end{bmatrix} +
\begin{bmatrix} 0 \\ 1 \end{bmatrix} =
$
$
\begin{bmatrix} x+2y \\ -y +1 \end{bmatrix}$
.
So we have
$\begin{bmatrix} x' \\ y' \end{bmatrix} =
\begin{bmatrix} x+2y \\ -y +1 \end{bmatrix}$
.
Now we multiply both sides by the integrating factor $e^{-tA}$ .
$
e^{-tA} \begin{bmatrix} x' \\ y' \end{bmatrix} =
e^{-tA} \begin{bmatrix} x+2y \\ -y +1 \end{bmatrix}
$
Now, let's explain $e^{-tA}$ , which is called a matrix exponential. I'll give a general definition:
Let M be a an n x n matrix. $e^M$ is the matrix exponential of M , which is expressed by the power series
$e^{M} = \sum_{k=0}^\infty \frac{1}{k!} M^k$ .
Note that the exponential of M is well-defined, as the series always converges.
Are you following so far? | 2,859 | 8,327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 50, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-50 | longest | en | 0.785051 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=RegularChains/MatrixTools | 1,498,479,561,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00416.warc.gz | 573,665,539 | 41,883 | Overview of the RegularChains[MatrixTools] Subpackage - Maple Programming Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : MatrixTools Subpackage : RegularChains/MatrixTools
Overview of the RegularChains[MatrixTools] Subpackage
Calling Sequence RegularChains[MatrixTools][command](arguments) command(arguments)
Description
• The RegularChains[MatrixTools] subpackage is a collection of commands to manipulate matrices of polynomials modulo regular chains.
• The commands of the RegularChains[MatrixTools] subpackage are quite standard in terms of their goals: multiplication of matrices, computation of inverse, or lower echelon of a matrix. However, these commands are considered here in a non-standard context. Indeed, the coefficients of these matrices are polynomials and the computations are performed modulo (the saturated ideal of) a regular chain.
• The saturated ideal of a regular chain is not necessarily prime and working modulo such ideals may involve zero-divisors. However, computing with these zero-divisors is possible following the celebrated D5 Principle, after J. Della Dora, C. Discrescenzo, and D. Duval. The idea is the following. When a zero-divisor is encountered during a computation modulo a regular chain, you can split the computations into several cases (or branches) such that in each case this zero-divisor becomes either zero or a regular element (that is, an element that is not a zero-divisor). Therefore, this zero-divisor is not an obstacle for the computations in any of these branches.
• The main purpose of the commands of the RegularChains[MatrixTools] subpackage is to compute the inverse of the Jacobian matrix of a polynomial systems modulo (the saturated ideal of) a regular chain. This question arises, for example, in Hensel lifting techniques for triangular sets. See the paper Degree bounds and lifting techniques for triangular sets by E. Schost.
List of RegularChains[MatrixTools] Subpackage Commands
The following is a list of available commands.
Examples
> $\mathrm{with}\left(\mathrm{RegularChains}\right):$
> $\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$$\mathrm{with}\left(\mathrm{MatrixTools}\right);$$\mathrm{with}\left(\mathrm{ChainTools}\right):$
$\left[{\mathrm{IsZeroMatrix}}{,}{\mathrm{JacobianMatrix}}{,}{\mathrm{LowerEchelonForm}}{,}{\mathrm{MatrixInverse}}{,}{\mathrm{MatrixMultiply}}{,}{\mathrm{MatrixOverChain}}\right]$ (1)
Consider the following polynomial ring.
> $R≔\mathrm{PolynomialRing}\left(\left[x,y,z\right],101\right)$
${R}{≔}{\mathrm{polynomial_ring}}$ (2)
Define a regular chain T with a saturated ideal that is clearly non-prime.
> $E≔\mathrm{Empty}\left(R\right):$
> $T≔\mathrm{Chain}\left(\left[\left(z+1\right)\left(z+2\right),{y}^{2}+z,\left(x-z\right)\left(x-y\right)\right],E,R\right)$
${T}{≔}{\mathrm{regular_chain}}$ (3)
> $\mathrm{Equations}\left(T,R\right)$
$\left[{{x}}^{{2}}{+}\left({100}{}{y}{+}{100}{}{z}\right){}{x}{+}{z}{}{y}{,}{{y}}^{{2}}{+}{z}{,}{{z}}^{{2}}{+}{3}{}{z}{+}{2}\right]$ (4)
Since T possesses as many variables as polynomials, the saturated ideal of T is the ideal generated by T. Clearly, the following numbers are zero-divisors modulo this ideal: z+1, z+2, x-z, x-y, and also y-1 and y+1. Observe that if the command ListConstruct is used instead of Chain, then this ideal splits into four.
> $\mathrm{split}≔\mathrm{ListConstruct}\left(\left[\left(z+1\right)\left(z+2\right),{y}^{2}+z,\left(x-z\right)\left(x-y\right)\right],E,R\right)$
${\mathrm{split}}{≔}\left[{\mathrm{regular_chain}}{,}{\mathrm{regular_chain}}{,}{\mathrm{regular_chain}}{,}{\mathrm{regular_chain}}\right]$ (5)
> $\mathrm{map}\left(\mathrm{Equations},\mathrm{split},R\right)$
$\left[\left[{x}{+}{1}{,}{y}{+}{1}{,}{z}{+}{1}\right]{,}\left[{x}{+}{1}{,}{y}{+}{100}{,}{z}{+}{1}\right]{,}\left[{x}{+}{100}{,}{y}{+}{100}{,}{z}{+}{1}\right]{,}\left[{{x}}^{{2}}{+}\left({100}{}{y}{+}{2}\right){}{x}{+}{99}{}{y}{,}{{y}}^{{2}}{+}{99}{,}{z}{+}{2}\right]\right]$ (6)
Now construct a random matrix with coefficients in R.
> $m≔\mathrm{Matrix}\left(\left[\left[87{z}^{2}-56z,-73-62{z}^{2}+97z,-10-4{z}^{2}-83z,80+62{z}^{2}-82z\right],\left[-44z+71,-17z-75,-10z-7,-40z+42\right],\left[-92-50{z}^{3}+23{z}^{2}+75z,37+6{z}^{3}+74{z}^{2}+72z,29-23{z}^{3}+87{z}^{2}+44z,-61+98{z}^{3}-23{z}^{2}+10z\right],\left[11-8{z}^{3}-29{z}^{2}+95z,-81-49{z}^{3}-47{z}^{2}+40z,31+91{z}^{3}+68{z}^{2}-10z,1-51{z}^{3}+77{z}^{2}+95z\right]\right]\right)$
${m}{≔}\left[\begin{array}{cccc}{87}{}{{z}}^{{2}}{-}{56}{}{z}& {-}{62}{}{{z}}^{{2}}{+}{97}{}{z}{-}{73}& {-}{4}{}{{z}}^{{2}}{-}{83}{}{z}{-}{10}& {62}{}{{z}}^{{2}}{-}{82}{}{z}{+}{80}\\ {-}{44}{}{z}{+}{71}& {-}{17}{}{z}{-}{75}& {-}{10}{}{z}{-}{7}& {-}{40}{}{z}{+}{42}\\ {-}{50}{}{{z}}^{{3}}{+}{23}{}{{z}}^{{2}}{+}{75}{}{z}{-}{92}& {6}{}{{z}}^{{3}}{+}{74}{}{{z}}^{{2}}{+}{72}{}{z}{+}{37}& {-}{23}{}{{z}}^{{3}}{+}{87}{}{{z}}^{{2}}{+}{44}{}{z}{+}{29}& {98}{}{{z}}^{{3}}{-}{23}{}{{z}}^{{2}}{+}{10}{}{z}{-}{61}\\ {-}{8}{}{{z}}^{{3}}{-}{29}{}{{z}}^{{2}}{+}{95}{}{z}{+}{11}& {-}{49}{}{{z}}^{{3}}{-}{47}{}{{z}}^{{2}}{+}{40}{}{z}{-}{81}& {91}{}{{z}}^{{3}}{+}{68}{}{{z}}^{{2}}{-}{10}{}{z}{+}{31}& {-}{51}{}{{z}}^{{3}}{+}{77}{}{{z}}^{{2}}{+}{95}{}{z}{+}{1}\end{array}\right]$ (7)
> $d≔\mathrm{Determinant}\left(m\right)$
${d}{≔}{58241707}{}{{z}}^{{9}}{+}{19263260}{}{{z}}^{{8}}{-}{145068342}{}{{z}}^{{7}}{+}{169363187}{}{{z}}^{{6}}{+}{98898633}{}{{z}}^{{5}}{-}{333772968}{}{{z}}^{{4}}{+}{57745075}{}{{z}}^{{3}}{+}{54218412}{}{{z}}^{{2}}{+}{4321491}{}{z}{+}{3727553}$ (8)
First, compute the inverse of this determinant modulo T.
> $\mathrm{inv}≔\mathrm{Inverse}\left(d,T,R\right)$
${\mathrm{inv}}{≔}\left[\left[\left[{89}{}{z}{+}{55}{,}{1}{,}{\mathrm{regular_chain}}\right]{,}\left[{89}{}{z}{+}{55}{,}{1}{,}{\mathrm{regular_chain}}\right]{,}\left[{89}{}{z}{+}{55}{,}{1}{,}{\mathrm{regular_chain}}\right]{,}\left[{89}{}{z}{+}{55}{,}{1}{,}{\mathrm{regular_chain}}\right]\right]{,}\left[{}\right]\right]$ (9)
> $\mathrm{seq}\left(\mathrm{Equations}\left({{{\mathrm{inv}}_{1}}_{i}}_{3},R\right),i=1..\mathrm{nops}\left({\mathrm{inv}}_{1}\right)\right)$
$\left[{x}{+}{1}{,}{y}{+}{1}{,}{z}{+}{1}\right]{,}\left[{x}{+}{1}{,}{y}{+}{100}{,}{z}{+}{1}\right]{,}\left[{x}{+}{100}{,}{y}{+}{100}{,}{z}{+}{1}\right]{,}\left[{{x}}^{{2}}{+}\left({100}{}{y}{+}{2}\right){}{x}{+}{99}{}{y}{,}{{y}}^{{2}}{+}{99}{,}{z}{+}{2}\right]$ (10)
You can also reduce the matrix m in the first place.
> $\mathrm{MatrixOverChain}\left(m,T,R\right)$
$\left[\left[\begin{array}{cccc}{87}{}{z}{+}{28}& {81}{}{z}{+}{51}& {30}{}{z}{+}{99}& {35}{}{z}{+}{57}\\ {57}{}{z}{+}{71}& {84}{}{z}{+}{26}& {91}{}{z}{+}{94}& {61}{}{z}{+}{42}\\ {60}{}{z}{+}{67}& {94}{}{z}{+}{26}& {26}{}{z}{+}{20}& {58}{}{z}{+}{68}\\ {25}{}{z}{+}{21}& {40}{}{z}{+}{22}& {19}{}{z}{+}{37}& {12}{}{z}{+}{46}\end{array}\right]{,}{\mathrm{regular_chain}}\right]$ (11)
Then, compute the inverse of the reduced matrix modulo T. You can see that the computation splits in the same manner as the inverse of the determinant below.
> $\mathrm{invm}≔\mathrm{MatrixInverse}\left(m,T,R\right)$
${\mathrm{invm}}{≔}\left[\left[\left[\left[\begin{array}{rrrr}{87}& {82}& {90}& {70}\\ {11}& {13}& {63}& {44}\\ {34}& {44}& {63}& {91}\\ {4}& {17}& {70}& {22}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{rrrr}{87}& {82}& {90}& {70}\\ {11}& {13}& {63}& {44}\\ {34}& {44}& {63}& {91}\\ {4}& {17}& {70}& {22}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{rrrr}{87}& {82}& {90}& {70}\\ {11}& {13}& {63}& {44}\\ {34}& {44}& {63}& {91}\\ {4}& {17}& {70}& {22}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{rrrr}{39}& {47}& {34}& {34}\\ {46}& {86}& {32}& {70}\\ {81}& {75}& {84}& {31}\\ {57}& {35}& {32}& {38}\end{array}\right]{,}{\mathrm{regular_chain}}\right]\right]{,}\left[{}\right]\right]$ (12)
> $\mathrm{seq}\left(\mathrm{Equations}\left({{{\mathrm{invm}}_{1}}_{i}}_{2},R\right),i=1..\mathrm{nops}\left({\mathrm{inm}}_{1}\right)\right)$
$\left[{x}{+}{1}{,}{y}{+}{1}{,}{z}{+}{1}\right]$ (13)
Finally, compute the low echelon form of m modulo T.
> $\mathrm{lem}≔\mathrm{LowerEchelonForm}\left(m,T,R\right)$
${\mathrm{lem}}{≔}\left[\left[\left[\begin{array}{cccc}{98}& {0}& {0}& {0}\\ {33}& {42}& {0}& {0}\\ {76}& {90}& {20}& {0}\\ {-}{8}{}{{z}}^{{3}}{-}{29}{}{{z}}^{{2}}{+}{95}{}{z}{+}{11}& {-}{49}{}{{z}}^{{3}}{-}{47}{}{{z}}^{{2}}{+}{40}{}{z}{-}{81}& {91}{}{{z}}^{{3}}{+}{68}{}{{z}}^{{2}}{-}{10}{}{z}{+}{31}& {-}{51}{}{{z}}^{{3}}{+}{77}{}{{z}}^{{2}}{+}{95}{}{z}{+}{1}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{cccc}{98}& {0}& {0}& {0}\\ {33}& {42}& {0}& {0}\\ {76}& {90}& {20}& {0}\\ {-}{8}{}{{z}}^{{3}}{-}{29}{}{{z}}^{{2}}{+}{95}{}{z}{+}{11}& {-}{49}{}{{z}}^{{3}}{-}{47}{}{{z}}^{{2}}{+}{40}{}{z}{-}{81}& {91}{}{{z}}^{{3}}{+}{68}{}{{z}}^{{2}}{-}{10}{}{z}{+}{31}& {-}{51}{}{{z}}^{{3}}{+}{77}{}{{z}}^{{2}}{+}{95}{}{z}{+}{1}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{cccc}{98}& {0}& {0}& {0}\\ {33}& {42}& {0}& {0}\\ {76}& {90}& {20}& {0}\\ {-}{8}{}{{z}}^{{3}}{-}{29}{}{{z}}^{{2}}{+}{95}{}{z}{+}{11}& {-}{49}{}{{z}}^{{3}}{-}{47}{}{{z}}^{{2}}{+}{40}{}{z}{-}{81}& {91}{}{{z}}^{{3}}{+}{68}{}{{z}}^{{2}}{-}{10}{}{z}{+}{31}& {-}{51}{}{{z}}^{{3}}{+}{77}{}{{z}}^{{2}}{+}{95}{}{z}{+}{1}\end{array}\right]{,}{\mathrm{regular_chain}}\right]{,}\left[\left[\begin{array}{cccc}{78}& {0}& {0}& {0}\\ {48}& {12}& {0}& {0}\\ {68}& {15}& {56}& {0}\\ {-}{8}{}{{z}}^{{3}}{-}{29}{}{{z}}^{{2}}{+}{95}{}{z}{+}{11}& {-}{49}{}{{z}}^{{3}}{-}{47}{}{{z}}^{{2}}{+}{40}{}{z}{-}{81}& {91}{}{{z}}^{{3}}{+}{68}{}{{z}}^{{2}}{-}{10}{}{z}{+}{31}& {-}{51}{}{{z}}^{{3}}{+}{77}{}{{z}}^{{2}}{+}{95}{}{z}{+}{1}\end{array}\right]{,}{\mathrm{regular_chain}}\right]\right]$ (14)
References
Aubry, P.; Lazard, D.; and Moreno Maza, M. "On the Theories of Triangular Sets." Journal of Symbolic Computation, (July/August 1999): 105-124.
Boulier, F. and Lemaire, F. "Computing Canonical Representatives of Regular Differential Ideals." Proceedings of ISSAC 2000, pp. 38-47. 2000.
Dahan, X.; Moreno Maza, M.; Schost, E.; Wu, W.; and Xie, Y. "Equiprojectable Decompositions of Zero-Dimensional Varieties." Proceedings of ICPSS 2004, pp. 69-71. 2004.
Della Dora, J.; Discrescenzo, C.; and Duval, D. "About a New Method for Computing in Algebraic Number Fields." Proceedings of EUROCAL 1985, pp. 289-290. 1985.
Hubert, E. "Notes on Triangular Sets and Triangulation-Decomposition Algorithms." Lecture Notes in Computer Science, Vol. 2630, (2003).
Kalkbrener, M. Three Contributions to Elimination Theory. PhD Thesis, University of Linz, Austria, 1991.
Lazard, D. "Solving zero-dimensional algebraic systems" J. Symb. Comp., Vol. 13, (1992): 117-133.
Moreno Maza, M. "On Triangular Decompositions of Algebraic Varieties." Proceedings of MEGA 2000. 2000.
Moreno Maza, M. and Rioboo, R. "Polynomial GCD computations over towers of algebraic extensions." In proc. of AAECC-11, Lecture Notes in Comput. Sci, Vol. 948, Springer, (1995). | 4,462 | 11,089 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-26 | latest | en | 0.83932 |
http://www.docstoc.com/docs/22735577/Discriminant-Analysis-Concepts | 1,386,801,675,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164061354/warc/CC-MAIN-20131204133421-00002-ip-10-33-133-15.ec2.internal.warc.gz | 327,663,681 | 18,294 | # Discriminant Analysis Concepts
Document Sample
``` Discriminant Analysis Concepts
• Used to predict group membership from a
set of continuous predictors
• Think of it as MANOVA in reverse – in
MANOVA we asked if groups are
significantly different on a set of linearly
combined responses.
• The same responses can be used to predict
group membership.
Discriminant Analysis Concepts
• Determine how can continuous variables be
linearly combined to best classify a subject
into a group.
• A better term may be “separation.”
• Slightly different is “classification” when
we seek rules that allocate new subjects into
established classes.
• Logistic regression is a competitor.
Classification
• Two populations π1 and π2
• We have measurements x' = [x1 x2 . . .xp]
on each of the individuals concerned.
• Given a new value of x for an unknown
individual our problem is how we can best
classify this individual.
Illustration
f1(x)
f2(x)
R1 R2
Probability of misclassifying Probability of misclassifying
Population 2 member in Population 1 member in
Population 1 Population 2
Misclassification
The probability an individual from π1 is wrongly
classified is
f (x)dx = P(2 | 1)
1
R2
and an individual from π2 is wrongly classified is:
f (x)dx
2 = P(1 | 2)
R1
Four Possibilities
• Assume p1 and p2 are the prior probabilities
of p1 and p2, respectively.
Classified
p1 p2
p1 P(1|1)p1 P(2|1)p1
Actual
p2 P(1|2)p2 P(2|2)p2
Costs
• In general there is a cost associated with
misclassification.
• Assume the cost is zero for correct classification.
• C(21) as the cost of misclassifying a π1
individual as a π2 individual.
• C(12) as the cost of misclassifying a π2
individual as a π1 individual.
Classified
p1 p2
p1 0 c(2|1)
Actual p c(1|2) 0
2
Expected Cost of
Misclassification (ECM)
ECM = c(2|1)P(2|1)p1 + c(1|2)P(1|2)p2
Goal: To minimize ECM
It can be shown that the ECM is minimized if R1
contains those values of for which
C(1|2)p2f2(x) -C(2|1)p1f1(x) 0
and excludes those x for which the above is > 0.
In other words R1 is the set of points x for
which:
f1(x) p2 C(1|2)
>
f2(x) p1 C(2|1)
So when x satisfies this inequality we would
classify the corresponding individual in p1.
Conversely since R2 is the complement of R1
R2 is the set such that:
f1(x) p2 C(1|2)
<
f2(x) p1 C(2|1)
and an individual whose x vector satisfied this
inequality would be allocated to p2.
Assuming x has a multivariate normal
distribution i.e.
x Np(µi , ) in population i (i =1,2)
(note that this implies the same covariance
matrix applies to each population) we have
f1(x) exp[-1/2(x-m1)‟-1(x-m1)]
=
f2(x) exp[-1/2(x-m2)‟-1(x-m2)]
and the general rule (1) after taking natural log‟s and
some rearrangement can be shown to be equivalent to
'x - '( µ1 + µ2 ) c
2
where = -1( µ1 - µ2 ) = 1
2
.
.
p say
(Correspondingly ‟ = [1,2….,p] = (m1-m2)‟-1 )
and c = ln C(1|2) p2
[ C(2|1) p1
]
Priors
• Typically, information is not available on the prior
probabilities p1 and p2.
• Typically taken to be the same making c a function
of only the ratio of the two costs.
• In addition, if the misclassification costs, C(1|2) and
C(2|1), are the same then c = 0.
Ordinarily , µ1, µ2 are not known and need to
be estimated from the data by S, x1 and x2
respectively and we therefore use:
S-1( x1 - x2 ) for etc
where S-1 is taken as the inverse of
Spooled = (n1-1)S1 +(n2-1)S2
(n1+n2-2)
Where S1 and S2 are the sample covariance
matrices for the each of the two groups
(populations) respectively.
Minimum ECN for Two Normals
Allocate xo to p1 for which:
'xo - 1 „(x1+ x2) c
2
Linear Discriminant Function
’ x = ( µ1 - µ2 ) -1x is called the linear
discriminant function of x.
This linear combination of x summarizes
all of the possible information in x that is
available for discriminating between the
two populations.
Unequal Covariance Matrices
Allocate xo to p1 for which:
-0.5xo’ (S1-1 - S2-1 ) xo + (x1S1-1 – x2S2-1 ) xo – k c
where k = 0.5ln(|S1| |S2|-1 ) xo + (x1’S1-1x1 – x2’S2-1 x2)
Fisher’s Discriminant Function
Allocate xo to p1 for which:
(x1-x2) SPooled-1 xo 0.5 (x1-x2) SPooled-1 (x1+ x2)
Note: The p-variate standard a ' x1 - x 2
X x1 , x 2 max
a a
distance between two vectors is 1/ 2
aR p t
defined as: a0
For this problem maximized at a = SPooled-1 (x1-x2)
Linear Discriminant Function,
Alternative View
The linear combination of x, say y x is called a linear
t
discriminant function if .
y β μ1, β' μ2 X μ1 , μ2
'
Example
4 0 1 - 1
Suppose , m1 0 , m 2 0 .
0 9
Example
Example of Linear Discriminant Function
4
The unscaled
3
0.5
2
β
1 0.0
0
-1
-2
-3
-4
-4 -3 -2 -1 0 1 2 3 4
Example
Example With Correlation
4
Correlation of 0.6
0.7813
3 The unscaled β .
2
- 0.3125
1
How separated are the “mean
0 scores” when projected onto
this line? (ans: 1.25 units)
-1
-2
How separated are the “mean
-3 scores” when projected onto
the x-axis? (ans: 1.0 units)
-4
-4 -3 -2 -1 0 1 2 3 4
More Than Two Groups
Sample variance/covariance matrix
g ni
1
SX
n-1 i 1 j 1
(x ij - x)(x ij - x)t
Among-Groups sums of squares and
cross-products matrix
g
H n i (xi - x)(xi - x) t
i 1
Pooled Within-Groups sums of squares and
cross-products matrix
g ni
E (x ij - x i )(x ij - x i )t
i 1 j 1
More Than Two Groups
(n - 1)S X H E
•Note: Same decomposition we
used with MANOVA
More Than Two Groups
Let X, Sx, E and H be defined as above. Suppose E is
positive definite and denote by A a1 a 2 a k -1 . Then,
(a t Ha)
arg max t ak ,
aR
p (a Ea)
a t S A 0 t
x
where a k is the eigenvector of E -1 H corresponding to
the kth largest eigenvalue.
Referred to as canonical discriminant analysis
Canonical Correlation Analysis
• A statistical technique to identify and measure
the association between two sets of variables.
• Multiple regression can be interpreted as a
special case of such an analysis.
• The “multiple correlation coefficient,” R, can
be thought of as the maximum correlation that
is attainable between the dependent variable
and a linear combination of the independent
variables
Canonical Correlation Analysis
• CCA is an extension of the multiple R in
multiple regression.
• In CCA, there can be multiple response
variables.
• Canonical correlations are the maximum
correlation between a linear combination of
the responses and a linear combination of the
predictor variables.
Canonical Correlations
Suppose
x1 m1 11 12
x N ,
x2 m2 21 22
where x1=(x11,…,x1q) and x2=(x21,…,x2,p-q).
Note that Var(x1) = 11 is qq, Var(x2) = 22 is
(p-q)(p-q), Cov(x1,x2) = 12 is q(p-q), Cov(x2,x1) = 21 is
(p-q)q, and 12 = 21
The First Canonical Correlation
• Find a1 and b1 (vectors of constants) such
that corr(a1x1,b1x2 ) is large as possible.
• Let U1= a1x1 and V1= b1x2 and call them
canonical variables.
• Then Var(U1) = a111a1 ,
Var(V1) = b122b1 ,
and Cov(U1,V1) = a112b1 .
The First Canonical Correlation
The correlation between U1 and V1 is
Cov(U1,V1 )
Corr(U1,V1 )
Var(U1 ) Var(V1 )
a112b1
a111a1 b122b1
Finding the Correlation
Let 1 = max corr(U1,V1 ) . It can be shown that
a1 0,b1 0
-1 -1
1 is the largest eigenvalue of 11122221
2
2
a1 is the eigenvector corresponding to1
b1 is the eigenvector corresponding to the largest eigenvalue of
-1 -1
22211112 - this largest eigenvalue also is 1
2 .
Note that 011.
```
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# GMAT Diagnostic Test Question 13
Author Message
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Joined: 20 Dec 2010
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Re: GMAT Diagnostic Test Question 12 [#permalink]
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04 Sep 2011, 08:03
sl750 wrote:
Can someone explain this step
(1) Suppose if 1/x > 1/y:
(1/x $1/y) = 1/x + 1/y = (x+y)/(xy). (1/y$ 1/x) = 1/x - 1/y = (y-x)/(xy). . Why does the sign change here? We are given that, for A>B A$B = A+B. So how does the sign change when we flip the quantities around the operator$
Thanks
Signs change because the stem presents a scenario for LHS<RHS as well. See the enlarged text below from the question stem.
If A>B, A$B = A+B and if A<B, A$B = B-A, then which of the followings is the highest for $$(\frac{1}{x}\frac{1}{y})(\frac{1}{y}\frac{1}{x})$$
BTW, We get the maximum value for the expression when,
$$\frac{1}{x}>\frac{1}{y}$$;-------Condition 1
For all the options in which Condition 1 hold true, we need to find the highest value for $$\frac{1}{x}$$.
D satisfies both of those conditions.
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Re: GMAT Diagnostic Test Question 12 [#permalink]
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04 Sep 2011, 11:18
A) (2$3)$(3$2) = 1$5 = 4
B) (3$4)$(4$3) = 1$7 = 6
c) (4$5)$(5$4) = 1$9 = 8
D) (5$4)$(4$5) = 9$1 = 10
E) (4$2)$(2$4) = 6$2 = 8
hence D
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Re: GMAT Diagnostic Test Question 12 [#permalink]
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29 Jun 2012, 07:52
LUGO wrote:
gmatoverduegirl wrote:
LUGO wrote:
Actually something that you could do for sure under two minutes is the following:
Possible combinations are:
+ + - = 2B
- + + = 2B
+ - - = 2A
- - + = - 2A
Clearly you can not have a negative answer for greatest? nor two choices with the same answer 2B? Hence the right answer has to be of the type 2A meaning => [(A>B) = A+ B] - [(A<B) = B - A] = [5 + 4] - [4 - 5] = 9 + 1 = 10.
It took me 18 seconds
Lugo can you please tell me how you came with the answers 2B, 2B, 2A and -2A? sorry but i still don't get it
All answers are positive real numbers, right? So there is no need to worry about adding/subtracting negative numbers. Also all answers can be converted to integer numbers since the formula to solve is made of inverse number of reals which are actually integer numbers i.e. 1 / (1/4) is 4 and 1 / (1/5) is 5 right?
Next, the formula to solve is (bracket_1) +/- (bracket_2) where either bracket is a combination of adding/subtracting 2 integer numbers and given to you in answers A,B,C,D and E. Think about all possible combinations of bracket_1 and bracket_2 in terms of sign. Given that you have two brackets and two possible signs (2^2) = 4 X 2 cases = 8 cases. However eliminate the ones where both bracket_1 and bracket_2 are equal, i.e. A>B and A>B or A<B and A<B since, from the answers provided and formula written in the question, no two real numbers can be the same, right?
This means that you are left with only 4 possible cases:
Bracket_1 +/- Bracket_2
+ + - = (A>B) + (A<B) = A+B+B-A = 2B
- + + = (A<B) + (A>B) = B-A+A+B = 2B
+ - - = (A>B) - (A<B) = A+B-B+A = 2A
- - + = (A<B) - (A>B) = B-A-A-B = -2A
Next the question ask you for greatest. Notice than from the cases above 2A and 2B can be greater than any other combination, correct? However 2B is duplicated twice so which one do you choose? It has to be 2A since each question can only have one possible answer. Besides try few cases from the answers provided and see what happens.
I hope it helps....
LUGO
Lugo, not sure everything is right here.
Some cases can never happen and you have to switch the order of the brackets when the second is negative.
Thus we are only left with (only possible cases as "ra011y" explained it the same way in page 1) :
A>B (i.e. B<A) :
==> Bracket_1 (A+B) > Bracket_2 (A-B) ==> Bracket_1 + Bracket_2
==> (+) + (-) = (A>B) $(B<A) = (A>B) + (B<A) = (A+B) + (A-B) = A+B+A-B = 2A A<B (i.e. B>A) : ==> Bracket_1 (B-A) < Bracket_2 (A+B) ==> Bracket_1 - Bracket_2 ==> (-) - (+) = (A<B)$ (B>A) = (B>A) - (A<B) = (B+A) - (B-A) = B+A-B+A = 2A
Hope it's clear.
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Re: GMAT Diagnostic Test Question 12 [#permalink]
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14 Aug 2013, 14:12
Is there an "intuitive" way to solve this problem quickly? Seems like a perfect candidate for that school of GMAT problem solving. While this problem isn't very difficult, I didn't really solve it intuitively and did it right at the two minute mark. Ended up utilizing the pattern with A, B, and C, which reverses with D and E.
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Re: GMAT Diagnostic Test Question 12 [#permalink] 14 Aug 2013, 14:12
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# GMAT Diagnostic Test Question 13
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Reduce the Problem: Permutations and Modulo Arithmetic
# Reduce the Problem: Permutations and Modulo Arithmetic
We’ve all seen permutations before. If you have ten distinct items, and rearrange them on a shelf, you’ve just performed a permutation. A permutation is actually a function that is performing the arrangement on a set of labeled objects. For simplicity, we can just number the objects and work with permuting the numbers.
If we just look at the object numbers $1,\ldots,n$, then we can just call the set of the first $n$ integers under the operation permutation forms the symmetric group $S_{n}$. These groups are so famous they get their own name. These groups are structurally similar to many other groups that the study of these runs throughout mathematics. Here we’ll take a look at permutations in a bit more depth and see that a certain type of permutation called a cycle behaves exactly the same as integers under modulo arithmetic.
## Notation: how do we look at permutations?
Permutations of integers are done by a function $\alpha$ that maps some $i$ in $\{1,2,\ldots,n\}$ to another $j$ in $\{1,2,\ldots,n\}$. Every number in the set must get mapped somewhere in the set, though the permutation can hold some, all, or no numbers fixed. All numbers that stay fixed gives us the identity permutation, which we will call $\epsilon$
We use a certain kind of notation for permutations that looks like this example of a permutation of the numbers 1, 2, and 3.
$$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}$$
The top row is the original set in order. The numbers in the bottom row beneath each top number tell you where the permutation sends each original number. So, the above permutation maps $1 \to 2$, $2 \to 3$, and $3 \to 1$
## Multiple permutations
Now, “products” of permutations work like compositions of functions. You perform the first permutation, and then the next permutation is performed on the result of the first. When we do this, we work right to left, just like in composition of functions.
Quick example: if $f(x)=x+5$ and $g(x) = 2x$, then $f(g(x))$ is calculated by first applying $g$, then inserting that result into $f$. So,
$$f(g(2)) = f(2\cdot2) = 2\cdot2 + 5 = 9$$
Now, let’s suppose we have two permutations on the integers $\{1,2,3\}$: $\alpha = \begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}$ and $\beta = \begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}$. What is $\beta\alpha$?
We can simply follow a chain. Take 1 for a start. If we apply the permutation $\alpha$, and then the permutation $\beta$, where does 1 end up?
Well, $\alpha(1) = 2$. Now we apply $\beta$ to the result of $\alpha$ acting on 1, which is $\alpha(1) = 2$. Then
$$\beta\alpha(1) = \beta(\alpha(1)) =\beta(2) = 3$$
So after applying these two permutations, 1 is sent to 3. We can do this for the other two numbers, and we get that
$$\beta\alpha =\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix} =\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$
## Cycles
Now we’ll introduce another concept and notation-that of cycles. A cycle is written just using parentheses instead of the larger notation we used before1.
The cycle $(123)$ denotes exactly that. Reading left to right, 1 maps to 2, 2 maps to 3, and 3 cycles back around to 1. So $(123)$ is actually just another way of expressing $\alpha = \begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}$
One other note about the shorthand: every permutation on $n$ integers can be written as the product of these disjoint cycles.2 (134) and (25) are disjoint cycles. We’re going to study powers of a single cycle. Powers of cycles work just like powers of numbers: just as $2^3$ is $2\cdot2\cdot2$, $(134)^{3} = (134)(134)(134)$3
## Examples of Powers of Cycles
Let’s take an example of a cycle: $\alpha = (123)$ and just compute some powers of it.
$$\alpha^{2} =(123)(123)=\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}=\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}=(132)$$
Now, $\alpha^{3} = \alpha\alpha^{2}$ (again note that we apply another “round” of permutations on the left), and can be calculated in a similar way:
$$\alpha^{3} = (123)(132)=\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}=\epsilon$$
Interesting notion here. Now we’re back to the identity permutation. That means if we multiply the cycle $(123)$ on the left of the identity permutation $\epsilon$, we just get $(123)$. In other words, $(123)^{4} = (123)$. Then that means $(123)^{5} = (123)(123)^{4} = (123)(123) = (123)^{2}$, and $(123)^{6} = \epsilon$. Our powers of this cycle form another “cycle” that’s starting to look awfully similar to something we’ve seen before: modulo addition.
## Modulo addition and our permutations
The great thing about studying algebra is that by studying the structure of things, we can observe that two groups that looked totally unrelated actually behave very similarly. Notice that the power of a cycle of length 3, like the ones we computed above, “start over” at powers that are multiples of 3: $(123)^{3} = (123)^{6} = \epsilon$. So if we paired the cycle $(123)$ with 1, $(123)^{2}$ with 2, and $(123)^{3}$ with 0, then the behavior of this cycle under its “permutation multiplication” is identical to the behavior of $\mathbb{Z}_{3}$, the integers modulo 3, that we have previously discussed.
To extend a bit further, since we can obvious take more powers of the cycle $(123)$ than just stopping at 3, if we wanted to make the correspondence to $\mathbb{Z}_{3}$, all powers of $(123)$ that are multiples of 3 (which evaluate to the identity permutation) would correspond to 0 (the identity element under modulo addition). The powers of $(123)$ that leave a remainder of 3 when you divide the power by 3 would behave like $1 \in \mathbb{Z}_{3}$, and the powers of $(123)$ that leave a remainder of 2 would match with $2 \in \mathbb{Z}_{3}$
To write this mathematically, we can actually map the set of all powers of $(123)$ or any cycle of length 34 to $\mathbb{Z}_{3}$
Let $n$ be the power of a cycle of length 3, which we will now call $(a_{1}a_{2}a_{3})$ to generalize. Then $(a_{1}a_{2}a_{3})^{n}$ maps to $n \bmod 3$. With this, we’ve connected the two groups and learned that powers of cycles of length 3 act just like integers modulo 3.
## One more extension
We just discovered that cycles of length 3 under powers (multiplying by itself) act just like integers modulo 3 under modulo addition. The next question we can ask is:
### Does this hold true for cycles of any length?
That is, if we have a cycle of length $s$, does it behave like integers modulo $s$? In fact, it does. We’re going to get a little more general here, so stick with me.
Take a generic cycle of length $s$, and call it $\beta = (b_{1}b_{2}\ldots b_{s})$ where each $b_{i}$ is some number or element in the permutation cycle. Then we’re going to start describing distinct powers of this $\beta$ and watch what happens.
\begin{aligned}\beta &=\begin{pmatrix}b_{1}&b_{2}&b_{3}&\ldots&b_{s}\\b_{2}&b_{3}&b_{4}&\ldots & b_{1}\end{pmatrix}\\\beta^{2} &=\begin{pmatrix}b_{1}&b_{2}&b_{3}&\ldots&b_{s}\\b_{3}&b_{4}&b_{5}&\ldots & b_{2}\end{pmatrix}\\\beta^{3} &=\begin{pmatrix}b_{1}&b_{2}&b_{3}&\ldots&b_{s}\\b_{4}&b_{5}&b_{6}&\ldots & b_{3}\end{pmatrix}\\&\vdots\\\beta^{s-1} &=\begin{pmatrix}b_{1}&b_{2}&b_{3}&\ldots&b_{s}\\b_{s}&b_{1}&b_{2}&\ldots & b_{s-1}\end{pmatrix}\\\beta^{s} &=\begin{pmatrix}b_{1}&b_{2}&b_{3}&\ldots&b_{s}\\b_{1}&b_{2}&b_{3}&\ldots & b_{s}\end{pmatrix}=\epsilon\end{aligned}
If this looks intimidating, just follow what $b_{1}$ gets permuted to as you take more powers. applying one $\beta$ to $b_{1}$ gives $\beta(b_{1}) = b_{2}$. Then $\beta^{2}(b_{1}) = \beta(\beta(b_{1})) = \beta(b_{2}) = b_{3}$. Next, $\beta^{3}(b_{1}) = b_{4}$. Now we start noticing the pattern, just as we did for the cycles of length 3: $\beta^{n}(b_{1}) = b_{1+n}$ while $n < s$, the length of our cycle. Then $\beta^{s}(b_{1}) = b_{1}$ again and we’re now starting over.
Do the same thing for any other element in the permutation. Follow it through the powers of $\beta$ and notice that $\beta^{n}(b_{j}) = b_{j+n}$ for any generic $j$. Now, when we take powers of the cycle that are larger than $s$, we just “start over again”. So $\beta^{s+1} = \beta$, $\beta^{s+2} = \beta^{2}$, and so forth. So $\beta^{n} = \beta^{n\mod s}$ in general, whether $n \geq s$ or $n < s$
Now that means we have made the same observation in general that we did for cycles of length 3. We can map powers of cycles of any length $s$ to integers modulo $s$, and both groups will behave the same way, with the same “starting over” property we observed in modulo arithmetic.
## Conclusion
What you’ve just discovered is a mathematical phenomenon called an isomorphism. We’ll discuss the formalities of this in another post. In a nutshell, an isomorphism is the mapping that lets us connect two groups that behave the same way, even if their elements and respective operations have nothing to do with each other. Picture it like two people running with the same stride length and form. Even if they are running in different locations or opposite directions, they’re still behaving the same way. Two things that are structurally and behaviorally equivalent, mathematically, are called isomorphic. We just proved (albeit a bit informally) that $\mathbb{Z}_{s}$, is isomorphic to permutation cycles of length $s$
Modulo arithmetic is much easier to compute than powers of permutations, so if we want to study how they work, we can actually study an isomorphic group that is simpler to get the same results. As a more tangible example, molecular structures and symmetries can get really icky to study and play with. There are simpler groups that are isomorphic to these structures and their symmetries that make their study much easier. For instance, instead of having to build a model and study what happens if you rotate it twice and reflect it across its vertical symmetry once, we can map those actions to a simpler algebraic structure with a simpler multiplication, calculate the result, then map it back using the isomorphism.
Abstract mathematics allows us to strip the more complex “fluff” from a problem and reduce it to its skeleton. By looking at the skeletons, we can study the problem’s mechanics more simply, and even reduce the problem to an isomorphic one that has been previously solved. | 2,986 | 10,399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 95, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-39 | latest | en | 0.854473 |
http://spotidoc.com/doc/152756/the-geometry-of-a-circle | 1,540,285,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516123.97/warc/CC-MAIN-20181023090235-20181023111735-00167.warc.gz | 325,214,417 | 9,889 | # The geometry of a circle
```The geometry of a
circle
mc-TY-circles-2009-1
In this unit we find the equation of a circle, when we are told its centre and its radius. There are
two different forms of the equation, and you should be able to recognise both of them. We also
look at some problems involving tangents to circles.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• find the equation of a circle, given its centre and radius;
• find the centre and radius of a circle, given its equation in standard form;
• find the equation of the tangent to a circle through a given point on its circumference;
• decide whether a given line is tangent to a given circle.
Contents
1. Introduction
2
2. The equation of a circle centred at the origin
2
3. The general equation of a circle
4
4. The equation of a tangent to a circle at a given point
7
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1. Introduction
The circle is a familiar shape and it has a host of geometric properties that can be proved using
the traditional Euclidean format. But it is sometimes useful to work in co-ordinates and this
requires us to know the standard equation of a circle, how to interpret that equation and how
to find the equation of a tangent to a circle. This video will explore these particular facets of a
circle, using co-ordinate geometry.
2. The equation of a circle centred at the origin
The simplest case is that of a circle whose centre is at the origin. Let us take an example. What
will be the equation centred on the origin with radius 5 units?
5
P(x, y)
5
y
x
O
−5
N
5
−5
If we take any point P (x, y) on the circle, then OP = 5 is the radius of the circle. But OP is also
the hypotenuse of the right-angled triangle OP N, formed when we drop a perpendicular from P
to the x-axis. Now in the right-angled triangle, ON = x and NP = y. Thus, using the theorem
of Pythagoras,
x2 + y 2 = 52 = 25 .
And this equation is true for any point on the circle. For instance, we could take a point Q(x1 , y1 )
5
Q(x1, y 1)
y1
−5
N
5
x1
O
5
−5
Once again, we can drop a perpendicular from Q to the x-axis. And now we can use the rightangled triangle OQN to see that x21 + y12 = 52 . So the co-ordinates (x1 , y1 ) of the point Q also
satisfy the equation x2 + y 2 = 25.
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We shall now take the radius of the circle to be r.
r
P(x, y)
r
y
x
−r
O
N
r
−r
If we take any point P (x, y) on the circle, then OP = r is the radius of the circle. But OP is also
the hypotenuse of the right-angled triangle OP N, formed when we drop a perpendicular from P
to the x-axis. In the right-angled triangle, ON = x and NP = y. Thus, using the theorem of
Pythagoras,
x2 + y 2 = r 2 ,
and this is the equation of a circle of radius r whose centre is the origin O(0, 0).
Key Point
The equation of a circle of radius r and centre the origin is
x2 + y 2 = r 2 .
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3. The general equation of a circle
What is the equation of a circle of radius r, centred at the point C(a, b)?
P(x, y)
r
y−b
C(a, b)
N
x−a
O
We shall take a horizontal line through the centre C and drop a perpendicular from P to meet
this horizontal line at N. Then again we have a right-angled triangle CP N, where CP = r is
the hypotenuse, and where we have CN = x − a and P N = y − b. Thus using Pythagoras again
we have
CN 2 + P N 2 = CP 2 ,
so that
Expanding the brackets gives
(x − a)2 + (y − b)2 = r 2 .
x2 − 2ax + a2 + y 2 − 2by + b2 = r 2 ,
and if we bring r 2 to the left-hand side and rearrange we get
x2 − 2ax + y 2 − 2by + a2 + b2 − r 2 = 0 .
It is a convention, at this point, to replace −a by g and −b by f . This gives
x2 + 2gx + y 2 + 2f y + g 2 + f 2 − r 2 = 0 .
Now look at the last three terms on the left-hand side, g 2 + f 2 − r 2 . These do not involve x or y
at all, so together they just represent a single number that we can call c. Substituting this into
the equation finally gives us
x2 + 2gx + y 2 + 2f y + c = 0 .
This is the general equation of a circle. We can recognise it, because it is quadratic in both x
and y, and it has two additional properties. First, there is no term in xy. And secondly, the
coefficient of x2 is the same as the coefficient of y 2. (In fact, our equation has both coefficients
equal to 1, but you can always multiply an equation by a non-zero constant to obtain another
valid equation, and so we must allow for this possibility.) The centre of the circle is then at
(a, b) = (−g, −f ) and, since c = g 2 + f 2 − r 2 , we have
r2 = g 2 + f 2 − c ,
so that the radius of the circle is given by
r=
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p
g2 + f 2 − c .
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Key Point
The general equation of a circle is
x2 + y 2 + 2gx + 2f y + c = 0 ,
p
where the centre is given by (−g, −f ) and the radius by r = g 2 + f 2 − c. The equation can
be recognised because it is given by a quadratic expression in both x and y with no xy term,
and where the coefficients of x2 and y 2 are equal.
Example
Find the centre and radius of the circle
x2 + y 2 − 6x + 4y − 12 = 0 .
Solution
First, we can check that the expression on the left-hand side is quadratic, that there is no term
involving xy, and that the coefficients of x2 and y 2 are equal. So this is the equation of a circle.
If we compare this equation with the standard equation
x2 + y 2 + 2gx + 2f y + c = 0 ,
we see that g = −3 and f = 2, so that the centre is (−g, −f ) = (3, −2). We also see that
c = −12, so we can find the radius by calculating
p
g2 + f 2 − c
r =
p
=
(−3)2 + 22 − (−12)
√
=
9 + 4 + 12
√
=
25
= 5.
An alternative method is to attempt to reconstruct the equation of the circle in the form
(x − a)2 + (y − b)2 = r 2
by completing the square. We start by collecting together the terms in x, and the terms in y.
So we rewrite our equation
x2 + y 2 − 6x + 4y − 12 = 0
as
x2 − 6x + y 2 + 4y − 12 = 0 .
Now the terms in x must come from (x − a)2 , a complete square, so we complete the square for
the x terms, and similarly for the y terms, to get
(x − 3)2 − 9 + (y + 2)2 − 4 − 12 = 0 .
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So we have
which we rewrite as
(x − 3)2 + (y + 2)2 − 25 = 0 ,
(x − 3)2 + (y + 2)2 = 25 = 52 .
We can now see that the centre of the circle is (3, −2) and the radius is 5.
Example
Find the centre and radius of the circle
2x2 + 2y 2 − 8x − 7y = 0 .
Solution
Notice that this is the equation of a circle, even though the coefficients of x2 and of y 2 are not
equal to 1. But we can divide throughout by 2, and we get
x2 + y 2 − 4x − 72 y = 0 .
If we compare this with the standard equation
x2 + y 2 + 2gx + 2f y + c = 0
we see that g = −2 and f = − 74 , so the centre of the circle is (−g, −f ) = (2, 74 ). We also see
that c = 0, so we find the radius by calculating
p
g2 + f 2 − c
r =
q
2
(−2)2 + − 47
=
q
49
4 + 16
=
q
113
=
16
√
1
= 4 113 .
Alternatively, we could try completing the square to regain the form (x − a)2 + (y − b)2 = r 2 . So
2x2 + 2y 2 − 8x − 7y = 0 ,
and again we divide by 2 to get
x2 + y 2 − 4x − 72 y = 0 .
Collecting the x terms together and the y terms together, we get
x2 − 4x + y 2 − 27 y = 0 ,
and then completing the square gives us
(x − 2)2 − 4 + y −
so that
(x − 2)2 + y −
7 2
4
7 2
4
−
=
49
16
113
16
=0
.
We can now see that the centre of the circle is (2, 74 ) and the radius is
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6
1
4
√
113.
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Example
Find the centre and radius of the circle
x2 + y 2 + 8x + 7 = 0 .
Solution
Notice that, in this example, there is no y term. If we compare our equation with the standard
equation
x2 + y 2 + 2gx + 2f y + c = 0 ,
we see that g = 4 and f = 0. So the centre of the circle is (−g, −f ) = (−4, 0). We also see
that c = 7, so we find the radius by calculating
p
g2 + f 2 − c
r =
√
=
42 + 02 − 7
√
=
16 − 7
√
=
9
= 3.
Exercises
1. Find the equation of the circle with given centre and radius:
(a) centre (3, 5), radius 3;
(b) centre (−2, 3), radius 1;
(d) centre (2, −2), radius 5; (e) centre (0, 5) radius 4.
2. Identify the centre and radius of the following circles:
(a) x2 + y 2 − 2x − 4y − 20 = 0,
(c) x2 + y 2 + 2x − 3 = 0,
(e) 3x2 + 3y 2 − 6x + 9y + 5 = 0.
(c) centre (−1, −3), radius 2;
(b) x2 + y 2 − 4x + 6y + 4 = 0,
(d) x2 + y 2 + 6x + 7y − 14 34 = 0,
4. The equation of a tangent to a circle at a given point
What is the equation of the tangent to the circle x2 + y 2 + 2x + 4y − 3 = 0 at the point (1, −4)
on the circle?
For a question like this, we should check first that the given point does indeed lie on the circle.
If we substitute x = 1 and y = −4 into the equation, we obtain
12 + (−4)2 + 2 × 1 + 4 × (−4) − 3 = 1 + 16 + 2 + (−16) − 3 = 0 ,
and so the equation is satisfied. In fact we can see this from a diagram.
tangent
(−1, −2)
(1, −4)
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We have also marked the centre of the circle on the diagram. To find the centre, we note that
g = 1 and f = 2, so that the centre is at (−1, −2).
A tangent is a straight line that just touches the circle. To find the equation of a straight line,
we need to know either two points on it, or one point on it together with its gradient. In this
example, we know one point on the line, the point (1, −4) where it is to touch the circle. But
we do not know another point. Nor do we know the gradient. So what should we do?
One fact we do know is that the tangent to a circle is perpendicular to the radius at the point
of contact. In this case, we know the point of contact (1, −4), and we also know the centre
(−1, −2). We can therefore calculate the gradient of the radius from the centre to the point of
contact, and hence the gradient of the tangent.
Now the gradient m of a straight line joining the points (x1 , y1 ) and (x2 , y2 ) is given by
m=
y2 − y1
.
x2 − x1
So if we take (x1 , y1) = (1, −4) and (x2 , y2 ) = (−1, −2), the gradient m1 of the radius is
m1 =
(−2) − (−4)
2
=
= −1 .
(−1) − 1
−2
We now use the result that, if two lines with gradients m1 and m2 are perpendicular, then
m1 m2 = −1. Here, the gradient of the radius is m1 = −1, and so the gradient of the tangent
must be m2 = 1.
Now we have enough information to find the equation of the tangent. We know that the equation
of a straight line with a given gradient m = 1 and containing a given point (x1 , y1) = (1, −4) can
be found from the formula
y − y1 = m(x − x1 ) ,
and so the equation of the tangent is given by
y − (−4) = 1 × (x − 1)
y+4 = x−1
y = x −5.
Key Point
To find the equation of the tangent to a circle through a given point of contact, you should first
find the centre of the circle and then calculate the gradient m1 of the line joining the centre to
the point of contact.
Having done this, you should find the gradient m2 of the tangent, using the formula m1 m2 = −1.
As you now know the gradient and one point on the tangent, you can find the equation of the
tangent.
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Example
Find the equation of the tangent at the point (0, 2) to the circle x2 + y 2 − 4x + 2y − 8 = 0.
Solution
We start by finding the centre of the circle. From the equation, we see that g = −2 and f = 1,
so the centre of the circle is at (2, −1).
(0, 2)
(2, −1)
Let us take (x1 , y1) = (0, 2) and (x2 , y2 ) = (2, −1). Then the gradient m1 of the radius joining
these two points is
(−1) − 2
3
y2 − y1
=
= − .
m1 =
x2 − x1
2−0
2
If the tangent has gradient m2 then we must have m1 m2 = −1 as the tangent and the radius
are perpendicular, and so m2 = 32 .
Now we can find the equation of the tangent. We know the gradient m2 = 32 , and we know a
point (x1 , y1 ) = (0, 2). So the tangent is given by
y − y1 = m2 (x − x1 )
y − 2 = 32 (x − 0)
y = 23 x + 2 .
Note that the y-intercept of this line is 2, as we would expect from the fact that it passes through
the given point (0, 2).
Exercises
3. Find the equation of the tangent to each circle at the point specified:
(a)
(b)
(c)
(d)
(e)
circle
circle
circle
circle
circle
x2 + y 2 − 2x − 4y − 20 = 0, point (4, −2);
x2 + y 2 + 4x + 2y − 20 = 0, point (1, 3);
x2 + y 2 − 6x + 4y − 87 = 0, point (−3, −10);
x2 + y 2 + 18x − 88 = 0, point (3, 5);
x2 + y 2 − 6y − 160 = 0, point (12, 8).
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4. Find the points of intersection of the line y = 2x + 1 and the circle x2 + y 2 − 2y + 4 = 0.
Show that the line y = 2x + 1 is a diameter of the circle. Find the equation of the tangent to
the circle at one of the points of intersection.
5. Find the points of intersection of the line y = x − 3 and the circle x2 + y 2 − 2x + 2y + 1 = 0.
What are the tangents at the points of intersection? Where do they intersect?
6. Find the points where the circle x2 + y 2 − 10x − 10y + 40 = 0 and the line y + 2x = 10
intersect. Find the equation of the tangent to the circle at each of the points of intersection.
Find the point of intersection of these two tangents.
7. Show that the equation of the tangent at the point (x1 , y1 ) on the circle
x2 + y 2 + 2gx + 2f y + c = 0
is given by
xx1 + yy1 + g(x + x1 ) + f (y + y1 ) + c = 0 .
1.
(a) x2 + y 2 − 6x − 10y + 25 = 0 (b)
(c) x2 + y 2 + 2x + 6y + 6 = 0
(d)
2
2
(e) x + y − 10y + 9 = 0
2.
(a)
(d) centre (−3, − 72 ), radius 6
(b)
(e)
x2 + y 2 + 4x − 6y + 12 = 0
x2 + y 2 − 4x + 4y − 17 = 0
centre (1, − 23 ), radius 19
12
3.
(a) 4y = 3x − 20 (b) 4y + 3x = 15
(c)
(d) 5y + 12x = 61 (e) 5y + 12x = 184
(c) centre (−1, 0), radius 2
4y + 3x + 49 = 0
4. The points of intersection are (1, 3) and (−1, −1). The mid-point of these is (0, 1) which is
the centre of the circle, and hence y = 2x + 1 is a diameter. The tangents are 2y + x = 7 and
2y + x = 3 respectively.
5. The points of intersection are (1, −2) and (2, −1). The tangents are y = −2 and x = 2
respectively. They intersect at the point (2, −2).
6. The points of intersection are (4, 2) and (2, 6). The tangents are 3y + x = 10 and y = 3x
respectively. They intersect at the point (1, 3).
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### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Polite Numbers
##### Stage: 5 Challenge Level:
A polite number can be written as the sum of two or more consecutive positive integers. Find the consecutive sums giving the polite numbers 544 and 424. What characterizes impolite numbers?
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### The Clue Is in the Question
##### Stage: 5 Challenge Level:
This problem is a sequence of linked mini-challenges leading up to the proof of a difficult final challenge, encouraging you to think mathematically. Starting with one of the mini-challenges, how. . . .
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Plus or Minus
##### Stage: 5 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Integral Sandwich
##### Stage: 5 Challenge Level:
Generalise this inequality involving integrals.
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Steve's Mapping
##### Stage: 5 Challenge Level:
Steve has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### Why Stop at Three by One
##### Stage: 5
Beautiful mathematics. Two 18 year old students gave eight different proofs of one result then generalised it from the 3 by 1 case to the n by 1 case and proved the general result.
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Fibonacci Factors
##### Stage: 5 Challenge Level:
For which values of n is the Fibonacci number fn even? Which Fibonnaci numbers are divisible by 3?
### Discrete Trends
##### Stage: 5 Challenge Level:
Find the maximum value of n to the power 1/n and prove that it is a maximum.
### 2^n -n Numbers
##### Stage: 5
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### Cyclic Triangles
##### Stage: 5 Challenge Level:
Make and prove a conjecture about the cyclic quadrilateral inscribed in a circle of radius r that has the maximum perimeter and the maximum area.
##### Stage: 4 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Multiplication Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Poly Fibs
##### Stage: 5 Challenge Level:
A sequence of polynomials starts 0, 1 and each poly is given by combining the two polys in the sequence just before it. Investigate and prove results about the roots of the polys.
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Sixty-seven Squared
##### Stage: 5 Challenge Level:
Evaluate these powers of 67. What do you notice? Can you convince someone what the answer would be to (a million sixes followed by a 7) squared?
### Few and Far Between?
##### Stage: 4 and 5 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Vecten
##### Stage: 5 Challenge Level:
Join in this ongoing research. Build squares on the sides of a triangle, join the outer vertices forming hexagons, build further rings of squares and quadrilaterals, investigate.
### Trig Rules OK
##### Stage: 5 Challenge Level:
Change the squares in this diagram and spot the property that stays the same for the triangles. Explain...
### Center Path
##### Stage: 3 and 4 Challenge Level:
Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . .
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Prime Sequences
##### Stage: 5 Challenge Level:
This group tasks allows you to search for arithmetic progressions in the prime numbers. How many of the challenges will you discover for yourself?
### Alison's Mapping
##### Stage: 4 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### Close to Triangular
##### Stage: 4 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Painting by Numbers
##### Stage: 5 Challenge Level:
How many different colours of paint would be needed to paint these pictures by numbers?
### Conjugate Tracker
##### Stage: 5 Challenge Level:
Make a conjecture about the curved track taken by the complex roots of a quadratic equation and use complex conjugates to prove your conjecture.
### OK! Now Prove It
##### Stage: 5 Challenge Level:
Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
##### Stage: 5 Challenge Level:
With red and blue beads on a circular wire; 'put a red bead between any two of the same colour and a blue between different colours then remove the original beads'. Keep repeating this. What happens?
### Binary Squares
##### Stage: 5 Challenge Level:
If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
### Summats Clear
##### Stage: 5 Challenge Level:
Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1, 2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a - b) = ab.
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
##### Stage: 4 Challenge Level:
Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles?
### Pythagorean Fibs
##### Stage: 5 Challenge Level:
What have Fibonacci numbers got to do with Pythagorean triples?
### The Kth Sum of N Numbers
##### Stage: 5
Yatir from Israel describes his method for summing a series of triangle numbers.
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Fibonacci Fashion
##### Stage: 5 Challenge Level:
What have Fibonacci numbers to do with solutions of the quadratic equation x^2 - x - 1 = 0 ? | 2,237 | 9,582 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-39 | latest | en | 0.825855 |
https://ininet.org/chapter-3-descriptive-statistics-numerical-measures-learning-o.html?page=2 | 1,716,002,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00542.warc.gz | 285,880,102 | 15,877 | # Chapter 3 Descriptive Statistics: Numerical Measures Learning Objectives
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For the Stivers mutual fund we have:
18000=10000, so =1.8 and
So the mean annual return for the Stivers mutual fund is (1.07624 – 1)100 = 7.624%
For the Trippi mutual fund we have:
10600=5000, so =2.12 and
So the mean annual return for the Trippi mutual fund is (1.09848 – 1)100 = 9.848%.
While the Stivers mutual fund has generated a nice annual return of 7.6%, the annual return of 9.8% earned by the Trippi mutual fund is far superior.
21. 5000=3500, so =1.428571, and so
So the mean annual growth rate is (1.040426 – 1)100 = 4.0404%
22. 25,000,000=10,000,000, so =2.50, and so
So the mean annual growth rate is (1.165 – 1)100 = 16.5%
23. Range 20 - 10 = 10
10, 12, 16, 17, 20
Q1 (2nd position) = 12
Q3 (4th position) = 17
IQR = Q3Q1 = 17 – 12 = 5
24.
25. 15, 20, 25, 25, 27, 28, 30, 34 Range = 34 – 15 = 19
IQR = Q3Q1 = 29 – 22.5 = 6.5
26. a. Range = 190 – 168 = 22
b.
c.
d.
27. a. The mean price for a round–trip flight into Atlanta is \$356.73, and the mean price for a round–trip flight into Salt Lake City is \$400.95. Flights into Atlanta are less expensive than flights into Salt Lake City. This possibly could be explained by the locations of these two cities relative to the 14 departure cities; Atlanta is generally closer than Salt Lake City to the departure cities.
b. For flights into Atlanta, the range is \$290.0, the variance is 5517.41, and the standard Deviation is \$74.28. For flights into Salt Lake City, the range is \$458.8, the variance is 18933.32, and the standard deviation is \$137.60.
The prices for round–trip flights into Atlanta are less variable than prices for round–trip flights into Salt Lake City. This could also be explained by Atlanta’s relative nearness to the 14 departure cities.
28. a. The mean serve speed is 180.95, the variance is 21.42, and the standard deviation is 4.63.
b. Although the mean serve speed for the twenty Women's Singles serve speed leaders for the 2011 Wimbledon tournament is slightly higher, the difference is very small. Furthermore, given the variation in the twenty Women's Singles serve speed leaders from the 2012 Australian Open and the twenty Women's Singles serve speed leaders from the 2011 Wimbledon tournament, the difference in the mean serve speeds is most likely due to random variation in the players’ performances.
29. a. Range = 60 – 28 = 32
IQR = Q3Q1 = 55 – 45 = 10
b.
c. The average air quality is about the same. But, the variability is greater in Anaheim.
30. Dawson Supply: Range = 11 – 9 = 2
J.C. Clark: Range = 15 – 7 = 8
31. a.
18–34 35–44 45+ mean 1368.0 1330.1 1070.4 median 1423.0 1382.5 1163.5 standard deviation 540.8 431.7 334.5
b. The 45+ group appears to spend less on coffee than the other two groups, and the 18–34 and 35–44 groups spend similar amounts of coffee.
32. a. Freshmen
Seniors
Freshmen spend almost three times as much on back-to-school items as seniors.
b. Freshmen Range = 2094 – 374 = 1720
Seniors Range = 632 – 280 = 352
c. Freshmen
Q1 = 1079 (7th item)
Q3 = 1475 (19th item)
IQR = Q3 Q1 = 1479 – 1075 = 404
Seniors
IQR = Q3 Q1 = 502 – 370.5 = 131.5
d.
Freshmen
Seniors
e. All measures of variability show freshmen have more variation in back-to-school expenditures.
33. a. For 2011
For 2012
b. The mean score is 76 for both years, but there is an increase in the standard deviation for the scores in 2012. The golfer is not as consistent in 2012 and shows a sizeable increase in the variation with golf scores ranging from 71 to 85. The increase in variation might be explained by the golfer trying to change or modify the golf swing. In general, a loss of consistency and an increase in the standard deviation could be viewed as a poorer performance in 2012. The optimism in 2012 is that three of the eight scores were better than any score reported for 2011. If the golfer can work for consistency, eliminate the high score rounds, and reduce the standard deviation, golf scores should show improvement.
34. Quarter milers
s = 0.0564
Coefficient of Variation = (s/)100% = (0.0564/0.966)100% = 5.8%
Milers
s = 0.1295
Coefficient of Variation = (s/)100% = (0.1295/4.534)100% = 2.9%
Yes; the coefficient of variation shows that as a percentage of the mean the quarter milers’ times show more variability.
35.
10
20
12
17
16
36.
37. a. At least 75%
b. At least 89%
c. At least 61%
d. At least 83%
e. At least 92%
38. a. Approximately 95%
b. Almost all
c. Approximately 68%
39. a. This is from 2 standard deviations below the mean to 2 standard deviations above the mean.
With z = 2, Chebyshev’s theorem gives:
Therefore, at least 75% of adults sleep between 4.5 and 9.3 hours per day.
b. This is from 2.5 standard deviations below the mean to 2.5 standard deviations above the mean.
With z = 2.5, Chebyshev’s theorem gives:
Therefore, at least 84% of adults sleep between 3.9 and 9.9 hours per day.
c. With z = 2, the empirical rule suggests that 95% of adults sleep between 4.5and 9.3 hours per day. The percentage obtained using the empirical rule is greater than the percentage obtained using Chebyshev’s theorem.
40. a. \$3.33 is one standard deviation below the mean and \$3.53 is one standard deviation above the mean. The empirical rule says that approximately 68% of gasoline sales are in this price range.
b. Part (a) shows that approximately 68% of the gasoline sales are between \$3.33 and \$3.53. Since the bell-shaped distribution is symmetric, approximately half of 68%, or 34%, of the gasoline sales should be between \$3.33 and the mean price of \$3.43. \$3.63 is two standard deviations above the mean price of \$3.43. The empirical rule says that approximately 95% of the gasoline sales should be within two standard deviations of the mean. Thus, approximately half of 95%, or 47.5%, of the gasoline sales should be between the mean price of \$3.43 and \$3.63. The percentage of gasoline sales between \$3.33 and \$3.63 should be approximately 34% + 47.5% = 81.5%.
c. \$3.63 is two standard deviations above the mean and the empirical rule says that approximately 95% of the gasoline sales should be within two standard deviations of the mean. Thus, 1 – 95% = 5% of the gasoline sales should be more than two standard deviations from the mean. Since the bell-shaped distribution is symmetric, we expected half of 5%, or 2.5%, would be more than \$3.63.
41. a. 615 is one standard deviation above the mean. Approximately 68% of the scores are between 415 and 615 with half of 68%, or 34%, of the scores between the mean of 515 and 615. Also, since the distribution is symmetric, 50% of the scores are above the mean of 515. With 50% of the scores above 515 and with 34% of the scores between 515 and 615, 50% – 34% = 16% of the scores are above 615.
b. 715 is two standard deviations above the mean. Approximately 95% of the scores are between 315 and 715 with half of 95%, or 47.5%, of the scores between the mean of 515 and 715. Also, since the distribution is symmetric, 50% of the scores are above the mean of 515. With 50% of the scores above 515 and with 47.5% of the scores between 515 and 715, 50%– 47.5% = 2.5% of the scores are above 715.
c. Approximately 68% of the scores are between 415 and 615 with half of 68%, or 34%, of the scores between 415 and the mean of 515.
d. Approximately 95% of the scores are between 315 and 715 with half of 95%, or 47.5%, of the scores between 315 and the mean of 515. Approximately 68% of the scores are between 415 and 615 with half of 68%, or 34%, of the scores between the mean of 515 and 615. Thus, 47.5% + 34% = 81.5% of the scores are between 315 and 615.
42. a.
b.
c. \$2300 is .67 standard deviations below the mean. \$4900 is 1.50 standard deviations above the mean. Neither is an outlier.
d.
\$13,000 is 8.25 standard deviations above the mean. This cost is an outlier.
43. a. days
Median: with n = 7, use 4th position
2, 3, 8, 8, 12, 13, 18
Median = 8 days
Mode: 8 days (occurred twice)
b. Range = Largest value – Smallest value
= 18 – 2 = 16
c.
The 18 days required to restore service after hurricane Wilma is not an outlier.
d. Yes, FP&L should consider ways to improve its emergency repair procedures. The mean, median and mode show repairs requiring an average of 8 to 9 days can be expected if similar hurricanes are encountered in the future. The 18 days required to restore service after hurricane Wilma should not be considered unusual if FP&L continues to use its current emergency repair procedures. With the number of customers affected running into the millions, plans to shorten the number of days to restore service should be undertaken by the company.
44. a.
b.
Approximately one standard deviation above the mean. Approximately 68% of the scores are within one standard deviation. Thus, half of (100–68), or 16%, of the games should have a winning score of 84 or more points.
Approximately two standard deviations above the mean. Approximately 95% of the scores are within two standard deviations. Thus, half of (100–95), or 2.5%, of the games should have a winning score of more than 90 points.
c.
Largest margin 24: . No outliers.
45. a.
b. \$75.00 – \$72.20 = \$2.80
\$2.80/\$72.20 = .0388 Ticket price increased 3.88% during the one-year period.
c. 7th position – Green Bay Packers 63
8th position – Pittsburgh Steelers 67
Median =
d. Use 4th position
Q1 = 61 (Tennessee Titans)
Use 11th position
Q3 = 83 (Indianapolis Colts)
e.
f. Dallas Cowboys:
With z> 3, this is an outlier. The Dallas Cowboys have an unusually high ticket price compared to the other NFL teams.
46. 15, 20, 25, 25, 27, 28, 30, 34
Smallest = 15
Largest = 34
47.
48. 5, 6, 8, 10, 10, 12, 15, 16, 18
Smallest = 5
Q1 = 8 (3rd position)
Median = 10
Q3 = 15 (7th position)
Largest = 18
49. IQR = 50 – 42 = 8
Lower Limit: Q1 – 1.5 IQR = 42 – 12 = 30
Upper Limit: Q3 + 1.5 IQR = 50 + 12 = 62
65 is an outlier
50. a. The first place runner in the men’s group finished minutes ahead of the first place runner in the women’s group. Lauren Wald would have finished in 11th place for the combined groups.
b. Men: . Use the 11th and 12th place finishes.
Median =
Women: . Use the 16th place finish. Median = 131.67.
Using the median finish times, the men’s group finished minutes ahead of the women’s group.
Also note that the fastest time for a woman runner, 109.03 minutes, is approximately equal to the median time of 109.64 minutes for the men’s group.
c. Men: Lowest time = 65.30; Highest time = 148.70
Q1: Use 6th position. Q1 = 87.18
Q3: Use 17th position. Q3 = 128.40
Five number summary for men: 65.30, 87.18, 109.64, 128.40, 148.70
Women: Lowest time = 109.03; Highest time = 189.28
Q1: Use 8th position. Q1 = 122.08
Q3: Use 24th position. Q3 = 147.18
Five number summary for women: 109.03, 122.08, 131.67, 147.18, 189.28
d. Men: IQR =
Lower Limit =
Upper Limit =
There are no outliers in the men’s group.
Women: IQR =
Lower Limit =
Upper Limit =
The two slowest women runners with times of 189.27 and 189.28 minutes are outliers in the women’s group.
e.
The box plots show the men runners with the faster or lower finish times. However, the box plots show the women runners with the lower variation in finish times. The interquartile ranges of 41.22 minutes for men and 25.10 minutes for women support this conclusion.
51. a. Median (11th position) = 4019
Q1 (6th position) = 1872
Q3 (16th position) = 8305
608, 1872, 4019, 8305, 14138
b. Limits:
IQR = Q3Q1 = 8305 – 1872 = 6433
Lower Limit: Q1 – 1.5 (IQR) = –7777
Upper Limit: Q3 + 1.5 (IQR) = 17955
c. There are no outliers, all data are within the limits.
d. Yes, if the first two digits in Johnson and Johnson's sales were transposed to 41,138, sales would have shown up as an outlier. A review of the data would have enabled the correction of the data.
e.
52. a. Median n = 20; 10th and 11th positions
Median =
b. Smallest 68
Q1: ; 5th and 6th positions
Q3: ; 15th and 16th positions
Largest 77
5- number summary: 68, 71.5, 73.5, 74.5, 77
c. IQR = Q3Q1 = 74.5 – 71.5 = 3
Lower Limit = Q1 – 1.5(IQR)
= 71.5 – 1.5(3) = 67
Upper Limit = Q3 + 1.5(IQR)
= 74.5 + 1.5(3) = 79
All ratings are between 67 and 79. There are no outliers for the T-Mobile service.
d. Using the solution procedures shown in parts a, b, and c, the five number summaries and outlier limits for the other three cell-phone services are as follows.
AT&T 66, 68, 71, 73, 75 Limits: 60.5 and 80.5
Sprint 63, 65, 66, 67.5, 69 Limits: 61.25 and 71.25
Verizon 75, 77, 78.5, 79.5, 81 Limits: 73.25 and 83.25
There are no outliers for any of the cell-phone services.
e.
The box plots show that Verizon is the best cell-phone service provider in terms of overall customer satisfaction. Verizon’s lowest rating is better than the highest AT&T and Sprint ratings and is better than 75% of the T-Mobile ratings. Sprint shows the lowest customer satisfaction ratings among the four services.
53. a. Total Salary for the Philadelphia Phillies = \$96,870,000
Median n = 28; 14th and 15th positions
Median =
Smallest 390
Q1: ; 7th and 8th positions
Q3: ; 21st and 22nd positions
Largest 14250
5– number summary for the Philadelphia Phillies: 390, 432.5, 1300, 6175, 14250
Using the 5-number summary, the lower quartile shows salaries closely bunched between 390 and 432.5. The median is 1300. The most variation is in the upper quartile where the salaries are spread between 6175 and 14250, or between \$6,175,000 and \$14,250,000.
b. IQR = Q3Q1 = 6175 – 432.5 = 5742.5
Lower Limit = Q1 – 1.5(IQR)
= 432.5 –1.5(5742.5) = – 8181.25; Use 0
Upper Limit = Q3 + 1.5(IQR)
= 6175 + 1.5(5742.5) = 14788.75
All salaries are between 0 and 14788.75. There are no salary outliers for the Philadelphia Phillies.
c. Using the solution procedures shown in parts a and b, the total salary, the five-number summaries, and the outlier limits for the other teams are as follows.
Los Angeles Dodgers \$136,373,000
390, 403, 857.5, 9125, 19000 Limits: 0 and 22208
Tampa Bay Rays \$ 42,334,000
390, 399, 415, 2350, 6000 Limits: 0 and 5276.5
Boston Red Sox \$120,460,000
396, 439.5, 2500, 8166.5, 14000 Limits: 0 and 19757
The Los Angeles Dodgers had the highest payroll while the Tampa Bay Rays clearly had the lowest payroll among the four teams. With the lower salaries, the Rays had two outlier salaries compared to other salaries on the team. But these top two salaries are substantially below the top salaries for the other three teams. There are no outliers for the Phillies, Dodgers and Red Sox.
d.
The box plots show that the lowest salaries for the four teams are very similar. The Red Sox have the highest median salary. Of the four teams the Dodgers have the highest upper end salaries and highest total payroll, while the Rays are clearly the lowest paid team.
For this data, we would conclude that paying higher salaries do not always bring championships. In the National League Championship, the lower paid Phillies beat the higher paid Dodgers. In the American League Championship, the lower paid Rays beat the higher paid Red Sox. The biggest surprise was how the Tampa Bay Rays over achieved based on their salaries and made it to the World Series. Teams with the highest salaries do not always win the championships.
54. a.
Median 23rd position 15.1
24th position 15.6
Median =
b. Q1:
12th position: Q1 = 11.7
Q3:
35th position: Q3 = 23.5
c. 3.4, 11.7, 15.35, 23.5, 41.3
d. IQR = 23.5 – 11.7 = 11.8
Lower Limit = Q1 – 1.5(IQR)
= 11.7 – 1.5(11.8) = –6 Use 0
Upper Limit = Q3 + 1.5(IQR)
= 23.5 + 1.5(11.8) = 41.2
Yes, one: Alger Small Cap 41.3
55. a.
b. Negative relationship
c/d.
There is a strong negative linear relationship.
56. a.
b. Positive relationship
c/d.
A positive linear relationship
57. a.
b. The scatter diagram shows a positive relationship with higher predicted point margins associated with higher actual point margins.
c. Let x = predicted point margin and y = actual point margin
A positive covariance shows a positive relationship between predicted point margins and actual point margins.
d.
The modest positive correlation shows that the Las Vegas predicted point margin is a general, but not a perfect, indicator of the actual point margin in college football bowl games.
Note: The Las Vegas odds makers set the point margins so that someone betting on a favored team has to have the team win by more than the point margin to win the bet. For example, someone betting on Auburn to win the Outback Bowl would have to have Auburn win by more than five points to win the bet. Since Auburn beat Northwestern by only three points, the person betting on Auburn would have lost the bet.
A review of the predicted and actual point margins shows that the favorites won by more than the predicted point margin in five bowl games: Gator, Sugar, Cotton, Alamo, and the Championship bowl game. The underdog either won its game or kept the actual point margin less than the predicted point margin in the other five bowl games. In this case, betting on the underdog would have provided winners in the Outback, Capital One, Rose, Fiesta and Orange bowls. In this example, the Las Vegas odds point margins made betting on the favored team a 50-50 probability of winning the bet.
58. Let x = miles per hour and y = miles per gallon
A strong negative linear relationship exists. For driving speeds between 25 and 60 miles per hour,
higher speeds are associated with lower miles per gallon.
59. a.
7.1 7.02 0.2852 0.6893 0.0813 0.4751 0.1966 5.2 5.31 -1.6148 -1.0207 2.6076 1.0419 1.6483 7.8 5.38 0.9852 -0.9507 0.9706 0.9039 -0.9367 7.8 5.40 0.9852 -0.9307 0.9706 0.8663 -0.9170 5.8 5.00 -1.0148 -1.3307 1.0298 1.7709 1.3505 5.8 4.07 -1.0148 -2.2607 1.0298 5.1109 2.2942 9.3 6.53 2.4852 0.1993 6.1761 0.0397 0.4952 5.7 5.57 -1.1148 -0.7607 1.2428 0.5787 0.8481 7.3 6.99 0.4852 0.6593 0.2354 0.4346 0.3199 7.6 11.12 0.7852 4.7893 0.6165 22.9370 3.7605 8.2 7.56 1.3852 1.2293 1.9187 1.5111 1.7028 7.1 12.11 0.2852 5.7793 0.0813 33.3998 1.6482 6.3 4.39 -0.5148 -1.9407 0.2650 3.7665 0.9991 6.6 4.78 -0.2148 -1.5507 0.0461 2.4048 0.3331 6.2 5.78 -0.6148 -0.5507 0.3780 0.3033 0.3386 6.3 6.08 -0.5148 -0.2507 0.2650 0.0629 0.1291 7.0 10.05 0.1852 3.7193 0.0343 13.8329 0.6888 6.2 4.75 -0.6148 -1.5807 0.3780 2.4987 0.9719 5.5 7.22 -1.3148 0.8893 1.7287 0.7908 -1.1692 6.5 3.79 -0.3148 -2.5407 0.0991 6.4554 0.7999 6.0 3.62 -0.8148 -2.7107 0.6639 7.3481 2.2088 8.3 9.24 1.4852 2.9093 2.2058 8.4638 4.3208 7.5 4.40 0.6852 -1.9307 0.4695 3.7278 -1.3229 7.1 6.91 0.2852 0.5793 0.0813 0.3355 0.1652 6.8 5.57 -0.0148 -0.7607 0.0002 0.5787 0.0113 5.5 3.87 -1.3148 -2.4607 1.7287 6.0552 3.2354 7.5 8.42 0.6852 2.0893 0.4695 4.3650 1.4315 Total 25.77407 130.0594 25.5517
There is evidence of a modest positive linear association between the jobless rate and the delinquent housing loan percentage. If the jobless rate were to increase, it is likely that an increase in the percentage of delinquent housing loans would also occur.
b.
60. a.
b.
0.20 0.24 0.04 0.11 0.0016 0.0121 0.0044 0.82 0.19 0.66 0.06 0.4356 0.0036 0.0396 -0.99 -0.91 -1.15 -1.04 1.3225 1.0816 1.1960 0.04 0.08 -0.12 -0.05 0.0144 0.0025 0.0060 -0.24 -0.33 -0.40 -0.46 0.1600 0.2166 0.1840 1.01 0.87 0.85 0.74 0.7225 0.5476 0.6290 0.30 0.36 0.14 0.23 0.0196 0.0529 0.0322 0.55 0.83 0.39 0.70 0.1521 0.4900 0.2730 -0.25 -0.16 -0.41 -0.29 0.1681 0.0841 0.1189 Total 2.9964 2.4860 2.4831
c. There is a strong positive linear association between DJIA and S&P 500. If you know the change in either, you will have a good idea of the stock market performance for the day.
61. a.
b.
c.
68 50 .5 -.4286 .25 .1837 -.2143 70 49 2.5 -1.4286 6.25 2.0408 -3.5714 65 44 -2.5 -6.4286 6.25 41.3265 16.0714 96 64 28.5 13.5714 812.25 184.1837 386.7857 57 46 -10.5 -4.4286 110.25 19.6122 46.5000 70 45 2.5 -5.4286 6.25 29.4694 -13.5714 80 73 12.5 22.5714 156.25 509.4694 282.1429 67 45 -.5 -5.4286 .25 29.4694 2.7143 44 29 -23.5 -21.4286 552.25 459.1837 503.5714 69 44 1.5 -6.4286 2.25 41.3265 -9.6429 76 69 8.5 18.5714 72.25 344.8980 157.8571 69 51 1.5 .5714 2.25 .3265 .8571 70 58 2.5 7.5714 6.25 57.3265 18.9286 44 39 -23.5 -11.4286 552.25 130.6122 268.5714 Total 2285.5 1849.4286 1657.0000
High positive correlation as should be expected.
62. a. The mean is 2.95 and the median is 3.0.
b. The index for the first quartile is , so the first quartile is the mean of the values of the 5th and 6th observations in the sorted data, or .
The index for the third quartile is , so the third quartile is the mean of the values of the 15th and 16th observations in the sorted data, or .
c. The range is 7 and the interquartile range is 4.5 – 1 = 3.5.
d. The variance is 4.37 and standard deviation is 2.09.
e. Because most people dine out a relatively few times per week and a few families dine out very frequently, we would expect the data to be positively skewed. The skewness measure of 0.34 indicates the data are somewhat skewed to the right.
f. The lower limit is –4.25 and the upper limit is 9.75. No values in the data are less than the lower limit or greater than the upper limit, so the Minitab boxplot indicates there are no outliers.
63. a. Arrange the data in order
Men
21 23 24 25 25 26 26 27 27 27 27 28 28 29 30 30 32 35
Median i = .5(18) = 9
Use 9th and 10th positions
Median = 27
Women
19 20 22 22 23 23 24 25 25 26 26 27 28 29 30
Median i = .5(15) = 7.5
Use 8th position
Median = 25
b. Men Women
Q1 i = .25(18) = 4.5 i = .25(15) = 3.75 Use 5th position Use 4th position Q1 = 25 Q1 = 22 Q3 i = .75(18) = 13.5 i = .75(15) = 11.25 Use 14th position Use 12th position Q3 = 29 Q3 = 27 | 7,384 | 21,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-22 | latest | en | 0.899845 |
https://www.quizzes.cc/metric/percentof.php?percent=315&of=879 | 1,606,802,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00454.warc.gz | 761,699,990 | 4,403 | #### What is 315 percent of 879?
How much is 315 percent of 879? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 315% of 879 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 315% of 879 = 2768.85
Calculate another percentage below. Type into inputs
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Calculating three hundred and fifteen of eight hundred and seventy-nine How to calculate 315% of 879? Simply divide the percent by 100 and multiply by the number. For example, 315 /100 x 879 = 2768.85 or 3.15 x 879 = 2768.85
#### How much is 315 percent of the following numbers?
315% of 879.01 = 276888.15 315% of 879.02 = 276891.3 315% of 879.03 = 276894.45 315% of 879.04 = 276897.6 315% of 879.05 = 276900.75 315% of 879.06 = 276903.9 315% of 879.07 = 276907.05 315% of 879.08 = 276910.2 315% of 879.09 = 276913.35 315% of 879.1 = 276916.5 315% of 879.11 = 276919.65 315% of 879.12 = 276922.8 315% of 879.13 = 276925.95 315% of 879.14 = 276929.1 315% of 879.15 = 276932.25 315% of 879.16 = 276935.4 315% of 879.17 = 276938.55 315% of 879.18 = 276941.7 315% of 879.19 = 276944.85 315% of 879.2 = 276948 315% of 879.21 = 276951.15 315% of 879.22 = 276954.3 315% of 879.23 = 276957.45 315% of 879.24 = 276960.6 315% of 879.25 = 276963.75
315% of 879.26 = 276966.9 315% of 879.27 = 276970.05 315% of 879.28 = 276973.2 315% of 879.29 = 276976.35 315% of 879.3 = 276979.5 315% of 879.31 = 276982.65 315% of 879.32 = 276985.8 315% of 879.33 = 276988.95 315% of 879.34 = 276992.1 315% of 879.35 = 276995.25 315% of 879.36 = 276998.4 315% of 879.37 = 277001.55 315% of 879.38 = 277004.7 315% of 879.39 = 277007.85 315% of 879.4 = 277011 315% of 879.41 = 277014.15 315% of 879.42 = 277017.3 315% of 879.43 = 277020.45 315% of 879.44 = 277023.6 315% of 879.45 = 277026.75 315% of 879.46 = 277029.9 315% of 879.47 = 277033.05 315% of 879.48 = 277036.2 315% of 879.49 = 277039.35 315% of 879.5 = 277042.5
315% of 879.51 = 277045.65 315% of 879.52 = 277048.8 315% of 879.53 = 277051.95 315% of 879.54 = 277055.1 315% of 879.55 = 277058.25 315% of 879.56 = 277061.4 315% of 879.57 = 277064.55 315% of 879.58 = 277067.7 315% of 879.59 = 277070.85 315% of 879.6 = 277074 315% of 879.61 = 277077.15 315% of 879.62 = 277080.3 315% of 879.63 = 277083.45 315% of 879.64 = 277086.6 315% of 879.65 = 277089.75 315% of 879.66 = 277092.9 315% of 879.67 = 277096.05 315% of 879.68 = 277099.2 315% of 879.69 = 277102.35 315% of 879.7 = 277105.5 315% of 879.71 = 277108.65 315% of 879.72 = 277111.8 315% of 879.73 = 277114.95 315% of 879.74 = 277118.1 315% of 879.75 = 277121.25
315% of 879.76 = 277124.4 315% of 879.77 = 277127.55 315% of 879.78 = 277130.7 315% of 879.79 = 277133.85 315% of 879.8 = 277137 315% of 879.81 = 277140.15 315% of 879.82 = 277143.3 315% of 879.83 = 277146.45 315% of 879.84 = 277149.6 315% of 879.85 = 277152.75 315% of 879.86 = 277155.9 315% of 879.87 = 277159.05 315% of 879.88 = 277162.2 315% of 879.89 = 277165.35 315% of 879.9 = 277168.5 315% of 879.91 = 277171.65 315% of 879.92 = 277174.8 315% of 879.93 = 277177.95 315% of 879.94 = 277181.1 315% of 879.95 = 277184.25 315% of 879.96 = 277187.4 315% of 879.97 = 277190.55 315% of 879.98 = 277193.7 315% of 879.99 = 277196.85 315% of 880 = 277200
1% of 879 = 8.79 2% of 879 = 17.58 3% of 879 = 26.37 4% of 879 = 35.16 5% of 879 = 43.95 6% of 879 = 52.74 7% of 879 = 61.53 8% of 879 = 70.32 9% of 879 = 79.11 10% of 879 = 87.9 11% of 879 = 96.69 12% of 879 = 105.48 13% of 879 = 114.27 14% of 879 = 123.06 15% of 879 = 131.85 16% of 879 = 140.64 17% of 879 = 149.43 18% of 879 = 158.22 19% of 879 = 167.01 20% of 879 = 175.8 21% of 879 = 184.59 22% of 879 = 193.38 23% of 879 = 202.17 24% of 879 = 210.96 25% of 879 = 219.75
26% of 879 = 228.54 27% of 879 = 237.33 28% of 879 = 246.12 29% of 879 = 254.91 30% of 879 = 263.7 31% of 879 = 272.49 32% of 879 = 281.28 33% of 879 = 290.07 34% of 879 = 298.86 35% of 879 = 307.65 36% of 879 = 316.44 37% of 879 = 325.23 38% of 879 = 334.02 39% of 879 = 342.81 40% of 879 = 351.6 41% of 879 = 360.39 42% of 879 = 369.18 43% of 879 = 377.97 44% of 879 = 386.76 45% of 879 = 395.55 46% of 879 = 404.34 47% of 879 = 413.13 48% of 879 = 421.92 49% of 879 = 430.71 50% of 879 = 439.5
51% of 879 = 448.29 52% of 879 = 457.08 53% of 879 = 465.87 54% of 879 = 474.66 55% of 879 = 483.45 56% of 879 = 492.24 57% of 879 = 501.03 58% of 879 = 509.82 59% of 879 = 518.61 60% of 879 = 527.4 61% of 879 = 536.19 62% of 879 = 544.98 63% of 879 = 553.77 64% of 879 = 562.56 65% of 879 = 571.35 66% of 879 = 580.14 67% of 879 = 588.93 68% of 879 = 597.72 69% of 879 = 606.51 70% of 879 = 615.3 71% of 879 = 624.09 72% of 879 = 632.88 73% of 879 = 641.67 74% of 879 = 650.46 75% of 879 = 659.25
76% of 879 = 668.04 77% of 879 = 676.83 78% of 879 = 685.62 79% of 879 = 694.41 80% of 879 = 703.2 81% of 879 = 711.99 82% of 879 = 720.78 83% of 879 = 729.57 84% of 879 = 738.36 85% of 879 = 747.15 86% of 879 = 755.94 87% of 879 = 764.73 88% of 879 = 773.52 89% of 879 = 782.31 90% of 879 = 791.1 91% of 879 = 799.89 92% of 879 = 808.68 93% of 879 = 817.47 94% of 879 = 826.26 95% of 879 = 835.05 96% of 879 = 843.84 97% of 879 = 852.63 98% of 879 = 861.42 99% of 879 = 870.21 100% of 879 = 879 | 2,689 | 5,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-50 | latest | en | 0.727958 |
https://runestone.academy/ns/books/published/fcla/section-VO.html | 1,686,031,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00085.warc.gz | 531,621,268 | 18,071 | ## SectionVOVector Operations
In this section we define some new operations involving vectors, and collect some basic properties of these operations. Begin by recalling our definition of a column vector as an ordered list of complex numbers, written vertically (Definition CV). The collection of all possible vectors of a fixed size is a commonly used set, so we start with its definition.
### SubsectionCVColumn Vectors
#### DefinitionVSCV.Vector Space of Column Vectors.
The vector space $$\complex{m}$$ is the set of all column vectors (Definition CV) of size $$m$$ with entries from the set of complex numbers, $$\complexes\text{.}$$
When a set similar to this is defined using only column vectors where all the entries are from the real numbers, it is written as $${\mathbb R}^m$$ and is known as Euclidean $$m$$-space.
The term vector is used in a variety of different ways. We have defined it as an ordered list written vertically. It could simply be an ordered list of numbers, and perhaps written as $$\left\langle 2,\,3,\,-1,\,6\right\rangle\text{.}$$ Or it could be interpreted as a point in $$m$$ dimensions, such as $$\left(3,\,4,\,-2\right)$$ representing a point in three dimensions relative to $$x\text{,}$$ $$y$$ and $$z$$ axes. With an interpretation as a point, we can construct an arrow from the origin to the point which is consistent with the notion that a vector has direction and magnitude.
All of these ideas can be shown to be related and equivalent, so keep that in mind as you connect the ideas of this course with ideas from other disciplines. For now, we will stick with the idea that a vector is just a list of numbers, in some particular order.
#### SageVSCV.Vector Spaces of Column Vectors.
It is possible to construct vector spaces several ways in Sage. For now, we will show you two basic ways. Remember that while our theory is all developed over the complex numbers, $$\complexes\text{,}$$ it is better to initially illustrate these ideas in Sage using the rationals, QQ.
To create a vector space, we use the VectorSpace()constructor, which requires the name of the number system for the entries and the number of entries in each vector. We can display some information about the vector space, and with tab-completion you can see what functions are available. We will not do too much with these methods immediately, but instead learn about them as we progress through the theory.
V = VectorSpace(QQ, 8)
V
Notice that the word “dimension” is used to refer to the number of entries in a vector contained in the vector space, whereas we have used the word “degree” before. If the way you use Sage provides tab-completion, then press the TABkey while in the next cell to see the range of methods you can use on a vector space.
V.
We can easily create “random” elements of any vector space, much as we did earlier for the kernel of a matrix. Try executing the next compute cell several times.
w = V.random_element()
w # random
Vector spaces are fundamental objects in Sage and in mathematics, and Sage has a nice compact way to create them, mimicking the notation we use when working on paper. While the next example includes the Sage version of the complex numbers, remember to be sure to do all your calculations with the rationals, QQ.
U = CC^5
U
W = QQ^3
W
Sage can determine if two vector spaces are the same. Notice that we use two equals sign to test equality, since we use a single equals sign to make assignments.
X = VectorSpace(QQ, 3)
W = QQ^3
X == W
### SubsectionVEASMVector Equality, Addition, Scalar Multiplication
We start our study of this set by first defining what it means for two vectors to be the same.
#### DefinitionCVE.Column Vector Equality.
Suppose that $$\vect{u},\,\vect{v}\in\complex{m}\text{.}$$ Then $$\vect{u}$$ and $$\vect{v}$$ are equal, written $$\vect{u}=\vect{v}$$ if
\begin{gather*} \vectorentry{\vect{u}}{i}=\vectorentry{\vect{v}}{i} \end{gather*}
for all $$1\leq i\leq m\text{.}$$
Now this may seem like a silly (or even stupid) thing to say so carefully. Of course two vectors are equal if they are equal for each corresponding entry! Well, this is not as silly as it appears. We will see a few occasions later where the obvious definition is not the right one. And besides, in doing mathematics we need to be very careful about making all the necessary definitions and making them unambiguous. And we have done that here.
Notice now that the symbol “=” is now doing triple-duty. We know from our earlier education what it means for two numbers (real or complex) to be equal, and we take this for granted. In Definition SE we defined what it meant for two sets to be equal. Now we have defined what it means for two vectors to be equal, and this new definition builds on our prior definition for when two numbers are equal when we use the condition $$\vectorentry{\vect{u}}{i}=\vectorentry{\vect{v}}{i}$$ repeatedly, for all $$1\leq i\leq m\text{.}$$ So think carefully about your objects when you see an equal sign and think about just which notion of equality you have encountered. This will be especially important when you are asked to construct proofs whose conclusion states that two objects are equal. If you have an electronic copy of the book, such as the PDF version, searching on “Definition CVE” can be an instructive exercise. See how often, and where, the definition is employed.
OK, let us do an example of vector equality that begins to hint at the utility of this definition.
Consider the system of linear equations in Archetype B,
\begin{align*} -7x_1 -6 x_2 - 12x_3 &=-33\\ 5x_1 + 5x_2 + 7x_3 &=24\\ x_1 +4x_3 &=5 \end{align*}
Note the use of three equals signs — each indicates an equality of numbers (the linear expressions are numbers when we evaluate them with fixed values of the variable quantities). Now write the vector equality,
\begin{equation*} \colvector{-7x_1 -6 x_2 - 12x_3\\ 5x_1 + 5x_2 + 7x_3\\ x_1 +4x_3} = \colvector{-33\\24\\5}\text{.} \end{equation*}
By Definition CVE, this single equality (of two column vectors) translates into three simultaneous equalities of numbers that form the system of equations. So with this new notion of vector equality we can become less reliant on referring to systems of simultaneous equations. There is more to vector equality than just this, but this is a good example for starters and we will develop it further.
We will now define two operations on the set $$\complex{m}\text{.}$$ By this we mean well-defined procedures that somehow convert vectors into other vectors. Here are two of the most basic definitions of the entire course.
Suppose that $$\vect{u},\,\vect{v}\in\complex{m}\text{.}$$ The sum of $$\vect{u}$$ and $$\vect{v}$$ is the vector $$\vect{u}+\vect{v}$$ defined by
\begin{gather*} \vectorentry{\vect{u}+\vect{v}}{i}=\vectorentry{\vect{u}}{i}+\vectorentry{\vect{v}}{i} \end{gather*}
for $$1\leq i\leq m\text{.}$$
So vector addition takes two vectors of the same size and combines them (in a natural way!) to create a new vector of the same size. Notice that this definition is required, even if we agree that this is the obvious, right, natural or correct way to do it. Notice too that the symbol ‘+’ is being recycled. We all know how to add numbers, but now we have the same symbol extended to double-duty and we use it to indicate how to add two new objects, vectors. And this definition of our new addition of vectors is built on our prior meaning of addition of numbers in the expressions $$\vectorentry{\vect{u}}{i}+\vectorentry{\vect{v}}{i}\text{.}$$ Think about your objects, especially when doing proofs. Vector addition is easy, here is an example from $$\complex{4}\text{.}$$
For vectors $$\vect{u},\,\vect{v}\text{,}$$ their sum is
\begin{align*} \vect{u}&=\colvector{2\\-3\\4\\2} & \vect{v}&=\colvector{-1\\5\\2\\-7} & \vect{u}+\vect{v}&= \colvector{2\\-3\\4\\2}+\colvector{-1\\5\\2\\-7}= \colvector{2+(-1)\\-3+5\\4+2\\2+(-7)}= \colvector{1\\2\\6\\-5}\text{.} \end{align*}
Our second operation takes two objects of different types, specifically a number and a vector, and combines them to create another vector. In this context we call the number a scalar in order to emphasize that it is not a vector.
#### DefinitionCVSM.Column Vector Scalar Multiplication.
Suppose $$\vect{u}\in\complex{m}$$ and $$\alpha\in\complexes\text{,}$$ then the scalar multiple of $$\vect{u}$$ by $$\alpha$$ is the vector $$\alpha\vect{u}$$ defined by
\begin{gather*} \vectorentry{\alpha\vect{u}}{i}=\alpha\vectorentry{\vect{u}}{i} \end{gather*}
for $$1\leq i\leq m\text{.}$$
Notice that we are doing a kind of multiplication here, but we are defining a new type, perhaps in what appears to be a natural way. We use juxtaposition (smashing two symbols together side-by-side) to denote this operation rather than using a symbol like we did with vector addition. So this can be another source of confusion. When two symbols are next to each other, are we doing regular old multiplication, the kind we have done for years, or are we doing scalar vector multiplication, the operation we just defined? Think about your objects — if the first object is a scalar, and the second is a vector, then it must be that we are doing our new operation, and the result of this operation will be another vector.
Notice how consistency in notation can be an aid here. If we write scalars as lower case Greek letters from the start of the alphabet (such as $$\alpha\text{,}$$ $$\beta\text{,}$$ …) and write vectors in bold Latin letters from the end of the alphabet ($$\vect{u}\text{,}$$ $$\vect{v}\text{,}$$ …), then we have some hints about what type of objects we are working with. This can be a blessing and a curse, since when we go read another book about linear algebra, or read an application in another discipline (physics, economics, …) the types of notation employed may be very different and hence unfamiliar.
Again, computationally, vector scalar multiplication is very easy.
For $$\vect{u}$$ and $$\alpha=6$$ we have
\begin{align*} \vect{u}&=\colvector{3\\1\\-2\\4\\-1} & \alpha\vect{u}&= 6\colvector{3\\1\\-2\\4\\-1}= \colvector{6(3)\\6(1)\\6(-2)\\6(4)\\6(-1)}= \colvector{18\\6\\-12\\24\\-6} \end{align*}
#### SageVO.Vector Operations.
Sage can easily perform the two basic operations with vectors, vector addition, +, and scalar vector multiplication, *. Notice that Sage is not confused by an ambiguity due to multiple meanings for the symbols +and *— for example, Sage knows that 3 + 12is different than the vector additions below.
x = vector(QQ, [1, 2, 3])
y = vector(QQ, [10, 20, 30])
5*x
x + y
3*x + 4*y
-y
w = (-4/3)*x - (1/10)*y
w
### SubsectionVSPVector Space Properties
With definitions of vector addition and scalar multiplication we can state, and prove, several properties of each operation, and some properties that involve their interplay. We now collect ten of them here for later reference.
While some of these properties seem very obvious, they all require proof. However, the proofs are not very interesting, and border on tedious. We will prove one version of distributivity very carefully, and you can test your proof-building skills on some of the others. Exercise VO.T13 has a proof in the solution, while Exercises VO.T17 and VO.T18 are excellent practice.
We need to establish an equality, so we will do so by beginning with one side of the equality, apply various definitions and theorems (listed to the right of each step) to massage the expression from the left into the expression on the right. Here we go with a proof of Property DSAC.
For $$1\leq i\leq m\text{,}$$
\begin{align*} \vectorentry{(\alpha+\beta)\vect{u}}{i}&=(\alpha+\beta)\vectorentry{\vect{u}}{i}&& \knowl{./knowl/definition-CVSM.html}{\text{Definition CVSM}}\\ &=\alpha\vectorentry{\vect{u}}{i}+\beta\vectorentry{\vect{u}}{i}&& \knowl{./knowl/property-DCN.html}{\text{Property DCN}}\\ &=\vectorentry{\alpha\vect{u}}{i}+\vectorentry{\beta\vect{u}}{i}&& \knowl{./knowl/definition-CVSM.html}{\text{Definition CVSM}}\\ &=\vectorentry{\alpha\vect{u}+\beta\vect{u}}{i}&& \knowl{./knowl/definition-CVA.html}{\text{Definition CVA}}\text{.} \end{align*}
Since the individual components of the vectors $$(\alpha+\beta)\vect{u}$$ and $$\alpha\vect{u}+\beta\vect{u}$$ are equal for all $$i\text{,}$$ $$1\leq i\leq m\text{,}$$ Definition CVE tells us the vectors are equal.
Many of the conclusions of our theorems can be characterized as “identities,” especially when we are establishing basic properties of operations such as those in this section. Most of the properties listed in Theorem VSPCV are examples. So some advice about the style we use for proving identities is appropriate right now. Have a look at Proof Technique PI.
Be careful with the notion of the vector $$\vect{-u}\text{.}$$ This is a vector that we add to $$\vect{u}$$ so that the result is the particular vector $$\zerovector\text{.}$$ This is basically a property of vector addition. It happens that we can compute $$\vect{-u}$$ using the other operation, scalar multiplication. We can prove this directly by writing that
\begin{gather*} \vectorentry{\vect{-u}}{i} =-\vectorentry{\vect{u}}{i} =(-1)\vectorentry{\vect{u}}{i} =\vectorentry{(-1)\vect{u}}{i} \end{gather*}
We will see later how to derive this property as a consequence of several of the ten properties listed in Theorem VSPCV.
Similarly, we will often write something you would immediately recognize as vector subtraction. This could be placed on a firm theoretical foundation — as you can do yourself with Exercise VO.T30.
A final note. Property AAC implies that we do not have to be careful about how we “parenthesize” the addition of vectors. In other words, there is nothing to be gained by writing$$\left(\vect{u}+\vect{v}\right)+\left(\vect{w}+\left(\vect{x}+\vect{y}\right)\right)$$ rather than $$\vect{u}+\vect{v}+\vect{w}+\vect{x}+\vect{y}\text{,}$$ since we get the same result no matter which order we choose to perform the four additions. So we will not be careful about using parentheses this way.
#### SageANC.A Note on Coercion.
Study the following sequence of commands, while cognizant of the failure to specify a number system for x.
x = vector([1, 2, 3])
u = 3*x
u
v = (1/3)*x
v
y = vector(QQ, [4, 5, 6])
w = 8*y
w
z = x + y
z
None of this should be too much of a surprise, and the results should be what we would have expected. Though for xwe never specified if 1, 2, 3are integers, rationals, reals, complexes, or …? Let us dig a little deeper and examine the parents of the five vectors involved.
x.parent()
u.parent()
v.parent()
y.parent()
w.parent()
z.parent()
So xand ubelong to something called an “ambient free module,” whatever that is. What is important here is that the parent of xuses the integers as its number system. How about u, v, y, w, z? All but the first has a parent that uses the rationals for its number system.
Three of the final four vectors are examples of a process that Sage calls “coercion.” Mathematical elements get converted to a new parent, as necessary, when the conversion is totally unambiguous. In the examples above:
• uis the result of scalar multiplication by an integer, so the computation and result can all be accommodated within the integers as the number system.
• vinvolves scalar multiplication by a scalar that is not an integer, and which could be construed as a rational number. So the result needs to have a parent whose number system is the rationals.
• yis created explicitly as a vector whose entries are rational numbers.
• Even though wis created only with products of integers, the fact that yhas entries considered as rational numbers, so too does the result.
• The creation of zis the result of adding a vector of integers to a vector of rationals. This is the best example of coercion — Sage promotes xto a vector of rationals and therefore returns a result that is a vector of rationals. Notice that there is no ambiguity and no argument about how to promote x, and the same would be true for any vector full of integers.
The coercion above is automatic, but we can also usually force it to happen without employing an operation.
t = vector([10, 20, 30])
t.parent()
V = QQ^3
t_rational = V(t)
t_rational
t_rational.parent()
W = CC^3
t_complex = W(t)
t_complex
t_complex.parent()
So the syntax is to use the name of the parent like a function and coerce the element into the new parent. This can fail if there is no natural way to make the conversion.
u = vector(CC, [5*I, 4-I])
u
V = QQ^2
V(u)
Coercion is one of the more mysterious aspects of Sage, and the above discussion may not be very clear the first time though. But if you get an error (like the one above) talking about coercion, you know to come back here and have another read through. For now, be sure to create all your vectors and matrices over QQand you should not have any difficulties.
#### 1.
Where have you seen vectors used before in other courses? How were they different?
#### 2.
In words only, when are two vectors equal?
#### 3.
Perform the following computation with vector operations
\begin{equation*} 2\colvector{1\\5\\0} + (-3)\colvector{7\\6\\5} \end{equation*}
### ExercisesVOExercises
#### C10.
Compute
\begin{equation*} 4\colvector{2\\-3\\4\\1\\0}+ (-2)\colvector{1\\2\\-5\\2\\4}+ \colvector{-1\\3\\0\\1\\2}\text{.} \end{equation*}
Solution.
$$\colvector{5\\-13\\26\\1\\-6}\text{.}$$
#### C11.
Solve the given vector equation for $$x\text{,}$$ or explain why no solution exists.
\begin{equation*} 3\colvector{1\\2\\-1}+ 4\colvector{2\\0\\x}= \colvector{11\\6\\17} \end{equation*}
Solution.
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
\begin{align*} \colvector{11\\6\\17} &= 3\colvector{1\\2\\-1} + 4\colvector{2 \\ 0 \\ x} = \colvector{11\\6\\ -3 + 4x}\text{.} \end{align*}
Since the entries of the vectors must be equal by Definition CVE, we have $$-3 + 4x = 17\text{,}$$ which leads to $$x = 5\text{.}$$
#### C12.
Solve the given vector equation for $$\alpha\text{,}$$ or explain why no solution exists.
\begin{equation*} \alpha\colvector{1\\2\\-1}+ 4\colvector{3\\4\\2} = \colvector{-1\\0\\4} \end{equation*}
Solution.
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
\begin{align*} \colvector{\alpha \\ 2\alpha\\ -\alpha} + \colvector{12\\16\\8} &=\colvector{\alpha + 12 \\ 2\alpha + 16\\ -\alpha + 8} =\colvector{-1\\0\\4}\text{.} \end{align*}
Thus, if a solution $$\alpha$$ exists, by Definition CVE then $$\alpha$$ must satisfy the three equations
\begin{align*} \alpha + 12 &= -1\\ 2\alpha + 16 &= 0\\ -\alpha + 8 &= 4 \end{align*}
which leads to $$\alpha = -13\text{,}$$ $$\alpha = -8$$ and $$\alpha = 4\text{.}$$ Since $$\alpha$$ cannot simultaneously have three different values, there is no solution to the original vector equation.
#### C13.
Solve the given vector equation for $$\alpha\text{,}$$ or explain why no solution exists.
\begin{equation*} \alpha\colvector{3\\2\\-2}+ \colvector{6\\1\\2} = \colvector{0\\-3\\6} \end{equation*}
Solution.
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
\begin{align*} \colvector{3\alpha \\ 2\alpha\\ - 2\alpha} + \colvector{6\\1\\2} &= \colvector{3\alpha + 6 \\ 2\alpha + 1\\ -2\alpha + 2} = \colvector{0\\-3\\6}\text{.} \end{align*}
Thus, if a solution $$\alpha$$ exists, by Definition CVE then $$\alpha$$ must satisfy the three equations
\begin{align*} 3\alpha + 6 &= 0\\ 2\alpha + 1 &= -3\\ -2 \alpha + 2 &= 6 \end{align*}
which leads to $$3\alpha = -6\text{,}$$ $$2\alpha = -4$$ and $$-2\alpha = 4\text{.}$$ And thus, the solution to the given vector equation is $$\alpha = -2\text{.}$$
#### C14.
Find $$\alpha$$ and $$\beta$$ that solve the vector equation
\begin{equation*} \alpha\colvector{1\\0}+\beta\colvector{0\\1} = \colvector{3\\2}\text{.} \end{equation*}
Solution.
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
\begin{align*} \colvector{3\\2} &=\alpha\colvector{1\\0} + \beta\colvector{0\\1} =\colvector{\alpha + 0 \\ 0 + \beta} =\colvector{\alpha\\ \beta}\text{.} \end{align*}
Since the entries of the vectors must be equal by Definition CVE, we have $$\alpha = 3$$ and $$\beta = 2\text{.}$$
#### C15.
Find $$\alpha$$ and $$\beta$$ that solve the vector equation.
\begin{equation*} \alpha\colvector{2\\1}+ \beta\colvector{1\\3 }= \colvector{5\\0} \end{equation*}
Solution.
Performing the indicated operations (Definition CVA, Definition CVSM), we obtain the vector equations
\begin{align*} \colvector{5\\0} &= \alpha\colvector{2\\1} + \beta\colvector{1\\3} = \colvector{2\alpha + \beta \\ \alpha + 3\beta}\text{.} \end{align*}
Since the entries of the vectors must be equal by Definition CVE, we obtain the system of equations
\begin{align*} 2\alpha + \beta &= 5\\ \alpha + 3\beta &= 0 \end{align*}
which we can solve by row-reducing the augmented matrix of the system
\begin{align*} \begin{bmatrix} 2 & 1 & 5\\ 1 & 3 & 0 \end{bmatrix} & \rref \begin{bmatrix} \leading{1} & 0 & 3\\ 0 & \leading{1} & -1 \end{bmatrix}\text{.} \end{align*}
Thus, the only solution is $$\alpha = 3\text{,}$$ $$\beta = -1\text{.}$$
#### T05.
Provide reasons (mostly vector space properties) as justification for each of the seven steps of the proof of the following theorem.
For any vectors $$\vect{u},\,\vect{v},\,\vect{w}\in\complex{m}\text{,}$$ if $$\vect{u} + \vect{v} = \vect{u} + \vect{w}\text{,}$$ then $$\vect{v} = \vect{w}\text{.}$$
Proof: Let $$\vect{u},\,\vect{v},\,\vect{w}\in\complex{m}\text{,}$$ and suppose $$\vect{u} + \vect{v} = \vect{u} + \vect{w}\text{.}$$
\begin{align*} \vect{v}&=\zerovector + \vect{v}&&\underline{\hspace{9.090909090909092em}}\\ &=(-\vect{u} + \vect{u}) + \vect{v}&&\underline{\hspace{9.090909090909092em}}\\ &=-\vect{u} + (\vect{u} + \vect{v})&&\underline{\hspace{9.090909090909092em}}\\ &=-\vect{u} + (\vect{u} + \vect{w})&&\underline{\hspace{9.090909090909092em}}\\ &=(-\vect{u} + \vect{u}) + \vect{w}&&\underline{\hspace{9.090909090909092em}}\\ &=\zerovector + \vect{w}&&\underline{\hspace{9.090909090909092em}}\\ &=\vect{w}&&\underline{\hspace{9.090909090909092em}} \end{align*}
Solution.
\begin{align*} \vect{v} &= \zerovector + \vect{v} &&\knowl{./knowl/property-ZC.html}{\text{Property ZC}}\\ &= (-\vect{u} + \vect{u}) + \vect{v}&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= -\vect{u} + (\vect{u} + \vect{v})&& \knowl{./knowl/property-AAC.html}{\text{Property AAC}}\\ &= -\vect{u} + (\vect{u} + \vect{w})&& \text{Hypothesis}\\ &= (-\vect{u} + \vect{u}) + \vect{w}&& \knowl{./knowl/property-AAC.html}{\text{Property AAC}}\\ &= \zerovector + \vect{w}&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= \vect{w}&& \knowl{./knowl/property-ZC.html}{\text{Property ZC}}\ \end{align*}
#### T06.
Provide reasons (mostly vector space properties) as justification for each of the six steps of the proof of the following theorem.
For any vector $$\vect{u}\in\complex{m}\text{,}$$ $$0\vect{u}=\zerovector\text{.}$$
Proof: Let $$\vect{u}\in\complex{m}\text{.}$$
\begin{align*} \zerovector&= 0\vect{u} +(-0\vect{u})&&\underline{\hspace{9.090909090909092em}}\\ &= (0+0)\vect{u} + (-0\vect{u})&&\underline{\hspace{9.090909090909092em}}\\ &= (0\vect{u}+0\vect{u}) + (-0\vect{u})&&\underline{\hspace{9.090909090909092em}}\\ &= 0\vect{u} + (0\vect{u} + (-0\vect{u}))&&\underline{\hspace{9.090909090909092em}}\\ &= 0\vect{u} + \zerovector&&\underline{\hspace{9.090909090909092em}}\\ &= 0\vect{u}&&\underline{\hspace{9.090909090909092em}} \end{align*}
Solution.
\begin{align*} \zerovector &= 0\vect{u} +(-0\vect{u})&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= (0+0)\vect{u} + (-0\vect{u})&& \knowl{./knowl/property-ZCN.html}{\text{Property ZCN}}\\ &= (0\vect{u}+0\vect{u}) + (-0\vect{u})&& \knowl{./knowl/property-DSAC.html}{\text{Property DSAC}}\\ &= 0\vect{u} + (0\vect{u} + (-0\vect{u}))&& \knowl{./knowl/property-AAC.html}{\text{Property AAC}}\\ &= 0\vect{u} + \zerovector&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= 0\vect{u}&& \knowl{./knowl/property-ZC.html}{\text{Property ZC}} \end{align*}
#### T07.
Provide reasons (mostly vector space properties) as justification for each of the six steps of the proof of the following theorem.
For any scalar $$c\text{,}$$ $$c\,\zerovector = \zerovector\text{.}$$
Proof: Let $$c$$ be an arbitrary scalar.
\begin{align*} \zerovector&= c\zerovector + (-c\zerovector)&&\underline{\hspace{9.090909090909092em}}\\ &= c(\zerovector + \zerovector) + (-c\zerovector)&&\underline{\hspace{9.090909090909092em}}\\ &= (c\zerovector + c\zerovector) + (-c\zerovector)&&\underline{\hspace{9.090909090909092em}}\\ &= c\zerovector + (c\zerovector + (-c\zerovector))&&\underline{\hspace{9.090909090909092em}}\\ &= c\zerovector + \zerovector&&\underline{\hspace{9.090909090909092em}}\\ &= c\zerovector&&\underline{\hspace{9.090909090909092em}} \end{align*}
Solution.
\begin{align*} \zerovector&= c\zerovector + (-c\zerovector)&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= c(\zerovector + \zerovector) + (-c\zerovector)&& \knowl{./knowl/property-ZC.html}{\text{Property ZC}}\\ &= (c\zerovector + c\zerovector) + (-c\zerovector)&& \knowl{./knowl/property-DVAC.html}{\text{Property DVAC}}\\ &= c\zerovector + (c\zerovector + (-c\zerovector))&& \knowl{./knowl/property-AAC.html}{\text{Property AAC}}\\ &= c\zerovector + \zerovector&& \knowl{./knowl/property-AIC.html}{\text{Property AIC}}\\ &= c\zerovector&& \knowl{./knowl/property-ZC.html}{\text{Property ZC}} \end{align*}
#### T13.
Prove Property CC of Theorem VSPCV. Write your proof in the style of the proof of Property DSAC given in this section.
Solution.
For all $$1\leq i\leq m\text{,}$$
\begin{align*} \vectorentry{\vect{u}+\vect{v}}{i}& =\vectorentry{\vect{u}}{i}+\vectorentry{\vect{v}}{i}&& \knowl{./knowl/definition-CVA.html}{\text{Definition CVA}}\\ &=\vectorentry{\vect{v}}{i}+\vectorentry{\vect{u}}{i}&& \knowl{./knowl/property-CACN.html}{\text{Property CACN}}\\ &=\vectorentry{\vect{v}+\vect{u}}{i} && \knowl{./knowl/definition-CVA.html}{\text{Definition CVA}}\text{.} \end{align*}
With equality of each component of the vectors $$\vect{u}+\vect{v}$$ and $$\vect{v}+\vect{u}$$ being equal Definition CVE tells us the two vectors are equal.
#### Exercise Group.
Exercises T30, T31 and T32 are about making a careful definition of vector subtraction.
##### T30.
Suppose $$\vect{u}$$ and $$\vect{v}$$ are two vectors in $$\complex{m}\text{.}$$ Define a new operation, called “subtraction,” as the new vector denoted $$\vect{u}-\vect{v}$$ and defined by
\begin{gather*} \vectorentry{\vect{u}-\vect{v}}{i}=\vectorentry{\vect{u}}{i} + \left(-\vectorentry{\vect{v}}{i}\right) \end{gather*}
for $$1\leq i\leq m\text{.}$$ Prove that we can express the subtraction of two vectors in terms of our two basic operations. More precisely, prove that $$\vect{u}-\vect{v}=\vect{u}+(-1)\vect{v}\text{.}$$ So in a sense, subtraction is not something new and different, but is just a convenience. Mimic the style of similar proofs in this section.
##### T31.
Prove, by giving counterexamples, that vector subtraction is not commutative and not associative.
##### T32.
Prove that vector subtraction obeys a distributive property. Specifically, prove that $$\alpha\left(\vect{u}-\vect{v}\right)=\alpha\vect{u}-\alpha\vect{v}\text{.}$$
Can you give two different proofs? Distinguish your two proofs by using the alternate descriptions of vector subtraction provided by Exercise VO.T30. | 8,407 | 27,669 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | latest | en | 0.92133 |
http://math.stackexchange.com/questions/76193/apostol-section-13-25-13-conic-sections | 1,469,379,162,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00185-ip-10-185-27-174.ec2.internal.warc.gz | 159,209,824 | 22,293 | # Apostol Section 13.25 #13 - Conic Sections
Question: Prove that a similarity transformation (replacing $x$ by $tx$ and $y$ by $ty$) carries an ellipse with center at the origin into another ellipse with the same eccentricity.
(The next questions go on to prove the same result (and converses) for hyperbolas.)
Please don't feel the need to read through all my attempts, I just wanted to show what I had tried and give some background on what has been covered in the book.
I think my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?
My Attempt
This is easy to prove for an ellipse (or hyperbola) in standard Cartesian form:
$$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$$
Replacing $x$ and $y$ by $tx$ and $ty$ respectively, we have
$$\frac{(tx)^2}{a^2}+\frac{(ty)^2}{a^2(1-e^2)}=\frac{x^2}{\left(\frac a t\right)^2}+\frac{(y)^2}{\left(\frac a t\right)^2(1-e^2)}=1$$
Which which is another ellipse with a center at the origin with the same eccentricity. The problem is that the standard form only describes ellipses and hyperbolas with vertical directrices. Intuitively I understand that you could choose the coordinate axes so that any ellipse is described by this equation, but this has not been made explicit in the book.
I suppose one question worth asking, at this point, is if this is all Apostol wanted me to prove? Taking the original question literally, it seems my argument for choosing the coordinate axes is a bit of hand-waving to get at the other cases of ellipses.
EDIT: After discussing this with a professor at the university I attended, his interpretation was that Apostol's (high-level) intention was just to show that scaling the shape of an ellipse does not change the eccentricity, because eccentricity is a description of the shape and not the size. He thought, after reading the chapter, that I should consider only those curves in the ideal standard form for all questions such as this, and if necessary remark (although it is yet unproven in the book) that any conic is congruent to one in standard form. In this case, my original solution is adequate.
EDIT #2: It now seems clear that Apostol's intention was to consider only the standard cases, as the next question is Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y'=(e^2-1)x/y$. Indeed, it seems that the integral curves of that differential equation are given by $\frac{x^2}{C}+\frac{y^2}{C(1-e^2)}=1$, which are really only the central conics at the origin with horizontal directrices, not all conics at the origin. Since, in that question, he does not make a distinction, I think he is just being atypically loose with his terminology.
-
Note that the next question is "If two concentric ellipses have the same eccentricity and major axes on the same line, then they are related by a similarity transformation.". Perhaps Apostol does expect me to just accept a bit of hand waving and say that I place the origin of my coordinate axes on the center of the ellipses, and the x axis in the direction of the major axes. If this is acceptable, it certainly speeds up the proof. – process91 Oct 26 '11 at 21:53
This doesn't have to do with your question, but one of the earliest and cruelest jokes of coordinate geometry is how replacing $x$ and $y$ with $tx$ and $ty$ in an equation for a curve does not give the equation for the image of that curve under the transformation of $\mathbb{R}^2$ given by $(u,v) \mapsto (tu, tv)$. (Try the unit circle and $t=2$. One must replace $x$ and $y$ with formulas for the inverse transformation.) For your problem, the result follows if you can prove that any ellipse can be rotated to one that your argument applies to, and that rotation does not change eccentricity. – leslie townes Oct 26 '11 at 22:02
That's interesting. The idea of rotation hasn't really been broached in the book yet, although I see how that proves my argument. One of the next questions after this is Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y'=(e^2-1)x/y$. This is easy if we're allowed to start by saying "Take the coordinate axes to have origin at the center of the conic, and the x axis parallel to the main axes of the conic." Otherwise I'm not sure what the Cartesian equation would be. – process91 Oct 26 '11 at 22:10
Here are two approaches. Since I suspect neither suits your needs, it will have to be done in comments. (a) The eccentricity is defined as a ratio. The major, minor axes are stretched by same factor. So eccentricity is unchanged. (b) Transform, Solve, Transform Back. Specifically, rotate about the center until major, minor axes are in the standard place. Eccentricity does not change. Dilate by factor $t$. You showed eccentricity does not change. Rotate back. Eccentricity does not change. – André Nicolas Oct 26 '11 at 22:21
@AndréNicolas While I understand both approaches, I agree that they probably won't work for me in this problem if I want to stay on course with the book (which I do). I'm starting to grow more comfortable with my "generalized attempt", can you comment on whether it is definitely wrong? – process91 Oct 26 '11 at 22:27
A central conic with center at the origin takes the general form
$$ax^2+bxy+cy^2=1$$
(missing linear terms; why?). We are being asked to prove that
$$t^2(ax^2+bxy+cy^2)=1$$
also has the same eccentricity.
The rotation that will zero out the cross-term will be the same for both, a fair bit of algebra shows that for the rotation
\begin{align*}x&=p\cos\,\varphi-q\sin\,\varphi\\y&=p\sin\,\varphi+q\cos\,\varphi\end{align*}
taking the $\varphi$ given by
$$\tan\;\varphi=\frac{b}{a-c+\sqrt{b^2+(a-c)^2}}$$
does the trick.
Applying that transformation yields the conic
$$\frac{a+c+\sqrt{(a-c)^2+b^2}}{2}p^2+\frac{a+c-\sqrt{(a-c)^2+b^2}}{2}q^2=1$$
in the "unscaled" case, and with the pendant factor $t^2$ in the scaled case, which now looks a lot like the standard form you're accustomed to...
-
From the linear algebra viewpoint, what we have actually done here is a special case of Jacobi's method for diagonalizing a symmetric matrix... – J. M. Oct 26 '11 at 22:57
Thanks, this is obviously correct. I'm still pretty sure that this isn't the method that the book was looking for, normally he leads the proofs a little more if they require this many jumps from currently developed material, but I will go with it for now. – process91 Oct 26 '11 at 23:10
At the very least, there is an easy way to derive the $\tan\,\varphi$ expression I gave: substitute the rotation transformations into the original conic, and equate the coefficient of the cross-term $pq$ to $0$. This yields an expression for $\tan\,2\varphi$; the double-angle formula + judicious use of the quadratic equation yields the $\tan\,\varphi$ expression. – J. M. Oct 26 '11 at 23:21
That's true, but the two big steps are - he has not introduced rotations yet (at all), and he has not had us show that a central conic with center at the origin takes that form (missing the linear terms), which I think he would have had as a separate question. I do really appreciate your answer though. – process91 Oct 26 '11 at 23:49
I did take your points. (apologies if it did not seem that way) :) I'll see if I can offer up a neater way that what I know to show that a conic with linear terms missing has to be central; the proof I am accustomed to will double the length of this answer... – J. M. Oct 26 '11 at 23:53
The general Equation of a ellipse whose axes are parallel to $X-Y$ axes is
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ and the eccentricity is given by $e=\sqrt{1-\dfrac{b^2}{a^2}}$ (I am assuming $a>b$).
If a similarity transformation occurs i.e, $(x,y)\to(tx,ty)$, then new equation of ellipse is $$\dfrac{(tx)^2}{a^2}+\dfrac{(ty)^2}{b^2}=1\ or \dfrac{x^2}{(a/t)^2}+\dfrac{y^2}{(b/t)^2}=1$$ So the eccentrity of the new ellipse $e'=\sqrt{1-\dfrac{(b/t)^2}{(a/t)^2}}=e$
In case Hyperbola just replace $b^2$ with $-b^2$ and remove the condition $(a>b)$.
Hence proved.
-
This is my first attempt above. However, this doesn't address ellipses whose directrices are not parallel to the coordinate axes. – process91 Oct 27 '11 at 11:56
You can always rotate the axes and write the can write thet equation in standard form. – Ramana Venkata Oct 27 '11 at 11:59
Yes, intuitively that makes sense to me. The question is related to rigorously proving that fact, or (at the very least) rigorously proving that the eccentricity doesn't change due to rotation. – process91 Oct 27 '11 at 12:23
my question boils down to this: am I allowed to say "given some conic, align the coordinate axes so that the origin is at the center of the conic and the x-axis is aligned with the main axes of the conic" without any further justification?
If eccentricity is invariant under the operation of "align the coordinate axes so that ..." then calculations involving the eccentricity will not be affected by the choice of a conveniently aligned coordinate system. The invariance can be seen by defining eccentricity in terms that are invariant and do not refer to position in a coordinate system. For that it is enough to define major and minor axis invariantly.
The major axis is the longest chord of the ellipse, and the minor axis is the longest chord perpendicular to the major axis. The lengths of these axes are invariant under rigid motion or reflection of the plane. The ratio of major to minor axes is also invariant under similarity transformations such as $(x,y) \to (tx,ty)$. The eccentricity can be calculated from this ratio, so that it is both scale-invariant (because the ratio is) and isometry-invariant (because it is a function of the axes).
-
The statement - "The lengths of these axes are invariant under rigid motion or reflection of the plane" - seems to be difficult to prove using the tools so far introduced. In particular, concepts such as "rigid motion" and "reflection" have not yet been introduced. – process91 Oct 26 '11 at 22:44
For this question it is only necessary to check that all distances change by a factor of $t$ under the similarity transformation, which can be done using the formula for distance. – zyx Oct 27 '11 at 0:26 | 2,738 | 10,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-30 | latest | en | 0.944886 |
http://gams.cam.nist.gov/1.2 | 1,516,438,461,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00062.warc.gz | 141,287,336 | 13,450 | # §1.2 Elementary Algebra
## §1.2(i) Binomial Coefficients
In (1.2.1) and (1.2.3) $k$ and $n$ are nonnegative integers and $k\leq n$. In (1.2.2), (1.2.4), and (1.2.5) $n$ is a positive integer. See also §26.3(i).
1.2.1 $\genfrac{(}{)}{0.0pt}{}{n}{k}=\frac{n!}{(n-k)!k!}=\genfrac{(}{)}{0.0pt}{}{n}{n% -k}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $!$: factorial (as in $n!$), $k$: integer and $n$: nonnegative integer A&S Ref: 3.1.2 Referenced by: §1.2(i), §1.2(i), Other Changes Permalink: http://dlmf.nist.gov/1.2.E1 Encodings: TeX, pMML, png See also: Annotations for 1.2(i), 1.2 and 1
For complex $z$ the binomial coefficient $\genfrac{(}{)}{0.0pt}{}{z}{k}$ is defined via (1.2.6).
### Binomial Theorem
1.2.2 $(a+b)^{n}=a^{n}+\genfrac{(}{)}{0.0pt}{}{n}{1}a^{n-1}b+\genfrac{(}{)}{0.0pt}{}{% n}{2}a^{n-2}b^{2}+\dots+\genfrac{(}{)}{0.0pt}{}{n}{n-1}ab^{n-1}+b^{n}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient and $n$: nonnegative integer A&S Ref: 3.1.1 Referenced by: §1.10(i), §1.10(i), §1.2(i), §1.2(i), §4.6(ii), §4.6(ii), Subsection 1.2(i) Permalink: http://dlmf.nist.gov/1.2.E2 Encodings: TeX, pMML, png See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.3 $\genfrac{(}{)}{0.0pt}{}{n}{0}+\genfrac{(}{)}{0.0pt}{}{n}{1}+\dots+\genfrac{(}{% )}{0.0pt}{}{n}{n}=2^{n}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient and $n$: nonnegative integer A&S Ref: 3.1.6 Referenced by: §1.2(i) Permalink: http://dlmf.nist.gov/1.2.E3 Encodings: TeX, pMML, png See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.4 $\genfrac{(}{)}{0.0pt}{}{n}{0}-\genfrac{(}{)}{0.0pt}{}{n}{1}+\dots+(-1)^{n}% \genfrac{(}{)}{0.0pt}{}{n}{n}=0.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient and $n$: nonnegative integer A&S Ref: 3.1.7 Referenced by: §1.2(i), §1.2(i), Subsection 1.2(i) Permalink: http://dlmf.nist.gov/1.2.E4 Encodings: TeX, pMML, png See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.5 $\genfrac{(}{)}{0.0pt}{}{n}{0}+\genfrac{(}{)}{0.0pt}{}{n}{2}+\genfrac{(}{)}{0.0% pt}{}{n}{4}+\dots+\genfrac{(}{)}{0.0pt}{}{n}{\ell}=2^{n-1},$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $\ell$: integer and $n$: nonnegative integer Referenced by: §1.2(i), §1.2(i), Subsection 1.2(i) Permalink: http://dlmf.nist.gov/1.2.E5 Encodings: TeX, pMML, png See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
where $\ell$ is $n$ or $n-1$ according as $n$ is even or odd.
In (1.2.6)–(1.2.9) $k$ and $m$ are nonnegative integers and $z$ is complex.
1.2.6 $\genfrac{(}{)}{0.0pt}{}{z}{k}=\frac{z(z-1)\cdots(z-k+1)}{k!}=\frac{(-1)^{k}{% \left(-z\right)_{k}}}{k!}=(-1)^{k}\genfrac{(}{)}{0.0pt}{}{k-z-1}{k}.$ ⓘ Symbols: ${\left(\NVar{a}\right)_{\NVar{n}}}$: Pochhammer’s symbol (or shifted factorial), $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $!$: factorial (as in $n!$), $z$: variable, $k$: integer and $n$: nonnegative integer A&S Ref: 3.1.3 Referenced by: §1.10(i), §1.10(i), §1.2(i), §1.2(i), §1.2(i), §4.6(ii), §4.6(ii), Other Changes Permalink: http://dlmf.nist.gov/1.2.E6 Encodings: TeX, pMML, png Clarification (effective with 1.0.11): As a notational clarification, wherever $n$ appeared originally in in this equation, it has been replaced by $z$. See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.7 $\genfrac{(}{)}{0.0pt}{}{z+1}{k}=\genfrac{(}{)}{0.0pt}{}{z}{k}+\genfrac{(}{)}{0% .0pt}{}{z}{k-1}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $z$: variable, $k$: integer and $n$: nonnegative integer A&S Ref: 3.1.4 Referenced by: §1.2(i) Permalink: http://dlmf.nist.gov/1.2.E7 Encodings: TeX, pMML, png Clarification (effective with 1.0.11): As a notational clarification, wherever $n$ appeared originally in in this equation, it has been replaced by $z$. See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.8 $\sum^{m}_{k=0}\genfrac{(}{)}{0.0pt}{}{z+k}{k}=\genfrac{(}{)}{0.0pt}{}{z+m+1}{m}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $z$: variable, $k$: integer, $m$: nonnegative integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.2.E8 Encodings: TeX, pMML, png Clarification (effective with 1.0.11): As a notational clarification, wherever $n$ appeared originally in in this equation, it has been replaced by $z$. See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
1.2.9 $\genfrac{(}{)}{0.0pt}{}{z}{0}-\genfrac{(}{)}{0.0pt}{}{z}{1}+\dots+(-1)^{m}% \genfrac{(}{)}{0.0pt}{}{z}{m}=(-1)^{m}\genfrac{(}{)}{0.0pt}{}{z-1}{m}.$ ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $z$: variable, $m$: nonnegative integer and $n$: nonnegative integer Referenced by: §1.2(i), §1.2(i), Other Changes Permalink: http://dlmf.nist.gov/1.2.E9 Encodings: TeX, pMML, png Clarification (effective with 1.0.11): As a notational clarification, wherever $n$ appeared originally in in this equation, it has been replaced by $z$. See also: Annotations for 1.2(i), 1.2(i), 1.2 and 1
## §1.2(ii) Finite Series
### Arithmetic Progression
1.2.10 $a+(a+d)+(a+2d)+\dots+(a+(n-1)d)=na+\tfrac{1}{2}n(n-1)d=\tfrac{1}{2}n(a+\ell),$ ⓘ Symbols: $n$: nonnegative integer A&S Ref: 3.1.9 Permalink: http://dlmf.nist.gov/1.2.E10 Encodings: TeX, pMML, png See also: Annotations for 1.2(ii), 1.2(ii), 1.2 and 1
where $\ell$ = last term of the series = $a+(n-1)d$.
### Geometric Progression
1.2.11 $a+ax+ax^{2}+\dots+ax^{n-1}=\frac{a(1-x^{n})}{1-x},$ $x\not=1$. ⓘ Symbols: $n$: nonnegative integer A&S Ref: 3.1.10 Permalink: http://dlmf.nist.gov/1.2.E11 Encodings: TeX, pMML, png See also: Annotations for 1.2(ii), 1.2(ii), 1.2 and 1
## §1.2(iii) Partial Fractions
Let $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ be distinct constants, and $f(x)$ be a polynomial of degree less than $n$. Then
1.2.12 $\frac{f(x)}{(x-\alpha_{1})(x-\alpha_{2})\cdots(x-\alpha_{n})}=\frac{A_{1}}{x-% \alpha_{1}}+\frac{A_{2}}{x-\alpha_{2}}+\dots+\frac{A_{n}}{x-\alpha_{n}},$ ⓘ Symbols: $n$: nonnegative integer, $f(x)$: polynomial of degree less than $n$ and $A_{j}$: coefficient Permalink: http://dlmf.nist.gov/1.2.E12 Encodings: TeX, pMML, png See also: Annotations for 1.2(iii), 1.2 and 1
where
1.2.13 $A_{j}=\frac{f(\alpha_{j})}{\prod\limits_{k\not=j}(\alpha_{j}-\alpha_{k})}.$ ⓘ Defines: $A_{j}$: coefficient (locally) Symbols: $j$: integer, $k$: integer and $f(x)$: polynomial of degree less than $n$ Permalink: http://dlmf.nist.gov/1.2.E13 Encodings: TeX, pMML, png See also: Annotations for 1.2(iii), 1.2 and 1
Also,
1.2.14 $\frac{f(x)}{(x-\alpha_{1})^{n}}=\frac{B_{1}}{x-\alpha_{1}}+\frac{B_{2}}{(x-% \alpha_{1})^{2}}+\dots+\frac{B_{n}}{(x-\alpha_{1})^{n}},$ ⓘ Symbols: $n$: nonnegative integer, $f(x)$: polynomial of degree less than $n$ and $B_{j}$: coefficient Permalink: http://dlmf.nist.gov/1.2.E14 Encodings: TeX, pMML, png See also: Annotations for 1.2(iii), 1.2 and 1
where
1.2.15 $B_{j}=\frac{f^{(n-j)}(\alpha_{1})}{(n-j)!},$ ⓘ Defines: $B_{j}$: coefficient (locally) Symbols: $!$: factorial (as in $n!$), $j$: integer, $n$: nonnegative integer and $f(x)$: polynomial of degree less than $n$ Permalink: http://dlmf.nist.gov/1.2.E15 Encodings: TeX, pMML, png See also: Annotations for 1.2(iii), 1.2 and 1
and $f^{(k)}$ is the $k$-th derivative of $f$1.4(iii)).
If $m_{1},m_{2},\dots,m_{n}$ are positive integers and $\deg f<\sum_{j=1}^{n}m_{j}$, then there exist polynomials $f_{j}(x)$, $\deg f_{j}, such that
1.2.16 $\frac{f(x)}{(x-\alpha_{1})^{m_{1}}(x-\alpha_{2})^{m_{2}}\cdots(x-\alpha_{n})^{% m_{n}}}=\frac{f_{1}(x)}{(x-\alpha_{1})^{m_{1}}}+\frac{f_{2}(x)}{(x-\alpha_{2})% ^{m_{2}}}+\cdots+\frac{f_{n}(x)}{(x-\alpha_{n})^{m_{n}}}.$ ⓘ Symbols: $m$: nonnegative integer, $n$: nonnegative integer and $f(x)$: polynomial of degree less than $n$ Permalink: http://dlmf.nist.gov/1.2.E16 Encodings: TeX, pMML, png See also: Annotations for 1.2(iii), 1.2 and 1
To find the polynomials $f_{j}(x)$, $j=1,2,\dots,n$, multiply both sides by the denominator of the left-hand side and equate coefficients. See Chrystal (1959a, pp. 151–159).
## §1.2(iv) Means
The arithmetic mean of $n$ numbers $a_{1},a_{2},\dots,a_{n}$ is
1.2.17 $A=\frac{a_{1}+a_{2}+\dots+a_{n}}{n}.$ ⓘ Defines: $A$: arithmetic mean (locally) Symbols: $n$: nonnegative integer A&S Ref: 3.1.11 Permalink: http://dlmf.nist.gov/1.2.E17 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
The geometric mean $G$ and harmonic mean $H$ of $n$ positive numbers $a_{1},a_{2},\dots,a_{n}$ are given by
1.2.18 $G=(a_{1}a_{2}\cdots a_{n})^{1/n},$ ⓘ Defines: $G$: geometric mean (locally) Symbols: $n$: nonnegative integer A&S Ref: 3.1.12 Permalink: http://dlmf.nist.gov/1.2.E18 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
1.2.19 $\frac{1}{H}=\frac{1}{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\dots+\frac{1}{a_% {n}}\right).$ ⓘ Defines: $H$: harmonic mean (locally) Symbols: $n$: nonnegative integer A&S Ref: 3.1.13 Permalink: http://dlmf.nist.gov/1.2.E19 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
If $r$ is a nonzero real number, then the weighted mean $M(r)$ of $n$ nonnegative numbers $a_{1},a_{2},\dots,a_{n}$, and $n$ positive numbers $p_{1},p_{2},\dots,p_{n}$ with
1.2.20 $p_{1}+p_{2}+\dots+p_{n}=1,$ ⓘ Symbols: $n$: nonnegative integer and $p_{j}$; positive numbers Permalink: http://dlmf.nist.gov/1.2.E20 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
is defined by
1.2.21 $M(r)=(p_{1}a_{1}^{r}+p_{2}a_{2}^{r}+\dots+p_{n}a_{n}^{r})^{1/r},$ ⓘ Defines: $M(r)$: weighted mean (locally) Symbols: $n$: nonnegative integer and $p_{j}$; positive numbers Permalink: http://dlmf.nist.gov/1.2.E21 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
with the exception
1.2.22 $M(r)=0,$ $r<0$ and $a_{1}a_{2}\dots a_{n}=0$. ⓘ Symbols: $n$: nonnegative integer and $M(r)$: weighted mean A&S Ref: 3.1.15 Permalink: http://dlmf.nist.gov/1.2.E22 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
1.2.23 $\displaystyle\lim_{r\to\infty}M(r)$ $\displaystyle=\max(a_{1},a_{2},\dots,a_{n}),$ ⓘ Symbols: $n$: nonnegative integer and $M(r)$: weighted mean A&S Ref: 3.1.16 Permalink: http://dlmf.nist.gov/1.2.E23 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1 1.2.24 $\displaystyle\lim_{r\to-\infty}M(r)$ $\displaystyle=\min(a_{1},a_{2},\dots,a_{n}).$ ⓘ Symbols: $n$: nonnegative integer and $M(r)$: weighted mean A&S Ref: 3.1.17 Permalink: http://dlmf.nist.gov/1.2.E24 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
For $p_{j}=1/n$, $j=1,2,\dots,n$,
1.2.25 $\displaystyle M(1)$ $\displaystyle=A,$ $\displaystyle M(-1)$ $\displaystyle=H,$ ⓘ Symbols: $A$: arithmetic mean, $H$: harmonic mean and $M(r)$: weighted mean A&S Ref: 3.1.19 3.1.20 Permalink: http://dlmf.nist.gov/1.2.E25 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.2(iv), 1.2 and 1
and
1.2.26 $\lim_{r\to 0}M(r)=G.$ ⓘ Symbols: $G$: geometric mean and $M(r)$: weighted mean A&S Ref: 3.1.18 Permalink: http://dlmf.nist.gov/1.2.E26 Encodings: TeX, pMML, png See also: Annotations for 1.2(iv), 1.2 and 1
The last two equations require $a_{j}>0$ for all $j$. | 4,762 | 11,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 174, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-05 | latest | en | 0.495037 |
https://www.omnicalculator.com/health/babys-milk-intake | 1,709,340,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00461.warc.gz | 916,526,261 | 101,538 | Last updated:
# Baby's Milk Intake Calculator
Baby's milk intake calculatorHow to calculate milk intake for a baby?Baby's milk intake chartBaby's milk intake by weightFAQs
The baby's milk intake calculator is a worry-free tool to help all the mothers who want to calculate the milk intake for their baby according to the baby's age, weight, and feeding habits.
If you are a parent to a newborn, you have definitely wondered, "How much milk does my baby need?", "How to calculate milk intake for my baby?" or maybe you know all that and are just looking to make a baby's milk intake chart. Worry no more in any of these cases because we have it all for you.
If you are a parent, some other noteworthy calculators for you to look at are:
## Baby's milk intake calculator
The baby's milk intake calculator is a hassle-free tool that helps you estimate the amount of milk needed by your baby in their early months.
Babies need proper feeding, whether it's breast milk or formula milk, and this intake depends on their age, weight, and the number of feedings they require per day.
Our calculator requires the weight of the baby, preferably in kilograms, but if you have it in any other unit (say, pounds or ounces), worry not. You can always use our pounds and ounces calculator to convert to the unit of your convenience. The next important thing you need is your baby's age in months, and, lastly, you need to record the number of feedings your baby requires during a day.
Based on these variables, you will find out the amount of milk the baby needs during the whole day and the amount per feeding.
## How to calculate milk intake for a baby?
The baby's milk intake calculator performs two different calculations. The first one finds the total amount of milk needed by the baby during the day and the second one computes the amount of milk per serving.
Since we are using both age and weight, the formula to evaluate a baby's milk intake is:
$\text {Required milk} = \text {Baby's age equivalence} \times \text{Baby's weight}$
where:
• $\text {Baby's age equivalence}$ - An estimated series of numbers used as constants for various age groups.
The formula to determine the milk amount per serving is:
$\text {Milk per serving = Required milk / Number of servings}$
So, all that you need is your baby's weight, age, and the number of feedings that he requires in a day. For this purpose, you might already be maintaining a chart, and, if not, it's never too late to start keeping your baby's milk intake chart.
Once you have this information, please place them in the formula to get your results.
For instance, your baby is $\text {2 months old}$, weighs $\text {4 kg}$, and needs $8$ feedings per day. The total amount of milk required during the whole day is $480 \text{ ml}$, which means $60 \text { ml}$ of milk per feeding.
The baby's age equivalence in terms of milk requirement is given in the table below:
Age
Required milk
Newborn
60 ml per kilogram per day
< 1 month
180 ml per kilogram per day
1 - 3 months
150 ml per kilogram per day
3 - 6 months
120 ml per kilogram per day
6 - 9 months
100 ml per kilogram per day
9 - 12 months
90 ml per kilogram per day
One important thing to keep in mind is that the default units used in the formula to calculate baby's milk intake needs are kilograms (kg) for weight, months for age, and milliliters (ml) for the amount of milk. Even though you have the freedom to choose any unit, remember to select appropriate units for all variables if you want to perform the computations by yourself. For instance, if you have your baby's weight in ounces, then it would be wise to use fluid ounces for milk quantity.
Generally, an infant requires at least 8-12 feedings daily, after every two to three hours, and this amount decreases with the passage of time as the baby develops a proper sleeping schedule.
## Baby's milk intake chart
We need to monitor infants that rely exclusively on breastfeeding. It's worth maintaining a journal or chart in case of the mother's absence due to work or other outdoor chores.
The important things to keep in the baby's milk intake chart are:
• The number of feedings per day for the baby;
• The weekly weight of the baby;
• Age of the baby;
• Sleeping habits of the baby; and
• Amount of milk required during a day and per serving.
Some research suggests that . Still, each baby has his own needs, so we could estimate the milk intake based on the baby's weight as well as his age.
Therefore, these charts can be useful in maintaining a baby's dietary and calorie records.
You can expect a decline in the milk intake any time after six months, especially when the baby starts eating solid food. If you maintain the baby's milk intake chart, you will not have to stress about slight changes in the baby's milk intake.
## Baby's milk intake by weight
As a parent, you might be concerned with your baby's milk intake needs, especially if you are a first-time parent, but our baby's milk intake calculator ensures you remain stress-free.
The baby's milk intake varies a lot during the baby's first month. After that, there should be a gradual decrease when the baby starts consuming foods other than milk. There could also be variation in the milk intake based on the weight of the baby, which is what our calculator bases its calculations on.
The variables that influence the milk intake of a baby are:
• Growth rate;
• Age;
• Metabolic rate;
• Activity level;
• Amount of solid/semi-solid food;
• Feeding habit/pattern; and
• Sleeping habits.
FAQs
### How much milk does my baby need?
The baby's milk intake depends on his weight, age, and other factors like growth rate and genetics.
But if you want to estimate how much milk is required, you need to multiply your baby's weight by the age equivalence number, which is the average amount of milk required at various ages.
Please refer to the list below:
• A newborn needs 60 ml per kg per day.
• A baby less than 1-month-old needs 180ml per kg per day.
• 1 - 3 months old baby needs 150 ml per kg per day.
• 3 - 6 months old baby needs 120 ml per kg per day.
• 6 - 9 months old baby needs 100 ml per kg per day.
• 9 - 12 months old baby needs 90 ml per kg per day.
### When does the baby's milk intake decrease?
The baby's milk intake starts to decrease when he starts taking solid or semi-solid food.
The decline in the baby's milk intake should not be a matter of major concern if he has started eating other foods like fruit or vegetable puree and should be considered a sign of good health.
### How much milk does my 1 month old baby need?
Your baby's milk intake should be around 2-4 ounces or 60 - 110 ml, which is comparable to the stomach capacity of a one-month-old baby (typically 2.5 to 5 ounces).
The average milk intake for a 1-month-old baby is 150 ml per kilogram per day. This means whatever the weight of your baby is, all you have to do is multiply it by 150 and you will have the total milk intake during the whole day for your beloved baby.
### How much milk does my 7 kg, 5 months old baby girl need?
Your baby needs 840 ml of milk during the whole day.
The required amount for a baby between 3 - 6 months old is 120 ml per kilogram per day. So, if you want to calculate the milk intake by yourself, all you have to do is multiply your baby's weight by 120 and you will get the milk quantity required during the whole day.
You can further divide this amount into amounts per feeding, the formula for which is:
Milk per serving = Required milk / Number of servings
People also viewed | 1,730 | 7,596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-10 | longest | en | 0.950346 |
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