text: # Thread: Pre-test 6b Help Hurry
  
  1. ## Pre-test 6b Help Hurry
  
  questions are on attachment!!!
  
  2. Hey Brookester:
  
  Let's do #4. The bacteria.
  
  What you need to do is use the given data and solve for k.
  
  Once you know k, you can use that to solve the problem.
  
  Given:
  
  $\displaystyle 125=50e^{3k}$
  
  Solve for k:
  
  $\displaystyle \frac{5}{2}=e^{3k}$
  
  $\displaystyle ln(\frac{5}{2})=3k$
  
  $\displaystyle k=\frac{ln(\frac{5}{2})}{3}$
  
  Now use that in the problem to solve. The initial population is 50. Triple that is 150.
  
  $\displaystyle 150=50e^{\frac{ln(\frac{5}{2})}{3}t}$
  
  $\displaystyle 3=e^{\frac{ln(\frac{5}{2})}{3}t}$
  
  $\displaystyle ln(3)=\frac{ln(\frac{5}{2})}{3}t$
  
  $\displaystyle t=\frac{3ln(3)}{ln(\frac{5}{2})}\approx{3.60}$
  
  3. Hello, Brooke!
  
  I believe we're suppose to use: $\displaystyle A \:= \:P\left(1 + \frac{i}{n}\right)^{nt}\quad\Rightarrow\quad P \;=\;\frac{A}{\left(1 + \frac{i}{n}\right)^{nt}}$
  
  where $\displaystyle P$ is the principal investment,
  $\displaystyle i$ is the annual interest rate,
  $\displaystyle n$ is the number of compounding periods per year,
  $\displaystyle t$ is the number of years.
  
  2. An investment with an interest rate of 7% per annum yields $4000 in 3.5 years. What is the present value for each compounding period? . . A: annually . . B: semi-annually . . C: quarterly . . D: monthly (A) annually:$\displaystyle n = 1\displaystyle P \;= \;\frac{4000}{(1 + 0.07)^{3.5}} \;=\;3156.579774 \:\approx\; \$3,156.58$
  
  (B) semi-annually: $\displaystyle n = 2$
  
  $\displaystyle P \;= \;\frac{4000}{\left(1 + \frac{0.07}{2}\right)^{(2)(3.5)}} \;= \; 3143.963843 \:\approx\:\$3143.96\$
  
  You can do the other two . . .
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