8 c. d. Using the LINear REGresssion program, the line of best fit is: = 3.8613x + 180.2920 y −5 5 160 e. −8 d. Using the LINear REGression program, the line of best fit is: = −1.8x + 3.6 y 0 –25 90 e. 8 18. a. 25 3 –3 −40 0 –2 0 Selection of points will vary. We select 17. a. (–30, 10) and (–14, 18). The of the –25 line containing these points is: 0 90 = 18 − 10 = 8 = 1 m (−30) 160 −14 − 16 2 b The equation of the line is: − − = ) y y1 m( x x1 − 10 = 1 y − (−30) x 2 b. Answers will vary. We select (–20,100) − 10 = 1 + y x and (–15,118). The slope of the line 15 containing these points is: 2 = 118 − 100 = 18 = 3.6 m 1 ( ) (−20) −15 − = + 25 y x 2 5 The equation of the line is: = 18 + 172 y x − = − y y1 m( x x1 ) c. 25 − = − 100 18 ( y x (−20)) 5 − 100 = 18 + 72 y x 5 −40 0 0 5 d. Using the LINear REGression program, c. 160 the line of best fit is: = + 0.4421x 23.4559 y
c. 8 d. Using the LINear REGresssion program, the line of best fit is: y = 3.8613x + 180.2920 −5 5 e. 160 −8 d. Using the LINear REGression program, the line of best fit is: y = −1.8x + 3.6 –25 0 90 e. 8 18. a. 25 –3 3 –2 −40 0 0 17. a. –25 160 0 90 Selection of points will vary. We select (–30, 10) and (–14, 18). The of the line containing these points is: m = 18 − 10 = 8 = 1 −14 − (−30) 16 2 The equation of the line is: y − y1 = m( x − x1 ) y − 10 = 1 x − (−30) b. Answers will vary. We select (–20,100) and (–15,118). The slope of the line containing these points is: m = 118 − 100 = 18 = 3.6 ( ) 2 y − 10 = 1 x + 15 2 1 −15 − (−20) 5 y = x + 25 2 The equation of the line is: y − y1 = m( x − x1 ) y − 100 = 18 ( x − (−20)) 5 y − 100 = 18 x + 72 5 y = 18 x + 172 c. 25 −40 0 0 5 c. 160 d. Using the LINear REGression program, the line of best fit is: y = 0.4421x + 23.4559 b