IAT SOLUTIONS - C_7.pdf

Chapter 7, continued4. x 1 2 3 4 5 6 7lny 2.71 3.14 3.69 3.95 4.38 4.65 4.94The line appears to pass through (1, 2.80) and(6, 4.65).4.65 2 2.80M 5 } ø 0.376 2 1ln yln y 2 2.80 5 0.37(x 2 1)ln y 5 0.37x 1 2.430.37x 1 2.43y 5 ey 5 e 2.43 (e 0.37 ) xy 5 11.36(1.45) x1(1, 2.80)1(6, 4.65)Using a scatter plot of (x, ln y), an exponential model forthe data (x, y) is y 5 11.36(1.45) x . Using the exponentialregression feature of a graphing calculator, a model isy 5 11.39(1.45) x .x7. (5, 8): 8 5 a p 5 b → a 5 }85 b(10, 34): 34 5 a p 10 b34 5 1 8 }5 b 2 p 10b34 5 8 p 2 b17}4 5 2blog 217}4 5 blog }174}log 2 5 b2.09 ø ba 5 }85 ø 8b } ø 0.2772.095So, an equation is y 5 0.277x 2.09 .Copyright © by McDougal Littell, a division of Houghton Mifflin Company.5. (2, 1): 1 5 a p 2 b → a 5 }12 b(7, 6): 6 5 a p 7 b6 5 1 1 }2 b 2 p 7b6 5 1 p 1 }7 22log 7/26 5 blog 65 b}log 7 }21.43 ø ba 5 }12 ø 1b } ø 0.3711.432So, an equation is y 5 0.371x 1.43 .6. (3, 4): 4 5 a p 3 b → a 5 }43 b(6, 15): 15 5 a p 6 bb15 5 1 4 }3 b 2 p 6b15 5 4 p 2 b15}4 5 2blog 215}4 5 blog }154}log 2 5 b1.91 ø ba 5 }43 ø 4b } ø 0.4911.913So, an equation is y 5 0.491x 1.91 .8. (3, 5): 5 5 a p 3 b → a 5 }53 b(3, 7): 7 5 a p 3 b7 5 1 5 }3 b 2 p 3b7 Þ 5 p 1Sample answer: The two points cannot form a powerfunction.9.ln x 1.099 2.398 2.944 3.612 4.007ln y 5.196 4.583 4.154 3.661 3.3501ln y(1.099, 5.196)1(4.007, 3.350)ln x3.350 2 5.196M 5 }}4.007 2 1.099 ø 20.635ln y 2 3.350 5 20.635(ln x 2 4.007)ln y 5 20.635 ln x 1 5.894ln y 5 ln x 20.635 1 5.894y 5 e ln x20.635 1 5.894y 5 e 5.894 p e lnx20.635y 5 362.85x 20.6357.7 Exercises (pp. 533 – 536)Skill Practice1. Given a set of more than two data pairs (x, y), you candecide whether an exponential function fits the data wellby making a scatterplot of the points (x, ln y).Algebra 2Worked-Out Solution Key451