Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.2 Mathematical Models: A Catalog of Essential Functions - 1.2 Exercises - Page 34: 17

Answer

( a.) T=$\frac{1}{6}$ N +$\frac{307}{6}$ ( b.) The slope of the equation is $\frac{1}{6}$. The slope represents a $\frac{1}{6}$ change in degrees Fahrenheit for every chirp per minute change. ( c.) 76 degrees Fahrenheit

Work Step by Step

( a.) Using the two sets of data given, 113 chirps per minute at 70 degrees Fahrenheit and 173 chirps per minute at 80 degrees Fahrenheit, calculate slope using the change in y over the change in x. Slope = $\frac{80-70}{173-113}$ = $\frac{1}{6}$. Next you find the equation using the equation y = mx+ b, solving for b. 70=$\frac{1}{6}$(113) + b. b=$\frac{307}{6}$ The finished equation is T= $\frac{1}{6}$N + $\frac{307}{6}$ ( b.) Slope is calculated by using the change of temperature over the change in chirps per minute, Slope = $\frac{80-70}{173-113}$ = $\frac{1}{6}$. Because of the way slope is found, you can rationalize that the change in Fahrenheit is proportionate to the change in chirps per minute. ( c.) Substitute 150 chirps per minute for the N variable. T = $\frac{1}{6}$(150) + $\frac{307}{6}$ T= 76 degrees Fahrenheit
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