# Chapter 1 - Section 1.2 - Mathematical Models: A Catalog of Essential Functions. - 1.2 Exercises - Page 34: 17

(a) $T = \frac{N}{6}+\frac{307}{6}$ (b) The slope of the graph is $\frac{1}{6}$. The slope represents the temperature increase for each additional chirp per minute. (c) $T = 76.2^{\circ}~F$

#### Work Step by Step

(a) We can use the data pairs $(113,70)$ and $(173,80)$ to find the slope $m$ of the function. $m = \frac{\Delta T}{\Delta N}$ $m = \frac{80-70}{173-113}$ $m = \frac{1}{6}$ We can use the slope and the data pair (113,70) to find the equation of the function. $T - 70 = \frac{1}{6}(N-113)$ $T = \frac{N}{6}+\frac{307}{6}$ (b) The slope of the graph is $\frac{1}{6}$. The slope represents the temperature increase for each additional chirp per minute. (c) We can find the temperature when the crickets are chirping at 150 chirps per minute. $T = \frac{N}{6}+\frac{307}{6}$ $T = \frac{150}{6}+\frac{307}{6}$ $T = \frac{457}{6}$ $T = 76.2^{\circ}~F$

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