Answer
(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.
$T = \frac{N}{6}+\frac{307}{6}$
$T = \frac{150}{6}+\frac{307}{6}$
$T = \frac{457}{6}$
$T = 76.2^{\circ}~F$
