Question

The chirping rate of crickets varies linearly with outdoor temperature. one evening, nicki counted 44 cricket chirps in one 15-second span, and noted that the outdoor temperature was 29c. on another evening, she counted 32 chirps in 15 seconds, and noted that the temperature was 22c. which equation can be used to find the temperature, T, in degree celsius, given the number of cricket chirps, c, that are counted in a 15-second span?
A. T=12/7c-230/7
B. T=12/7c-160/7
C. T=7/12c+115/6
D. T=7/12c+10/3

Answer 1

Answer:

Option D. T=7/12c+10/3

Step-by-step explanation:

Let

T -----> the temperature in degree Celsius (y-axis)

c ---> the number of cricket chirps,that are counted in a 15-second span (x-axis)

we have the ordered pairs

(32,22) and (44,29)

step 1

Find the slope of the linear equation

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{29-22}{44-32}$

$m=\frac{7}{12}\ \frac{^oC}{chirp}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{7}{12}$

$point\ (32,22)$

substitute

$y-22=\frac{7}{12}(x-32)$

step 3

Convert to slope intercept form

$y=mx+b$

isolate the variable y

$y-22=\frac{7}{12}(x-32)\\\\y-22=\frac{7}{12}x-\frac{224}{12}\\\\y=\frac{7}{12}x-\frac{224}{12}+22\\\\y=\frac{7}{12}x+\frac{40}{12}$

simplify

$y=\frac{7}{12}x+\frac{10}{3}$