Answer:
a) [tex]y=4x-160[/tex]
b) [tex]y=4(102)-160=248\frac{chirps}{min}[/tex]
Step-by-step explanation:
We know that theres a linear relationship between the rate of the chirping of crickets and the air temperature.
The equation of a line is:
[tex]y=mx+b[/tex]
So, let's name our variables
x= Temperature
y=Rate of the chirping
First of all, we need to find the slope with the two given points
[tex]x_{1}=[/tex]60ºF , [tex]y_{1} =80\frac{chirps}{min}[/tex]
[tex]x_{2} =[/tex]80ºF, [tex]y_{2} =160\frac{chirps}{min}[/tex]
By,
[tex]m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} } =\frac{160-80}{80-60}[/tex]
[tex]m=4[/tex]
Now, the equation between the air temperature and the number of chirps is:
[tex]y-y_{1} =m(x-x_{1} )[/tex]
[tex]y-80=4(x-60)[/tex]
Solving for y,
a) [tex]y=4x-160[/tex]
b) To calculate the rate at which the crickets chirp when the temperature is 102 ºF we need to evaluate y(102)
[tex]y=4(102)-160=248\frac{chirps}{min}[/tex]
