Question 107512
Linear equation means you need two points to calculate the equation. 
You have two points. 
Let C=chirp rate and T=temperature. 
The equation of a line is given by,
{{{C=mT+b}}}
where m is the slope of the line and b is the intercept. 
The slope is given by 
{{{m=(T[2]-T[1])/(C[2]-C[1])}}}
Using the two data points, you gave,
{{{m=(C[2]-C[1])/(T[2]-T[1])}}}
{{{m=(126-70)/(69-55)}}}
{{{m=56/14}}}
{{{m=4}}} and the units are chirps per minute/degree F.
You can solve for b by using m you just calculated and one data point. 
{{{C=mT+b}}}
{{{C=4T+b}}}
{{{126=4(69)+b}}}
{{{126=276+b}}}
{{{126-276=276-276+b}}}
{{{b=-150}}}
The equation of your line is then
{{{C(T)=4T-150}}}
You can verify with your two points. 
{{{C(55)=4T-150}}}
{{{C(55)=4(55)-150}}}
{{{C(55)=220-150}}}
{{{C(55)=70}}}
70 chirps at 55 degrees F. That checks out. 
{{{C(69)=4T-150}}}
{{{C(69)=4(69)-150}}}
{{{C(69)=276-150}}}
{{{C(69)=126}}}
126 chirps at 69 degrees F. That checks out. 
Your equation matches the data.
a.)C(T)=4T-150 where C is chirps per minute, T is degrees F. 
Chirping rate at 90F.
C(90)=4(90)-150
C(90)=360-150
C(90)=210
b.)210 chirps per minute at 90 degrees F.
C(T)=4T-150=120
4T-150=120
4T-150+150=120+150
4T=270
T=270/4
c.)T=67.5 degrees F for 120 chirps per minute.

d.) The temperature intercept is where the chirping rate is equal to zero. 
4T-120=0
4T-120+120=120
4T=120
T=30 degrees F. 
In the real world, that says chirping doesn't start until 30 degrees.
The birds are probably conserving their energy because its so cold. 
e.) The chirping intercept is where the temperature is equal to zero. 
C(0)=4(0)-120=-120.
This doesn't really have a real world interpretation since there is no such thing as a negative chirp. Chirps need to be positive, that is not defined below T=30 degrees F. 


{{{ graph( 300, 300, -10, 90, -10, 190, 4x-120) }}}