Answer: The mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.
Step-by-step explanation:
if there is a linear relationship between two variables x and y , then we represent this relation in the form of equation as :[tex]y= mx+c[/tex] (*)
, where m = rate of change of y with respect to x
c= Constant or the value of y when x=0.
We assume there is a linear relationship between the curb weight of a car (in thousands of kilograms) and its combined mileage (in miles per gallon).
Let y= combined mileage (in miles per gallon)
x= curb weight of a car (in thousands of kilograms)
When x= 3 thousand kg , y = 9 mpg
⇒ [tex]9= m(3)+c[/tex] [Put values in (*)]
⇒ [tex]9= 3m+c--------(1)[/tex]
When x= 1.9 thousand kg , y = 21 mpg
⇒ [tex]21= m(1.9)+c[/tex] [Put values in (*)]
⇒ [tex]21= 1.9m+c--------(2)[/tex]
Eliminate the equation (1) from (2) , we get
[tex]1.9m-3m=21-9[/tex]
[tex]-1.1m=12\\\Rightarrow\ m=\dfrac{12}{-1.1}=\dfrac{-120}{11}[/tex]
Put value of m in (1) , we gte
[tex]9= 3(\dfrac{-120}{11})+c[/tex]
[tex]9= \dfrac{-360}{11}+c\\\Rightarrow\ c=9+\dfrac{360}{11}=\dfrac{459}{11}[/tex]
Substitute the value of m and c in (*) , we get
[tex]y= \dfrac{-120}{11}x+\dfrac{459}{11}[/tex]
When x= 1.1
[tex]y= \dfrac{-120}{11}(1.1)+\dfrac{459}{11}[/tex]
[tex]y= \dfrac{-120}{11}(\dfrac{11}{10})+\dfrac{459}{11}[/tex]
[tex]y=-12+\dfrac{459}{11}=\dfrac{327}{11}\approx29.72[/tex]
Hence, the mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.
