Answer-
[tex]\boxed{\boxed{y=9000(1.0592)^x}}[/tex]
And cost of a car in the year 2000 will be $20134.18
Solution-
Let's assume,
x = number of years after 1986
y = average cost of sedan in dollar
The exponential model that will model the scenario will be in the form of,
[tex]y=ab^x[/tex]
where a and b are constants
As given that, in 1986 average cost of sedan was $9000 and in 1991 average cost of sedan was $12000
So, the points (0, 9000) and (5, 12000) will satisfy or lie on the exponential curve.
Putting (0, 9000) in the equation,
[tex]\Rightarrow 9000=ab^0[/tex]
[tex]\Rightarrow a\times 1=9000[/tex]
[tex]\Rightarrow a=9000[/tex]
Now, the equation becomes [tex]y=9000b^x[/tex]
Putting (5, 12000) in this equation,
[tex]\Rightarrow 12000=9000b^5[/tex]
[tex]\Rightarrow b^5=\dfrac{12000}{9000}[/tex]
[tex]\Rightarrow b=\sqrt[5]{\dfrac{4}{3}}[/tex]
[tex]\Rightarrow b=1.0592[/tex]
Putting the values,
[tex]y=9000(1.0592)^x[/tex]
As we have to calculate the cost of sedan in 2000, so putting x=14(as 2000-1986=14),
[tex]y=9000(1.0592)^{14}=20134.18[/tex]
Therefore, cost of a car in the year 2000 will be $20134.18
