Let's assume we have a base value of K and we are told that it depreciates an average of p% each year. The value V(t) after t years is given by the following exponential function:
[tex]V(t)=K\cdot(1-\frac{p}{100})^t[/tex]In this case K is the original value of the car $39367, p is the percentage of depreciation i.e. 22. Then the value of the car bought by Brianna t years after she bought it is:
[tex]V(t)=39367\cdot(1-\frac{22}{100})^t=39367\cdot0.78^t[/tex]Then the approximate value in 9 years will be V(9):
[tex]V(9)=39367\cdot0.78^9=4207.11[/tex]We need to round it to the nearest dollar so the answer is $4207.
