Answer:
A) [tex]V =12500[1-\frac{17.2 \times t}{100} ][/tex]
B) [tex]V = 12500[1-\frac{19}{100} ]^{t}[/tex]
Step-by-step explanation:
In two years i.e. from 2013 to 2015 the car value decreases from $12500 to $8200.
a) If the rate of decrease is constant and it is r% per year, then
[tex]8200 = 12500[1-\frac{r \times 2}{100}}][/tex]
⇒ r = 17.2%
Therefore, the value of the car is given by [tex]V =12500[1-\frac{17.2 \times t}{100} ][/tex], where, t is in years since 2013. (Answer)
b) If the rate of decrease is exponential and it is r%, then
[tex]8200 = 12500[1-\frac{r}{100} ]^{2}[/tex]
⇒ r = 19%
Therefore, the value of the car is given by [tex]V = 12500[1-\frac{19}{100} ]^{t}[/tex], where, t is in years since 2013. (Answer)
