Question

In 2013 your car was worth $12,500. In 2015 your car was worth $8,200.

Answer 1

Answer:

A) $V =12500[1-\frac{17.2 \times t}{100} ]$

B)  $V = 12500[1-\frac{19}{100} ]^{t}$

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

$8200 = 12500[1-\frac{r \times 2}{100}}]$

⇒ r = 17.2%

Therefore, the value of the car is given by $V =12500[1-\frac{17.2 \times t}{100} ]$, where, t is in years since 2013. (Answer)

b) If the rate of decrease is exponential and it is r%, then

$8200 = 12500[1-\frac{r}{100} ]^{2}$

⇒ r = 19%

Therefore, the value of the car is given by $V = 12500[1-\frac{19}{100} ]^{t}$, where, t is in years since 2013. (Answer)