Question

question
mav purchased a new car in 2001 for $27,900. the value of the car has been depreciating exponentially at a constant rate. if the value of the car was $13,600 in the year 2008, then what would be the predicted value of the car in the year 2011, to the nearest dollar?
answer

Explanation:

Step1: Define exponential decay formula

The exponential decay model is $V(t) = V_0(1-r)^t$, where $V_0$ is initial value, $r$ is decay rate, $t$ is time in years.

Step2: Calculate time elapsed to 2008

Time from 2001 to 2008: $t=2008-2001=7$ years.

Step3: Solve for decay rate $r$

Substitute $V(7)=13600$, $V_0=27900$, $t=7$:
$$13600 = 27900(1-r)^7$$
$$(1-r)^7 = \frac{13600}{27900} \approx 0.487455$$
$$1-r = 0.487455^{\frac{1}{7}} \approx 0.9072$$
$$r \approx 1-0.9072=0.0928$$

Step4: Calculate time to 2011

Time from 2001 to 2011: $t=2011-2001=10$ years.

Step5: Compute 2011 car value

Substitute $V_0=27900$, $1-r\approx0.9072$, $t=10$:
$$V(10)=27900(0.9072)^{10} \approx 27900 \times 0.3829 \approx 10683$$

Answer:

$\boldsymbol{10683}$