Question

in the year 2005, a person bought a new car for $24500. for each consecutive year after that, the value of the car depreciated by 9%. how much would the car be worth in the year 2008, to the nearest hundred dollars?

Explanation:

Step1: Define depreciation formula

The value of the car after $n$ years follows the formula:
$$V = V_0(1 - r)^n$$
where $V_0 = 24500$, $r = 0.09$, and $n$ is the number of years since 2005.

Step2: Calculate years elapsed

From 2005 to 2008, $n = 2008 - 2005 = 3$

Step3: Substitute values into formula

$$V = 24500(1 - 0.09)^3 = 24500(0.91)^3$$

Step4: Compute $(0.91)^3$

$$0.91^3 = 0.91 \times 0.91 \times 0.91 = 0.753571$$

Step5: Calculate final value

$$V = 24500 \times 0.753571 \approx 18462.49$$

Step6: Round to nearest hundred

Round 18462.49 to the nearest hundred: $\approx 18500$

Answer:

$\$18500$