Question

a new car is purchased for 15300 dollars. the value of the car depreciates at 14.25% per year. what will the value of the car be, to the nearest cent, after 6 years?

Explanation:

Step1: Recall the formula for exponential depreciation

The formula for the value of an asset after \( t \) years with an initial value \( P \), a depreciation rate \( r \) (expressed as a decimal), is \( V = P(1 - r)^t \).
Here, \( P = 15300 \) dollars, \( r = 0.1425 \) (since \( 14.25\%= \frac{14.25}{100}=0.1425 \)), and \( t = 6 \) years.

Step2: Substitute the values into the formula

Substitute \( P = 15300 \), \( r = 0.1425 \), and \( t = 6 \) into the formula:
\( V=15300\times(1 - 0.1425)^{6}\)
First, calculate \( 1- 0.1425=0.8575 \)
Then, calculate \( 0.8575^{6}\). Let's compute this step by step:
\( 0.8575^{2}=0.8575\times0.8575 = 0.73530625\)
\( 0.8575^{3}=0.73530625\times0.8575\approx0.630525\)
\( 0.8575^{4}=0.630525\times0.8575\approx0.5407\)
\( 0.8575^{5}=0.5407\times0.8575\approx0.4637\)
\( 0.8575^{6}=0.4637\times0.8575\approx0.3976\) (A more accurate way is to use a calculator: \( 0.8575^{6}\approx0.397579\))

Step3: Calculate the value of \( V \)

Now, multiply \( 15300 \) by \( 0.397579 \):
\( V = 15300\times0.397579\)
\( 15300\times0.397579 = 15300\times0.397579\)
\( 15300\times0.397579=6082.9587\)

Answer:

The value of the car after 6 years, to the nearest cent, is \(\$6082.96\) (Note: If we use a more precise calculation of \( 0.8575^{6}\) using a calculator directly: \( 0.8575^{6}\approx0.397579\), \( 15300\times0.397579 = 15300\times0.397579 = 6082.9587\), which rounds to \( 6082.96\) when rounded to the nearest cent).