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?
Answer
Attempt 1 out of 4
Answer
Attempt 1 out of 4
Answers
GPT-4o mini
To calculate the value of a car after a certain number of years given a certain rate of depreciation, we can use the formula for exponential decay:
\[
V = P \times (1 - r)^t
\]
Where:
- \( V \) is the value of the car after \( t \) years,
- \( P \) is the initial purchase price of the car,
- \( r \) is the depreciation rate (as a decimal),
- \( t \) is the number of years.
Given:
- \( P = 15300 \) dollars,
- \( r = 14.25\% = 0.1425 \),
- \( t = 6 \) years.
Now, plug these values into the formula:
\[
V = 15300 \times (1 - 0.1425)^6
\]
Calculate \( (1 - 0.1425) \):
\[
1 - 0.1425 = 0.8575
\]
Now, raise \( 0.8575 \) to the power of \( 6 \):
\[
0.8575^6 \approx 0.3875 \quad (\text{rounded to four decimal places})
\]
Now multiply by the initial value:
\[
V \approx 15300 \times 0.3875 \approx 5921.25
\]
Thus, the value of the car after 6 years is approximately **$5921.25**.
\[
V = P \times (1 - r)^t
\]
Where:
- \( V \) is the value of the car after \( t \) years,
- \( P \) is the initial purchase price of the car,
- \( r \) is the depreciation rate (as a decimal),
- \( t \) is the number of years.
Given:
- \( P = 15300 \) dollars,
- \( r = 14.25\% = 0.1425 \),
- \( t = 6 \) years.
Now, plug these values into the formula:
\[
V = 15300 \times (1 - 0.1425)^6
\]
Calculate \( (1 - 0.1425) \):
\[
1 - 0.1425 = 0.8575
\]
Now, raise \( 0.8575 \) to the power of \( 6 \):
\[
0.8575^6 \approx 0.3875 \quad (\text{rounded to four decimal places})
\]
Now multiply by the initial value:
\[
V \approx 15300 \times 0.3875 \approx 5921.25
\]
Thus, the value of the car after 6 years is approximately **$5921.25**.