Question
6.1.2 Exam: Semester 2 Exam Question 14 of 40 Prakash bought a new car at the dealership for \( \$ 27,000 \). It is estimated that the value of the car will decrease \( 7 \% \) each year. Which exponential function models the value \( v \) of the car after \( t \) years? A. \( v=27,000(0.3)^{t} \) B. \( v=27,000(1.3)^{t} \) c. \( v=27,000(1.03)^{t} \) D. \( v=27,000(0.93)^{t} \)
Ask by Cervantes Ramirez. in the United States
Jan 31,2025
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Answer
The correct exponential function is \( v = 27,000(0.93)^t \).
Solution
To model the depreciation of Prakash's new car, we use an exponential decay function. Here's how to determine the correct function:
1. **Initial Value (\( v_0 \))**: The car's initial value is \$27,000.
2. **Depreciation Rate**: The car loses 7% of its value each year. This means it retains \( 100\% - 7\% = 93\% \) of its value annually.
3. **Exponential Decay Formula**: The general formula for exponential decay is:
\[
v = v_0 \times (1 - r)^t
\]
where:
- \( v \) is the value after \( t \) years,
- \( v_0 \) is the initial value,
- \( r \) is the depreciation rate.
Applying the values:
\[
v = 27,000 \times (0.93)^t
\]
**Option D** correctly represents this formula:
\[
\boxed{D.\ v=27,000(0.93)^{t}}
\]
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When a car depreciates in value by a percentage each year, we can model it using an exponential decay function. Since the value decreases by 7%, the remaining value after depreciation each year is \( 100\% - 7\% = 93\% \) or \( 0.93 \). Thus, the correct exponential function is \( v = 27,000(0.93)^{t} \), which corresponds to option D. This kind of modeling is important in finance as it helps buyers understand how their assets will lose value over time. Having a grasp on depreciation can inform decisions on buying, selling, or trading in vehicles, ensuring you make the best financial choice possible with your hard-earned money.
