Question
The table lists the value of a car each year after it is purchased.
Year 0 1 2 3 4
Value ($) 54,000 51,840 49,766 47,776 45,865
Determine whether the decreasing value can be modeled by a linear decay function or an exponential decay function, and why.
(1 point)
Responses
exponential decay function; the value decreases by 4% per year, which is a constant percent rate of change
exponential decay function; the value decreases by 4% per year, which is a constant percent rate of change
linear decay function; the value decreases by 96% per year, which is a constant percent rate of change
linear decay function; the value decreases by 96% per year, which is a constant percent rate of change
linear decay function; the value decreases by 0.96 per year, which is a constant percent rate of change
linear decay function; the value decreases by 0.96 per year, which is a constant percent rate of change
exponential decay function; the value decreases by $6,000 per year, which is a constant percent rate of change
Answers
GPT-4o mini
To determine whether the decreasing value of the car can be modeled by a linear decay function or an exponential decay function, we should analyze the yearly changes in the value of the car.
From the table, we can observe the changes in value each year:
- Year 0 to Year 1:
\( 54,000 - 51,840 = 2,160 \)
- Year 1 to Year 2:
\( 51,840 - 49,766 = 2,074 \)
- Year 2 to Year 3:
\( 49,766 - 47,776 = 1,990 \)
- Year 3 to Year 4:
\( 47,776 - 45,865 = 1,911 \)
The differences in the car's value are decreasing each year, indicating that the absolute decrease is not constant.
Now, let's analyze the percentage change:
- From Year 0 to Year 1:
\( \frac{2,160}{54,000} \approx 0.04 \) or \( 4\% \)
- From Year 1 to Year 2:
\( \frac{2,074}{51,840} \approx 0.040 \) or \( 4\% \)
- From Year 2 to Year 3:
\( \frac{1,990}{49,766} \approx 0.040 \) or \( 4\% \)
- From Year 3 to Year 4:
\( \frac{1,911}{47,776} \approx 0.040 \) or \( 4\% \)
The percentage decrease appears to remain roughly consistent around 4% per year. This indicates a constant percent rate of change, which is characteristic of exponential decay, rather than a constant amount which would characterize linear decay.
Based on this analysis, the correct response is:
**exponential decay function; the value decreases by 4% per year, which is a constant percent rate of change.**
From the table, we can observe the changes in value each year:
- Year 0 to Year 1:
\( 54,000 - 51,840 = 2,160 \)
- Year 1 to Year 2:
\( 51,840 - 49,766 = 2,074 \)
- Year 2 to Year 3:
\( 49,766 - 47,776 = 1,990 \)
- Year 3 to Year 4:
\( 47,776 - 45,865 = 1,911 \)
The differences in the car's value are decreasing each year, indicating that the absolute decrease is not constant.
Now, let's analyze the percentage change:
- From Year 0 to Year 1:
\( \frac{2,160}{54,000} \approx 0.04 \) or \( 4\% \)
- From Year 1 to Year 2:
\( \frac{2,074}{51,840} \approx 0.040 \) or \( 4\% \)
- From Year 2 to Year 3:
\( \frac{1,990}{49,766} \approx 0.040 \) or \( 4\% \)
- From Year 3 to Year 4:
\( \frac{1,911}{47,776} \approx 0.040 \) or \( 4\% \)
The percentage decrease appears to remain roughly consistent around 4% per year. This indicates a constant percent rate of change, which is characteristic of exponential decay, rather than a constant amount which would characterize linear decay.
Based on this analysis, the correct response is:
**exponential decay function; the value decreases by 4% per year, which is a constant percent rate of change.**