Asked by Lemy
Analyzing a Data Set
The value of a car decreases as shown in the table below.
A 2-column table with 6 rows. The first column is labeled years after purchase with entries 0, 1, 2, 3, 4, 5. The second column is labeled value (dollar sign) with entries 25,000; 18,000; 14,000; 10,500; 8,000; 6,000.
Which statements are true? Check all that apply.
The function that best represents the data is f(x) = 24,512(0.755)x.
The function that best represents the data is f(x) = 554x2 – 5,439x + 24,600.
The function decreases indefinitely.
It is reasonable to interpolate to the value of the car at 4.5 years.
It is reasonable to extrapolate to 40 years.
All Answers 1
Answered by
GPT-5 mini
AI
Answers: 1 True, 2 False, 3 True, 4 True, 5 False.
Brief reasons:
- f(x) = 24,512(0.755)^x (exponential) fits the points well (e.g., f(0) ≈ 24,512 ≈ 25,000 and f(5) ≈ 6,038 ≈ 6,000), so it is a good model.
- The quadratic 554x^2 – 5,439x + 24,600 gives values far from the data (e.g., f(5) ≈ 11,255), so it does not represent the data well.
- The exponential model has base 0.755 < 1, so it keeps decreasing as x increases (approaches 0), i.e., it decreases indefinitely.
- Interpolating at 4.5 years is reasonable because 4.5 is inside the measured range (between 4 and 5 years).
- Extrapolating to 40 years is not reasonable — it’s far beyond the data range and ignores realistic limits (wear, scrap value, model changes), so predictions that far out would be unreliable.
Brief reasons:
- f(x) = 24,512(0.755)^x (exponential) fits the points well (e.g., f(0) ≈ 24,512 ≈ 25,000 and f(5) ≈ 6,038 ≈ 6,000), so it is a good model.
- The quadratic 554x^2 – 5,439x + 24,600 gives values far from the data (e.g., f(5) ≈ 11,255), so it does not represent the data well.
- The exponential model has base 0.755 < 1, so it keeps decreasing as x increases (approaches 0), i.e., it decreases indefinitely.
- Interpolating at 4.5 years is reasonable because 4.5 is inside the measured range (between 4 and 5 years).
- Extrapolating to 40 years is not reasonable — it’s far beyond the data range and ignores realistic limits (wear, scrap value, model changes), so predictions that far out would be unreliable.
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