To complete the table representing the value of a car after x years, typically, we would use a formula for depreciation. A common model for car depreciation is that a car loses a fixed percentage of its value each year. For this example, let's assume the car starts with an initial value of $20,000 and depreciates at a rate of 15% per year.
Using the formula for depreciation:
\[ f(x) = P(1 - r)^x \]
where:
- \( P \) is the initial value of the car (here, \( 20000 \)),
- \( r \) is the rate of depreciation (15% or 0.15),
- \( x \) is the number of years.
Let's calculate \( f(x) \) for different values of \( x \):
-
For \( x = 0 \): \[ f(0) = 20000 \times (1 - 0.15)^0 = 20000 \times 1 = 20000.00 \]
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For \( x = 1 \): \[ f(1) = 20000 \times (1 - 0.15)^1 = 20000 \times 0.85 = 17000.00 \]
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For \( x = 3 \): \[ f(3) = 20000 \times (1 - 0.15)^3 = 20000 \times 0.85^3 \approx 20000 \times 0.614125 = 12282.50 \]
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For \( x = 5 \): \[ f(5) = 20000 \times (1 - 0.15)^5 = 20000 \times 0.85^5 \approx 20000 \times 0.443705 = 8874.10 \]
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For \( x = 8 \): \[ f(8) = 20000 \times (1 - 0.15)^8 = 20000 \times 0.85^8 \approx 20000 \times 0.272490 = 5449.80 \]
Now we can fill out the table:
\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 0 & 20000.00 \ 1 & 17000.00 \ 3 & 12282.50 \ 5 & 8874.10 \ 8 & 5449.80 \ \hline \end{array} \]
The completed values are:
- For \( x = 0 \): $20000.00
- For \( x = 1 \): $17000.00
- For \( x = 3 \): $12282.50
- For \( x = 5 \): $8874.10
- For \( x = 8 \): $5449.80