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A new car is purchased for 24600 dollars. The value of the car depreciates at
13.75% per year. What will the value of the car be, to the nearest cent, after 15 years?
Answer:

A new car is purchased for $24600$ dollars. The value of the car depreciates at $13.75 \%$ per year. What will the value of the car be, to the nearest cent, after $15$ years? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

24600

dollars. The value of the car depreciates at

13.75 \%

per year. What will the value of the car be, to the nearest cent, after

15

years?

\newline

Answer:

Identify values: First, identify the initial value $P_0$ , the rate of depreciation $r$ , and the time $t$ . $P_0 = \$24,600$ $r = 13.75\%$ per year $= 0.1375$ (as a decimal) $t = 15$ years
Plug into formula: Next, plug these values into the exponential decay formula to calculate the future value of the car. $\newline$ $P(t) = P_0 \times (1 - r)^t$ $\newline$ $P(t) = (\$24,600) \times (1 - 0.1375)^{15}$
Calculate inside parentheses: Now, calculate the value inside the parentheses first. $1 - 0.1375 = 0.8625$
Calculate decay factor: Raise $0.8625$ to the power of $15$ to find the decay factor. $\newline$ $(0.8625)^{15}$
Multiply initial value: Using a calculator, we find that $(0.8625)^{15} \approx 0.1663$ (rounded to four decimal places for precision in intermediate steps).
Find final value: Now, multiply the initial value of the car by the decay factor to find the value of the car after $15$ years. $P(t) = \$24,600 \times 0.1663$
Find final value: Now, multiply the initial value of the car by the decay factor to find the value of the car after $15$ years. $\newline$ $P(t) = \$24,600 \times 0.1663$ Perform the multiplication to find the final value. $\newline$ $P(t) \approx \$24,600 \times 0.1663 \approx \$4090.98$

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