Respuesta :
Answer:
To solve this problem, we can use the formula for exponential decay:
\[ A = P \times (1 - r)^t \]
Where:
- \( A \) is the final amount (the value of the car in this case)
- \( P \) is the initial amount (the cost of the car)
- \( r \) is the annual rate of decrease (as a decimal)
- \( t \) is the time in years
Given:
- \( P = $25000 \)
- \( A = $10000 \)
- \( r = 9.8\% = 0.098 \) (since it's a decrease, we use \( 1 - 0.098 \))
- We need to find \( t \)
Plugging these values into the formula, we get:
\[ 10000 = 25000 \times (1 - 0.098)^t \]
Now, let's solve for \( t \). We'll start by dividing both sides by \( 25000 \):
\[ \frac{10000}{25000} = (1 - 0.098)^t \]
Simplify:
\[ 0.4 = (0.902)^t \]
Now, take the natural logarithm (ln) of both sides to solve for \( t \):
\[ \ln(0.4) = \ln(0.902)^t \]
\[ \ln(0.4) = t \times \ln(0.902) \]
\[ t = \frac{\ln(0.4)}{\ln(0.902)} \]
Now, calculate:
\[ t \approx \frac{-0.916291}{-0.102982} \]
\[ t \approx 8.913 \]
Rounding to the nearest tenth, \( t \approx 8.9 \) years.
So, it will take approximately 8.9 years for the car to be worth $10000.
Step-by-step explanation:
