\begin{tabular}{|l}\hline Question 25 \\ Jana invests a sum of money in a retirement account with a fixed annual interest rate of \\ 2.15\% compounded continuously. After 10 years, the balance reaches \( \$ 1,912.41 \). What was \\ the amount of the initial investment? \\ 4 pts \\ \hline\end{tabular}

To find the initial investment, we can use the formula for continuous compounding, which is given by: \[ A = Pe^{rt} \] where: - \( A \) is the amount of money accumulated after n years, including interest, - \( P \) is the principal amount (the initial investment), - \( r \) is the annual interest rate (decimal), - \( t \) is the time the money is invested for in years, - \( e \) is the base of the natural logarithm, approximately equal to 2.71828. In this case: - \( A = 1912.41 \) - \( r = 0.0215 \) - \( t = 10 \) We need to solve for \( P \): \[ 1912.41 = P e^{0.0215 \times 10} \] Calculating the exponent: \[ e^{0.215} \approx 1.24076 \] Now we can substitute this value back into the equation: \[ 1912.41 = P \times 1.24076 \] Now, we solve for \( P \): \[ P = \frac{1912.41}{1.24076} \approx 1540.60 \] So, the initial investment was approximately \( \$ 1,540.60 \).