\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 amount \( P \) using the formula for continuous compounding, we can use the equation: \[ A = Pe^{rt} \] where: - \( A \) is the amount after time \( t \) (which is \( \$ 1,912.41 \)), - \( r \) is the annual interest rate (which is \( 0.0215 \)), - \( t \) is the time in years (which is \( 10 \)), - \( e \) is the base of the natural logarithm. Rearranging the formula to solve for \( P \): \[ P = \frac{A}{e^{rt}} \] Substituting the known values in: \[ P = \frac{1912.41}{e^{0.0215 \times 10}} = \frac{1912.41}{e^{0.215}} \approx \frac{1912.41}{1.24076} \approx 1541.79 \] Therefore, the initial investment amount was approximately \( \$ 1,541.79 \).