How much would you have to deposit in an account with a \( 7.5 \% \) interest rate, compounded continuously, to have \( \$ 2500 \) in your account 8 years later? \[ P=\$[?] \] Round to the nearest cent.

To find out how much you need to deposit now to reach \$2500 in 8 years with a \( 7.5\% \) interest rate compounded continuously, you can use the formula: \[ A = Pe^{rt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) 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). Given: - \( A = 2500 \) - \( r = 0.075 \) - \( t = 8 \) Rearranging to solve for \( P \): \[ P = \frac{A}{e^{rt}} \] Plugging in the numbers: \[ P = \frac{2500}{e^{0.075 \times 8}} \] Calculating the exponent: \[ e^{0.6} \approx 1.8221 \] Now, calculate \( P \): \[ P = \frac{2500}{1.8221} \approx 1371.47 \] So, you would need to deposit approximately **$1371.47**.