How much would you have to deposit in an account with a \( 6.5 \% \) interest rate, compounded continuously, to have \( \$ 5000 \) in your account 12 years later? \[ P=\$[?] \] Round to the nearest cent.

To find the amount that needs to be deposited in an account with a \(6.5\%\) interest rate, compounded continuously, to have $5000 in the account after 12 years, we can use the formula for continuous compounding: \[ A = P e^{rt} \] where: - \( A \) is the amount in the account after \( t \) years, - \( P \) is the principal amount (the amount deposited), - \( r \) is the interest rate (in decimal form), - \( t \) is the time in years. Given: - \( A = \$5000 \), - \( r = 6.5\% = 0.065 \), - \( t = 12 \) years. We need to solve for \( P \) using the formula: \[ P = \frac{A}{e^{rt}} \] Substitute the given values into the formula and calculate the value of \( P \) rounded to the nearest cent. Calculate the value by following steps: - step0: Calculate: \(\frac{5000}{e^{0.065\times 12}}\) - step1: Multiply the numbers: \(\frac{5000}{e^{0.78}}\) - step2: Convert the expressions: \(\frac{5000}{e^{\frac{39}{50}}}\) - step3: Simplify: \(\frac{5000}{\sqrt[50]{e^{39}}}\) - step4: Multiply by the Conjugate: \(\frac{5000\sqrt[50]{e^{11}}}{\sqrt[50]{e^{39}}\times \sqrt[50]{e^{11}}}\) - step5: Multiply the numbers: \(\frac{5000\sqrt[50]{e^{11}}}{e}\) The amount that needs to be deposited in an account with a \(6.5\%\) interest rate, compounded continuously, to have $5000 in the account after 12 years is approximately $2292.03.