\( \$ 3000 \) are deposited in an account with \( 9 \% \) interest rate, compounded continuously. What is the balance after 14 years? \( \mathrm{F}=\$[?] \) Round to the nearest cent.

To find the balance after 14 years with continuous compounding, we can use the formula for continuous compounding: \[ A = P e^{rt} \] Where: - \( A \) is the amount after \( t \) years, - \( P \) is the principal amount (initial deposit), - \( r \) is the annual interest rate (in decimal form), - \( t \) is the time in years. Given: - Principal amount \( P = \$3000 \), - Annual interest rate \( r = 9\% = 0.09 \), - Time \( t = 14 \) years. Substitute the values into the formula: \[ A = 3000 e^{0.09 \times 14} \] Now, we can calculate the balance after 14 years. Calculate the value by following steps: - step0: Calculate: \(3000e^{0.09\times 14}\) - step1: Multiply the numbers: \(3000e^{1.26}\) - step2: Convert the expressions: \(3000e^{\frac{63}{50}}\) The balance after 14 years with continuous compounding is approximately $10,576.26. Rounding to the nearest cent, the balance is $10,576.26.