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

To calculate the balance after 14 years with continuous compounding, you can use the formula: \[ F = Pe^{rt} \] where: - \( F \) is the future value of the investment/loan, including interest, - \( P \) is the principal investment amount (\$3000), - \( r \) is the annual interest rate (decimal) (0.09 for 9%), - \( t \) is the time the money is invested or borrowed for, in years (14), - \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Plugging the values into the formula: \[ F = 3000 \times e^{0.09 \times 14} \] Calculating \( 0.09 \times 14 \): \[ 0.09 \times 14 = 1.26 \] Now, calculating \( e^{1.26} \): \[ e^{1.26} \approx 3.5291 \] Now plug back into the equation: \[ F \approx 3000 \times 3.5291 \approx 10587.30 \] So, the balance after 14 years would be approximately: \[ F = \$10587.30 \]