Question
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\( \$ 3000 \) 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.

Ask by Cervantes Zimmerman. in the United States
Jan 21,2025

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Tutor-Verified Answer

Answer

The balance after 14 years is approximately \$10,576.26.

Solution

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.

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To find the balance after 14 years for an investment compounded continuously, we use the formula \( F = Pe^{rt} \), where \( F \) is the final amount, \( P \) is the principal amount (initial deposit), \( r \) is the interest rate (as a decimal), and \( t \) is the time in years. Given \( P = 3000 \), \( r = 0.09 \), and \( t = 14 \): \[ F = 3000 \times e^{0.09 \times 14} \] Calculating \( 0.09 \times 14 = 1.26 \), and \( e^{1.26} \approx 3.535 \): \[ F \approx 3000 \times 3.535 \approx 10605.00 \] So, the balance after 14 years is approximately \( F = \$10605.00 \).

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