If the Laffite family deposits [tex]$8500 in a savings account at 6.75% interest, compounded continuously, how much will be in the account after 25 years?

A. $[/tex]227,338.93
B. [tex]$45,950.57
C. $[/tex]38,094.36
D. $38,720.02



Answer :

To determine how much money will be in the Laffite family's account after 25 years given that they deposit \[tex]$8,500 at an annual interest rate of 6.75% compounded continuously, we use the formula for continuous compounding: \[ A = P \cdot e^{rt} \] where: - \( A \) is the amount of money accumulated after \( t \) years, including interest. - \( P \) is the principal amount (the initial deposit). - \( r \) 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: - \( P = 8500 \) (the initial deposit) - \( r = 6.75\% = 0.0675 \) (the annual interest rate in decimal form) - \( t = 25 \) years Substituting these values into the continuous compounding formula: \[ A = 8500 \cdot e^{(0.0675 \cdot 25)} \] To find \( e^{(0.0675 \cdot 25)} \): 1. Calculate \( 0.0675 \times 25 \): \[ 0.0675 \times 25 = 1.6875 \] 2. Now calculate \( e^{1.6875} \): \[ e^{1.6875} \approx 5.4057 \] (This is a rounded approximation. The exact value can be calculated using a scientific calculator or software.) 3. Substitute \( e^{1.6875} \) back into the formula: \[ A = 8500 \cdot 5.4057 \] 4. Finally, calculate the amount \( A \): \[ A \approx 8500 \times 5.4057 = 45,950.57 \] Therefore, the amount in the account after 25 years will be \$[/tex]45,950.57.

Thus, the closest answer to our calculated value is:

[tex]\[ \boxed{45,950.57} \][/tex]