Answer :
To find the amount of money in the account after 5 years with continuous compounding, we can use the formula for continuous compound interest:
\[ A = P e^{rt} \]
Where:
- \( A \) is the amount of money accumulated after \( t \) years
- \( P \) is the principal amount (the initial deposit)
- \( r \) is the annual interest rate (in 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:
- \( P = \$2000 \)
- \( r = 0.08 \) (8% interest rate in decimal)
- \( t = 5 \) years
Substitute the values into the formula:
\[ A = 2000 \times e^{0.08 \times 5} \]
\[ A = 2000 \times e^{0.4} \]
Now, calculate \( e^{0.4} \):
\[ e^{0.4} \approx 2.71828^{0.4} \approx 1.49182469764 \]
Finally, multiply by the initial deposit:
\[ A \approx 2000 \times 1.49182469764 \approx \$2983.65 \]
Therefore, you will have approximately $2983.65 in the account after 5 years with continuous compounding.
Hope this helps!
\[ A = P e^{rt} \]
Where:
- \( A \) is the amount of money accumulated after \( t \) years
- \( P \) is the principal amount (the initial deposit)
- \( r \) is the annual interest rate (in 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:
- \( P = \$2000 \)
- \( r = 0.08 \) (8% interest rate in decimal)
- \( t = 5 \) years
Substitute the values into the formula:
\[ A = 2000 \times e^{0.08 \times 5} \]
\[ A = 2000 \times e^{0.4} \]
Now, calculate \( e^{0.4} \):
\[ e^{0.4} \approx 2.71828^{0.4} \approx 1.49182469764 \]
Finally, multiply by the initial deposit:
\[ A \approx 2000 \times 1.49182469764 \approx \$2983.65 \]
Therefore, you will have approximately $2983.65 in the account after 5 years with continuous compounding.
Hope this helps!
