Question
Alexandra opened a savings account and deposited $800.00 as principal. The account earns 14% interest, compounded continuously. What is the balance after 4 years?
Round your answer to the nearest cent.
Round your answer to the nearest cent.
Answers
GPT-4o mini
To calculate the balance of a savings account with continuous compounding, we use the formula:
\[
A = Pe^{rt}
\]
where:
- \(A\) is the amount of money accumulated after time \(t\),
- \(P\) is the principal amount (the initial deposit),
- \(r\) is the annual interest rate (decimal),
- \(t\) is the time in years,
- \(e\) is the base of the natural logarithm (approximately equal to 2.71828).
In your case:
- \(P = 800.00\),
- \(r = 0.14\), (14% as a decimal),
- \(t = 4\) years.
Substituting these values into the formula gives:
\[
A = 800 \cdot e^{0.14 \cdot 4}
\]
Now calculate \(0.14 \cdot 4\):
\[
0.14 \cdot 4 = 0.56
\]
Next, we need to calculate \(e^{0.56}\):
Using a calculator, we find:
\[
e^{0.56} \approx 1.751
\]
Now substitute this value back into the formula:
\[
A = 800 \cdot 1.751
\]
Calculating the final amount:
\[
A \approx 800 \cdot 1.751 = 1400.80
\]
Therefore, the balance after 4 years is approximately **$1400.80**.
\[
A = Pe^{rt}
\]
where:
- \(A\) is the amount of money accumulated after time \(t\),
- \(P\) is the principal amount (the initial deposit),
- \(r\) is the annual interest rate (decimal),
- \(t\) is the time in years,
- \(e\) is the base of the natural logarithm (approximately equal to 2.71828).
In your case:
- \(P = 800.00\),
- \(r = 0.14\), (14% as a decimal),
- \(t = 4\) years.
Substituting these values into the formula gives:
\[
A = 800 \cdot e^{0.14 \cdot 4}
\]
Now calculate \(0.14 \cdot 4\):
\[
0.14 \cdot 4 = 0.56
\]
Next, we need to calculate \(e^{0.56}\):
Using a calculator, we find:
\[
e^{0.56} \approx 1.751
\]
Now substitute this value back into the formula:
\[
A = 800 \cdot 1.751
\]
Calculating the final amount:
\[
A \approx 800 \cdot 1.751 = 1400.80
\]
Therefore, the balance after 4 years is approximately **$1400.80**.