To calculate the future value of an investment with daily compounding interest, we can use the formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.
Given:
- \( P = 3800 \)
- \( r = 3.6% = 0.036 \)
- \( n = 365 \) (compounded daily)
- \( t = 15 \)
Now, substitute these values into the formula:
\[ A = 3800 \left(1 + \frac{0.036}{365}\right)^{365 \times 15} \]
First, calculate \( \frac{0.036}{365} \):
\[ \frac{0.036}{365} \approx 0.000098630137 \]
Now add 1 to that value:
\[ 1 + 0.000098630137 \approx 1.000098630137 \]
Next, calculate \( 365 \times 15 \):
\[ 365 \times 15 = 5475 \]
Now we raise the result from the previous step to the power of 5475:
\[ A = 3800 \times (1.000098630137)^{5475} \]
Calculating \( (1.000098630137)^{5475} \):
\[ (1.000098630137)^{5475} \approx 1.668051402 \]
Now we can calculate \( A \):
\[ A \approx 3800 \times 1.668051402 \approx 6349.59 \]
Rounding this amount to the nearest hundred dollars gives:
\[ A \approx 6300 \]
Therefore, the amount of money in the account after 15 years would be approximately $6,300.