To find the total amount of an investment after a certain period 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 time the money is invested for in years.
Given:
- \( P = 6000 \) (the initial investment)
- \( r = 0.0743 \) (7.43% as a decimal)
- \( n = 365 \) (interest is compounded daily)
- \( t = 10 \) (the investment is for 10 years)
Now, plug these values into the formula:
\[ A = 6000 \left(1 + \frac{0.0743}{365}\right)^{365 \times 10} \]
Calculating \( \frac{0.0743}{365} \):
\[ \frac{0.0743}{365} \approx 0.0002030151 \]
Now, we plug this back into the equation:
\[ A = 6000 \left(1 + 0.0002030151\right)^{3650} \]
Calculating \( 1 + 0.0002030151 \):
\[ 1 + 0.0002030151 \approx 1.0002030151 \]
Now raising this to the power of \( 3650 \):
\[ A = 6000 \left(1.0002030151\right)^{3650} \]
Using a calculator or software to compute \( (1.0002030151)^{3650} \):
\[ (1.0002030151)^{3650} \approx 2.095378644 \]
Now, we can calculate \( A \):
\[ A \approx 6000 \times 2.095378644 \]
\[ A \approx 12572.27 \]
Therefore, the total amount of the investment after 10 years is approximately:
\[ \boxed{12572.27} \]
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