To determine the account balance, we will use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( P \)[/tex] is the principal amount,
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal),
- [tex]\( n \)[/tex] is the number of compounding periods per year,
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given values:
- [tex]\( P = 18,483 \)[/tex]
- [tex]\( r = 5.7\% = 0.057 \)[/tex] (as a decimal)
- [tex]\( n = 365 \)[/tex] (compounded daily)
- [tex]\( t = 3 \)[/tex] years
Substitute these values into the formula:
[tex]\[ A = 18,483 \left(1 + \frac{0.057}{365}\right)^{365 \times 3} \][/tex]
First, calculate [tex]\( \frac{0.057}{365} \)[/tex]:
[tex]\[ \frac{0.057}{365} \approx 0.000156164 \][/tex]
Add 1 to this result:
[tex]\[ 1 + 0.000156164 \approx 1.000156164 \][/tex]
Then, raise this to the power of [tex]\( 365 \times 3 \)[/tex]:
[tex]\[ 1.000156164^{1095} \approx 1.234259515 \][/tex]
Finally, multiply this result by the principal [tex]\( 18,483 \)[/tex]:
[tex]\[ A = 18,483 \times 1.234259515 \approx 22,929.615737264525 \][/tex]
Rounding to two decimal places, the account balance [tex]\( A \)[/tex] is approximately:
[tex]\[ \$ 22,929.62 \][/tex]
So, the account balance after 3 years is approximately \$22,929.62.
