Sure, let's go through the compound interest formula step-by-step to find the account balance [tex]\( A \)[/tex]:
The compound interest formula is given by:
[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 (in decimal form).
- [tex]\( n \)[/tex] is the number of compounding periods per year.
- [tex]\( t \)[/tex] is the time in years.
Given:
- [tex]\( P = \$17,547 \)[/tex]
- [tex]\( r = 3.6\% = 0.036 \)[/tex]
- [tex]\( n = 365 \)[/tex] (compounded daily)
- [tex]\( t = 3 \)[/tex] years
Let's substitute the given values into the formula:
1. [tex]\( r/n \)[/tex]:
[tex]\[ \frac{r}{n} = \frac{0.036}{365} \approx 0.00009863 \][/tex]
2. [tex]\( 1 + r/n \)[/tex]:
[tex]\[ 1 + \frac{r}{n} \approx 1 + 0.00009863 = 1.00009863 \][/tex]
3. [tex]\( nt \)[/tex]:
[tex]\[ nt = 365 \times 3 = 1095 \][/tex]
4. Finally, raising to the power of [tex]\( nt \)[/tex] and multiplying by [tex]\( P \)[/tex]:
[tex]\[ A = 17547 \times (1.00009863)^{1095} \][/tex]
First, calculate the exponentiation part:
[tex]\[ (1.00009863)^{1095} \approx 1.1143 \][/tex]
Then multiply by [tex]\( P \)[/tex]:
[tex]\[ A \approx 17547 \times 1.1143 = 19548.09 \][/tex]
Thus, the account balance after 3 years, compounded daily, at an annual interest rate of 3.6%, is approximately:
[tex]\[ A \approx \$19,548.09 \][/tex]
So, the final account balance is approximately \$19,548.09.
