Use the compound-interest formula to find the account balance [tex]A[/tex], where [tex]P[/tex] is the principal, [tex]r[/tex] is the interest rate, [tex]n[/tex] is the number of compounding periods per year, and [tex]t[/tex] is the time in years.

[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
P & r & n & t \\
\hline
\$17,547 & 3.6\% & \text{Daily} & 3 \\
\hline
\end{tabular}
\][/tex]

The account balance is approximately [tex] \boxed{\$}[/tex].

(Simplify your answer. Do not round until the final answer. Then round to two decimal places as needed.)



Answer :

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.