Sure, let's solve this step-by-step.
We are given:
- Initial investment (Principal), [tex]\( P = \$ 10,000 \)[/tex]
- Annual interest rate, [tex]\( r = 0.03 \)[/tex] (3%)
- Number of times the interest is compounded per year, [tex]\( m = 12 \)[/tex]
- Number of years, [tex]\( t = 10 \)[/tex]
The formula to calculate the amount [tex]\( A \)[/tex] in the account after [tex]\( t \)[/tex] years with interest compounded [tex]\( m \)[/tex] times per year is:
[tex]\[ A = P \left( 1 + \frac{r}{m} \right)^{m \cdot t} \][/tex]
Now, let's break down the steps to compute this:
1. Calculate the monthly interest rate:
[tex]\[ \frac{r}{m} = \frac{0.03}{12} = 0.0025 \][/tex]
2. Calculate the total number of compounding periods:
[tex]\[ m \cdot t = 12 \cdot 10 = 120 \][/tex]
3. Calculate the growth factor:
[tex]\[ 1 + \frac{r}{m} = 1 + 0.0025 = 1.0025 \][/tex]
4. Raise the growth factor to the power of the total number of compounding periods:
[tex]\[ (1.0025)^{120} \approx 1.3493535471908246 \][/tex]
5. Multiply this result by the principal amount to find [tex]\( A \)[/tex]:
[tex]\[ A = 10000 \cdot 1.3493535471908246 \approx 13493.54 \][/tex]
Therefore, the amount of money in the savings account after 10 years will be approximately [tex]$\$[/tex]13,493.54[tex]$.
Among the provided choices, $[/tex]\[tex]$ 13,493.54$[/tex] is the correct answer.
