To calculate the balance in the given savings account after one year, we will use the compound interest formula. The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (initial deposit),
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal),
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year,
- [tex]\( t \)[/tex] is the time the money is invested for, in years,
- and [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
Given:
- The principal amount ([tex]\( P \)[/tex]) is \[tex]$1,450.
- The annual interest rate (\( r \)) is 62%, which as a decimal is \( 0.62 \).
- The interest is compounded semi-annually, so \( n = 2 \) (twice a year).
- The time period (\( t \)) is 1 year.
Now, let's substitute these values into the formula:
\[ A = 1450 \left(1 + \frac{0.62}{2}\right)^{2 \cdot 1} \]
First, we calculate the value inside the parentheses:
\[ 1 + \frac{0.62}{2} = 1 + 0.31 = 1.31 \]
Now, we raise this result to the power of \( 2 \cdot 1 \) (which is 2):
\[ 1.31^2 = 1.7161 \]
Finally, we multiply by the principal amount:
\[ A = 1450 \times 1.7161 \]
\[ A \approx 2488.345 \]
Therefore, after one year, Sing's balance in the savings account will be approximately \$[/tex]2,488.35.
