To find the balance after 2 years in a savings account with an initial investment of [tex]$1,400 and a 2% annual compound interest rate, we will use the compound interest formula. The formula for compound interest is:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the future value of the investment/loan, including interest.
- \( P \) is the principal investment amount (initial deposit), which is $[/tex]1,400.
- [tex]\( r \)[/tex] is the annual interest rate (decimal), which is 0.02.
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year. Since the interest is compounded annually, [tex]\( n \)[/tex] is 1.
- [tex]\( t \)[/tex] is the number of years the money is invested, which is 2 years.
Now we substitute the given values into the formula:
[tex]\[ A = 1400 \times \left(1 + \frac{0.02}{1}\right)^{1 \times 2} \][/tex]
First, simplify the fraction inside the parentheses:
[tex]\[ 1 + \frac{0.02}{1} = 1 + 0.02 = 1.02 \][/tex]
Now raise this to the power of [tex]\( 1 \times 2 \)[/tex]:
[tex]\[ 1.02^2 \][/tex]
Calculate [tex]\( 1.02^2 \)[/tex]:
[tex]\[ 1.02^2 = 1.0404 \][/tex]
Next, multiply the principal [tex]\( P \)[/tex] by this result:
[tex]\[ 1400 \times 1.0404 = 1456.56 \][/tex]
Thus, the balance after 2 years is:
[tex]\[ \text{Balance} = \$1456.56 \][/tex]
