To determine the balance after 2 years in a savings account with an initial investment of \[tex]$1,250 and an annual compound interest rate of 3.5%, follow these steps:
1. Identify the initial investment and the interest rate:
- Initial investment (\( P \)): \$[/tex]1,250
- Annual interest rate ([tex]\( r \)[/tex]): 3.5% or 0.035 (expressed as a decimal)
2. Identify the time period:
- Number of years ([tex]\( t \)[/tex]): 2 years
3. Use the compound interest formula:
The formula for compound interest is given by:
[tex]\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest
- [tex]\( P \)[/tex] is the principal amount (initial investment)
- [tex]\( r \)[/tex] is the annual interest rate (decimal)
- [tex]\( n \)[/tex] is the number of times interest is compounded per year
- [tex]\( t \)[/tex] is the time the money is invested or borrowed for, in years
In this case, since the interest is compounded annually, [tex]\( n = 1 \)[/tex], so the formula simplifies to:
[tex]\[
A = P (1 + r)^t
\][/tex]
4. Plug in the values:
[tex]\[
A = 1250 (1 + 0.035)^2
\][/tex]
5. Calculate the balance:
[tex]\[
A = 1250 \times (1.035)^2
\][/tex]
[tex]\[
A = 1250 \times 1.071225
\][/tex]
[tex]\[
A \approx 1339.03
\][/tex]
6. Round to the nearest hundredth:
The balance after 2 years, rounded to the nearest hundredth, is \[tex]$1339.03.
So, the balance after 2 years in the savings account is \$[/tex]1339.03.
