To determine the balance of a savings account after 2 years with an initial investment of [tex]$800 and an annual compound interest rate of 2%, we will use the compound interest formula:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
where:
- \(A\) is the amount of money accumulated after \(t\) years, including interest.
- \(P\) is the principal amount (the initial amount of money).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times interest is compounded per year.
- \(t\) is the time the money is invested for in years.
For this problem:
- The principal amount \(P\) is $[/tex]800.
- The annual interest rate [tex]\(r\)[/tex] is 0.02 (2% expressed as a decimal).
- The interest is compounded annually, so [tex]\(n = 1\)[/tex].
- The number of years [tex]\(t\)[/tex] is 2.
We can plug these values into the formula:
[tex]\[
A = 800 \left(1 + \frac{0.02}{1}\right)^{1 \times 2}
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
A = 800 \left(1 + 0.02\right)^{2}
\][/tex]
[tex]\[
A = 800 \left(1.02\right)^{2}
\][/tex]
Next, we calculate [tex]\( \left(1.02\right)^{2} \)[/tex]:
[tex]\[
1.02 \times 1.02 = 1.0404
\][/tex]
Now, multiply this result by the principal amount to find [tex]\(A\)[/tex]:
[tex]\[
A = 800 \times 1.0404
\][/tex]
[tex]\[
A = 832.3199999999999
\][/tex]
Thus, the balance after 2 years will be:
[tex]\[
\text{Balance} = \$832.3199999999999
\][/tex]
