To find the balance after 3 years in a savings account with an initial investment of \[tex]$1,350 and a 3% annual compound interest rate, follow these steps:
1. Identify the initial principal (P):
\[
P = \$[/tex]1,350
\]
2. Identify the annual interest rate (r):
[tex]\[
r = 0.03 \quad \text{(3% as a decimal)}
\][/tex]
3. Determine the number of years (t):
[tex]\[
t = 3
\][/tex]
4. Use the compound interest formula:
The compound interest formula for annual compounding is:
[tex]\[
A = P(1 + r)^t
\][/tex]
where:
- [tex]\(A\)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial amount of money).
- [tex]\(r\)[/tex] is the annual interest rate (decimal).
- [tex]\(t\)[/tex] is the number of years the money is invested or borrowed for.
- Since the interest is compounded annually, [tex]\(n\)[/tex] is 1.
5. Substitute the values into the formula:
[tex]\[
A = 1350 \times (1 + 0.03)^3
\][/tex]
6. Calculate inside the parentheses first:
[tex]\[
1 + 0.03 = 1.03
\][/tex]
7. Raise this result to the power of 3:
[tex]\[
(1.03)^3 \approx 1.0930909
\][/tex]
8. Multiply this result by the principal:
[tex]\[
1350 \times 1.0930909 \approx 1475.18145
\][/tex]
9. Round the result to the nearest hundredth:
[tex]\[
1475.18145 \approx 1475.18
\][/tex]
Therefore, the balance after 3 years is:
[tex]\[
\text{Balance} = \$1475.18
\][/tex]
