To determine the balance in a savings account after 3 years with an initial investment of [tex]$1,350 and an annual compound interest rate of 4.6%, we use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money), which is \$[/tex]1,350.
- [tex]\( r \)[/tex] is the annual interest rate (decimal), which is 4.6% or 0.046.
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year. Here, we assume it is compounded annually, so n = 1.
- [tex]\( t \)[/tex] is the number of years the money is invested or borrowed, in this case, 3 years.
Given that the interest is compounded annually, the formula simplifies to:
[tex]\[ A = P (1 + r)^t \][/tex]
Now, let's substitute the given values into the formula:
[tex]\[ A = 1350 \left(1 + 0.046\right)^3 \][/tex]
[tex]\[ A = 1350 \left(1.046\right)^3 \][/tex]
First, we compute [tex]\( 1.046^3 \)[/tex]:
[tex]\[ 1.046^3 \approx 1.217568456 \][/tex]
Next, we multiply this result by the initial investment amount:
[tex]\[ A = 1350 \times 1.217568456 \][/tex]
[tex]\[ A \approx 1643.6354126 \][/tex]
Finally, we round the balance to the nearest hundredth:
[tex]\[ A \approx 1545.00 \][/tex]
Therefore, the balance after 3 years in the savings account is approximately [tex]\(\$1545.00\)[/tex], rounded to the nearest hundredth.
