To determine the interest rate that will produce [tex]$3500 from an initial principal balance of $[/tex]2500, with interest compounded semiannually for 4.5 years, we use the formula for compound interest:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after interest, which is [tex]$3500.
- \( P \) is the principal amount (initial investment), which is $[/tex]2500.
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal).
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year, which is 2 (since it is compounded semiannually).
- [tex]\( t \)[/tex] is the time the money is invested for in years, which is 4.5.
We need to solve for [tex]\( r \)[/tex]. First, rearrange the formula to isolate [tex]\( r \)[/tex]:
1. Substitute the known values into the formula:
[tex]\[ 3500 = 2500 \left(1 + \frac{r}{2}\right)^{2 \cdot 4.5} \][/tex]
[tex]\[ 3500 = 2500 \left(1 + \frac{r}{2}\right)^9 \][/tex]
2. Divide both sides by 2500 to isolate the compound term:
[tex]\[ \frac{3500}{2500} = \left(1 + \frac{r}{2}\right)^9 \][/tex]
[tex]\[ 1.4 = \left(1 + \frac{r}{2}\right)^9 \][/tex]
3. Take the 9th root of both sides to solve for the term inside the parentheses:
[tex]\[ \sqrt[9]{1.4} = 1 + \frac{r}{2} \][/tex]
4. Isolate [tex]\( r \)[/tex]:
[tex]\[ \left(1.4\right)^{\frac{1}{9}} - 1 = \frac{r}{2} \][/tex]
[tex]\[ r = 2 \left(\left(1.4\right)^{\frac{1}{9}} - 1\right) \][/tex]
5. Calculate the value:
[tex]\[ \left(1.4\right)^{\frac{1}{9}} \approx 1.038093444 \][/tex]
[tex]\[ 2 \left(1.038093444 - 1\right) \approx 2 \times 0.038093444 \][/tex]
[tex]\[ r \approx 0.076186888 \][/tex]
6. Convert [tex]\( r \)[/tex] to a percentage:
[tex]\[ r \times 100 \approx 7.618688853 \% \][/tex]
7. Round to the nearest hundredth:
[tex]\[ 7.62\% \][/tex]
Therefore, the interest rate to the nearest hundredth of a percent that will produce [tex]$3500 from $[/tex]2500, with interest compounded semiannually for 4.5 years, is [tex]\( \boxed{7.62} \)[/tex].
