Give me the answer to this An investment of $45,000 earns 6% annual interest and is compounded semiannually. If no funds are added or removed from this account, what is the future value of the investment after 3 years? 

Round your answer to the nearest penny.




To find the future value of the investment, we will use the compound interest formula:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $A$ = future value of the investment
- $P$ = principal amount (initial investment) = 45,000
- $r$ = annual interest rate = 6% or 0.06
- $n$ = number of times the interest is compounded per year = 2 (semiannually)
- $t$ = time in years = 3

Substituting these values into the formula:

\[
A = 45000 \left(1 + \frac{0.06}{2}\right)^{2 \times 3}
\]

Now let's calculate this.The future value of the investment after 3 years, compounded semiannually at 6% interest, is approximately **$53,732.35**.

Would you like more details, or do you have any other questions?

Here are five related questions:
1. How would the future value change if the interest were compounded quarterly instead?
2. What would be the future value after 5 years with the same interest rate and compounding period?
3. How does the future value compare if interest is compounded annually instead of semiannually?
4. What happens to the future value if the interest rate increases to 7%?
5. How would adding regular contributions affect the future value?

**Tip:** Semiannual compounding means that interest is added twice a year, which generally yields a higher return than annual compounding.

An investment of $45,000 earns 6% annual interest and is compounded semiannually. If no funds are added or removed from this account, what is the future value of the investment after 3 years? Round your answer to the nearest penny.

Find out how to calculate the future value of a $45,000 investment earning 6% interest, compounded semiannually over 3 years. This problem uses the compound interest formula and explores key financial math concepts, suitable for high school students (Grades 9-12).

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