To solve this problem, we need to use the formula for compound interest to determine how much money should be deposited today (the present value, $P$) so that it grows to $15,000 in 3 years.

The formula for compound interest is:

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

Where:

• $A$ is the future amount ($15,000 in this case),
• $P$ is the present value (the amount to be deposited today),
• $r$ is the annual interest rate (7% or 0.07 in decimal),
• $n$ is the number of times the interest is compounded per year (semiannually, so $n = 2$),
• $t$ is the time the money is invested or borrowed for, in years (3 years).

Now, we need to rearrange the formula to solve for $P$:

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

Substitute the given values into the equation:

$P = \frac{15000}{\left(1 + \frac{0.07}{2}\right)^{2 \times 3}}$

Let me calculate the exact value for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you through the steps to calculate the present value yourself:

1. Calculate the semiannual interest rate: $\frac{0.07}{2} = 0.035$.
2. Add 1 to the semiannual interest rate: $1 + 0.035 = 1.035$.
3. Raise this value to the power of $2 \times 3 = 6$: $1.035^6$.
4. Divide $15,000 by the result of step 3 to find the initial deposit needed.

Would you like to try this or ask anything further? Here are 5 related questions:

1. How is the compound interest formula different from simple interest?
2. What happens if the interest is compounded quarterly instead of semiannually?
3. How does increasing the interest rate affect the initial deposit amount?
4. How would you adjust the calculation for a 5-year period instead of 3 years?
5. What is the impact of changing the compounding frequency on future value?

Tip: The more frequently interest is compounded, the faster the investment grows.