To solve this problem, we can use the compound interest formula:

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

Where:

• $A$ is the amount of money accumulated (future value) = $11,000
• $P$ is the principal (initial deposit) — this is what we need to find
• $r$ is the annual interest rate = 4% or 0.04
• $n$ is the number of times the interest is compounded per year = 2 (since it's compounded semiannually)
• $t$ is the time in years = 3

Step 1: Plug the known values into the formula

$11,000 = P \left( 1 + \frac{0.04}{2} \right)^{2 \times 3}$

Step 2: Simplify the equation

$11,000 = P \left( 1 + 0.02 \right)^6$ $11,000 = P \left( 1.02 \right)^6$ $11,000 = P \times 1.126162$

Step 3: Solve for $P$

$P = \frac{11,000}{1.126162} = 9,782.01$

So, the amount of money that should be deposited today is $9,782.01 (rounded to the nearest cent).

---

Would you like further clarification on any part of the solution? Here are some related questions:

1. What if the interest rate were different, say 5%? How would that affect the deposit amount?
2. How does compounding frequency (quarterly vs. semiannually) impact the amount to be deposited?
3. What is the difference between simple interest and compound interest in such problems?
4. If you increase the deposit duration to 5 years, how would that change the amount of money needed today?
5. How would the final amount change if the principal is deposited annually instead of semiannually?

Tip: When working with compound interest problems, always double-check the compounding frequency, as it can significantly affect the outcome.