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

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

Where:

• $A$ is the future amount ($11,000 in this case),
• $P$ is the present value (the amount to be deposited),
• $r$ is the annual interest rate (5% or 0.05),
• $n$ is the number of compounding periods per year (2 for semiannual compounding),
• $t$ is the time in years (3 years).

We solve for $P$:

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

Step-by-Step Calculation:

1. Substitute the known values: $P = \frac{11,000}{\left(1 + \frac{0.05}{2}\right)^{2 \cdot 3}}$

2. Simplify the terms: $P = \frac{11,000}{\left(1 + 0.025\right)^6}$ $P = \frac{11,000}{1.025^6}$

3. Calculate $1.025^6$:
   $1.025^6 \approx 1.159693$

4. Divide $11,000$ by $1.159693$:
   $P \approx \frac{11,000}{1.159693} \approx 9,482.68$

Final Answer:

Approximately $9,482.68 should be deposited today.

---

Would you like a detailed explanation or have additional questions? Here are 5 related questions to explore:

1. What is the formula for continuous compounding, and how would the result differ?
2. How would the amount change if the interest rate increased to 6%?
3. What if the compounding frequency were quarterly instead of semiannually?
4. How is compound interest different from simple interest, and how are their formulas used?
5. How would the result change if the time period were extended to 5 years?

Tip: Always double-check your compounding frequency ($n$) since it significantly affects the final result.