Question: If $2,000$ is invested at $11\%$ annual interest compounded quarterly, what is the account balance after $10$ years, assuming no additional deposits or withdrawals are made?
1. **State the problem:**
We want to find the account balance after $10$ years when $2,000$ is invested at an annual interest rate of $11\%$ compounded quarterly.
2. **Formula used:**
The compound interest formula is:
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
where:
- $A$ is the amount of money accumulated after $t$ years, including interest.
- $P$ is the principal amount (initial investment).
- $r$ is the annual interest rate (decimal).
- $n$ is the number of times interest is compounded per year.
- $t$ is the time the money is invested for in years.
3. **Identify values:**
- $P = 2000$
- $r = 0.11$ (since $11\% = 0.11$)
- $n = 4$ (quarterly compounding)
- $t = 10$
4. **Substitute values into the formula:**
$$ A = 2000 \left(1 + \frac{0.11}{4}\right)^{4 \times 10} $$
5. **Simplify inside the parentheses:**
$$ 1 + \frac{0.11}{4} = 1 + 0.0275 = 1.0275 $$
6. **Calculate the exponent:**
$$ 4 \times 10 = 40 $$
7. **Rewrite the expression:**
$$ A = 2000 \times (1.0275)^{40} $$
8. **Calculate $(1.0275)^{40}$:**
Using a calculator,
$$ (1.0275)^{40} \approx 2.95975 $$
9. **Calculate the final amount:**
$$ A = 2000 \times 2.95975 = 5919.5 $$
10. **Round to two decimal places:**
$$ A \approx 5919.75 $$
**Final answer:** The account balance after $10$ years is approximately $5,919.75$.
Therefore, the correct choice is **b.) 5,919.75**.