Interest rate (with changing years). Keiko is looking at the following investment choices and wants to know what annual rate of return each choice produces. a. Invest \( \$ 450.00 \) and receive \( \$ 746.23 \) in 11 years. b. Invest \( \$ 3,600.00 \) and receive \( \$ 10,465.82 \) in 17 years. c. Invest \( \$ 32,914.85 \) and receive \( \$ 120,000.00 \) in 20 years. d. Invest \( \$ 33,619.78 \) and receive \( \$ 1,300,000.00 \) in 40 years. c. What annual rate of reth will Keiko earn if she invests \( \$ 32,914.85 \) today and receives \( \$ 120,000.00 \) in 20 years? \% (Round to two decimal places.)

To determine the annual rate of return (\( r \)) Keiko earns by investing \( \$32,914.85 \) today and receiving \( \$120,000.00 \) in 20 years, we can use the formula for compound interest: \[ FV = PV \times (1 + r)^n \] Where: - \( FV \) = Future Value (\$120,000.00) - \( PV \) = Present Value (\$32,914.85) - \( n \) = Number of years (20) - \( r \) = Annual rate of return **Step 1: Set Up the Equation** \[ 120,000 = 32,914.85 \times (1 + r)^{20} \] **Step 2: Solve for \( (1 + r)^{20} \)** \[ (1 + r)^{20} = \frac{120,000}{32,914.85} \approx 3.64528 \] **Step 3: Take the Natural Logarithm of Both Sides** \[ \ln((1 + r)^{20}) = \ln(3.64528) \] \[ 20 \ln(1 + r) = 1.2953 \] \[ \ln(1 + r) = \frac{1.2953}{20} \approx 0.064765 \] **Step 4: Exponentiate to Solve for \( r \)** \[ 1 + r = e^{0.064765} \approx 1.066905 \] \[ r \approx 1.066905 - 1 = 0.066905 \text{ or } 6.69\% \] **Conclusion:** Keiko will earn an **annual rate of return of approximately 6.69%**. **Answer:** Keiko earns an annual rate of return of approximately 6.69 %.