If Sandy can afford car payments of \( \$ 300 \) per month for 6 years, what is the price of a car that she can afford now? Assume an interest rate of 8.4 percent HW Score: \( 49.2 \%, 33.95 \) of 69 points Sandy can afford a car that costs \( \$ \square \) or less. (Round to the nearest dollar as needed.)

Given: - Monthly payment, \( PMT = \$300 \) - Loan term, \( n = 6 \text{ years} \times 12 \text{ months/year} = 72 \text{ months} \) - Annual interest rate, \( 8.4\% \); monthly rate, \( r = \frac{0.084}{12} = 0.007 \) We use the present value formula for an annuity (which represents the car price): \[ PV = PMT \times \frac{1 - (1 + r)^{-n}}{r} \] Substitute the given values: \[ PV = 300 \times \frac{1 - (1 + 0.007)^{-72}}{0.007} \] Step 1. Calculate the term \( (1 + 0.007)^{-72} \): \[ (1.007)^{-72} = \frac{1}{(1.007)^{72}} \] Step 2. Compute \( (1.007)^{72} \). Taking the natural logarithm: \[ (1.007)^{72} = e^{72 \ln(1.007)} \] Approximating \( \ln(1.007) \approx 0.006975 \): \[ 72 \ln(1.007) \approx 72 \times 0.006975 = 0.5022 \] \[ (1.007)^{72} \approx e^{0.5022} \approx 1.652 \] Thus, \[ (1.007)^{-72} \approx \frac{1}{1.652} \approx 0.6056 \] Step 3. Substitute back into the PV formula: \[ PV = 300 \times \frac{1 - 0.6056}{0.007} \] \[ PV = 300 \times \frac{0.3944}{0.007} \] Step 4. Compute the fraction: \[ \frac{0.3944}{0.007} \approx 56.34 \] Step 5. Finally, calculate the price: \[ PV \approx 300 \times 56.34 \approx 16,902 \] Rounding to the nearest dollar, Sandy can afford a car priced at approximately: \[ \$16,902 \]