Latoya is taking out an amortized loan for \( \$ 37,000 \) to buy a new car and is deciding between the offers from two lenders. She wants to know which one would be the better deal over the life of the car loan, and by how much. Answer each part. Do not round intermediate computations, and round your answers to the nearest cent. If necessary, refer to the list of financial formulas. (a) Her credit union has offered her a 6 -year car loan at an annual interest rate of \( 8.9 \% \). Find the monthly payment. \( \$ \square \) (b) A bank has offered her a 7 -year car loan at an annual interest rate of \( 7.6 \% \). Find the monthly payment. \( \$ \square \) (c) Suppose Latoya pays the monthly payment each month for the full term. Which lender's car loan would have the lowest total amount to pay off, and by how much? Credit union The total amount paid would be \( \$ \square \) less than to the bank. Bank The total amount paid would be \( \$ \square \) less than to the credit union.

To solve this problem, we will use the formula for the monthly payment on an amortized loan, which is given by: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] where: - \( M \) is the monthly payment, - \( P \) is the loan principal (amount borrowed), - \( r \) is the monthly interest rate (annual interest rate divided by 12), - \( n \) is the total number of payments (loan term in months). ### Part (a): Credit Union Offer 1. **Loan Amount**: \( P = 37000 \) 2. **Annual Interest Rate**: \( 8.9\% \) or \( 0.089 \) 3. **Monthly Interest Rate**: \( r = \frac{0.089}{12} \) 4. **Loan Term**: \( 6 \) years or \( n = 6 \times 12 \) Now, let's calculate the monthly payment \( M \) for the credit union offer. ### Part (b): Bank Offer 1. **Loan Amount**: \( P = 37000 \) 2. **Annual Interest Rate**: \( 7.6\% \) or \( 0.076 \) 3. **Monthly Interest Rate**: \( r = \frac{0.076}{12} \) 4. **Loan Term**: \( 7 \) years or \( n = 7 \times 12 \) Now, let's calculate the monthly payment \( M \) for the bank offer. ### Part (c): Total Amount Paid To find out which lender's car loan would have the lowest total amount to pay off, we will calculate the total amount paid for each loan: \[ \text{Total Amount Paid} = M \times n \] Now, let's perform the calculations for both parts (a) and (b) and then compare the total amounts. Calculate the value by following steps: - step0: Calculate: \(\frac{37000\left(\frac{0.076}{12}\right)\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}}{\left(\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}-1\right)}\) - step1: Remove the parentheses: \(\frac{37000\left(\frac{0.076}{12}\right)\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}}{\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}-1}\) - step2: Divide the terms: \(\frac{37000\left(\frac{0.076}{12}\right)\left(1+\frac{19}{3000}\right)^{7\times 12}}{\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}-1}\) - step3: Add the numbers: \(\frac{37000\left(\frac{0.076}{12}\right)\left(\frac{3019}{3000}\right)^{7\times 12}}{\left(1+\left(\frac{0.076}{12}\right)\right)^{7\times 12}-1}\) - step4: Divide the terms: \(\frac{37000\left(\frac{0.076}{12}\right)\left(\frac{3019}{3000}\right)^{7\times 12}}{\left(1+\frac{19}{3000}\right)^{7\times 12}-1}\) - step5: Add the numbers: \(\frac{37000\left(\frac{0.076}{12}\right)\left(\frac{3019}{3000}\right)^{7\times 12}}{\left(\frac{3019}{3000}\right)^{7\times 12}-1}\) - step6: Divide the terms: \(\frac{37000\times \frac{19}{3000}\left(\frac{3019}{3000}\right)^{7\times 12}}{\left(\frac{3019}{3000}\right)^{7\times 12}-1}\) - step7: Multiply the numbers: \(\frac{37000\times \frac{19}{3000}\left(\frac{3019}{3000}\right)^{84}}{\left(\frac{3019}{3000}\right)^{7\times 12}-1}\) - step8: Multiply the numbers: \(\frac{37000\times \frac{19}{3000}\left(\frac{3019}{3000}\right)^{84}}{\left(\frac{3019}{3000}\right)^{84}-1}\) - step9: Multiply: \(\frac{\frac{703\times 3019^{84}}{3\times 3000^{84}}}{\left(\frac{3019}{3000}\right)^{84}-1}\) - step10: Subtract the numbers: \(\frac{\frac{703\times 3019^{84}}{3\times 3000^{84}}}{\frac{3019^{84}-3000^{84}}{3000^{84}}}\) - step11: Multiply by the reciprocal: \(\frac{703\times 3019^{84}}{3\times 3000^{84}}\times \frac{3000^{84}}{3019^{84}-3000^{84}}\) - step12: Rewrite the expression: \(\frac{703\times 3019^{84}}{3\times 3000^{84}}\times \frac{3^{84}\times 1000^{84}}{3019^{84}-3000^{84}}\) - step13: Reduce the numbers: \(\frac{703\times 3019^{84}}{3000^{84}}\times \frac{3^{83}\times 1000^{84}}{3019^{84}-3000^{84}}\) - step14: Rewrite the expression: \(\frac{703\times 3019^{84}}{3^{84}\times 1000^{84}}\times \frac{3^{83}\times 1000^{84}}{3019^{84}-3000^{84}}\) - step15: Reduce the numbers: \(\frac{703\times 3019^{84}}{3}\times \frac{1}{3019^{84}-3000^{84}}\) - step16: Multiply the fractions: \(\frac{703\times 3019^{84}}{3\left(3019^{84}-3000^{84}\right)}\) - step17: Multiply: \(\frac{703\times 3019^{84}}{3\times 3019^{84}-3\times 3000^{84}}\) Calculate or simplify the expression \( 37000 * (0.089/12) * (1 + (0.089/12))^(6*12) / ((1 + (0.089/12))^(6*12) - 1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{37000\left(\frac{0.089}{12}\right)\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}}{\left(\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}-1\right)}\) - step1: Remove the parentheses: \(\frac{37000\left(\frac{0.089}{12}\right)\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}}{\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}-1}\) - step2: Divide the terms: \(\frac{37000\left(\frac{0.089}{12}\right)\left(1+\frac{89}{12000}\right)^{6\times 12}}{\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}-1}\) - step3: Add the numbers: \(\frac{37000\left(\frac{0.089}{12}\right)\left(\frac{12089}{12000}\right)^{6\times 12}}{\left(1+\left(\frac{0.089}{12}\right)\right)^{6\times 12}-1}\) - step4: Divide the terms: \(\frac{37000\left(\frac{0.089}{12}\right)\left(\frac{12089}{12000}\right)^{6\times 12}}{\left(1+\frac{89}{12000}\right)^{6\times 12}-1}\) - step5: Add the numbers: \(\frac{37000\left(\frac{0.089}{12}\right)\left(\frac{12089}{12000}\right)^{6\times 12}}{\left(\frac{12089}{12000}\right)^{6\times 12}-1}\) - step6: Divide the terms: \(\frac{37000\times \frac{89}{12000}\left(\frac{12089}{12000}\right)^{6\times 12}}{\left(\frac{12089}{12000}\right)^{6\times 12}-1}\) - step7: Multiply the numbers: \(\frac{37000\times \frac{89}{12000}\left(\frac{12089}{12000}\right)^{72}}{\left(\frac{12089}{12000}\right)^{6\times 12}-1}\) - step8: Multiply the numbers: \(\frac{37000\times \frac{89}{12000}\left(\frac{12089}{12000}\right)^{72}}{\left(\frac{12089}{12000}\right)^{72}-1}\) - step9: Multiply: \(\frac{\frac{3293\times 12089^{72}}{12\times 12000^{72}}}{\left(\frac{12089}{12000}\right)^{72}-1}\) - step10: Subtract the numbers: \(\frac{\frac{3293\times 12089^{72}}{12\times 12000^{72}}}{\frac{12089^{72}-12000^{72}}{12000^{72}}}\) - step11: Multiply by the reciprocal: \(\frac{3293\times 12089^{72}}{12\times 12000^{72}}\times \frac{12000^{72}}{12089^{72}-12000^{72}}\) - step12: Rewrite the expression: \(\frac{3293\times 12089^{72}}{12\times 12000^{72}}\times \frac{12^{72}\times 1000^{72}}{12089^{72}-12000^{72}}\) - step13: Reduce the numbers: \(\frac{3293\times 12089^{72}}{12000^{72}}\times \frac{12^{71}\times 1000^{72}}{12089^{72}-12000^{72}}\) - step14: Rewrite the expression: \(\frac{3293\times 12089^{72}}{12^{72}\times 1000^{72}}\times \frac{12^{71}\times 1000^{72}}{12089^{72}-12000^{72}}\) - step15: Reduce the numbers: \(\frac{3293\times 12089^{72}}{12\times 1000^{72}}\times \frac{1000^{72}}{12089^{72}-12000^{72}}\) - step16: Rewrite the expression: \(\frac{3293\times 12089^{72}}{12\times 1000^{72}}\times \frac{8^{72}\times 125^{72}}{12089^{72}-12000^{72}}\) - step17: Rewrite the expression: \(\frac{3293\times 12089^{72}}{4\times 3\times 1000^{72}}\times \frac{8^{72}\times 125^{72}}{12089^{72}-12000^{72}}\) - step18: Rewrite the expression: \(\frac{3293\times 12089^{72}}{2^{2}\times 3\times 1000^{72}}\times \frac{2^{216}\times 125^{72}}{12089^{72}-12000^{72}}\) - step19: Reduce the numbers: \(\frac{3293\times 12089^{72}}{3\times 1000^{72}}\times \frac{2^{214}\times 125^{72}}{12089^{72}-12000^{72}}\) - step20: Rewrite the expression: \(\frac{3293\times 12089^{72}}{3\times 8^{72}\times 125^{72}}\times \frac{2^{214}\times 125^{72}}{12089^{72}-12000^{72}}\) - step21: Rewrite the expression: \(\frac{3293\times 12089^{72}}{3\times 2^{216}\times 125^{72}}\times \frac{2^{214}\times 125^{72}}{12089^{72}-12000^{72}}\) - step22: Reduce the numbers: \(\frac{3293\times 12089^{72}}{3\times 2^{2}}\times \frac{1}{12089^{72}-12000^{72}}\) - step23: Multiply the fractions: \(\frac{3293\times 12089^{72}}{12\left(12089^{72}-12000^{72}\right)}\) - step24: Multiply: \(\frac{3293\times 12089^{72}}{12\times 12089^{72}-12\times 12000^{72}}\) Calculate or simplify the expression \( 665.110014 * (6 * 12) \). Calculate the value by following steps: - step0: Calculate: \(665.110014\left(6\times 12\right)\) - step1: Remove the parentheses: \(665.110014\times 6\times 12\) - step2: Multiply the terms: \(3990.660084\times 12\) - step3: Multiply the numbers: \(47887.921008\) Calculate or simplify the expression \( 569.344029 * (7 * 12) \). Calculate the value by following steps: - step0: Calculate: \(569.344029\left(7\times 12\right)\) - step1: Remove the parentheses: \(569.344029\times 7\times 12\) - step2: Multiply the terms: \(3985.408203\times 12\) - step3: Multiply the numbers: \(47824.898436\) ### Part (a): Monthly Payment for Credit Union Using the formula, we calculated the monthly payment for the credit union: \[ M = 665.11 \] Thus, the monthly payment for the credit union is: \[ \text{Monthly Payment (Credit Union)} = \$665.11 \] ### Part (b): Monthly Payment for Bank Using the formula, we calculated the monthly payment for the bank: \[ M = 569.34 \] Thus, the monthly payment for the bank is: \[ \text{Monthly Payment (Bank)} = \$569.34 \] ### Part (c): Total Amount Paid Now, we calculate the total amount paid for each loan: 1. **Total Amount Paid to Credit Union**: \[ \text{Total Amount (Credit Union)} = 665.11 \times (6 \times 12) = 47887.92 \] 2. **Total Amount Paid to Bank**: \[ \text{Total Amount (Bank)} = 569.34 \times (7 \times 12) = 47824.90 \] ### Comparison of Total Amounts - Total amount paid to the credit union: \( \$47887.92 \) - Total amount paid to the bank: \( \$47824.90 \) Now, we find the difference: \[ \text{Difference} = 47887.92 - 47824.90 = 63.02 \] ### Final Answers (a) The monthly payment for the credit union is \( \$665.11 \). (b) The monthly payment for the bank is \( \$569.34 \). (c) The bank's car loan would have the lowest total amount to pay off, and the total amount paid would be \( \$63.02 \) less than to the credit union.