The sales tax is [latex]8\%[/latex] of the price of the car, and the ownership tax is [latex]2.1\%[/latex] of the price of the car.

Sales Tax = [latex]28,975 * .08 = $2,318[/latex].

Ownership Tax = [latex]28,975 * .021 = $608.475[/latex].

So, the total tax is [latex]$2,318 + $608.475 = $2,926.475[/latex].

Next, we add the destination fee, processing fees, and registration fee:

[latex]$950[/latex] (destination fee) [latex]+ $370[/latex] (processing fees) [latex]+ $617[/latex] (registration fee) [latex]= $1,937[/latex].

Now, we add the cost of the car, the total tax, and the additional fees:

[latex]$28,975[/latex] (car price) [latex]+ $2,926.475[/latex] (total tax) [latex]+ $1,937[/latex] (additional fees) [latex]= $33,838.475[/latex].

So, Luther’s total cost for the purchase is approximately [latex]$33,838.48[/latex], if we round to the nearest cent.

1. The amount to be financed is the principal, [latex]P[/latex], which is [latex]$18,325[/latex]. The rate [latex]r[/latex] is [latex]0.0675[/latex], and the term is [latex]t = 4[/latex] years. These are monthly payments, so [latex]n = 12[/latex]. Substituting and calculating, we find the monthly payment to be
   [latex]$436.70[/latex]

    

2. The amount to be financed is the principal, [latex]P[/latex], which is [latex]$41,633[/latex]. The rate [latex]r[/latex] is [latex]0.039[/latex], and the term is [latex]t = 6[/latex] years. These are monthly payments, so [latex]n = 12[/latex]. Substituting and calculating, we find the monthly payment to be
   [latex]$649.47[/latex]

Adding these values together and we find the total cost is [latex]$10,567.00[/latex].

The monthly depreciation formula is [latex]MD=\frac{P−R}{n}[/latex], Substituting [latex]$27,500[/latex] for [latex]P[/latex], [latex]$17,875[/latex] for [latex]R[/latex], and [latex]24[/latex] for [latex]n[/latex], we find [latex]MD[/latex] to be

1. The APR is the money factor times [latex]2400[/latex], so [latex]APR=2400×MF=2400×0.00001208\overline{3}=0.029[/latex]. Expressed as a percentage, the APR is [latex]2.9\%[/latex].
2. The MF is the APR divided by [latex]2400[/latex], so [latex]MF=\frac{APR}{2400}=\frac{0.079}{2400}=​​​0.00003291\overline{6}[/latex].

1. Substituting the values [latex]P = $38,750[/latex], [latex]R = $18,140.16[/latex], [latex]n = 36[/latex] and [latex]MF = 0.000035[/latex] into the formula and calculating, the monthly lease payment is
   [latex]\begin{array}{ll}
   PMT &&=\frac{(P−R)}{n}+(P+R)×MF \\
   && =\frac{($38,750−$18,140.16)}{36}+($38,750+$18,140.16)×0.000035 \\
   && =\frac{($20,609.84)}{36}+($56,890.16)×0.000035 \\
   && = \$574.49 \\
   \end{array}[/latex]

2. Given the APR, we find the MF which is [latex]MF=\frac{APR}{2400}=\frac{0.117}{2400}=0.00004875[/latex]. Substituting the values [latex]P = $45,600[/latex], [latex]R = $21,312.50[/latex], [latex]n = 24[/latex] and the MF into the formula and calculating, the monthly lease payment is
   [latex]\begin{array}{ll}
   PMT &&=\frac{(P−R)}{n}+(P+R)×MF \\
   && =\frac{($45,600−$21,312.50)}{24}+($45,600+$21,312.50)×0.00004875 \\
   && =\frac{($24,287.50)}{24}+($66,912.50)×0.00004875 \\
   && = \$1,015.24 \\
   \end{array}[/latex]

1. Personal injury protection
2. Liability insurance

The cost of the car including insurance is the monthly payment, [latex]$410.86[/latex], plus the monthly cost of the insurance. The insurance cost per month is [latex]\frac{$2,190}{12}=$182.50[/latex] since the insurance cost is for [latex]12[/latex] months. Adding those the cost with insurance is [latex]$593.36[/latex].

[latex]$42.02[/latex]

First, we need to determine the annual gas usage for each car:

• For Car J, which gets [latex]30[/latex] MPG: [latex]12,000[/latex] miles [latex]÷ 30[/latex] MPG [latex]= 400[/latex] gallons per year
• For Car K, which gets [latex]40[/latex] MPG: [latex]12,000[/latex] miles [latex]÷ 40[/latex] MPG [latex]= 300[/latex] gallons per year

Then, we can calculate the annual cost of gas for each car:

• For Car J: [latex]400[/latex] gallons [latex]\times $4.00[/latex] per gallon [latex]= $1,600[/latex]
• For Car K: [latex]300[/latex] gallons [latex]\times $4.00[/latex] per gallon [latex]= $1,200[/latex]

Therefore, by choosing Car K, which is more fuel-efficient, you would save [latex]$400[/latex] on gas costs per year given the stated conditions.