Charles is going to purchase a new car that has a list price of $21,450. he is planning on trading in his good-condition 2004 dodge neon and financing the rest of the cost over three years, paying monthly. his finance plan has an interest rate of 12.28%, compounded monthly. charles will also be responsible for 6.88% sales tax, a $1,089 vehicle registration fee, and a $124 documentation fee. if the dealer gives charles 80% of the listed trade-in price on his car, once the financing is paid off, what percent of the total amount paid will the interest be? (consider the trade-in to be a reduction in the amount paid.)

dodge cars in good condition
model/year
2004
2005
2006
2007
2008
viper
$7,068
$7,225
$7,626
$7,901
$8,116
neon
$6,591
$6,777
$6,822
$7,191
$7,440
intrepid
$8,285
$8,579
$8,699
$9,030
$9,121
dakota
$7,578
$7,763
$7,945
$8,313
$8,581
a.
17.64%
b.
15.67%
c.
16.70%
d.
12.86%

is C

Step-by-step explanation:

16.70% XD OWO -W- ;$

The answer is C

Step-by-step explanation:

The after-tax amount Charles from saling his 2004 Neon is : 6,591 x 80% x ( 1 - 6.88%) = $4,910

The total amount Charles has to pay for the new car : $ 21,450 + 1,089 + $124 = $22,663

The total amount Charles has to get from finance plan : $ 22,663 - 4,910 = $17,753

For the finance plan, we have: Monthly IR = 12.28% /12 = 1.023%; Number of payment(N) = 12 x 3 = 36; Present value (PV) = Financing amount = $17,753

We apply the present value formular for annuity to find the equal payment Charles needs to pay monthly : C = (PV x Monthly IR) : ( 1 - (1+i)^(-n) ) = $592

So the total repayment needs to be make in Charles' finance plan = $592 x 36 = $21,312

The total interest expenses is: Total repayment - Financing Amount = $21,312 - $17,753 = $3,559 => percentage of total interest expenses in the amount paid = 3,559/21,312 = 16.70%

The measure distances for the bus drivers is 5 miles less than that of distances for the teachers

Step-by-step explanation:

The range of the distance for the bus drivers live from school with that of the teachers is estimated to be 5 miles apart

The data that was generated in the school by the principal did an estimation of the distances of both the bus driver and that of the teachers, it was found out that the distances was five miles.