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4 years ago

What is the finance charge on $13,300 financed at 7.9 percent for 4 years if the monthly payment per $100 is $2.44?

Mathematics

1 answer:

luda_lava [24]4 years ago

5 0

If you borrowed $100, then your monthly payment is $2.44

If you borrowed $200, then your monthly payment is 2*2.44 = 4.88

etc etc

We can set up a proportion

2.44/100 = x/13300

to figure out the monthly payment x. Cross multiply and solve for x

2.44*13300 = 100*x

100x = 2.44*13300

100x = 32452

x = 32452/100

x = 324.52

So the monthly payment is $324.52

An alternative way to get this monthly payment is to apply 2.44% to 13300, which is another way to view the phrase "monthly payment per $100 is 2.44"

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There are 48 months in 4 years (start with 12 mon = 1 yr, then multiply both sides by 4) so we multiply 48 by the monthly payment to get the result 48*324.52 = 15,576.96. This is the total amount you have to pay back which is the principal plus interest.

Subtract off the principal (amount borrowed) to find the interest or finance charge: 15,576.96 - 13,300 = 2,276.96

Answer: Choice B

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3 years ago

Word problem plz help pt.1

shusha [124]

equation csn be represented as

money left = 500 - (n-1)×4.65

so we can see in first week money left

= 500 - (1-1)×4.65= 500-0=500

which is true because first week chaya got gift card and does not spend any money.

now lets assume that on week n money left is 327.95

so by equation

500 -(n-1)×4.65 = 327.95

500- 327.95 =( n-1)× 4.65

n -1 = 172.05/4.65=37

n = 37+1 = 38

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3 years ago

How many real solutions if any, does 2x^2-3x+8=0

fenix001 [56]

Answer:

The quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ has no real solution.

Step-by-step explanation:

Rewrite the quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ in the standard form $a\, x^{2} + b\, x + c = 0$ :

$2\, x^{2} + (-3)\, x + 8 = 0$ , for which:

$a = 2$ .
$b = (-3)$ .
$c = 8$ .

The quadratic discriminant of $a\, x^{2} + b\, x + c = 0$ is $(b^{2} - 4\, a\, c)$ . The quadratic discriminant of $2\, x^{2} + (-3)\, x + 8 = 0$ would be: