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3 years ago

What is the amount paid for the loan for $4800 at 9.9% for 4 years

Mathematics

2 answers:

lara31 [8.8K]3 years ago

7 0

Answer:

$1900.80 in interest on this $4800.00 loan

Step-by-step explanation:

By "amount paid" you most likely mean "amount of interest owed."

This would be i = p * r * t, and here,

i = $4800*0.099*4 = $1900.80.

ElenaW [278]3 years ago

5 0

Answer:

Total amount paid = 5,832.48

Total Cost = [rate * principal * # payments] / (1-(1+rate)^ -# pmts)

rate = 9.9 / 1,200 = 0.00825

Total Cost = [0.00825 * 4,800 * 48] / (1 -(1.00825)^-48)

Total Cost = 1,900.80 / 1 - 0.6741009317

Total Cost = 1,900.80 / 0.3258990683

Total Cost = 5,832.48

Source: 1728.com/loanfrm4.htm

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

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2 years ago

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2 years ago

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olga2289 [7]

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3 years ago

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