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a_sh-v [17]
2 years ago
12

Niki makes the same payment every two months to pay off his $61,600 loan. The loan has an interest rate of 9.84%, compounded eve

ry two months. If Niki pays off his loan after exactly eleven years, how much interest will he have paid in total? Round all dollar values to the nearest cent. a.$39,695.48b.$10,294.26c.$3,126.29d.$39,467.12
Mathematics
2 answers:
Umnica [9.8K]2 years ago
8 0
<h2>Answer:</h2>

The correct option is A: $39,695.48

Step-by-step explanation:

The formula to be used is :

PV = P(1 - (1 + r/t)^-nt) / (r/t)

P is the periodic payment.

Here, PV = $61,60

r = 0.0984

t = 6

n = 11 years  

Putting values in the formula we get,

61600 = P(1 - (1 + 0.0984/6)^-(11 x 6)) / (0.0984 / 6)

61600 = P(1 - (1 + 0.0164)^-66) / 0.0164

61600*0.0164 = P(1 - (1.0164)^-66)

1010.24=P(1-0.341769)=0.658231*P

P = \frac{1010.24}{0.658231}=1534.78

So, Niki pays $1534.78 every two months for eleven years.

Now the total payment made by Niki = 11*6*1534.78=101295.48

Hence, interest paid is = 101295.48-61600=39695.48

So, the correct option is A: $39,695.48

saw5 [17]2 years ago
7 0
The correct answer for this problem is a. $39695. Hope this helps!
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Answer:

0; 10; 20

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x is the independent variable

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We have to make it to be in the form r=\frac{ep}{1\pm e*cos\theta}:

r=\frac{12}{3-10cos\theta}\\\\multiply\ both\ sides\ by\ \frac{1}{3} \\\\r=\frac{12*\frac{1}{3}}{(3-10cos\theta)*\frac{1}{3}}\\\\r=\frac{12*\frac{1}{3}}{3*\frac{1}{3}-10cos\theta*\frac{1}{3}}\\\\r=\frac{4}{1-\frac{10}{3}cos\theta } \\\\r=\frac{\frac{10}{3}(\frac{6}{5} ) }{1-\frac{10}{3}cos\theta }

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