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MakcuM [25]
3 years ago
11

A borrower has a loan of $ 50,000.00 at 6 % compounded annually with 9 annual payments. Suppose the borrower paid off the loan a

fter 5 years. Calculate the amount needed to pay off the loan.
Mathematics
1 answer:
disa [49]3 years ago
5 0

Answer:

We say that $16911.28 is compound interest of 50,000 for 5 year  from  now ,at 6% compound annually. The total amount needed to pay off the loan $66,911.28

Step-by-step explanation:

Given that,

P=50,000

r=6%

t=5 years

A=?

According to formula  A=P\left ( 1+\frac{r}{100} \right )^{t}

A=50,000\left ( 1+\frac{6}{100} \right )^{5}

A=50,000\left ( 1+0.06 \right )^{5}

A=50,000\left ( 1.06 \right )^{5}

A=66911.28

CI = A-P  ( CI = compound interest)

CI=66,911.28-50,000.00

CI=16,911.28

Total amount needed to pay off loan= \left ( 50,000.00+16,911.28 \right ) =  $66,911.28

Compound interest  $16911.28

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0/1 For positive integer n, n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?. What is the value of 4# · 3# · 2# · 1#?
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Answer:

331776

Step-by-step explanation:

Since n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?

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Now, n? = n! · (n − 1)! · … · 1!

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Also, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

and 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 4# = 4? · 3? · 2?· 1? = 288 × 12 × 2 × 1 = 6912

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3# = 3? · (3 − 1)? · 1? = 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

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and, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 3# = 3? · 2?· 1? = 12 × 2 × 1 = 24

We now find 2#

2# = 2? · (2 − 1)? · 1? = 2? · 1?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 2# = 2?· 1? = 2 × 1 = 2

We now find 1#

1# = 1? · 1? = 1? · 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

And, 1# = 1? · 1? = 1 × 1 = 1

So,  4# · 3# · 2# · 1#? =  6912 · 24 · 2 · 1? = 331776

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