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VLD [36.1K]
4 years ago
15

Aphrodite took out a 30-year loan from her bank for $170,000 at an APR of

Mathematics
1 answer:
horsena [70]4 years ago
8 0

Answer:

  A. $3246.74

Step-by-step explanation:

The monthly payment can be found from the amortization formula.

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the principal amount, r is the annual rate compounded n times per year for t years.

Filling in the values, we compute the monthly payment to be ...

  A = $170,000(.072/12)/(1 -(1 +.072/12)^(-12·30)) = $1153.94

__

The remaining balance after t years will be ...

  B = P(1 +r/n)^(nt) -A((1 +r/n)^(nt) -1)/(r/n)

For the given initial principal and the computed payment, after 18 years, the balance will be ...

  B = $170000(1 +.072/12)^(12·18) -$1153.94((1 +.072/12)^(12·18) -1)/(.072/12)

  B = $111,054.71

The prepayment penalty appears to be ...

  (r/2)(0.80B) = (.072/2)(0.80)($111,054.71) = $3,198.38

The closest listed answer choice is ...

  A.  $3246.74

_____

Please ask your teacher how to get the answer, since none of the offered choices appear to be correct.

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3 years ago
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Answer:

In radian :

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In degree :

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Step-by-step explanation:

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2 years ago
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\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &499\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.10\\ t=\textit{elapsed time}\dotfill &3\\ \end{cases} \\\\\\ A=499(1-0.10)^3\implies A=499(0.9)^3\implies A=363.771 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{total depreciation 499 - 363.771}}{135.229}~\hfill

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Answer:

Step-by-step explanation:

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