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butalik [34]
3 years ago
8

Find the amount of each payment necessary to amortize the following loan. A company borrows $84,700 for new equipment. The compa

ny agrees to make quarterly payments for 9 years at 10% per year. Find the amount of the quarterly payment.
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
5 0
This problem can be approached using the present value of annuity formula.

The present value of annuity is given by:

PV=P \left(\frac{1-\left(1+\frac{r}{t}\right)^{-nt}}{\frac{r}{t}} \right)

Where:: P is equal periodic payment, r is the annual interest rate, t is the number of payments in a year and n is the number of years for the loan to be paid.

Given that a<span> company borrows $84,700 for new equipment and that the company agrees to make quarterly payments for 9 years at 10% per year.

Thus, P = $84,700; n = 9 years and r = 10% = 0.1.

Since the payment is to be made quarterly, thus, in one year, there will be 4 payments. i.e. t = 4.

Thus, we have:

</span>84,700=P \left(\frac{1-\left(1+\frac{0.1}{4}\right)^{-9\times4}}{\frac{0.1}{4}} \right) \\  \\ =P\left( \frac{1-(1+0.025)^{-36}}{0.025} \right)=P\left(\frac{1-1.025^{-36}}{0.025}\right) \\  \\ =P\left(\frac{1-0.4111}{0.025}\right)=P\left(\frac{0.5889}{0.025}\right)=23.56P \\  \\ \therefore P= \frac{84,700}{23.56} =\$3,595.65<span>
</span>
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Answer:

x = 0.954

Step-by-step explanation:

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(4-x)² = 1.2² + 2.8²

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(Please heart and rate if you find it helpful, it's a motivation for me to help more people)

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Answer:

x-4 is a factor

Step-by-step explanation:

To know the factor, we simply equate the binomial to zero

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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The
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Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that n = 387, \pi = \frac{298}{387} = 0.77

84% confidence level

So \alpha = 0.16, z is the value of Z that has a p-value of 1 - \frac{0.16}{2} = 0.92, so Z = 1.405.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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3 years ago
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