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Talja [164]
1 year ago
12

A company is considering the purchase of a new machine for $75,660. management predicts that the machine can produce sales of $2

0,000 each year for the next 10 years. expenses are expected to include direct materials, direct labor, and factory overhead totaling $16,800 per year, including depreciation of $4,600 per year. what is the payback period for the new machine?
Mathematics
1 answer:
nekit [7.7K]1 year ago
3 0

The company is considering the purchase of a new machine for $75,660 (based on the available data), and the payback period is <u>24 years</u>.

<h3>What is the payback period?</h3>

The payback period is the time the company requires to recoup its investment for the new machine.

The payback period can be computed by dividing the investment cash outflows by the annual net cash inflows.

<h3>Data and Calculations:</h3>

Initial investment in new machine = $75,660

Annual depreciation expense = $4,600

Investment period = 10 years

Annual sales revenue = $20,000

Annual expenses = $16,800

Ne annual cash inflow = $3,200 ($20,000 - $16,800)

Payback period = 24 years ($75,660/$3,200)

Thus, since the payback period is <u>24 years</u>, while the investment period is 10 years, it sounds unwise for the company to continue the investment.

Learn more about the payback period at brainly.com/question/23149718

#SPJ1

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a. The amount that is saved at the expiration of the 5 year period is $22,769.20¢

b. The amount of interest is $2,769.20¢

Step-by-step explanation:

Since the amount that is deposited every year for a period of five years is $4,000 and the rate of the interest is 6.5%. We can always calculate the amount that is saved at the expiration of the five years.

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                      P[\frac{(1 + r)^{t}-1 }{r}]

   Where P is the amount deposited per year.

   r is the rate of interest

   t is the time or period

 

    and in this case, the actual value of P = $4,000

      rate of interest, r is 6.5% = 0.065

      time, t is 5 years.

   Substituting e, we have:

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                          4,000[\frac{(1 + 0.065)^{5}-1 }{0.065 }]

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  Since I deposited 4,000 every year for five years, the total amount of deposit I made at the period =

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  The amount of interest is then = $22,769.20¢ - $20,000 = $2,769.20¢

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