Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with the symbol whose value is to be determined. Thompson Mechanical Products is planning to set aside $150,000 now for possibly replacing its large synchronous refiner motors whenever it becomes necessary. If the replacement is not needed for 7 years, how much will the company have in its investment set-aside account, provided it achieves a rate of return of 11% per year?

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Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

The amount of money that the company will have in its investment set-aside account is; $311424

What is the future value?

Formula for the future value is;

F = P(1 + i)ⁿ

Where;

F is Future value

P is Present value

i is interest rate

n is Time

The four engineering economy symbols and their values from the question are;

F = ?

P = $150000

i = 11% = 0.11

n = 7 hours

Thus;

F = 150000(1 + 0.11)⁷

F = $311424

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