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KatRina [158]
3 years ago
5

The manager of an orchard expects about 70% of his apples to exceed the weight requirement for ""Grade A"" designation. At least

how many apples must he sample to be 90% confident of estimating the true proportion within ±4%? A) 19 B) 23 C) 89 D) 356 E) 505
Business
1 answer:
kenny6666 [7]3 years ago
6 0

Answer:

D) 356

Explanation:

ME = Z x √[(P x Q) / N]  

  • margin of error (ME) = 4%
  • 90% confidence level (Z) = 1.645 (by convention)
  • P = 70% of apples exceed Grade A
  • Q = 30% of apples do not exceed Grade A
  • N = sample size = ?  

0.04 = 1.645 x √[(0.7 x 0.3) / N]

0.04 = 1.645 x √(0.21 / N)

0.04 = 1.645 x 0.458 / √N

0.04 = 0.7538 / √N

√N = 0.7538 / 0.04 = 18.84

N = 18.84² = 355.2 ≈ 356 (there is no 0.2 apples, you must round up)

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If a borrower can afford to make monthly principal and interest payments of 1000 and the lender will make a 30 year loan at 5 1/
Alexus [3.1K]

Answer:

The the largest loan this buyer can afford is 14,533.75.

Explanation:

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PV30 = Present value or the loan the buyer can afford for a 30 year loan at 5 1/2% =?

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Step 2: Calculation of the present value or the loan the buyer can afford for a 20 year loan at 4 1/2%

PV20 = P * ((1 - (1 / (1 + r))^n) / r) …………………………………. (2)

Where;

PV30 = Present value or the loan the buyer can afford for a 20 year loan at 4 1/2% =?

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PV20 = 1000 * 13.0079364514537

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Conclusion

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