Gerry's Tire Service provides each employee with a clean, sharp-looking uniform. It also instructs employees to put all tools ba
1 answer:
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Answer:
The answer is below
Explanation:
a)
Given that mean (μ) = $1500, standard deviation (σ) = $200, sample size (n) = 100
confidence (C) = 95% = 0.95
α = 1 - C = 1 - 0.95 = 0.05
α/2 = 0.05 / 2 = 0.025
The z score that corresponds with 0.475 (0.5 - 0.025) is 1.96. Therefore the margin of error (E) is:
The confidence interval = (μ ± E) = (1500 ± 39.2) = (1500 - 39.2, 1500 + 39.2) = (1460.8, 1539.2)
The confidence interval is between $1460.8 and $1539.2.
b) Given that mean (μ) = $1500, standard deviation for 100 samples = σ /√n = $200,
confidence (C) = 95% = 0.95
The confidence interval = (μ ± E) = (1500 ± 392) = (1500 - 392, 1500 + 392) = (1108, 1892)
The confidence interval is between $1108 and $1892.
Answer:
option (D) loss, $3,000
Explanation:
Given:
price of the truck = $48,000
estimated residual value = $8,000
Exchange price of the truck = $60,000
Trade allowance = $35,000
Since, straight line depreciation is given, thus,
Total depreciation =
or
Total depreciation = $5,000 per year
Therefore,
the book value after two years
= Price of truck - total depreciation in two years
or
= $48,000 − ($5,000 × 2 years)
= $38,000
Now,
a trade allowance received ( i.e $35,000 ) is less than the book value
therefore a loss is recorded
The amount of loss = (Book value - trade allowance received)
or
The amount of loss = $38,000 - $35,000 = $3,000
Hence, correct answer is option (D) loss, $3,000