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Maurinko [17]
3 years ago
6

When originally purchased, a vehicle costing $25,560 had an estimated useful life of 8 years and an estimated salvage value of $

3,000. after 4 years of straight-line depreciation, the asset's total estimated useful life was revised from 8 years to 6 years and there was no change in the estimated salvage value. the depreciation expense in year 5 equals:?
Mathematics
1 answer:
e-lub [12.9K]3 years ago
5 0
If straight-line depreciation is used, the value being depreciated over 8 years is
  25,560 -3,000 = 22,560
so the depreciation in 4 years is
  22,560*4/8 = 11,280
This is also the remaining value to be depreciated, now in 2 years. The depreciation in year 5 is
  11,280/2 = 5,640
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Answer:

a = -5/3

Step-by-step explanation:

3/a × -4 = 20

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3a = -5

a = -5/3

Check:

3/(-5/3) × - 4 = 20

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Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
katrin2010 [14]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31.4 or higher are significantly high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 20.8, \sigma = 5.3

Identify the test scores that are significantly low or significantly high.

Significantly low

Z = -2 and lower.

So the significantly low scores are thoses values that are lower or equal than X when Z = -2. So

Z = \frac{X - \mu}{\sigma}

-2 = \frac{X - 20.8}{5.3}

X - 20.8 = -2*5.3

X = 10.2

Test scores of 10.2 or lower are significantly low.

Significantly high

Z = 2 and higher.

So the significantly high scores are thoses values that are higherr or equal than X when Z = 2. So

Z = \frac{X - \mu}{\sigma}

2 = \frac{X - 20.8}{5.3}

X - 20.8 = 2*5.3

X = 31.4

Test scores of 31.4 or higher are significantly high.

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3 years ago
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Answer:

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