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CaHeK987 [17]
3 years ago
11

Equipment was purchased at a cost of $52,000. It had an estimated useful life of seven years and a residual value of $3,000. Ass

uming the equipment was sold at the end of Year 6 for $14,000, which of the following will be included in the journal entry? (Assume the straight-line depreciation method.)
Mathematics
1 answer:
Kipish [7]3 years ago
4 0

Answer:

a debit to Accumulated Depreciation—Equipment

Step-by-step explanation:

we know that

Depreciation/year=(Cost-Residual value)/Useful life

put values

=(52000-3000)/7=$7000

Hence book value as on date of sale=52000-(7000*6)=$10000

Hence gain on sale=(14000-10000)=$4000

Hence the journal is :

Cash a/c..Dr$14000

Accumulated Depreciation..Dr$42000

To equipment $52000

To gain on sale $4000

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In principle, the Harris and the Arlen family should cover the same distance. Since distance is speed x time,

45 (6+y) = 58y

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3 years ago
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

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3 years ago
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Answer : AB = 12

Explanation:

2x + 6 + 5x - 3 = 24
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x = 3

AB = 2x + 6 = 2(3) + 6 = 12
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The expression that does not represent the sum of n and 6 is 6n
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