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Romashka [77]
2 years ago
7

On april 1 of the current year, a company purchased and placed in service a machine with a cost of $240,000. the company estimat

ed the machine's useful life to be four years or 60,000 units of output with an estimated salvage value of $60,000. during the current year, 12,000 units were produced. prepare the necessary december 31 adjusting journal entry to record depreciation for the current year assuming the company uses: a. the straight-line method of depreciation b. the units-of-production method of depreciation c. the double-declining balance method of depreciation
Mathematics
1 answer:
Viktor [21]2 years ago
8 0

The answers to the questions are:

  • a. $33,750
  • b. $36000
  • c. $90000

a. <u>the straight-line method of depreciation</u>

\frac{cost of machine - salvage}{useful life} * period

= (240000 - 60000) / 4 * (9 / 12)

= $33,750

The debit is 33750 as well as the credit

b. <u> the units-of-production method of depreciation</u>

(240000 - 60000) / 60000 * 12000

= $36000

c. <u>the double-declining balance method of depreciation</u>

(24000 *2 /4) * (9/12)

= $90000

Read more on depreciation here: brainly.com/question/1203926

#SPJ1

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Please help! Correct answers only please!
Marrrta [24]

Answer:

D. is your answer

Step-by-step explanation:

V=whl=7*6.5*7=318.5unit^3

3 0
3 years ago
Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the
Troyanec [42]

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}

Now, we calculate the sum

\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\

Therefore, we get

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98

So, the sample standard deviation is 6.98.

3 0
3 years ago
PLEASE HELO ASAP, 12 POINTS!!!
never [62]

Answer:b

Step-by-step explanation:l

3 0
3 years ago
Read 2 more answers
How much money is borrowed if the interest rate is 9.25% simple interest and the loan is made for 3.5 years and has 904.88 inter
Mars2501 [29]

The amount borrowed is 2795

<h3>What is Principal?</h3>

Principal is the amount loaned or invested in a bank or a financial institution that gives a certain interest on a specific rate over a period of time.

Analysis:

P = principal = unknown

R = rate = 9.25%

T = time = 3.5 years

I  = interest = 904.88

simple interest(I) = PRT/100

P = 100I/RT

P = 100 x 904.88/3.5 x 9.25 = 2795

In conclusion, the amount borrowed is 2795.

Learn more about simple interest: brainly.com/question/20690803

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6 0
2 years ago
Can someone help me with this problem
expeople1 [14]
Increasing
Increasing
Decreasing
t=3
t=14
V=4
7 0
3 years ago
Read 2 more answers
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