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# the efficiency of simple machine is 80%when a load of 60N is moved by 20N by How to find out how muchdoes the load move when 20N moves 10CM​

Option (A) is correct.

Given that,

overhead to each unit produced = \$7.50

Department 1:

Direct labor hours (DLH) = 6,610

Machine hours = 700 MH

Department 2 :

Machine hours = 800 MH

= \$74,358 + \$49,572

= \$123,930

Total Direct Labor Hours:

= 16,524

Direct Labor Hours for Department 2:

= Total Direct Labor Hours - Direct labor hours of department 1

= 16,524 - 6,610

= 9,914 DLH

1) \$ 11 per hour

2) Work in Process (Debit)                                      660,000

3) Manufacturing overhead is over-allocated by \$ 69,500

4) Manufacturing overhead clearing (debit)         69,500

Cost of good sold                                               69,500

Yes, the entry decreases the cost of goods sold as the actual overhead expenses are less than the applied over head cots.

1) Overhead rate = Expected manufacturing overhead / Estimated number of machine hours

Overhead rate: 880,000 / 80,000 = \$ 11 / machine hour

2) Journal entry is recorded based on every hour worked during the period. Since 60,000 hours are worked so total of 660,000 manufacturing overheads are recorded using following entry.

Work in Process (Debit)                                      660,000

3) Under-application results when actual manufacturing overheads are higher than expectation and vice versa. In question provided total hours were 60,000 instead of 80,000 budgeted hours, further actual cost amounted to \$590,500 (530,000 for depreciation, 36,500 for property taxes, 24,000 plant janitor), thus there is over application by 660,000 - 590,500 = 69,500.

Note: Sales salaries and delivery wages are not included in overhead expenses as they are not part of manufacturing overheads.

4) Manufacturing overhead clearing (debit)         69,500

Cost of good sold                                               69,500

As there is over application of overhead expenses so there is a credit in manufacturing overhead account, so in order to clear the clearing account balance is transferred to cost of goods sold resulting in net profit to bottom line.

13,200

2 hour = 120 miniteas

3/4 hours = 45 minutes

120 + 45 = 165 minutes

165 minutes × 80 boxes = 13,200 boxes

8385

9,600 boxes

60 mins in 1 hr = so 120 minutes in 2 hrs

80 boxes/ per minute

so 80 x 120 minutes = 9,600 boxes

a) To determine the minimum sample size we need to use the formula shown in the picture 1.

E is the margin of error, which is the distance from the limits to the middle (the mean) of the confidence interval. This means that we have to divide the range of the interval by 2 to find this distance.

E = 0.5/2 = 0.25

Now we apply the formula

n = (1.645*0.80/0.25)^2 = 27.7 = 28

The minimum sample size would be 28.

b) To answer the question we are going to make a 90% confidence interval. The formula is:

(μ - E, μ + E)

μ is the mean which is 127. The formula for E is shown in the picture.

E = 0.80*1.645/√8 = 0.47

(126.5, 127.5)

This means that the true mean is going to be contained in this interval 90% of the time. This is why it doesn't seem possible that the population mean is exactly 128.

D. 400 m

I just took the test, so I know that is right : )

Find copies per minute:

80 / 2.5 = 32 copies per minute.

Multiply copies per minute by total minutes:

32 x 0.25 = 8 copies

The distance Mr Franklin applied the force is 400 m

The efficiency of the machine = 80%

The force applied by the machine = 800 N

The distance over which the force is applied by the machine = 20 m

The force applied by Mr. Franklin = 50 N

The work done by the machine = 800 × 20 = 16000 J

The ideal work at 100% efficiency = X

80% of X = 16000 J

X = 16000 N/0.8 = 20,000 J

Therefore, we have have;

Work = Force × Distance;

20,000 J = 50 N × Distance Mr Franklin applied the force

Distance Mr Franklin applied the force = 20,000 J/(50 N) = 400 m

The distance Mr Franklin applied the force = 400 m.

Efficiency of a machine is 80% this means that the total energy produced to the machine only 80% is useful and given as a output.ლ(´ ❥ `ლ)
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