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3 years ago

Please answer this question...

Business

2 answers:

konstantin123 [22]3 years ago

6 0

Answer:yes i did thank you for caringgggggggggg have a good day^.^

Explanation:

oee [108]3 years ago

5 0

Answer: yes! Aww thanks, you too :))

Explanation: have a great day!!

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Direct Materials, Direct Labor, and Factory Overhead Cost Variance Analysis

seropon [69]

Answer:

Instructions are listed below.

Explanation:

Giving the following information:

Mackinaw Inc. processes a base chemical into plastic. Standard costs and actual costs for direct materials, direct labor, and factory overhead incurred for the manufacture of 40,000 units of product were as follows:

Standard Costs - Actual Costs

Direct materials 120,000 lb. at $3.20 118,500 lb. at $3.25

Direct labor 12,000 hrs. at $24.40 11,700 hrs. at $25.00

Factory overhead Rates per direct labor hr., based on 100% of normal capacity of 15,000 direct labor hrs.:

Variable cost, $8.00 $91,200 variable cost

Fixed cost, $10.00 $150,000 fixed cost

Each unit requires 0.3 hour of direct labor.

A) Direct material price variance= (standard price - actual price)*actual quantity

Direct material price variance= (3.20 - 3.25)*118,500= $5925 unfavorable

Direct material quantity variance= (standard quantity - actual quantity)*standard price

Direct material quantity variance= (120,000 - 118,500)*3.20=-$4,800 favorable

Total direct material variance= 5,925 - 4,800= 1,125 unfavorable

B)Direct labor efficiency variance= (SQ - AQ)*standard rate

Direct labor efficiency variance= (12,000 - 11,700)*24.40= -$7,320 favorable

Direct labor price variance= (SR - AR)*AQ

Direct labor price variance= (24.40 - 25)*11,700= $7,020 unfavorable

Total direct labor variance= $300 favorable

C) Variable factory overhead controllable variance= (8*15,000 - 92,100)= -$27,900 favorable

Fixed factory overhead volume variance= (10*15,000 - 150,000)= 0

Total factory overhead variance= 27,900 favorable

3 0

4 years ago

How does a home inspection help buyers?

Katen [24]

A, the purpose of inspections is to verify there are not problems within the house itself whether those problems arise in the form of plumbing, electric, or the foundation to warn possible buyers.

3 0

3 years ago

Read 2 more answers

An initial time study was done on a process with the following results (in minutes). Based on the data obtained so far, assumi

kiruha [24]

Answer: 24.60 minutes (2dp)

Explanation:

The question is essentially asking that we find the Standard Time it takes to serve a customer.

To do this we would have to calculate the averages of the different elements and then use this to find the normal time which we can then use to find the Standard Time.

Element 1

= (3 + 4 + 4 + 3 + 3)/5

= 3.4

Element 2

= ( 9 + 8 + 10 + 11 + 10)/5

= 9.6

Element 3

= ( 7 + 8 + 6 + 7 + 8)/5

= 7.2

We then calculate the normal times by multiplying each of the individual means with their performance ratings.

That would be,

Element A

= 3.4 * 70%

= 2.38

Element B

= 9.6 * 110%

= 10.56

Element C

= 7.2 * 120%

= 8.64

We then add up the normal times to get the total normal time

= 2.38 + 10.56 + 8.64

= 21.58 minutes

Now we can solve for the standard time using this formula,

Standard time = Normal time (1+Allowance factor)

= 21.58 ( 1 + 0.14)

= 24.6012 minutes

= 24.60 minutes (2dp)

The time per unit customer served is 24.60 minutes.

If you need any clarification please do comment. Cheers.

6 0

3 years ago

The ability to read, understand, and analyze accounting reports and financial statements is critical in understanding ______.

Pie

Answer:

business operations

Explanation:

7 0

2 years ago

Before the last presidential debates, 50% of registered votes indicated they were planning to vote for the incumbent president.

MAVERICK [17]

Answer:

a) H₀ : p = 0.5

$H_{a} : p > 0.5$

b) z = 2.0785

c) p - value = 0.0188

d) Our conclusion is that there has been a significant increase in the proportion of registered voters who are planning to vote for the incumbent president

Explanation:

Probability of those planning to vote for the incumbent president before the debate, P₀ = 50/100 = 0.5

a) The null and the alternative hypothesis

Let the null hypothesis be represented by $H_{0}$

and the alternative hypothesis be represented by $H_{a}$

H₀ : p = 0.5

$H_{a} : p > 0.5$

b) compute the test statistics

sample size, n = 1200

number of voters planning to vote for the incumbent president, x = 636

Probability of those planning to vote for the incumbent president in the selected sample:

$\bar{p} = x/n\\\bar{p} = 636/1200\\\bar{p} = 0.53$

Formula for the test statistic:

$z = \frac{\bar{p} - p_{0} }{\sqrt{\frac{p_{0}(1-p_{0} ) }{n} } }$

$z = \frac{0.53 - 0.5 }{\sqrt{\frac{0.5(1-0.5 ) }{1200} } }$

z = 2.0785

For $\alpha = 0.05\\$ , the critical value as checked from the normal table will be:

$z_{\alpha} = 1.645$

The P value for z > 2.0785

$P-value = P(z > 2.0785) \\P - value = \phi (-2.0785)\\P - value = 0.0188$

d) What do you conclude?

since $( z = 2.0785) > (z_{\alpha} = 1.645)$ , we will reject the null hypothesis.

Our conclusion is that there has been a significant increase in the proportion of registered voters who are planning to vote for the incumbent president

6 0

4 years ago