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vodka [1.7K]
4 years ago
10

Transland Company has recently tried to improve its analysis for its manufacturing process. Units started into production equale

d 3,200 and ending work in process equaled 1,700 units. Transland had no beginning work in process inventory. Conversion costs are applied equally throughout production, and materials are applied at the beginning of the process. How much is the materials cost per unit if ending work in process was 25% complete and total materials costs equaled $42,880?
Business
1 answer:
Maksim231197 [3]4 years ago
6 0

Answer:

Materials cost per unit is $13.40

Explanation:

Equivalent Unit of Material = (3,200-1,700)*100% + 1,700*100%

Equivalent Unit of Material = 3,200

Materials cost per unit = total materials costs/Equivalent Unit of Material

Materials cost per unit = 42,880/3,200

Materials cost per unit = $ 13.40

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A university officer wants to know the proportion of registered students that spend more than 20 minutes to get to school. He se
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Answer:

1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.

2) We need to satisfy the following inequalities:

n\hat p =25*0.52= 13 \geq 10

n(1-\hat p) = 25*(1-0.52) =12 \geq 10

So then we satisfy this condition

3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.

So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So then the answer for this case would be:

a. Yes.

Explanation:

For this case we assume that the question is: If in the experiment described we can use the normal approximation for the proportion of interest.

For this case we have a sample of n =25

And we are interested in the proportion of registered students that spend more than 20 minutes to get to school.

X = 13 represent the number of students in the sample selected that have a time more than 20 min.

And then the estimated proportion of interest would be:

\hat p = \frac{X}{n}= \frac{13}{25}= 0.52

And we want to check if we can use the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So in order to do this approximation we need to satisfy some conditions listed below:

1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.

2) We need to satisfy the following inequalities:

n\hat p =25*0.52= 13 \geq 10

n(1-\hat p) = 25*(1-0.52) =12 \geq 10

So then we satisfy this condition:

3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.

So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So then the answer for this case would be:

a. Yes.

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3 years ago
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