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Outdoor Company expects to sell 7 comma 500 units for $ 175 each for a total of $ 1 comma 312 comma 500 in January and 2 comma 5

nikitadnepr [17]

Answer and Explanation:

The preparation is presented below:

Outdoor Company

Inventory, Purchases, and Cost of Goods Sold Budget

Two months Ended January 31 and February 28

Particulars January February March

Sales in units 7,500 units 2,500 units 4,700 units

Sales price $175 $195 $270

Sales in dollars $1,312,500 $487,500 $1,269,000

Percentage of cost of goods sold 60% 60% 60%

Cost of goods sold $787,500 $292,500 $761,400

Add: Desired ending merchandise inventory $185,500 466,840

($292,500 × 60% + $10,000) ($761,400 × 60% + $10,000)

Total merchandise inventory required $973,000 $759,340

Less: Beginning merchandise inventory $482,500 $185,500

($787,500 × 60% + $10,000)

Budgeted purchases $490,500 $573,840

The ending inventory of month of Jan should be beginning inventory of Feb and the same is shown above

6 0

4 years ago

If the demand for bread is more elastic than the supply of bread, which group will bear more of the incidence of a tax on bread?

MatroZZZ [7]

The supplier because the tax will decrease demand bcause it is elastic.

5 0

3 years ago

True or False: All insurance policies have a cash or loan value.

Anvisha [2.4K]

The answer is false please rate me most brainleyst

5 0

3 years ago

Which of the following statements is true of raw materials?

STatiana [176]

Can you put the complete question, please?

7 0

3 years ago

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

vlada-n [284]

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.

3 0

3 years ago