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Orlov [11]
3 years ago
15

Bramble Corp. reported net sales of $248,700, cost of goods sold of $146,900, operating expenses of $58,000, net income of $39,9

00, beginning total assets of $473,900, and ending total assets of $635,400. Calculate profit margin and gross profit rate. (Round answers to 1 decimal place, e.g. 10.5%.) Profit margin enter percentages rounded to 1 decimal place % Gross profit rate
Business
1 answer:
juin [17]3 years ago
5 0

Answer:

profit margin is 16.0 %

gross profit rate  is 39.6 %

Explanation:

given data

net sales = $248,700

cost of goods sold = $146,900

operating expenses = $58,000

net income = $39,900

beginning total assets = $473,900

ending total assets of $635,400

to find out

profit margin and gross profit rate

solution

we will apply here profit margin formula that is

profit margin = \frac{net income}{sale} * 100      ..............1

put here value

profit margin = \frac{39900}{248700} * 100  

profit margin = 16.04 = 16.0 %

and

gross profit rate formula is

gross profit rate  = \frac{sales - cost of good }{sale} * 100    ..............2

put here value

gross profit rate  = \frac{245700 - 146900}{248700} * 100

gross profit rate   is 39.72 = 39.6 %

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The following information pertains to Crane Video Company:1. Cash balance per bank, July 31, $7,263.2. July bank service charge
My name is Ann [436]

Answer:

A. Prepare a bank reconciliation at July 31.

                                        Crane Video Company

                                             Bank Reconciliation

                                                        July 31

Cash balance per bank statement                             $7,263

Add: Deposits in transit                                                 1300

                                                                                      8,563

Less: Outstanding checks                                              591

Adjusted cash balance per bank                                $7,972

Cash balance per books                                             $7,284

Add: Collection of N/R ($700 plus

accrued interest $36 less collection

fee $20)                                                                          716

                                                                                       8,000

Less: Bank service charge                                               28

Adjusted cash balance per books                               $7,972

B. Journalize the adjusting entries at July 31 on the books of Crane Video Company.

The adjusting entry would be,

Date           Account Title                         Debit        Credit

Jul 31         Cash                                         716

                 Miscellaneous Expense           20

                 Notes Receivable                                         700

                 Interest Revenue                                            36

(to record Collection of N/R ($700 plus accrued interest $36 less collection fee $20)

31              Miscellaneous Expense            28

                 Cash                                                                28

(to record bank service charge)

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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
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.

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