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zalisa [80]
3 years ago
7

corey’s campus store has $4,000 of inventory on hand at the beginning of the month. during the month, the company buys $41,000 o

f merchandise and sells merchandise that had cost $30,000. at the end of the month, $13,000 of inventory is on hand. how much shrinkage occurred during the month?
Mathematics
1 answer:
Scrat [10]3 years ago
6 0

If Corey’s campus store has $4,000 of inventory on hand at the beginning of the month. during the month, the company buys $41,000 of merchandise and sells merchandise that had cost $30,000. at the end of the month, $13,000 of inventory is on hand. how much shrinkage occurred during the month will be $2,000

Beginning inventory on hand $4,000

Add purchase $41,000

Cost of goods sold $45,000

Less Ending Merchandise sold ($30,000)

Less Ending Inventory on hand  ($13,000)

Shrinkage $2,000

Inconclusion if Corey’s campus store has $4,000 of inventory on hand at the beginning of the month. during the month, the company buys $41,000 of merchandise and sells merchandise that had cost $30,000. at the end of the month, $13,000 of inventory is on hand. how much shrinkage occurred during the month will be $2,000

Learn more here:

brainly.com/question/18685380

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Answer:

The answer is \sqrt{\frac{6}{5}}

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

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Solving the double integral with these new limits we have:

\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy

This part is a little bit tricky so let's solve the integral first for dy:

\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx

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\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx

Solving now for dx:

[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.

Replacing the limits:

\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}

As I mentioned before, this volume is equal to 1, hence:

\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }

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