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lubasha [3.4K]
2 years ago
14

A buyer needs 150 blouses to retail at $15 each and 80 shirts to retail at $20 each. She needs to average a 50.5% markup. If she

pays $7 for each blouse, how much can she pay for each shirt in order to achieve the planned markup percent?
Business
1 answer:
kumpel [21]2 years ago
7 0

Answer:

To achieve the planned markup percent, she can pay for each shirt, the price of:

$18.85

Explanation:

a) Data and Calculations:

Number of blouses to retail at $15 each = 150

Total revenue from blouses = $2,250 (150 * $15)

Number of shirts to retail at $20 each = 80

Total revenue from shirts = $1,600 (80 * $20)

Total sales revenue = $3,850

Average markup = 50.5%

Therefore, the cost of the blouses and shirts = $3,850/1.505 = $2,558

If the cost of blouse = $7 each, the total cost of blouses = $1,050 (150 * $7)

Therefore, the cost of shirts will be equal to $1,508 (2,558 - $1,050)

Then, the cost of each shirt will be equal to $18.85 ($1,508/80).

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4 0
1 year ago
Does GAAP routinely require companies to disclosure forecasts of financial variables to external users? Indicate yes or no and e
Andrei [34K]

Answer:

No, they don´t.

Explanation:

Forecast is not required by GAAP, as the <u>Relevance</u> and the <u>Faithful</u> <u>Representation</u> are concepts that are not compatible with data projection.  Forecast implies estimates, and subjective interpretations that do not fulfill financial statements aim and are difficult to verify.

4 0
3 years ago
The Beef-up ranch feeds cattle for midwestern farmers and delivers them to processing plants in Topeka,Kansas and Tulsa, Oklahom
AfilCa [17]

Answer:

a. Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b. x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c. The optimal solution is Z = 79.25

Explanation:

Given - The table is as follows :

  • Nutrient           Feed 1                    Feed 2                      Feed 3
  •    A                      3                              2                                4
  •    B                      3                               1                                 3
  •    C                      1                                0                                2
  •    D                      6                               8                                4

The minimum requirement per cow each month is 4 pounds of nutrient A, 5 pounds of nutrient B, 1 pound of nutrient C, and 8 pounds of nutrient D. However, cows should not be fed more than twice the minimum requirement for any nutrient each month. Additionally, the ranch can only obtain 1,500 pounds of each type of feed each month. Because there are usually 100 cows at the beef-up ranch at any given time, this means that no more than 15 pounds of each type of feed can be used per cow each month.

To find - a. Formulate a linear programming problem to determine how  

                  much of each type of feed a cow should be fed each month.

              b. Create a spreadsheet model for this problem, and solve it using

                   Solver.

              c. What is the optimal solution?

Proof -

a.

Let feed 1 per cow per month = x₁

     feed 2 per cow per month = x₂

     feed 3 per cow per month = x₃

Now,

As given, The cost per pound of feeds 1,2, and 3 are $2.00, $2.50, and $3.00, respectively.

So, we have to minimize the cost , Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 4(32)

3x₁ + x₂ + 3x₃ ≤ 5(32)

x₁ + 0x₂ + 2x₃ ≤ 1(32)

6x₁ + 8x₂ + 4x₃ ≤ 8(32)

∴ we get

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

Now, as given

However, cows should not be fed more than twice the minimum requirement for any nutrient each month.

∴ we have

3x₁ + 2x₂ + 4x₃ ≥ \frac{128}{2}

3x₁ + x₂ + 3x₃ ≥ \frac{160}{2}

x₁ + 0x₂ + 2x₃ ≥ \frac{32}{2}

6x₁ + 8x₂ + 4x₃ ≥ \frac{256}{2}

and also

No more than 15 pounds of each type of feed can be used per cow each month.

⇒x₁ , x₂, x₃ ≤ 15

So,

The LPP model becomes

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b.)

We use simplex method calculator  to solve this LPP Problem

we get

x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c.)

The optimal solution is Z = 79.25

4 0
2 years ago
Alex and Jane are the best economists that FlexMoney Inc., a consulting firm, has. However, FlexMoney is in urgent need to hire
REY [17]

Answer:

Job elements method

Explanation:

Job elements methodbis defined as a work oriented job analysis method. It is more focused on human attributes that are needed to ensure superior performance on the job.

Job elements method is used to match employees with the activities that best suits their abilities.

Alex and Jane work as economists, and asked to provide criteria that are instrumental to success in their field. This is done to help with hiring of more economists. This emphasis on human attributes that will give success in the job role is called Job elements method.

5 0
3 years ago
Read 2 more answers
OJ's Orange Juice produces orange juice to sell in a competitive market.Given uncertainty in weather patterns, OJ has to determi
sweet-ann [11.9K]

Answer:

c- 1.15 units.

Explanation:

This can be calculated as follows:

Expected price at 10 percent = $5 * 10% = $0.5

Expected price at 90 percent = $2 * 90% = $1.80

Total expected price (EP) = $0.5 + $1.80 = $2.3

Since profit is maximized when EP = MC, we have:

2.3 = 2Q

Q = 2.3 / 2 = 1.15

Therefore, OJ should produce 1.15 units to maximize expected profit. The correction is therefore c- 1.15 units.

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3 years ago
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