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Aleksandr-060686 [28]
3 years ago
13

The following information pertains to the January operating budget for Casey Corporation. ∙ Budgeted sales for January $200,000

and February $107,000. ∙ Collections for sales are 40% in the month of sale and 60% the next month. ∙ Gross margin is 25% of sales. ∙ Administrative costs are $11,000 each month. ∙ Beginning accounts receivable is $26,000. ∙ Beginning inventory is $15,000. ∙ Beginning accounts payable is $68,000. (All from inventory purchases.) ∙ Purchases are paid in full the following month. ∙ Desired ending inventory is 25% of next month's cost of goods sold (COGS).
Engineering
1 answer:
stich3 [128]3 years ago
5 0

Answer:

For January, budgeted cost of goods sold is $150,000

Explanation:

Sales is the consists of Gross income and Cost of goods sold. We can derive the gross income value after deducting cost of goods sold from sales value.

Budgeted Sales for next month = $200,000

Gross Margin = 25%

Gross Margin = Gross Income / Sales

25% = Gross Income / $200,000

Gross Income is the net value of sales and cost of goods sold.

Gross Income = $200,000 x 25%

Gross Income = $50,000

As we know,

Gross Income = Sales - Cost of Goods sold

$50,000 = $200,000 - Cost of Goods sold

Cost of Goods sold = $200,000 - $50,000

Cost of Goods sold = $150,000

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A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to
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3 0
3 years ago
The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher
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Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or  since A = π×D²/4 we have

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or 260 kPa-18.82 m × 9.81 m/s²×ρ=  P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

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