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frutty [35]
2 years ago
12

Data concerning Sinisi Corporation's single product appear below: Selling price per unit $ 200.00 Variable expense per unit $ 58

.00 Fixed expense per month $ 407,540 The break-even in monthly dollar sales is closest to
Business
1 answer:
Finger [1]2 years ago
7 0

Answer:

Break-even point (dollars)= $574,000

Explanation:

Giving the following information:

Selling price per unit $ 200.00

Variable expense per unit $ 58.00

Fixed expense per month $ 407,540

<u>To calculate the break-even point in dollars, we need to use the following formula:</u>

Break-even point (dollars)= fixed costs/ contribution margin ratio

Break-even point (dollars)= 407,540 / [(200 - 58)/200]

Break-even point (dollars)= $574,000

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You just stuffed yourself with a hot dog, a large tub of popcorn, and a box of milk duds while watching a movie. when you come o
marishachu [46]

Answer:

Incentive Theory

Explanation:

Reason behind would be because how many things you ate your brain and taste are processing that all at the same time making it taste like a completely different substance.

7 0
2 years ago
Big Dig LLC makes an offer to perform an excavation and related tasks for Commercial Development Corporation, but due to a subst
valkas [14]

Answer:C. A mistake of value support the cancellation of a contract.

Explanation:

The law of equity says ' he who comes to equity must come with a clean hand. Although the law requires the enforcement of a valid contract but the precensce of a substantial mathematics mistakes make the contract unenforceable.

It's not a bilateral mistake because it's from one the parties, though not all unilateral mistakes can cancel a contract especially when done with negligence.

The contract been below the price of contract of similar nature is not a valid excuse for non performance after agreement.

7 0
2 years ago
Organic Laboratories allocates research and development costs to its three research facilities based on each facility's total an
spayn [35]

Answer:

$21,000,000

Explanation:

Ratio is used in allocating the research and development cost

This is the expression of relationship between two or more data showing the number of times one data contains or is contained in another data

Total research and development cost = $60,000,000

Revenue

Kentucky = $56,000,000

Arizona -= $ 100,000,000

Illinois =    $84,000,000

Total =     $240,000,000

Illinois allocation of research and development cost=

84,000,000/240,000,000*60,000,000 =$21,000,000

3 0
2 years ago
A marketing consultant, Sofia, has been studying the effect of increasing advertising spending on product sales. Sofia conducts
UkoKoshka [18]

No, because 100,000 is much greater than the values used in the experiment

Explanation:

The advertisement budget is an estimation of the company's commercial spending for a specified amount of time. More specifically, it is the capital that a organisation is able to put aside to accomplish its marketing goals.

In developing an advertisement budget, a corporation must balance the importance of the promotional dollar against the value of the dollar as known revenue.

Better promotional budgets — and campaigns — focus on consumers' desires and address their challenges, not on business concerns such as overstock elimination.

5 0
2 years ago
Diogo has a utility function,U(q1, q2) = q1 0.8 q2 0.2,where q1 is chocolate candy and q2 is slices of pie. If the price of slic
guapka [62]

Answer:

(0.5 \times 8q_2)+q_2=100\\\\5q_2=100\\\\q_2=20

since q_2 = 20

q_1 = 8*20\\\\q_1=160

Explanation:

U(q₁ q₂)

q_1^{0.8}q_2^{0.2}\\\\P_1= \$0.5 \ P_2=\$1 \ Y=100

Budget law can be given by

P_1q_1+P_2q_2=Y\\\\0.5q_1+q_2=100

Lagrangian function can be given by

L=q_1^{0.8}q_2^{0.2}+ \lambda (100-0.5q_1-q_2)

First order condition csn be given by

\frac{dL}{dq} =0.8q_1^{-0.2}q_2^{0.2}-0.5 \lambda=0\\\\0.5 \lambda=0.8q_1^{-0.2}q_2^{0.2}---(i)

\frac{dL}{dq} =0.2q_1^{0.8}q_2^{-0.8}- \lambda=0\\\\ \lambda=0.2q_1^{0.8}q_2^{-0.8}---(ii)

\frac{dL}{d \lambda} =100-0.5q_1-q_2=0\\\\0.5q_1+q_2=100---(iii)

From eqn (i) and eqn (ii) we have

\frac{0.5 \lambda}{\lambda} =\frac{0.8q_1^{-0.2}q_2^{0.2}}{0.2q_1^{0.8}q_2^{-0.8}} \\\\0.5=\frac{4q_2}{q_1}\\\\q_1=8q_2}

Putting q_1=8q_2 in euqtion (iii) we have

(0.5 \times 8q_2)+q_2=100\\\\5q_2=100\\\\q_2=20

since q_2 = 20

q_1 = 8*20\\\\q_1=160

3 0
2 years ago
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