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riadik2000 [5.3K]
3 years ago
14

Edwin Edwards and Karen Davis owned EEE, Inc., which owned three convenience stores, all of which sold gasoline. Reid Ellis deli

vered to the three convenience stores $26,675.02 worth of gasoline for which he was not paid. Ellis proved that Edwards and Davis owned the business, ran it, and in fact personally ordered the gasoline. He claimed that they were personally liable for the debt owed him by EEE, Inc. Decide. [Ellis v.
Edwards, 348 S.E.2d 764 (Ga. App.)]
Mathematics
1 answer:
Ulleksa [173]3 years ago
5 0

Question Completion:

Note: This is a Law question, not Mathematics.

Answer:

Decision:

Edwards and Davis are not personally liable for the debts of EEE, Inc.

Step-by-step explanation:

That Edwin Edwards and Karen Davis owned EEE Inc. does not force them to be personally liable for the debts of EEE Inc.  EEE Inc. is a separate legal entity.  There is the legal separation of ownership and the liabilities of Edwards and Davis are limited to the capital they contributed or undertook to contribute to the corporation.  The corporate veil is only lifted when it is ascertained that EEE Inc. was not run as a legal entity separate from the owners.  We can also conclude that there is no evidence of abuse of the corporate form of EEE Inc. with comingling of assets or for the purpose of promoting fraud or injustice or evasion of tort or contractual responsibility on the part of Edwards and Davis.  Therefore, Edwin Edwards and Karen Davis are not personally liable for the debts of EEE Inc. as claimed by Reid Ellis.

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Consider the numberless roulette game at a casino. On a spin of the wheel, the ball lands in a space with color red (r), green (
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Answer:

a. Probability of Event A = 2.724037476607E−24

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Step-by-step explanation:

Given

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This is calculated as follows;

Let P(R) = Probability of Red

P(R) = 19/40

Let P(G) = Probability of Green

P(G) = 19/40

Let P(B) = Probability of Black

P(B) = 2/40

Total number of arrangement = 40!/(19!19!2!) = 27,569,305,764,000

Probability of Event A = 27,569,305,764,000 * (19/40)^19 * (19/40)^19 * (2/40)^19

Probability of Event A = 2.724037476607E−24

b. In 40 spins of the wheel, find the probability of the event G19 = {19 greens}.

Let P(G) = Probability of Green

P(G) = 19/40

Let P(Other) = Probability of any colour other than green = (2+19)/40

P(Other) = 21/40

Total = 40C19

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c. Given that you randomly choose to bet red and green only, what is the probability p that you bet a winner?

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P(R) = 19/40

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P(G) = 19/40

Let P(Winning) = Probability of Winning

P(Winning) = ½ * P(G) + ½ * P(R)

P(Winning) = ½ * 19/40 + ½ * 19/40

P(Winning) = ½(19/40 + 19/40)

P(Winning) = ½(38/40)

P(Winning) = 19/40

P(Winning) = 0.475

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