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kodGreya [7K]
3 years ago
8

When budgeted and actual results are not the same amount, there is a budget

Business
1 answer:
torisob [31]3 years ago
7 0

Answer:

a) difference.

Explanation:

As we know the budget represents the difference between the expected and the actual results

So as per the given situation, in the case when the amount of the expected and the actual results are not same or similar so it should be the budget difference

hence, the option a is correct

And, the rest of the options are incorrect

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because they just are buddy

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Fore Farms reported a pretax operating loss of $137 million for financial reporting purposes in 2021. Contributing to the loss w
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answer:

1) attached below

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4 0
3 years ago
A fast-food restaurant has determined that the chance a customer will order a soft drink is 0.90. The proba- bility that a custo
azamat

Answer:

(a) The probability that the order will include a soft drink and no fries is 0.45.

(b) The probability that the order will include a hamburger and fries is 0.48.

Explanation:

Let the events be denoted as follows:

S = an order of soft drink

H = an order of hamburger

F = an order of french fries.

Given:

P (S) = 0.90

P (H) = 0.60

P (F) = 0.50

(a)

It is provided that the event of ordering a soft drink and fries are independent.

If events A and B are independent then the probability of event (A ∩ B) is:

P(A\cap B)=P(A)\times P(B)

Compute the probability that the order will include a soft drink and no fries as follows:

P(S\cap \bar F)=P(S)\times P(\bar F)\\=P(S)\times[1-P(F)]\\=0.90\times (1-0.50)\\=0.45

Thus, the probability that the order will include a soft drink and no fries is 0.45.

(b)

It is provided that the conditional probability that a customer will order fries given that he/she has already ordered a hamburger as, P (F|H) = 0.80.

The conditional probability of an event B given another event A has already occurred is:

P(B|A)=\frac{P(A\cap B}{P(A)}

Compute the probability that the order will include a hamburger and fries as follows:

P(F|H)=\frac{P(H\cap F)}{P(H)}\\P(H\cap F)=P(F|H)\times P(H)\\=0.80\times 0.60\\=0.48

Thus, the probability that the order will include a hamburger and fries is 0.48.

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