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Softa [21]
3 years ago
8

Dinham Kennel uses tenant-days as its measure of activity; an animal housed in the kennel for one day is counted as one tenant-d

ay. During March, the kennel budgeted for 4,700 tenant-days, but its actual level of activity was 4,780 tenant-days. The kennel has provided the following data concerning the formulas used in its budgeting and its actual results for March: Data used in budgeting: Fixed element per month Variable element per tenant-day Revenue - $ 35.20 Wages and salaries $ 3,600 $ 8.60 Food and supplies 1,100 15.10 Facility expenses 7,500 4.10 Administrative expenses 7,600 0.50 Total expenses $ 19,800 $ 28.30 Actual results for March: Revenue $ 144,230 Wages and salaries $ 28,660 Food and supplies $ 74,070 Facility expenses $ 27,220 Administrative expenses $ 7,106 The spending variance for food and supplies in March would be closest to:
Business
1 answer:
djverab [1.8K]3 years ago
4 0

Answer:

The spending variance for food and supplies in March would be closest to $122 U

Explanation:

In order to calculate the spending variance for food and supplies in March we would have to use the following formula:

Spending variance = Budgeted cost - Actual cost

Budgeted facility expenses for actual tenant days = $7,500 + (4780*$4.10) Budgeted facility expenses for actual tenant days =$27,098

Actual facility expenses = $27,220

Therefore, Spending variance = $27,098 - $27,220 = $122 U

Spending variance = $122 U

The spending variance for food and supplies in March would be closest to $122 U

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Answer:

The answer is below

Explanation:

A What is the probability that all 4 selected workers will be the day shift?

B What is the probability that all 4 selected workers will be the same shift?

C What is the probability that at least two different shifts will be represented among the selected workers.

A)

The total number of workers = 10 + 8 + 6 = 24

The probability that all 4 selected workers will be the day shift is given as:

P_a=\frac{C(10,4)}{C(24,4)}= \frac{210}{10626}=0.0198

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Hence:

P_B=\frac{C(10,4)}{C(24,4)}+\frac{C(8,4)}{C(24,4)}+\frac{C(6,4)}{C(24,4)}=0.0198+0.0066+0.0014=0.0278

C) The probability that at least two different shifts will be represented among the selected workers (P_C)= 1 - the probability that all 4 selected workers will be the same shift(P_B)

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Answer:

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