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bagirrra123 [75]
2 years ago
5

At the beginning of the period, the Cutting Department budgeted direct labor of $46,300 and supervisor salaries of $37,200 for 4

,630 hours of production. The department actually completed 5,000 hours of production.
Mathematics
1 answer:
Korvikt [17]2 years ago
4 0

Answer:

$87200

Step-by-step explanation:

Here is the complete question:

At the beginning of the period, the Cutting Department budgeted direct labor of $46,300 and supervisor salaries of $37,200 for 4,630 hours of production. The department actually completed 5,000 hours of production.

Determine the budget of the department assuming that it uses flexible budgeting?

Given: Budget for direct labour= $46300

           Supervisor salaries= $37200

           Expected production hours= 4630 hours

           Completed production hours= 5000 hours

Now, we know that company budget include both fixed and variable cost.

∴ Direct labour cost is a variable cost and Supervisor salaries are fixed cost.

Using flexible budgeting for determining the budget of department, we will pro rate the direct labour cost on the basis of production hours.

Direct labour= Budget\times \frac{completed\ production\ hours}{expected\ production\ hours}

Direct labour= 46300\times \frac{5000}{4630}

∴ Direct labour= $50000

we know the department budget = Fixed cost+variable cost

∴ Department budget= \$ 37200+\$ 50000 = \$ 87200

∴ The department budget is $87200.

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Which equation represents the line with a slope of 5/6 that passes through the point (-2, 1)?
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(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

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                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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