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Ivahew [28]
3 years ago
12

Peak Industries saw a 20% increase in activity levels over the fourth quarter. Which of the following changes should the firm ex

pect to see in relation to this increase?
A : a 20% increase in total costs
B : a 20% increase in variable costs
C : a 20% increase in fixed costs
D : a 20% decrease in variable costs per unit
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

B : a 20% increase in variable costs

Step-by-step explanation:

Here the industry is seeing a 20% increase in activity levels over the fourth quarter. Since, there has been an increase in activity levels the fixed cost would remain the same but the variable cost being variable would increase with increase in production activity in same proportion as the increase in production.

B : a 20% increase in variable costs

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Leya [2.2K]

Answer:

the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

Step-by-step explanation:

Let assume that n should represent the number of the students

SO, \bar x can now be the sample mean of number of students  in GPA's

To obtain n such that P( \bar x \leq 2.3 ) \leq .04

⇒ P( \bar x \geq 2.3 ) \geq .96

However ;

E(x) = \int\limits^4_2 Dx (2+e^{-x} ) 4x = D  \\ \\ = D(e^{-x} (e^xx^2 - x-1 ) ) ^D_2 = 12.314 D

E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D

Similarly;

D\int\limits^4_2(2+ e^{-x}) dx = 1

⇒ D*(2x-e^{-x} ) |^4_2 = 1

⇒ D*4.117 = 1

⇒ D= \dfrac{1}{4.117}

\mu = E(x) = 2.991013 ; \\ \\ E(x^2) = 9.28103

∴  Var (x)  = E(x^2) - E^2(x) \\ \\  = .3348711

Now; P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega  \geq .691013)  \ \ \ \  \ \ \ \ \ \ (x = E(\bar x ) - \mu)

Using Chebysher one sided inequality ; we have:

P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}

So; (\omega = \bar x - \mu)

⇒ E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}

∴ P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}

To determine n; such that ;

\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}

⇒ n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}

n \geq 16.83125

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

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Answer:

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Step-by-step explanation:

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Choice C, it has the same slope as the given equation without being identical to it like choice a.
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