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Lina20 [59]
2 years ago
10

Gordon​ Miller's job shop has four work​ areas, A,​ B, C, and D. Distances in feet between centers of the work areas​ are: A B C

D A − 5 9 7 B − − 6 8 C − − − 11 D − − − − Workpieces moved per week between work areas​ are: A B C D A − 900 900 500 B − − 500 200 C − − − 600 D − − − − It costs Gordon ​$22 to move 1 work piece 1 foot.What is the weekly total material handling cost of the layout?

Mathematics
1 answer:
andreev551 [17]2 years ago
4 0

Answer: $600,600

Step-by-step explanation:

Total handling cost :

Workpiece moved * cost * distance

Work area A :

-, (5 × 22 × 900), (9 × 22 × 900), (7 × 22 × 500)

-, 99000, 178200, 77000

Work area B:

-, -, (6 × 22 × 500), (8 × 22 × 200)

-, -, 66000, 35200

Work area C:

-, -, -, (11 × 22 × 600)

-,-,-, 145200

Work area D:

-, -, -, -

Total weekly handling cost :

(99000 + 178200 + 77000 + 66000 + 35200 + 145200)

= $600,600

Kindly check attached picture for more explanation

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The question is incomplete. The complete question is :

Davison Electronics manufactures two LCD television monitors, identified as model A and model B. Each model has its lowest possible production cost when produced on Davison’s new production line. However, the new production line does not have the capacity to handle the total production of both models. As a result, at least some of the production must be routed to a higher-cost, old production line. The following table shows the minimum production requirements for next month, the production line capacities in units per month, and the production cost per unit for each production line:

                      Production per unit                         Minimum production

Model        New Line        Old Line                              requirements

A                   $30                $50                                          50,000

B                   $25                $40                                           70,000

Production  80000          60000

line capacity.

Let

AN- Units of model A produced on the new production line

AO Units of model A produced on the old production line

BN Units of model B produced on the new production line

BO Units of model B produced on the old production line

Davison's objective is to determine the minimum cost production plan. The computer solution is shown in Figure 3.21.

a. Formulate the linear programming model for this problem using the following four constraints:

Constraint 1: Minimum production for model A

Constraint 2: Minimum production for model B

Constraint 3: Capacity of the new production line

Constraint 4: Capacity of the old production line  

Solution :

<u>Linear programming model</u>:

Linear programming is defined as a mathematical model where the linear function is either minimize or maximize when they are subjected to some constraints.

The linear programming model is determined as follows :

Minimum : 30 AN + 50 AO + 25 BN + 40 BO

This is subject to the constraints as :

AN+AO \geq 60,000

BN+BO \geq 70,000

AN+BN \leq 80,000

AO+BO \leq 60,000

Learn more :

https://brainly.in/question/15044395

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Answer:

The following points are not arranged in a parallelogram or rectangle order.

Step-by-step explanation:

Well first we need to graph the following.

A(1,1) B(2,2) C(3,3) D(4,4)

By looking at the image below we can tell it is not any shape, it’s not a parallelogram or a rectangle.

It is a line with a slope of 1 or x.

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Suppose a horticulturist measures the aboveground height growth rate of four different ornamental shrub species grown in a green
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The decision is to <u>reject</u> the <u> null hypothesis</u> at a significant level of <u>significance </u> \alpha  = 0.05 There is <u>sufficient </u> evidence to conclude that <u>at least one of the population mean</u>  is <u>different from</u>  <u>at least of the population</u>  

Step-by-step explanation:

From the question we are told that the claim is

     The mean growth rates of all four species are equal.

The  null hypothesis is  

             H_o  :  \mu _1 =  \mu_2 = \mu_3  =  \mu_4

Th alternative hypothesis is    

             H_a: at \ least \ one \ of \  the \  means \ is \not\ equal

From question the p-value is p-value  =  0.015

  And since the p-value <  \alpha so the null hypothesis will be rejected

So  

   The decision is to <u>reject</u> the <u> null hypothesis</u> at a significant level of <u>significance </u> \alpha  = 0.05 There is <u>sufficient </u> evidence to conclude that <u>at least one of the population mean</u>  is <u>different from</u>  <u>at least of the population</u>  

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2 years ago
The number of pits in a corroded steel coupon follows a Poisson distribution with a mean of 6 pits per cm2. Let X represent the
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Answer:

P(X=2)=0.0446

Step-by-step explanation:

If X follows a Poisson distribution, the probability p to have x pits in 1 cm2 is calculated as:

p(x)=\frac{e^{-m}*m^x}{x!}

Where m is the mean of pits per cm2, so, replacing m by 6 pits per cm2, we get that the probability is equal to:

p(x)=\frac{e^{-6}*6^x}{x!}

Now, the probability P(x=2) that there are 2 pits in a 1 cm2 is calculated as:

p(x=2)=\frac{e^{-6}*6^2}{2!}\\p(x=2)=0.0446

4 0
2 years ago
The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to ap
MAXImum [283]

Answer:

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.

150 calls. So:

\mu = 150*0.02 = 3

Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Either there are less than two calls from wrong numbers, or there are at least two calls from wrong numbers. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

We want to find P(X \geq 2). So

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992

Then

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.2008

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

6 0
2 years ago
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