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Sindrei [870]
3 years ago
12

Find all optimal solutions to the following LP using the Simplex Algorithm:

Mathematics
1 answer:
IgorLugansk [536]3 years ago
6 0

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

First Step convert your constraints in standard equations

so we have

x1 + 2x2 + 3x3+x4 = 10

x1 + x2 +x5= 5

x1 +x6= 1

Now we pass it all to the simplex table

Remember that we choose the column with the most negative value

Pivot Element=3

Divide all elements on Pivot Line by Pivot Element

Line x5= 0*Pivot Line +Line x5

Line x6= 0*Pivot Line+ Line X6

Line Z= 3* Pivot Line + Line Z

We finish when all the elements from the line Z are positive

Hence we have that x3=3.33 and x1=0, x2=0 and the max of z is 10

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Degger [83]

Answer:

17 square yards

Step-by-step explanation:

split it so you have a big rectangle and a small one. find the areas of both

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25t means "25 times t" where t is some unknown number. It is a placeholder for a number.

To find what the number is, we undo what is happening to t. So we divide both sides by 25 to undo the operation "multiply by 25"

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25*t = 1125

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Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding no
baherus [9]

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.

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4 years ago
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