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Ede4ka [16]
3 years ago
15

Which equation could have been used to create this function table?

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0
A) y=3x
You can plug in the given values (inputs) of x to see if it creates the number in y (output). 
3*1=3 
3*2=6
3*3=9
3*4=12
3*5=15
3*6=18

All these values are true, therefore the equation y=3x is the correct answer.

:)
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The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management
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Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1=  13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the  number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting  X4= 0 in

X1 + X4 ≥ 12

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X1 ≥ 12

Now Putting the value X1 ≥ 12  in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

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this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

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X2  ≥ 2

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A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is
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Answer:

A. P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336

Step-by-step explanation:

Working with an ordinary Normal Distribution of probability and trying to find the probabilities asked in it could be difficult, because there´s no easy method to find probabilities in a generic Normal Distribution (with mean μ=4 and STD σ=0.1). The recommended approach to this question is to use a process called "Normalize", this process let us translate the problem of any Normal Distribution to a Standard Normal Distribution (μ=0 and σ=1) where there´s easier ways to find probabilities in there. The "Normalization" goes as follows:

Suppose you want to know P(x<a) of the Normal Distribution you are working with:

P(x<a)=P( (x-μ)/σ < (a-μ)/σ )=P(z<b)   ( b=(a-μ)/σ )

Where μ is the mean and σ is the STD of your Normal Distribution. Notice P(z<b) now it´s a probability in a Standard Normal Distribution, now we can find it using the available method to do so. My favorite is a chart (It´s attached to this answer) that contains a lot of probabilities in a Standard Normal Distribution. Let´s solve this as an example

A. We want to find the probability of the cork being defective (P(x<3.85) + P(x>4.15)). Now we find those separated and, then, add them for our answer.

Let´s begin with P(x<3.85), we start by normalizing that probability:

P(x<3.85)= P( (x-μ)/σ < (3.85-4)/0.1 )= P(z<-1.5)

And now it´s time to use the chart, it works like this: If you want P(z<c) and the decimal expansion of c=a.bd... , then:

P(z<c)=(a.b , d)

Where (a.b , d) are the coordinates of the probability in the chart. Keep in mind that will only work with "<" (It won´t work directly with P(z>c)) and we will do some extra work in those cases.

P(z<-1.5) is in the coordinates (-1.5 , 0)

P(z<-1.5)= 0.0668

P(x<3.85)= 0.0668

Now we are looking for P(x>4.15), let´s Normalize it too:

P(x>4.15)=P( (x-μ)/σ < (4.15-4)/0.1 )=P(z>1.5)

But remember the chart only work with "<", so we need to use a property of probability:

P(z>1.5)= 1 - P(z<1.5)

Using the chart:

P(z<1.5)=0.9332                             (1.5 , 0)

P(z>1.5)= 1 - 0.9332

P(z>1.5)= 0.0668

P(x>4.15)= 0.0668

And our final answer will be:

P(x<3.85 or x>4.15)= P(x<3.85)+P(x>4.15) = 0.1336

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