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user100 [1]
3 years ago
15

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.

A quality control consultant is to select 7 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 7 workers has the same chance of being selected as does any other group (drawing 7 slips without replacement from among 45).
1. How many selections result in all 7 workers coming from the day shift?
2. What is the probability that all 7 selected workers will be from the day shift?
3. What is the probability that all 7 selected workers will be from the same shift?
4. What is the probability that at least two different shifts will be represented among the selected workers?
5. What is the probability that at least one of the shifts will be un-represented in that sample of workers?
Mathematics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

1. 77520

2. P_1 = 0.0017

3. P_2 = 0.0019

4. P_3 = 0.9981

5. P_4 = 0.2036

Step-by-step explanation:

The number of ways or combinations in which we can select x elements from a group of n can be calculated as:

nCx = \frac{n!}{x!(n-x)!}

So, there are 77520 selections that result in all 7 workers coming from the day shift. It is calculated as:

20C7 = \frac{20!}{7!(20-7)!}=77520

At the same way, the total number of selections of 7 workers from the 45 is 45C7, so the probability that all 7 selected workers will be from the day shift is:

P_1=\frac{20C7}{45C7} =0.0017

The probability that all 7 selected workers will be from the same shift is calculated as:

P_2=\frac{20C7+15C7+10C7}{45C7} =0.0019

Because the consultant can select all workers from the day shift (20C7) or can select all workers from the swing shift (15C7) or can select all workers from the graveyard shift (10C7).

On the other hand, the probability that at least two different shifts will be represented among the selected workers is the complement of the probability that all 7 selected workers will be from the same shift. So it is calculated as:

P_3 = 1- P_2=1 - 0.0019 = 0.9981

Finally, the probability that at least one of the shifts will be un-represented in that sample of workers is:

P_4=\frac{25C7+30C7+35C7}{45C7} =0.2036

Where 25C7 is the number of ways to select all 7 workers from swing or graveyard shift, 30C7 is the number of ways to select all 7 workers from day or graveyard shift and 35C7 is the number of ways to selects all 7 workers from day shift and swing shift.

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Step-by-step explanation:

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( 3x + 2) + (4x - 3) + ( x + 7) = 86 what is x
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Step-by-step explanation:

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2 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

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And finally, we can eliminate the second 1 by adding -(R₃):

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Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
2 years ago
Two functions are shown in the table below.
SVETLANKA909090 [29]

Answer:

(d)\ x = 4

Step-by-step explanation:

Given

f(x) = -x^2 + 4x + 12

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Required

Find x such that: f(x) = g(x)

This gives:

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Factorize

x(x - 4) - 1(x-4) = 0

Factor out x - 1

(x- 1)(x - 4)=0

Solve:

x =1\ or\ x = 4

8 0
3 years ago
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