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svp [43]
3 years ago
12

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A

quality control consultant is to select 4 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 4 workers has the same chance of being selected as does any other group (drawing 4 slips without replacement from among 24).
Required:
a. How many selections result in all 5 workers coming from the day shift?
b. What is the probability that all 5 selected workers will be from the day shift?
c. What is the probability that all 5 selected workers will be from the same shift?
d. What is the probability that at least two different shifts will be represented among the selected workers?
e. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?
Mathematics
1 answer:
Bezzdna [24]3 years ago
5 0
Yeah i think it’s A hope
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Step-by-step explanation:

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8 0
3 years ago
Simplify the expression shown. You must show your work to receive credit.
ZanzabumX [31]

ANSWER

12 \sqrt{5}

EXPLANATION

The given expression is

\sqrt{2}  \times  \sqrt{72}  \times  \sqrt{5}

The middle radical contains a perfect square.

\sqrt{2}  \times  \sqrt{36 \times 2}  \times  \sqrt{5}

\sqrt{2}  \times  \sqrt{36}   \times  \sqrt{2} \times  \sqrt{5}

6\sqrt{2}  \times  \sqrt{2} \times  \sqrt{5}

Note that:

\sqrt{a}  \times  \sqrt{a}  = a

6\times 2 \sqrt{5}  = 12 \sqrt{5}

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2 years ago
What is one half added to three quarters
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3 years ago
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2 years ago
Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.
valentina_108 [34]

Answer:

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with f^{-1}(x)

Given that:

I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = a^{-1}(x) = a(x)  Hence, I is true.

II. y =\dfrac{1}{x}

Replacing y with and x with y:

x =\dfrac{1}{y}

x=\dfrac{1}{a^{-1}(x)}

\Rightarrow a^{-1}(x) = \dfrac{1}{x}

a^{-1}(x) = a(x)  Hence, II is true.

III. y =x^{2}

Replacing y with and x with y:

x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)

 Hence, III is not true.

IV. y =x^{3}

Replacing y with and x with y:

x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)

Hence, IV is not true.

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

4 0
3 years ago
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