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Harrizon [31]
3 years ago
13

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 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 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? (Round your answer to four decimal places.)
c. What is the probability that all 5 selected workers will be from the same shift? (Round your answer to four decimal places.)
d. What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.)
c. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

252 ; 0.0059 ; 0.0074 ; 0.9926

Step-by-step explanation:

Day shift = 10

Swing shift = 8

Graveyard shift = 6

Total number of workers = 24

A.) Number of selections resulting in 5 workers coming from day shift :

10C5 = 10! ÷ (10-5)!5!

= (10*9*8*7*6) / (5*4*3*2*1)

= 252

B.) All 5 workers coming from day shift :

Required outcome = 10C5

Total possible outcomes = 24C5

10C5 ÷ 24C5

252 ÷ 42504

= 0.0059288

= 0.0059

C.) 5 selected workers are from the same shift :

[day shift + swing shift + graveyard shift] / total possible outcomes

[(10C5) + (8C5) + (6C5)] ÷ 24C5

(252 + 56 + 6) / 42504

= 0.0074

D.) What is the probability that at least two different shifts will be represented among the selected workers?

1 - [[(10C5) + (8C5) + (6C5)] ÷ 24C5]

1 - 0.0073875

= 0.9926124

= 0.9926

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
4 years ago
1/3(m-1)=-5<br> Please help
dezoksy [38]

Answer:

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Step-by-step explanation:


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hichkok12 [17]

Answer:

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Step-by-step explanation:

Input Values Form the Domain of the Function

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3 0
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