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Readme [11.4K]
3 years ago
10

Chapin Manufacturing Company operates 24 hours a day, five days a week. The workers rotate shifts each week. Management is inter

ested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift.
Units Produced

Employee Day Afternoon Night

Skaff 38 24 34

Lum 35 23 37

Clark 28 26 38

Treece 35 26 23

Morgan 22 23 24

1. At the 0.01 significance level, can we conclude there is a difference in the mean production rate by shift or by employee?
Mathematics
1 answer:
Nata [24]3 years ago
5 0

Answer:

The answer is there is a difference in the mean production rate by shift

Step-by-step explanation:

The level of significance is actually very low since the same production level is maintained for each of the 5 workers at the same time in the afternoon.

As for the other shifts, that is, the day shift and the night shift, the level of significance varies according to the production established by each worker.h

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A House needs to some maintenance one sink is dripping water every 10 minutes and the other sink is dripping every 12 minutes if
e-lub [12.9K]

Answer:

60 minutes

Step-by-step explanation:

We solve this question using Lowest Common Factor method

Find and list multiples of each number of minutes( 12 and 10 minutes) until the first common multiple is found. This is the lowest common multiple.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80

Multiples of 12:

12, 24, 36, 48, 60, 72, 84

Therefore,

LCM(10, 12) = 60

The number if minutes until they both drip again is 60 minutes

7 0
3 years ago
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Easyish PreCalculus; Find the exact values of the six trigonometric functions of the given angle. (you can use a calculator.) (D
Brut [27]

Answer:

there is no given angle here man

Step-by-step explanation:

5 0
3 years ago
Select the correct measurement of the angle shown on the protractor. a protractor showing an angle with one side lined up with t
Mumz [18]

Answer:

110°

Step-by-step explanation:

Each tick line in a protractor shows 10°. So the next is 110° inclined.

6 0
3 years ago
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Hi please help i will mark brain list giving all my points
DochEvi [55]

Answer:

B

Step-by-step explanation:

It looks like addition to me.

3 0
3 years ago
Read 2 more answers
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