A sample of employees of Worldwide Enterprises is to be surveyed about a new health
care plan. The employees are classified as follows:
Classification Event Number of Employees
Supervisors A 120
Maintenance B 50
Production C 1,460
Management D 302
Secretarial E 68
(a) What is the probability that the first person selected is:
(i) either in maintenance or a secretary?
(ii) not in management?

(c) Are the events in part (a)(i) complementary or mutually exclusive or both

Answers

Answer 1

Using the probability concept, it is found that:

a) i) There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.

ii) There is a 0.849 = 84.9% probability that the first person selected is not in management.

b) The events are neither complementary nor mutually exclusive.

A probability is the number of desired outcomes divided by the number of total outcomes.

Item a:

In this problem, there is a total of 120 + 50 + 1460 + 302 + 68 = 2000 people.

Of those, 50 + 68 = 118 are either in maintenance or a secretary, hence:

[tex]p = \frac{118}{2000} = 0.059[/tex]

There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.

Item ii:

2000 - 302 = 1698 people are not in management, hence:

[tex]p = \frac{1698}{2000} = 0.849[/tex]

There is a 0.849 = 84.9% probability that the first person selected is not in management.

Item b:

If a person is either in maintenance or a secretary, they are also not in management, hence the events are neither complementary nor mutually exclusive.

To learn more about the probability concept, you can take a look at https://brainly.com/question/24437717

Answer 2

Following are the solution to the given points:

For point a:

i)

[tex]\to P(B \cup E) = P(B) + P(E) - P( B\cap E)[/tex]

Where [tex]P(B \cap E) = 0[/tex] because they are both separate occasions It is impossible to be a member of both departments.  

[tex]\to P(B\cup E) =\frac{50}{120 + 50 + 1460 + 302 + 68} +\frac{68}{120 + 50 + 1460+ 302 +68}\\\\[/tex]

                    [tex]=\frac{50}{2000} +\frac{68}{2000}\\\\=\frac{50+68}{2000} \\\\=\frac{118}{2000} \\\\=\frac{ 59}{1000} \\\\=0.059\\[/tex]

ii)

[tex]\to P(D^c) = 1 - P(D)[/tex]

               [tex]=1- (\frac{302}{120+50+1460+302+68})\\\\=1- (\frac{302}{2000})\\\\=1- (\frac{ 151}{1000})\\\\= 1- 0.151 \\\\= 0.849\\\\[/tex]

For point b:

Please find the attached file.

For point c:

Events in a. I are mutually incompatible since [tex]P(B \cap E) = 0[/tex]. One would be a part of the both departments.

Learn more:

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A Sample Of Employees Of Worldwide Enterprises Is To Be Surveyed About A New Healthcare Plan. The Employees
A Sample Of Employees Of Worldwide Enterprises Is To Be Surveyed About A New Healthcare Plan. The Employees

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Answers

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Step-by-step explanation:

\text{A.S.A.}\rightarrow \text{Law of Sines}

A.S.A.→Law of Sines

\frac{a}{\sin A}=\frac{b}{\sin B}

sinA

a

=

sinB

b

From reference sheet.

\text{Find other angle:}

Find other angle:

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u = ?

71°

36°

73°

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sin71

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Let's break this down...

9.03+10.71 is 19.74 and subtracted by 6.8 it will be 12.94 not 13.94 so it is not correct

So the answer is No

Hope this helps!

Answer:

No, it is 12.94

Step-by-step explanation:

9.03 + 10.71 = 19.74, 19.74 - 6.8 = 12.94

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Answer:

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Step-by-step explanation:

correct on edg

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