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Aloiza [94]
3 years ago
11

Ashton surveyed some of the employees at his company about their cell phone habits. From the data, he concluded that most employ

ees at his company use cell phones primarily for business. For which sample could this generalization be valid?
A. all managers in the company
B. some managers and some sales executives selected at random
C. every fifth employee on the employee list
D. every other sales executive on the employee list
Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0
It can't be A. since if you only look at managers, you are missing all the sales executives.

It may be C. this option is more random but doesn't guarantee that you will represent both groups of employee's. Also, each time you would conduct the survey, you will receive the exact same results since it is the same people. 

It isn't D. for the exact same reason as A. but you're missing managers now. 

Therefore the answer is B. Some managers and some sales executives selected at random. This way you get a sample from both categories, and within those groups, it is randomly selected.

 I hope this helps!

 
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90.125 in expanded form
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Answer:

90 + 0.1 + 0.02 + 0.005

Step-by-step explanation:

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Step-by-step explanation:

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Help me pls !!!<br><br> answer all pls
polet [3.4K]
1. We assume, that the number 128 is 100% - because it's the output value of the task. 
<span>2. We assume, that x is the value we are looking for. </span>
<span>3. If 128 is 100%, so we can write it down as 128=100%. </span>
<span>4. We know, that x is 51% of the output value, so we can write it down as x=51%. </span>
5. Now we have two simple equations:
1) 128=100%
2) x=51%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
128/x=100%/51%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 51% of 128

128/x=100/51
<span>(128/x)*x=(100/51)*x       - </span>we multiply both sides of the equation by x
<span>128=1.96078431373*x       - </span>we divide both sides of the equation by (1.96078431373) to get x
<span>128/1.96078431373=x </span>
<span>65.28=x </span>
x=65.28

<span>now we have: </span>
<span>51% of 128=65.28</span>
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At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play
Mademuasel [1]

The probability that a student selected at random majors in engineering is 30% which is 0.3.

The probability  that the student both majors in engineering and play club sports is 10% which is 0.1.

For a student who is selected at random to be one who majors in engineering, there are two possible ways.

The student majors in engineering OR the Student both majors in engineering and plays club

The Probabilty that the student majors in Enginnering =

The probability that the student majors in engineering plus the  probability that Student both majors in engineering and plays club sport

= 0.3+0.1= 0.4

3 0
4 years ago
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