Mathematics: In a survey of 200 employees, 65% agreed that the management of the cafeteria should be changed. If the confidence interval is 95%, the margin of error for this survey is +/- 6.75% and the maximum number of employees who could have agreed is about? a. 144 b. 140 c. 136

In a survey of 200 employees, 65% agreed that the

Ответ:

amandasantiago2001

19.01.2022

Mathematics

Option A) 144

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 200

$\hat{p} = 65\% = 0.65$

Margin of error =

$\pm 6.75\% = \pm 0.0675$

Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$\hat{p}\pm \text{Margin of error}$

Putting values:

$(0.65 \pm 0.0675)\\=(0.5825,0.7175)$

Thus, maximum proportion is

$0.7175 = 71.75\%$

Maximum number of employees who could have agreed =

$71.75\% \times 200\\\\=\dfrac{71.75}{100}\times 200\\\\= 143.5$

This, approximately maximum 144 employees agreed that the management of the cafeteria should be changed.

Option A) 144

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Ответ:

nicolacorio02

19.01.2022

Mathematics

The answers are: in the first one +/- 6.75% and the second one: 144.

Step-by-step explanation:

I just did it on Edmentum.

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Ответ:

christian2510

10.01.2023

Mathematics

Let missing place be y

Solution has no solution means both sides are equal

-2(2x-y)+1=17-4x

$\\ \sf\longmapsto -4x+2y+1=17-4x$

$\\ \sf\longmapsto 17-4x=2y+1-4x$

On comparing

$\\ \sf\longmapsto 2y+1=17$

$\\ \sf\longmapsto 2y=17-1$

$\\ \sf\longmapsto 2y=16$

$\\ \sf\longmapsto y=\dfrac{16}{2}$

$\\ \sf\longmapsto y=8$

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In a survey of 200 employees, 65% agreed that the management of the cafeteria should be changed. If the confidence interval is 95%, the margin of error for this survey is +/- 6.75% and the maximum number of employees who could have agreed is about?

a. 144 b. 140 c. 136

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