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DanielleElmas [232]
4 years ago
5

could someone help with the steps! In a survey conducted in an office, 60% of the sample respondents agreed that employees shoul

d have flexible working hours. If the margin of error is ±2%, the expected population proportion that wants flexible working hours is between ? % and ? %
Mathematics
2 answers:
nignag [31]4 years ago
7 0

Answer: The expected population proportion that wants flexible working hours is between <u> 58%</u> and <u>62%.</u>

Step-by-step explanation:

The margin of error (E) is used to represent the level of accuracy of any random sample.

Confidence interval for population percentage (p):

 (p-E, p+E ) , where E is the margin of error ( in percentage.)

Given : The percentage of the sample respondents agreed that employees should have flexible working hours : p = 60%

The margin of error : E=  ± 2%

Then , the expected population proportion that wants flexible working hours is between : 60%-2% and 60% +2%

= 58% and 62%.

Hence, the expected population proportion that wants flexible working hours is between <u> 58%</u> and <u>62%</u>.

Ira Lisetskai [31]4 years ago
5 0
58%-62%
just add 2 to 60 
and subtract 2 from 60 to find the range of percents
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The inequality that models the given situation will be y*$22 + x*$38 ≤ $220 And some solutions are (1, 1), (1, 2), (2, 2).

<h3>What is inequality?</h3>

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