Could someone with the steps! in a survey conducted in an office, 60% of the sample respondents agreed that employees should have flexible working hours. if the margin of error is ±2%, the expected population proportion that wants flexible working hours is between ? % and ? %

The expected population proportion that wants flexible working hours is between 58% and 62%.

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 58% and 62%.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

answer:   14

step-by-step explanation:          

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