A simple random sample of size n=250 individuals who are currently employed is
asked if they work at home at least once per week. Of the 250 employed
individuals surveyed, 26 responded that they did work at home at least once per
week. Construct a 99% confidence interval for the population proportion of
employed individuals who work at home at least once per week.
The lower bound is

(Round to three decimal places as needed.)

The 99% confidence interval would be given by (0.054;0.154) . So we are confident at 99% that the true proportion of people that they did work at home at least once per  week is between 0.054 and 0.154

Step-by-step explanation:

For this case we can estimate the population proportion of people that they did work at home at least once per  week with this formula:

We need to find the critical value using the normal standard distribution the z distribution. Since our condifence interval is at 99%, our significance level would be given by and . And the critical value would be given by:

The confidence interval for the mean is given by the following formula:  

If we replace the values obtained we got:

The 99% confidence interval would be given by (0.054;0.154) . So we are confident at 99% that the true proportion of people that they did work at home at least once per  week is between 0.054 and 0.154

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation:

answer: 12x^2+6x

step-by-step explanation:

6x does not equal 24x. hope it works!