In March of 2018, a survey asked 801 US adults whether they had at least one subscription to a video-streaming service. Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service. What is the correct 95 percent confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service?

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service.

This means that

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).

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step-by-step explanation:

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2