Answer:
The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).
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 2809 people who responded to survey, 1634 stated that they currently use social media.
This means that 
98% 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 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).
20/10=2.
14/7=2.
2 dollars per pound.
The first one is D
The second one is true
Answer:
1
Explanation:
Radius = C/2π
C is 6.28 and 2*pi is also 6.28
So 6.28/6.28 = 1.