Zarina wants to determine a confidence interval based on a random poll she conducted about the percent of attendees at a conference who are vegetarian. she determined the margin of error for her poll results using the equation below. e = 1.96 /.3(1-.3)/540 based on zarina’s work, what can be concluded? with 90% confidence, between 4% and 30% of the attendees are vegetarian. with 90% confidence, between 26% and 34% of the attendees are vegetarian. with 95% confidence, between 4% and 30% of the attendees are vegetarian. with 95% confidence, between 26% and 34% of the attendees are vegetarian.

With 95% confidence, between 26% and 34% of the attendees are vegetarian.

Step-by-step explanation:

Given:

Zarina used 1.96 for confidence interval for the proportion

We know that a sample proportion will have standard error as

square root of pq/n

So from the information given,E = 1.96 /.3(1-.3)/540

we find that since 1.96 is used, 95% confidence level was done.

Sample size n =540

p = 0.3 and q = 0.7

Std error =

Margin of error = 1.96(std error) = 0.038

=0.040 (after rounding off)

=4%

Based on Zarina’s work, it can be concluded that: 

suitable option would be which as 4 as margin of error and 95% Conf level.

With 95% confidence, between 26% and 34% of the attendees are vegetarian.

is right answer.

ddd

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b