A survey of 1000 Californians finds reports that 48% are excited by the opportunity to take a statistics class. The margin of error in this report is 5%. Which answer choice best describes a reasonable range of values for the percent of
students who are excited to take a statistics class?

(0.449, 0.511) is the range for the percent of students who are excited to take a statistics class

Step-by-step explanation:

First of all, let's check that the sample size is large enough to use normal approximation.

Conditions are;

np > 10

n(1-p) > 10

Now, we have;

n = 1000

p = 48% = 0.48

Thus;

np = 1000 × 0.48 = 480 > 10

Also;

n(1 - p) = 1000 × (1 - 0.48) = 520 > 10. This means the conditions for normal approximation have been fulfilled. Hence we can use the confidence interval formula to get the range of values which is;

CI = p ± z√(pq/n)

We are given a margin of error and thus the confidence interval is at 95%.

Critical value of z at 95% CI is 1.96

Thus;

CI = 0.48 ± 1.96√(0.48 × 0.52/1000)

CI = 0.48 ± 0.031

CI = (0.449, 0.511) is the range for the percent of students who are excited to take a statistics class

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2