Answer:
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
Step-by-step explanation:
<u>Explanation:</u>
Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.
Given sample size 'n' = 1112
Sample proportion 'p' =
q = 1 - p = 1- 0.7940 = 0.206
<u>The 95% level of confidence intervals</u>
The confidence interval for the proportion of students supporting the fee increase
The Z-score at 95% level of significance =1.96
(0.7940-0.02376 , 0.7940+0.02376)
( 0.77024, 0.81776)
<u>Conclusion:</u>-
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
Answer:
288 in²
Step-by-step explanation:
The formula used to solve this question is :
Lateral Surface Area = s( P1 + P2)
Where s = slant height = 2 inches
P1 and P2 = Perimeter of the bases
Perimeter of base 1 = 4 × length of the end of the small square
4 × 17 inches = 68 inches
Perimeter of base 2 = 4 × length of the end of the large square
4 × 19 inches = 76 inches
Lateral Surface Area = 2 × (68 + 76)
= 2 × 144
= 288 in²
Answer:
2 hurry!
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Im just about to work one out for you and you just do the same on the rest