A I think I am not 100% right
Answer:
95% confidence interval for the proportion of students supporting the fee increase is [0.767, 0.815]. Option C
Step-by-step explanation:
The confidence interval for a proportion is given as [p +/- margin of error (E)]
p is sample proportion = 870/1,100 = 0.791
n is sample size = 1,100
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value (z) at 5% significance level is 1.96.
E = z × sqrt[p(1-p) ÷ n] = 1.96 × sqrt[0.791(1-0.791) ÷ 1,100] = 1.96 × 0.0123 = 0.024
Lower limit of proportion = p - E = 0.791 - 0.024 = 0.767
Upper limit of proportion = p + E = 0.791 + 0.024 = 0.815
95% confidence interval for the proportion of students supporting the fee increase is between a lower limit of 0.767 and an upper limit of 0.815.
Estimation:
23 --> rounded to 20
71 --> rounded to 70
70 + 20 = 90. I rounded both numbers down so the exact sum must be only a little greater than my estimate.
You equal each to zero, so -6,0,3,7
Not enough info to answer the question it should have 2 trapezoids to compare if your talking about transformations
