Answer:
<h2>Not sure!! I don't know this asnwer is wrong or right.</h2>
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.
<span>Standard Form of an Equation of an Hyperbola opening up and down is:
(y-k)^2 / b^2 - (x-h)^2 / a^2 = 1
</span><span>where Pt(h,k) is a center with vertices 'b' units up and down from center.
</span><span>vertices: vertices: (0,+-2) b = 2 AND Center is (0,-0)
(y)^2/4-(x)^2/a^2=1
</span><span>foci: (0,+-7)
a^2 = 24
square root of 4 + a^2 = 5.29
4 + 5.29 = 9.29
y^2 / 4 - x^2 / 24 = 1</span>
