A survey of an urban university (population of 25,450) showed that 870 of 1,100 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase? A. [0.714, 0.866] B. [0.759, 0.822] C. [0.767, 0.815] D. [0.771, 0.811]

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.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

3; 9 is 8 to -4

step-by-step explanation: