Answer: 0.1854
Step-by-step explanation:
Given : Suppose that a particular candidate for public office is in fact favored by 48% of all registered voters in the district.
Let 
be the sample proportion of voters in the district favored a particular candidate for public office .
A polling organization will take a random sample of n=500 voters .
Then, the probability that p will be greater than 0.5, causing the polling organization to incorrectly predict the result of the upcoming election :
![P(\hat{p}>0.5)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.5-0.48}{\sqrt{\dfrac{0.48(0.52)}{500}}})\\\\=P(z>0.8951)\ \ [\because\ z=(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]\\\\=1-P(z\leq0.8951)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\ = 1-0.8146=0.1854](https://tex.z-dn.net/?f=P%28%5Chat%7Bp%7D%3E0.5%29%3DP%28%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3E%5Cdfrac%7B0.5-0.48%7D%7B%5Csqrt%7B%5Cdfrac%7B0.48%280.52%29%7D%7B500%7D%7D%7D%29%5C%5C%5C%5C%3DP%28z%3E0.8951%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%28%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3D1-P%28z%5Cleq0.8951%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%3Ez%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%20%3D%201-0.8146%3D0.1854)
∴ Required probability = 0.1854
Answer:
No it is not correct.
Step-by-step explanation:
9.009-0.01=8.999
Answer:
a) Prepare a frequency table for the data
Step-by-step explanation:
I'm assuming that this is the complete question.
If f(x) = 3 – 2x and g(x)=1/(x+5), what is the value of (f/g)(8)? a) –169 b) –1 c) 13 d) 104
x = 8
f(x) = 3 -2xf(8) = 3 - 2(8) = 3 - 16 = -13
g(x) = 1/(x+5)g(8) = 1/(8+5) = 1/13
(f/g)(8)f(8)/g(8) = -13/ (1/13) = -13 * 13 = -169 Choice A :)
The answer to this question could be 44
