First we make the fractions homonym ( they must have the same denomitator)
0.6 = 6 / 10 7/9
6/10 = 6 * 9 / 10 * 9 = 54 / 90 7/9 = 7 * 10 / 9 * 10 = 70 / 90
The numbers which are between these numbers are ;
55 / 90 , 56 / 90 , 57 / 90 , 58 / 90 , 59 / 90, 60 / 90 , 61 / 90 , 62 / 90 , 63 / 90 , 64 / 90, 67 / 90 , 68 / 90 , 69 / 90
62 / 90 = 62 / 2 / 90 /2 = 31 / 45
So the right answer is B.
Hope that helps :)
Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:

Compute the probability that a person does not have the disease as follows:

The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:


Thus, the probability of the test indicating that the person does not have the disease is 0.1452.
I CALCULATED IT AND GOT
C) 4.29
1. set up the equation which makes these shapes equal, 1/2(2(x+1)) = 1x
2. solve, x+1=x, 1=0
3. since this is impossible, there is no solution and zero answers