Answer:
Step-by-step explanation:
94-89=5
60-45=15
5/3=1
15/5=3
so the solution is 1/3x+74=y
if it is not i'm sorry
We're told that
![P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995](https://tex.z-dn.net/?f=P%28A%29%3D%5Cdfrac1%7B200%7D%3D0.005%5Cimplies%20P%28A%5EC%29%3D0.995)
![P(B\mid A)=0.7](https://tex.z-dn.net/?f=P%28B%5Cmid%20A%29%3D0.7)
![P(B\mid A^C)=0.05](https://tex.z-dn.net/?f=P%28B%5Cmid%20A%5EC%29%3D0.05)
a. We want to find
. By definition of conditional probability,
![P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}](https://tex.z-dn.net/?f=P%28A%5Cmid%20B%29%3D%5Cdfrac%7BP%28A%5Ccap%20B%29%7D%7BP%28B%29%7D)
By the law of total probability,
![P(B)=P(B\cap A)+P(B\cap A^C)=P(B\mid A)P(A)+P(B\mid A^C)P(A^C)](https://tex.z-dn.net/?f=P%28B%29%3DP%28B%5Ccap%20A%29%2BP%28B%5Ccap%20A%5EC%29%3DP%28B%5Cmid%20A%29P%28A%29%2BP%28B%5Cmid%20A%5EC%29P%28A%5EC%29)
Then
![P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657](https://tex.z-dn.net/?f=P%28A%5Cmid%20B%29%3D%5Cdfrac%7BP%28B%5Cmid%20A%29P%28A%29%7D%7BP%28B%5Cmid%20A%29P%28A%29%2BP%28B%5Cmid%20A%5EC%29P%28A%5EC%29%7D%5Capprox0.0657)
(the first equality is Bayes' theorem)
b. We want to find
.
![P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984](https://tex.z-dn.net/?f=P%28A%5EC%5Cmid%20B%5EC%29%3D%5Cdfrac%7BP%28A%5EC%5Ccap%20B%5EC%29%7D%7BP%28B%5EC%29%7D%3D%5Cdfrac%7BP%28B%5EC%5Cmid%20A%5EC%29P%28A%5EC%29%7D%7B1-P%28B%29%7D%5Capprox0.9984)
since
.
Function would be f(x) = 5x+85
here, x= 15
5(15)+85 = 75+85 = $160
312/17
= (306+6)/ 17
= 306/17+ 6/17
= 18+ 6/17
= 18 6/17
The final answer is 18 6/17~