Answer:
0.665
Step-by-step explanation:
Given: 100 people are split into two groups 70 and 30. I group is given cough syrup treatment but second group did not.
Prob for a person to be in the cough medication group = 0.70
Out of people who received medication, 34% did not have cough
Prob for a person to be in cough medication and did not have cough
=![\frac{0.70(34)}{60}=0.397](https://tex.z-dn.net/?f=%5Cfrac%7B0.70%2834%29%7D%7B60%7D%3D0.397)
Prob for a person to be not in cough medication and did not have cough
=![\frac{0.3(20)}{30}=0.20](https://tex.z-dn.net/?f=%5Cfrac%7B0.3%2820%29%7D%7B30%7D%3D0.20)
Probability for a person not to have cough
= P(M1C')+P(M2C')
where M1 = event of having medication and M2 = not having medication and C' not having cough
This is because M1 and M2 are mutually exclusive and exhaustive
SO P(C') = 0.397+0.2=0.597
Hence required prob =P(M1/C') = ![\frac{0.397}{0.597}=0.665](https://tex.z-dn.net/?f=%5Cfrac%7B0.397%7D%7B0.597%7D%3D0.665)
Notice that
![7^{2014} - 7^{2012} = 7^{2012} \bigg(7^2 - 1\bigg) = 7^{2012} \times 48 = 2^4 \times 3 \times 7^{2012}](https://tex.z-dn.net/?f=7%5E%7B2014%7D%20-%207%5E%7B2012%7D%20%3D%207%5E%7B2012%7D%20%5Cbigg%287%5E2%20-%201%5Cbigg%29%20%3D%207%5E%7B2012%7D%20%5Ctimes%2048%20%3D%202%5E4%20%5Ctimes%203%20%5Ctimes%207%5E%7B2012%7D)
, so
![\dfrac{7^{2014}-7^{2012}}{12} = 2^2 \times 7^{2012}](https://tex.z-dn.net/?f=%5Cdfrac%7B7%5E%7B2014%7D-7%5E%7B2012%7D%7D%7B12%7D%20%3D%202%5E2%20%5Ctimes%207%5E%7B2012%7D)
Then taking the positive square root gives
![\sqrt{\dfrac{7^{2014}-7^{2012}}{12}} = \sqrt{2^2 \times 7^{2012}} = 2\times7^{1006}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7B7%5E%7B2014%7D-7%5E%7B2012%7D%7D%7B12%7D%7D%20%3D%20%5Csqrt%7B2%5E2%20%5Ctimes%207%5E%7B2012%7D%7D%20%3D%202%5Ctimes7%5E%7B1006%7D)
so
and
.
P(P|K) = 82.6%.
P(P|K) = P(K and P)/P(K) = 1.9%/2.3% = 0.019/0.023 = 0.8261 = 82.6%
Answer:
oop... glll
Step-by-step explanation:
have u uploaded any? Maybe I can help