Bobby is testing the effectiveness of a new cough medication. there are 100 people with a cough in the study. seventy patients received the cough medication, and 30 other patients did not receive treatment. thirty-four of the patients who received the medication reported no cough at the end of the study. twenty of the patients who did not receive medication reported no cough at the end of the study. what is the probability that a patient chosen at random from this study took the medication, given that they reported no cough?

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

=

Prob for a person to be not in cough medication and did not have cough

=

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') =

The answer is : 63%. 

Step-by-step explanation:

There are 100 people with a cough in the study.

70 patients received the cough medication, and 30 other patients did not receive treatment.

34 of the patients who received the medication reported no cough at the end of the study.

20 of the patients who did not receive medication reported no cough at the end of the study.

So, there are a total of 54 patients who reported no cough after the medication.

Out of this 54 patients, 34 took the medication.

So, the required probability is:

% ≈ 63%

The answer is : 63%. 

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.