Two methods, A and B, are available for teaching a certain industrial skill. The failure rate is 30% for method A and 10% for method B. Method B is more expensive, however, and hence is used only 20% of the time. (Method A is used the other 80% of the time.) A worker is taught the skill by one of the two methods, but he fails to learn it correctly. What is the probability that he was taught by using method A

Probability that he was taught by using method A is 14/17

Step-by-step explanation:

Following probabilities are given -  

P (F|A) = 0.2

P (F|B) = 0.1

P (A) = 0.7

P (B) = 0.3

Now, P(A|F) = { P (F|A)* P(A)}/{{ P (F|A)* P(A)} + P (F|B)* P(B)}}

P(A|F) = (0.2*0.7)/( (0.2*0.7)+(0.1*0.3)) = 14/17

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

(a) the volume of the cone is 20/3 in3

hence, option a is correct.