Astudent takes a multiple-choice exam. suppose for each question he either knows the answer or guesses and chooses an option at random. further suppose that if he knows the answer, the probability of a correct answer is 1, but if he guesses, the probability of a correct answer is 1/4. to pass, students need to answer at least 60% of the questions correctly. the student has "studied for a minimal pass," i. e., the student knows the answer to a question with probability 0.6. given that he answers a correctly, what is the probability that he actually knows the answer? group of answer choices

The probability that he knows the answer given his answer is correct is 0.8571

Step-by-step explanation:

The problem can be solved by Bayes theorem

Let us take the Following Events

1) P(A)= probability that the student's answer is correct

2) P(B) = probability that the student knows the correct answer

For 2 events A and B Bayes theorem states that

The probability that an event 'B' occurs provided that event A has already occurred is given by "P(B|A)" as

where

P(A|B) is the probability of event A to occur provided event B has occurred

from the given data we have

P(B) = 0.6

P(A) (probability that his answer is correct = probability that his answer is correct provided he knows the answer + Probability his answer is correct provided he gambles)

Thus

Thus

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet