Refer to the attachment please :-)
Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Fail the test.
- Event B: Unfit.
The probability of <u>failing the test</u> is composed by:
- 46% of 37%(are fit).
- 100% of 63%(not fit).
Hence:

The probability of both failing the test and being unfit is:

Hence, the conditional probability is:

0.7873 = 78.73% probability that Mona was justifiably dropped.
A similar problem is given at brainly.com/question/14398287
Answer: he payed $56 in total
Explanation:
50% of 100 is 50.
50 time 1.12 is 56
A:He should choose team Y because they have a Median, Low and High that are close in value and they have the lowest inter quartile range, IQR
B: He should choose team Z because they have the highest Mean, meaning that they have the highest average and the highest scores.
Answer:
−2k=k−7−8
−2k=k+−7+−8
−2k=(k)+(−7+−8)(Combine Like Terms)
−2k=k+−15
−2k=k−15
Step 2: Subtract k from both sides.
−2k−k=k−15−k
−3k=−15
Step 3: Divide both sides by -3.
−3k
−3
=
−15
−3
k=5
Step-by-step explanation: