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:
your answer would be 18 units
Step-by-step explanation:
Answer:
7x+7<u> </u><u>></u><u> </u>23-x
group like terms
7x+ x <u>></u><u> </u><u>2</u><u>3</u><u>-</u><u>7</u>
8x<u> </u><u>></u><u> </u>16
we divide both side by
<u>8</u><u>x</u><u> </u> <u>></u><u> </u> <u>1</u><u>6</u>
8 8
so 8 cancelled out 8 living
x <u>></u><u> </u> <u>1</u><u>6</u>
<u> </u> 8
x <u>></u><u> </u>2
8
we
Answer:
-3 is greater
Step-by-step explanation:
Have a good day :)