Please hurry up and answer i’ll mark brainliest
1 answer:
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Answer:
The correct option is (c).
Step-by-step explanation:
The complete question is:
The data for the student enrollment at a college in Southern California is:
Traditional Accelerated Total
Math-pathway Math-pathway
Female 1244 116 1360
Male 1054 54 1108
Total 2298 170 2468
We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the student is female. Which of the following pairs of probabilities is not a useful comparison?
a. 1360/2468 and 116/170
b. 170/2468 and 116/1360
c. 1360/2468 and 170/2468
Solution:
If two events <em>A</em> and <em>B</em> are independent then:
In this case we need to determine whether a student enrolled in an accelerated math pathway is independent of the student being a female.
Consider the following probabilities:
If the two events are independent then:
P (F|A) = P(F)
&
P (A|F) = P (A)
But what would not be a valid comparison is:
P (A) = P(F)
Thus, the correct option is (c).
Acc. to midpoint theorem,
15={(2x+9) + (4x-15)}/2
⇒ 30=2x+9+4x-15
⇒ 30-9+15=6x
⇒ 36/6=x
⇒6=x
WZ= 4x-15
=4*6-15
=24-15=9
15={(2x+9) + (4x-15)}/2
⇒ 30=2x+9+4x-15
⇒ 30-9+15=6x
⇒ 36/6=x
⇒6=x
WZ= 4x-15
=4*6-15
=24-15=9