Correction:
Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on
P(E∩E') = P(E)P(E').
Answer:
The case is not always true.
Step-by-step explanation:
Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.
And for any two mutually exclusive events, E and E',
P(E∩E') = 0
Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then
P(E)P(E') cannot be equal to zero.
So
P(E)P(E') ≠ 0
This makes P(E∩E') different from P(E)P(E')
Therefore,
P(E∩E') ≠ P(E)P(E') in this case.
The anwer is b because you can multiply or subraction from it
Step-by-step explanation:
1st expression = 3a - 6 = 3 ( a - 2 )
2nd expression = a² - 4 = a² - 2² = ( a + 2) (a - 2)
The HCF is (a -2).
Hope it helps :)❤
The point lies in quadrant 1 and the distance is 4.
Why?
Because if we subtract the x values which is 1 and 5 we get 4!
If helped mark me the brainiest!
