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.
You need to solve for one variable in equation 1 and substitute it in equation 2 to solve.
Equation 1: x+y=24
x= number of 3 pt questions
y= number of 5 pt questions
24= Total number of questions
Equation 2: 3x+5y=100
100= Total point value possible on test
3x= point value of 3 pt questions
5y= point value of 5 pt questions
x+y=24
Subtract y from both sides
x=24-y
Substitute in equation 2:
3x+5y=100
3(24-y) +5y=100
72-3y+5y=100
72+2y=100
Subtract 72 from both sides
2y=28
Divide both sides by 2
y=14
Substitute y=14 back in to solve for x:
3x+5y=100
3x+5(14)=100
3x+70=100
Subtract 70 from both sides
3x=30
Divide both sides by 3
x=10
So there are 10 three point questions
There are 14 five point questions.
Hope this helped! :)
Answer:
y = -2
Step-by-step explanation:
Any asymptotes of a rational function will be described by the quotient of the numerator and denominator (excluding any remainder).
The horizontal asymptote is ...
y = -2
