If we draw the contingency table of x (vertical) against y (horiz.), we have a square.
For n=4, we have (legend: < : x<y = : x=y > : x>y
y 1 2 3 4
x
1 = < < <
2 > = < <
3 > > = <
4 > > > =
We see that there are n(n-1)/2 cases of x<y out of n^2.
Therefore,
p(x<y)=n(n-1)/(2n^2)=(n-1)/(2n)
However, if the sample space is continuous, it will be simply p(x<y)=1/2.
I think the first one might be 18 second might be 10
Answer:
(3,0),(2,0)
Step-by-step explanation:
Here is your answer
A. 1,3,5,7,9,11,.....
REASON :
In an AP their is a common difference between two consecutive terms.
i.e. t2-t1=t3-t2=t4-t3=....= constant
The option A satisfies above condition.
HOPE IT IS USEFUL
1) No, because the line does not divide the figure into two mirrored images.
2)Yes, because the line divides the figure into two mirrored images.
3) Yes, because the line divides the figure into two mirrored images.
4)No, because the line does not divide the figure into two mirrored images.
5)One line, vertical down the middle.
6) Zero lines, because the figure can not be divided into mirrored images.
7)Four lines, horizontal down the middle, vertical down the middle and diagonal down from each top corner.
8) One line, vertical down the middle
