Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [In other words, the two points X and Y are independent random variables such that X is uniformly distributed over (0, L/2) and Y is uniformly distributed over (L/2, L).] Find the probability that the distance between the two points is greater than L/3.

P(Y-X>L/3)=7/9

Step-by-step explanation:

We know  that  X is uniformly distributed over (0, L/2) and Y is uniformly distributed over (L/2, L).

We know that:

f(x,y)=f(x)·f(y)=\frac{1}{L/2}·\frac{1}{L/2}=4/L²

Therefore, the probability is  

P(Y-X>L/3)=∫∫ f(x,y) dx dy

P(Y-X>L/3)=∫∫ 4/L²   dx dy

We conclude that is  

P(Y-X>L/3)=1-P(Y-X≤L/3)

We know  that for the region  (0, L/2)x(L/2, L) where is satisfied Y-X≤L/3

is a right triangle with bases x∈(L/6, L/2) and y∈(L/2, 5L/6).

We calculate the area of that triangle:

p=1/2 · (L/3)²= L²/18

Now, we have

P(Y-X≤L/3)=4/L² ·  L²/18=4/18=2/9

Therefore, the probability is    

P(Y-X>L/3)=1-P(Y-X≤L/3)=1 - 2/9

P(Y-X>L/3)=7/9

over the argument its not good to fight.

8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.