11. 9/21 6/x
9/21 = 3/7
6/x = 3/7
6/14 = 3/7
x = 14
5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10
And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.
Answer: x = -2 and y = 4
Step-by-step explanation:
We have given a system of equations
2x + 4y =12 ...........(1)
and
x + y = 2 .............(2)
Now to plot them in graph we need to find the points of the above linear equations .
From equation 1 ,we get
So get the points (0,3) and (6,0).
From equation 2 ,we get
So we get the points (0,2) and (2,0).
So after plotting these points we get two lines intersecting at (-2,4).
Therefore our solution is x=-2 and y= 4.
It can't be graphed because it's not a proper function