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.
7(y+2)=3
seven times the sum of a number (y) and 2
means y+2 multiplying by 7
Answer:
h = -0.3d^2 + 1.2d + 1.5
Step-by-step explanation:
Question 1:
F(x) and g(x) are like variables, just plug into the equation.
f(x) + g(x) = (x + 6) + (12x - 7)
x+6+12x-7 = 13x-1
Question 2: f(3) + g(-1)
You plug in the x-values into the equation, and then take the answer and add them together.
f(3) = 3+4
g(-1) = 12(-1)-6
f(3) = 7
g(-1) = -18
7 + (-18) = -11
Question 3:
This is similar to question 1, plug in the variables and simplify.
9x - (7x+3)
Remember to distribute the "-"
9x - 7x - 3
2x - 3