Well I don't know.
Let's think about it:
-- There are 6 possibilities for each role.
So 36 possibilities for 2 rolls.
Doesn't take us anywhere.
New direction:
-- If the first roll is odd, then you need another odd on the second one.
-- If the first roll is even, then you need another even on the second one.
This may be the key, right here !
-- The die has 3 odds and 3 evens.
-- Probability of an odd followed by another odd = (1/2) x (1/2) = 1/4
-- Probability of an even followed by another even = (1/2) x (1/2) = 1/4
I'm sure this is it. I'm a little shaky on how to combine those 2 probs.
Ah hah !
Try this:
Probability of either 1 sequence or the other one is (1/4) + (1/4) = 1/2 .
That means ... Regardless of what the first roll is, the probability of
the second roll matching it in oddness or evenness is 1/2 .
So the probability of 2 rolls that sum to an even number is 1/2 = 50% .
Is this reasonable, or sleazy ?
Domain: x² - 4 ≠ 0
+ 4 + 4
x² ≠ 4
√x² ≠ √4
x ≠ ±2
x ≠ -2 and x ≠ 2
(-∞, -2) ∨ (-2, 2) ∨ (2, ∞)
Range: y ≠ 1
(-∞, 1) ∨ (1, ∞)
Intervals: Increasing: (0.25 , ∞)
Decreasing: (-∞, 0.25)
Symmetry: X-axis: Not Symmetric
Y-axis: Not Symmetric
Origin: Not Symmteric
Extrema: Maximum Relative: x = 0
Minimum Relative: Nothing
The factor theorem says that for a function f(x), if f(a) = 0, then (x - a) is a factor of f(x).
For f(x) = 2x^4 + 22x^3 + 60x^2
f(-5) = 2(-5)^4 + 22(-5)^3 + 60(-5)^2 = 2(625) + 22(-125) + 60(25) = 1250 - 2750 + 1500 = 0
Therefore, x - (-5) = x + 5 is a factor of f(x).
