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 ?
Answer: huh ?
Step-by-step explanation:
Answer:
40 classrooms
Step-by-step explanation:
Divide 1200 by 30.
1200÷30=40
If 390 students are boys just subtract 390 from 1200.
1200-390= 810
There must be 810 girls if there are 390 boys
Answer:
The answer to your question is: third option is correct
Step-by-step explanation:
Data
F (5, -1)
directrix y = 1
Then, after getting the graph, the vertix = (5, 0)
and p = 1
Formula
(x - h) ² = -4p(y - k)
(x - 5)² = -4(1)(y - 0)
(x - 5)² = -4y
y = (x - 5)² / -4
Answer: 1 and 1/3 or 4/3
6/9 = 2/3
1 x 0.5 = 0.5
2/3 divided by 1/2 = 1 and 1/3 or 4/3
