Multiply 2x-2y=10 by 2,
4x-4y=20,
given:4x-5y=17
y=3
pluck it in
2x-6=10
move the 6 to the right
2x=16
divide
x=8
y=3
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:
5/4
Step-by-step explanation:
Using coordinates (10,0) and (90,100)
Slope m = (y2-y1)/(x2-x1)
m = (100 - 0)/(90 - 10) = 100/80 = 10/8 = 5/4
Answer:
B should be 3 inches long because the area of the square above is 48 inches squared, so that long side is probably 8. B is 3 inches long. As for A, this should be simple. since we know the longer side of the room has the area of 27, we can tell the length(longest side) is 9. 14 minus 9 is equal to 4. A's width is 4 inches long.
Step-by-step explanation:
for the rest, since we know one of the widths is 6 inches, we know that the length of the room with an area of 32 inches is 8. therefore, since 8 times 4 equals the number 32, C is 4 inches long. back to B. since we know that c is 4 inches long and that there is a length of 11 inches, we can tell the longest side of room A is 7. 4 Times 7 equals 28, so B is 28 inches squared.
4/6 with 2 throws left i think.
