Answer:
The probability of selecting a solid black marbles both times;
P = 9/100
Attached is the completed question;
Step-by-step explanation:
Number of solid black marbles = 3
Total number of marbles = 10
The probability of selecting a solid black marble;
P1 = 3/10
With the assumption that the marbles are replaced before next selection.
The probability of selecting a solid black marbles both times;
P = P1 × P1 = 3/10 × 3/10 = 9/100
P = 9/100
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 ?
The surface area of the cube is 600cm^2.
6(10^2)
6(100)
600
1200,000(1+.1)=(the answer) sorry I don't have a calculator with me.
Y=2/3x-2
to find intercepts, substitute 0 into y and solve for x = x intercept is (3,0) and repeat for y. substitute 0 into x to get y intercept of (0,-2)
