-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
9514 1404 393
Answer:
- 0.5
- 0.4444444... repeating
Step-by-step explanation:
Your calculator can do this for you.
1. 2/4 = 1/2 = 5/10 = 0.5
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2. The long division immediately becomes repetitive. The decimal equivalent is repeating 4s: 0.444...
_____
The long division is shown in the attachments. When the remainder is the same as the initial dividend, you know the quotient will repeat.
To find the remaining amount of money on the gift card you would substitute 8 in for the variable 't' and solve for g.
g= -9(8) + 100
g = -72 + 100
g = $28
The equation can be written as (2x-3)^2
Ans since it is square, then we can take the root of the equation to find on side of the square, which is (2x-3) and by multiplying it by4, we can find the perimeter.
