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sweet [91]
3 years ago
7

A bag contains 9 green marbles, 7 red marbles, and 5 white marbles. These are the only marbles in the bag. What is the ratio of

green marbles to red marbles?
Mathematics
1 answer:
MrRa [10]3 years ago
6 0

Step-by-step explanation:

A sack of marbles contains 9 red, 7 green, and 5 blue marbles. If we pull five marbles without replacement, what is the probability that exactly one is green?

The probability of an event occurring is the ratio of the number of ways it can occur to the total number of events in the sample space.

Let’s pretend for a moment that each of the marbles is labeled with a number in addition to its color, so each is distinct. Note that this doesn’t change the probability of the event in question.

First, how many ways are there for the set of marbles drawn to include exactly one green marble? Well, first we would need to choose 1 of the green marbles from the set of 7 , and then in each of those cases, we would need to choose 4 other marbles from the remaining 9+5=14 . Using binomial coefficients to represent these combinations[1] , this gives us

(71)(144)

ways to select exactly one green marble.

Now, how many total possibilities are there? This is just the number of combinations of 5 marbles that can be drawn from a set of 9+7+5=21 . Thus, there are

(215)

total possibilities.

With that, we have the probability as

Ways to draw exactly one green marble in a set of fiveWays to draw any five marbles

=(71)(144)(215)

=10012907≈34.43%

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Some body can help me with a geometric mean maze
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See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.

2. South-East point from the Start: By the 2nd theorem,

x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.

3. West point from the previous: By the 2nd theorem,

x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.

4. West point from the previous: By the 1st theorem,

9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.

5. West point from the previous: By the 2nd theorem,

10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.

6. North point from the previous: By the 1st theorem,

x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.

7. East point from the previous: By the 2nd theorem,

x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.

8. North point from the previous: By the 1st theorem,

x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.

8. West point from the previous: By the 2nd theorem,

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12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.

101. East point from the previous: By the 1st theorem,

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11. East point from the previous: By the 2nd theorem,

20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.

12. South-east point from the previous: By the 2nd theorem,

18^2=x\cdot 21.6\Rightarrow x=15.

13. North point=The end.

6 0
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