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Flura [38]
3 years ago
12

A bag contains 7 red marbles, 8 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the probabili

ty, to the nearest 10th of a percent, that both marbles drawn will be green?
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0

Answer: 0.053

Step-by-step explanation:

n(r) = 7

n(b) = 8

n ( g) = 5

Total number of marbles = 20

The probability that both marbles drawn will be green will be

The probability of the first being green = 5/20 and the probability that the second marble is green = 4/19.

We will multiply the two together , we have

5/20 x 4/19 = 20/380 = 0.05263157895

Therefore : the probability, to the nearest 10th of a percent, that both marbles drawn will be green will be 0.053

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Deffense [45]

The answer you are looking for is 7\frac{101}{14}.



Explanation:

4(4x−8)+4(3x−7)=7(2x+5)+6

Step 1: Simplify both sides of the equation.

4(4x−8)+4(3x−7)=7(2x+5)+6

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Step 2: Subtract 14x from both sides.

28x−60−14x=14x+41−14x

14x−60=41

Step 3: Add 60 to both sides.

14x−60+60=41+60

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Step 4: Divide both sides by 14.

14x

14

=

101

14

x=

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Help would be appreciated:)
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Answer:

<BCE=130°

Step-by-step explanation:

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3 years ago
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