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Roman55 [17]
3 years ago
13

Joshua has a bag of marbles. In the bag are 5 white marbles, 3 blue marbles, and 7 green marbles. Peter randomly draws one marbl

e, sets it aside, and then randomly draws another marble. What is the approximate probability of Peter drawing out two green marbles
Mathematics
1 answer:
Step2247 [10]3 years ago
6 0

Answer:

1 / 5

Step-by-step explanation:

Given that:

Number of white marbles = 5

Number of blue marbles = 3

Number of green marbles = 7

Required is the approximate probability of drawing 2 green marbles, Note that drawing is done without replacement :

Probability = required outcome / Total possible outcomes

Total possible outcomes = sum of all marbles = (5 + 3 + 7) = 15 marbles

First draw:

P(Green) = 7 / 15

Second draw:

Required outcome = 7 - 1 = 6

Possible outcomes = 15 - 1 = 14

P(green) = 6 / 14

Probability of drawing out two green marbles :

(7/15 * 6/14) = 42 / 210 = 1 / 5

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Rama09 [41]

Answer:

The wire is sufficient for making a cube of volume 2197 cm³

Step-by-step explanation:

Volume of cube = L³

Volume of the cube = 2197 cm³

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Let's see if the wire is sufficient for making a cube.

Volume of the cube that could be made from the wire is:

V = (L)³

V = (150)³

V = 3,375,000 cm³

So,

The volume of the cube using a 150 cm length wire is much more grater than 2197 cm³.

That is why, the wire is sufficient for making a cube of volume 2197 cm³.

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The answer is x = seven
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3 years ago
I don’t get this can someone explain this
LekaFEV [45]

Answer:

15.7 units

Step-by-step explanation:

AB = 6, BC = 3

Therefore by Pythagoras theorem

AC = \sqrt{45}

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