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3 years ago

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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

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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Step-by-step explanation:

Given:

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The probability that there are 8 occurrences in ten minutes = ?

Solution:

Let X be the number of occurrences of the event X

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Possion of distribution is given by ,

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Given

The shape above

Required

Determine PQ

Determine dimension of the garden

Calculating Length PQ

Represent the length of the inner rectangle as L

Represent the width of the inner rectangle as W

$W = x$

$L = x + 9$

The distance between the inner rectangle and the outer triangle is 4ft

This implies that;

$PQ = L + 4$

Substitute $x + 9$ for $L$

$PQ = x + 9 + 4$

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Also;

$QR = W + 4$

Substitute $x$ for $W$

$QR = x + 4$

Calculating The Dimension of The Garden

First, we need to determine the Area of the inner rectangle

$Area_1 = L * W$

Recall that $W = x$ and $L = x + 9$

So;

$Area_1 = x * (x + 9)$

For the bigger rectangle

$Area_2 = PQ * QR$

Recall that $PQ = x + 13$ and $QR = x + 4$

So;

$Area_2 = (x + 13)(x + 4)$

Given that the Area of the walkway is $400ft^2$

This implies that

$Area_2 = Area_1 + 400$

Substitute $Area_1 = x * (x + 9)$ and $Area_2 = (x + 13)(x + 4)$

$(x + 13)(x + 4) = x * (x + 9) + 400$

Open All Brackets

$x^2 + 13x + 4x + 52 = x^2 + 9x + 400$

$x^2 + 17x + 52 = x^2 + 9x + 400$

Collect Like Terms

$x^2 - x^2 + 17x - 9x = 400- 52$

$8x = 348$

Divide both sides by 8

$\frac{8x}{8} = \frac{348}{8}$

$x = \frac{348}{8}$

$x =43.5$

Since the dimensions of the garden is $W = x$ and $L = x + 9$

<em>Substitute 43.5 for x in both cases</em>

$L = 43.5 + 9 = 52.5$

$W = 43.5$

<em>Hence, the dimension of the garden is 52.5ft by 43.5ft</em>

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