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Nezavi [6.7K]
1 year ago
13

a couple has three children. assuming each child has an equal chance of being a boy or a girl, what is the probability that they

have at least one girl? a. 1 b. .125 c. .5 d. .875
Mathematics
1 answer:
yaroslaw [1]1 year ago
3 0

Answer:

d. 0.875

Step-by-step explanation:

Having at least 1 girl out of 3 children means that there can be 1, 2 or 3 girls out of the 3 children.

So having at lest 1 girl out of 3 children is the same as saying that all 3 children are not boys.

The probability that all 3 children are boys would be

chance that child 1 is a boy * chance that child 2 is a boy * chance that child 3 is a boy = 0.5 * 0.5 * 0.5 = 0.125

And the probability that NOT all 3 children are boys would be

100% - probability that all 3 children are boys = 1 - 0.125 = 0.875

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3 years ago
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Which statement is true about the equation? 3y-6x=-3
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<span>B.When x=-4, y=-9 and when x=2, y=3.

</span><span>3y-6x=-3
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8 0
3 years ago
X ⋅ x = ? (1 point) jjjj
liraira [26]

Answer:

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2 years ago
What did i do wrong?
irakobra [83]

Answer:

27x +42+ 3x²

Step-by-step explanation:

so im learning this rn in class and what you do is FOIL

So its stands for

First

Outside

Inside

Last

(3x+6)(x+7)

(3x+6)(x+7) -- 3x² (if you have x times x it is x to the second power)

(3x+6)(x+7) -- 21x

(3x+6)(x+7) -- 6x

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Now add 21x and 6x

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7 0
3 years ago
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stiv31 [10]

Answer:

Option C.

Step-by-step explanation:

Given information: RSTU is a parallelogram, Digonals RT and SU intersect each other at point V, UV=(x-3) and VS=(3x-13).

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Therefore, the correct option is C.

3 0
3 years ago
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