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ddd [48]
3 years ago
7

Suppose a couple decided to keep having children until they had a girl. Which

Mathematics
1 answer:
Serhud [2]3 years ago
8 0

Answer:

The couple having 3 children

Step-by-step explanation:

For each children, there is a 50% probability that it is a girl and a 50% probability that it is a boy.

Suppose a couple decided to keep having children until they have a girl.

The probability that they have 3 children, that is 2 boys before a girl is:

The desired outcome is two boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 4 children, that is 3 boys before a girl is:

The desired outcome is three boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 5 children, that is 4 boys before a girl is:

The desired outcome is four boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 6 children, that is 5 boys before a girl is:

The desired outcome is five boys then a girl. Each of these outcomes has a 50% probability. So

Which of these is most likely?

The largest probability is the one of them having three children, that is, two boys before a girl.

So the correct answer is:

The couple having 3 children

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3 years ago
Factor: x(3y-5)+3(3y-15)
alina1380 [7]

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Step-by-step explanation:

Step by Step Solution

STEP1:STEP2:Pulling out like terms

 2.1     Pull out like factors :

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1 year ago
1 - (8 to the second power + 6)<br><br> Which do I do first 8+6 the then exponent or Vice versa?
Svet_ta [14]
You would need to do it vice versa
6 0
3 years ago
Read 2 more answers
How do i solve this?
shusha [124]

Answer:

Step-by-step explanation:

You can use Pythagoras Theorem,

8^{2} + y^{2}  = (8\sqrt{6} )^{2} \\      y^{2}        = (8\sqrt{6} )^{2} - 8^{2} \\y^{2} = 384 - 64\\y^{2} = 320\\y  = \sqrt{320}

∴ y = 8\sqrt{5}

Hope that helps!!

Please mark as Brainliest!!

7 0
3 years ago
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