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bogdanovich [222]
3 years ago
7

Bonus: Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family

has twice as many sisters as he has brothers. How many sons and daughters are in the family? (There is exactly one correct answer to this, I promise.)
THIS IS FOR FUN...IF YOU CAN GUESS IT I WILL MARK YOUR ANSWER THE BRAINLIEST>>>>>>GOOD LUCK!!!
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
3 0
<h3>Answer: 3 sons, 4 daughters</h3>

===================================================

Explanation:

I used trial and error to get this answer.

B = boy

G = girl

I started with a family of 2 kids, so I wrote BG. But this doesn't work because neither child has any siblings of the same gender. So the daughter doesn't have any sisters, and the son doesn't have any brothers.

If we add a son, then we have BBG. We would have to add a daughter so that he has twice as many sisters compared to the number of brothers of any given son.

The family is now BBGG. This doesn't quite work because each girl has 1 sister, but 2 brothers.

-------------

If we add on a girl, then we get BBGGG

Now each daughter has the same number of sisters (2) as brothers (also 2).

But now each son has 1 brother, but 3 sisters. We want a ratio of 1 to 2, instead of 1 to 3.

We can fix this by adding a brother and sister to this family getting BBBGGGG

Each daughter has 3 sisters and 3 brothers

Each son has 2 brothers and 4 sisters

So this fits the criteria.

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Answer:

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z =\frac{(p'_1 -p'_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2} } }

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t₀ = \sqrt{\frac{n_1n_2(n_1+n_2-2)}{n_1+n_2} } *\frac{p_1'-p_2'}{\sqrt{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2} } =      1104.83

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