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alex41 [277]
3 years ago
11

A family plans to have three children. The wife and husband are trying to determine the probabilities of the different gender ou

tcomes for the children.
The husband thinks that the probability that the first child is a girl is greater than the probability that the first child is a girl and the second child is a girl. The wife disagrees. She thinks that the two probabilities are equal.
The sample space of possible outcomes is listed below. B represents a boy, and G represents a girl.
Who is correct, the husband or the wife?
Mathematics
1 answer:
nevsk [136]3 years ago
5 0
There is a higher chance that the dad is right because he is betting on a pretty much 50/50 chance while the mom is betting on a 25% chance
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24/25

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Heights in inches of several tomato plants : 16,18,18,20,17,10,18,17 mean absolute deviation :
AysviL [449]
18 would have to be the mean because of the average
7 0
3 years ago
2/3÷2⁴+(3/4+1/6)÷1/3<br> How do you figure this out
Alecsey [184]

Answer: \frac{67}{24}

Step-by-step explanation:

\frac{2}{3} /2^{4}+(\frac{3}{4} +\frac{1}{6})/\frac{1}{3} \\\\\frac{2}{3} /2^{4}+ \frac{11}{12} /\frac{1}{3} = \frac{2}{3} /16+ \frac{11}{12} /\frac{1}{3}

Use the rule a÷b/c=a*c/b

\frac{2}{3} *\frac{1}{16}+ \frac{11}{12}/ \frac{1}{3}

Use the rule a/b*c/d=ac/bd

\frac{2}{3*16}+ \frac{11}{12}/ \frac{1}{3} \\\\\frac{2}{48} +\frac{11}{12}/ \frac{1}{3} \\\\\frac{1}{24}+\frac{11}{12}/\frac{1}{3}

Use the rule a÷b/c=a*c/b

\frac{1}{24} +\frac{11}{12} *3

Use the rule a/b*c=ac/b

\frac{1}{24}+ \frac{11*3}{12}

Simplify three times then you're done.

\frac{1}{24}+ \frac{33}{12}= \frac{1}{24}+ \frac{11}{4} =\frac{67}{24}

Hope this helps, HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

- Cutiepatutie ☺❀❤

5 0
3 years ago
‼️‼️any help would be appreciated
never [62]
Because the triangle is equilateral we know that z=60, therefore:

60=\frac{1}{3}x-7
67=\frac{x}{3}
67 \times 3 = x
x=201 <span />
6 0
3 years ago
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