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murzikaleks [220]
3 years ago
10

The sample space listing the eight simple events that are possible when a couple has three children is {bbb, bbg, bgb, bgg, gbb,

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

bbbb,

bbbg, bbgb, bgbb, gbbb,

bbgg, bgbg, bggb, gbgb, gbbg, ggbb,

bggg, gbgg, ggbg, gggb,

gggg

Step-by-step explanation:

3 children: bbb, bbg, bgb, gbb, bgg, gbg, ggb, ggg

4 children: bbbb, bbbg, bbgb, bgbb, gbbb, bbgg, bgbg, bggb, gbgb, ggbb, bggg, gbgg, ggbg, gggb, gggg

You need to take a methodological approach;

The 2 easiest are the possibility of all boys and all girls;

Then consider 3 boys and 1 girl:

bbbg, bbgb, bgbb, gbbb

Then 2 boys and 2 girls:

bbgg, bgbg, bggb, gbgb, gbbg, ggbb

Lastly, 1 boy and 3 girls:

bggg, gbgg, ggbg, gggb

In total, there are 16 possibilities

P.S. interesting to note is that the number of possibilities here follows the pattern of Pascal's triangle:

       1

     1   1

   1  2  1

 1  3  3  1

1  4  6  4  1

The last row is relevant here;

There is 1 possibility where there are 4 boys;

There are 4 possibilities (in terms of the order of birth) where there are 3 boys and 1 girl;

There are 6 possibilities where there are 2 boys and 2 girls

There are 4 possibilities where there is 1 boy and 3 girls;

There is 1 possibility where there are 4 girls;

The pattern ∴ is 1 4 6 4 1, as the 5th row of Pascal's triangle reads;

If your talking about 3 children, it would match the 4th row of Pascal's triangle;

So, 1 possibility of 3 boys;

3 possibilities of 2 boys and 1 girl;

3 possibilities of 1 boy and 2 girls;

And 1 possibility of 3 girls;

If your talking about 10 children, it would match the 11th row of Pascal's triangle.

(Maths can be so cool XD)

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Melissa has more than 40 coins in her collection which inequality the nuber of coinsin melissas collection
nata0808 [166]

Answer:

X > 40

Step-by-step explanation:

Given that:

Let number of coins in her collection is more Than 40

To represent the number of coins in Melissa's collection as an inequality :

Let total number of items in her collection = X

Coins in her collection > 40

Hence, coins in her collection can be represented as :

X > 40

4 0
2 years ago
-3.4 + 11. 25=<br><br> Please help i'm horrible at math
jek_recluse [69]
The correct answer is 7. 85
8 0
2 years ago
Read 2 more answers
How do I solve this ?
timofeeve [1]

Answer:

\large\boxed{\{4,\ 7,\ 10\}}

Step-by-step explanation:

X\ \cup\ Y\\\text{The union of a collection of sets is the set of all elements in the collection.}\\\\X\ \cap\ Y\\\text{The intersection}\ X\cap Y\ \text{of two sets X and Y is the set that contains}\\\text{all elements of X that also belong to Y, but no other elements.}\\\\A'\\\text{When A is a subset of a given set U, the absolute complement A}\\\text{is the set of elements in U, but not in A.}

U=\{3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10\}\\A=\{4,\ 6,\ 8\}\\B=\{3,\ 4,\ 7,\ 10\}\\C=\{3,\ 5,\ 9\}\\\\A\ \cup\ B=\{4,\ 6,\ 8\}\ \cup\ \{3,\ 4,\ 7,\ 10\}=\{3,\ 4,\ 6,\ 7,\ 8,\ 10\}\\C'=\{3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10\}-\{3,\ 5,\ 9\}=\{4,\ 6,\ 7,\ 8,\ 10\}\\B\ \cap\ C'=\{3,\ \b4,\ \b7,\ \b1\b0\}\ \cap\ \{\b4,\ 6,\ \b7,\ 8,\ \b1\b0\}=\{\b4,\ \b7,\ \b1\b0\}\\\\(A\ \cup\ B)\ \cap\ (B\ \cap\ C')=\{3,\ \b4,\ 6,\ \b7,\ 8,\ \b1\b0\}\ \cap\ \{\b4,\ \b7,\ \b1\b0\}=\{\b4,\ \b7,\ \b1\b0\}

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3 years ago
I WILL GIVE BRAINLIEST!!!!
madreJ [45]
- 3.4 > - 4.2 is the only correct statement

> means larger than
< means smaller than

A negative number which closer to 0 than an other negative number, is larger. Therefore - 3.4 is larger than - 4.2, and therefore - 3.4 > - 4.2
5 0
2 years ago
Which diagram is NOT a good model of 3÷1/4?
Arisa [49]

Answer:

The diagram with three triangles is not a good representation.

The other diagrams have three parts divided into 4 places. The diagram of the triangles only has 3 parts in each of the three triangles. The equation for that would be 3 divided by 1/3.





Step-by-step explanation:

Have a great rest of your day
#TheWizzer

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