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bekas [8.4K]
3 years ago
10

6th grade math, due today. please help

Mathematics
2 answers:
Harman [31]3 years ago
7 0
6. 4x
7. 3y
8. y-4.5
9. 3.5x+2
10.4y+2
11. 7x
12. 2.2w-0.5
13. 1 2/3b + 6 2/3
14.3 3/4x + 1 1/2
15. 0.8x +2.5
lara31 [8.8K]3 years ago
4 0

Answer:

6.4x

7. 3y

8. y−4.5

9. 7/2 x+2

10. 4y+2

11. 7x

12.  2.2w−0.5

13. (1 2/3) (b) + 6 2/3

14. 15/4x + 3/2

15. 0.8x+2.5

Step-by-step explanation:

I GOTCHU FAM

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A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible o
Agata [3.3K]

Answer:

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

Step-by-step explanation:

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG.

Let X represent the number of children that are girls. Then

1. When X=0, there is one possible outcome BBBB. So

Pr(X=0)=\dfrac{1}{16}

2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so

Pr(X=1)=\dfrac{4}{16}=\dfrac{1}{4}

3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so

Pr(X=2)=\dfrac{6}{16}=\dfrac{3}{8}

4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so

Pr(X=3)=\dfrac{4}{16}=\dfrac{1}{4}

5. When X=4, then there is one possible outcome GGGG, so

Pr(X=4)=\dfrac{1}{16}

Now, the probability distribution table is

\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}

5 0
4 years ago
Read 2 more answers
Does -2 4/5 + 6 1/3= 2/5
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Answer:

yes.

Step-by-step explanation:

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Item 18
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Approximately 14%

14-12=2
2/14=1/7
1/7*100=14.29
14.29%
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3 years ago
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Answer:

Step-by-step explanation:

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X1=4x5 I think it is. Though this would then give x=20x with I think is impossible...
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