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soldier1979 [14.2K]
3 years ago
6

Solve: X/7-(4x-3)/2=1/2-x

Mathematics
1 answer:
Alex3 years ago
6 0
.................................................

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When 3/11 is written as a repeating decimal, which digits are repeating?
oksian1 [2.3K]

Answer:

I think it's 27 I may be wrong tho

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3 years ago
Simplify each expression by distributing and combining like terms <br> 4(x+5)+2
faltersainse [42]

Answer:

4x+22

Step-by-step explanation:

In pic

(Hope this helps can I pls have brainlist (crown) ☺️)

3 0
3 years ago
8. 10 ÷ 5 + 10 – 9 x 11 has to show work
guajiro [1.7K]

Answer:

<h2>-87</h2>

Step-by-step explanation:

(10 ÷ 5) + 10 – (9 x 11)

=(2) + 10 - (99)

=2 +10 - 99

= -87

hope \: this \: helps

There's "8. '' In the first, is this also included? Or it's the number of the expression?

8 0
3 years ago
Please I really need help! I will mark as brainliest for the correct answer!
IgorC [24]

Answer:

77.112

Step-by-step explanation:

circle area A=pirsquared so A= 3.142×6×6= 113.112

rectangle area A= 6×6=36

shaded area= 113.112-36=77.112

3 0
3 years ago
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
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