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Sergeeva-Olga [200]
3 years ago
6

John and Alan have a collection of x baseball cards. John has x/4 cards. What fraction of the cards does Alan have?

Mathematics
2 answers:
maw [93]3 years ago
8 0

Answer: B. \frac{3x}{4}

Step-by-step explanation:

Given: The total number of cards John and Alan has = x

The number of cards John has = \frac{x}{4}

Then to find the number of cards Alan has, we need to subtract the number of cards John has from the total number of cards.

Then, the number of cards John has =x-\frac{x}{4}

⇒ The number of cards John has =\frac{4x-x}{4}

⇒ The number of cards John has =\frac{3x}{4}

Furkat [3]3 years ago
3 0
The fraction of cards that Alan has is 3x/4. The correct answer is B. 

x-x/4
4x-x/4
3x/4
he has 3x/4 cards
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Can someone show me the workings out for each of these three questions
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3 0
3 years ago
Identify least to greatest<br><br> 4/25<br><br> .15<br><br> 7%<br><br> .28
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7%, .15, 4/25, .28

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4 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}

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4 years ago
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