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AnnyKZ [126]
2 years ago
5

If each point in the diagram can act as an endpoint, how many distinct line segments can be formed using the set of points

Mathematics
2 answers:
sdas [7]2 years ago
8 0

Answer:

Option D 6 is the answer.

Step-by-step explanation:

the given points are A, D, C and B. If each point in the diagram can act as end points, we have to calculate number of distinct line segments formed.

This ca be calculated in two ways.

(1) We will form the line segments

AB, AC, AD, BC, BD, DC

Therefore 6 segments can be formed.

(2) By combination method

Number of segments = ^{4}C_{2}

= \frac{4!}{2!2!}=\frac{4\times 3\times 2\times 1}{(2\times 1)\times (2\times 1)}

= \frac{4\times 3}{2\times 3}=\frac{12}{2}

= 6

Option D 6 is the answer.

Juli2301 [7.4K]2 years ago
6 0

Answer:

6

Step-by-step explanation:

possible points are:

ad

ac

ab

dc

db

cb

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Step-by-step explanation:

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P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

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  • A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed)=P(MW)+P(WM)

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(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

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P(at least one woman is appointed)=P(MW)+P(WM)+P(WW)

P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}

In Part B, P(MW)+P(WM)=\frac{8}{39}

Therefore:

P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}

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