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bezimeni [28]
3 years ago
7

Six people are sitting around a circular table, and each person has either blue eyes or green eyes. Let x be the number of peopl

e sitting next to at least one blue-eyed person, and let y be the number of people sitting next to at least one green-eyed person. How many possible ordered pairs (x,y) are there? (For example, (x,y) = (6,0) if all six people have blue eyes, since all six people are sitting next to a blue-eyed person, and zero people are sitting next to a green-eyed person.)

Mathematics
1 answer:
Slav-nsk [51]3 years ago
3 0

Answer:

10

Step-by-step explanation:

Given that:

Six people are sitting around a circular table

and each person has either blue or green eyes.

So; we are told to represent x as the number of people sitting next to at least one blue-eyed person

and y be the number of people sitting next to at least one green-eyed person.

The objective is to find how many possible ordered pairs (x,y) are there?

The ordered pairs (x,y) i.e ( blue , green)   are : ( (6, 0), (6, 2), (6, 4), (5, 3), (5, 5), (3, 3) ).

The above proves the ordered pairs for  at least as blue as green.

Now; to determine the ordered pairs for as many green as blue, we will require to go the opposite direction by reversing the values; so we can arrive at a new values of  10 possible outcomes which are as follows:

(6, 0), (6, 2), (6, 4), (5, 3), (5, 5), (3, 3), (3, 5), (4, 6), (2, 6), (0, 6)

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<h3>Answer: Mean = 218.9.</h3><h3>Median = 229</h3><h3>Mode = Zero mode.</h3>

Step-by-step explanation:

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<h3>Mean = 218.9.</h3>

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\mathrm{The\:mode\:is\:the\:term\:in\:the\:data\:set\:that\:appears\:the\:most.}

\mathrm{If\:there\:is\:more\:than\:one\:term\:that\:appears\:the\:most,\:then\:there\:is\:no\:mode.}

\mathrm{Count\:the\:number\:of\:times\:each\:element\:appears\:in\:the\:list}

\begin{pmatrix}148&160&191&212&246&262&264&268\\ 1&1&1&1&1&1&1&1\end{pmatrix}

\mathrm{The\:most\:common\:element\:is\:not\:unique,\:so\:there\:is\:no\:mode}

<h3>Therefore, Mode = Zero mode.</h3>
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