Problem is in the pictures

Mathematics

1 answer:

marta [7]3 years ago

3 0

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}6x+3y=-21&\text{divide both sides by (-3)}\\2x+5y=5\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-y=7\\2x+5y=5\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad4y=12\qquad\text{divide both sides by 4}\\.\qquad\qquad y=3\\\\\text{Put it to the second equation:}\\2x+5(3)=5\\2x+15=5\qquad\text{subtract 15 from both sides}\\2x=-10\qquad\text{divide both sides by 2}\\x=-5$

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Observe that the first and the second player have equal probability to get any of number. Using the principle of that symmetry, we have that

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<u>Event X = 2</u> means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence

$P(X=2)=\frac{1}{12}$

<u>Event X = 3</u> means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence

$P(X=3)=\frac{1}{20}$