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Neko [114]
2 years ago
6

If the average of 2 and p is equal to the average of 1,6, and p, what is the value of p

Mathematics
1 answer:
Vanyuwa [196]2 years ago
8 0

Answer:

p = 8

Step-by-step explanation:

\frac{2+p}{2}=\frac{1+6+p}{3}\\\\\frac{2+p}{2}=\frac{7+p}{3}\\\\

Cross multiply

 (2+p) * 3 = (7 + p) * 2          {Use distributive rule}

2*3 + p*3 = 7*2 + p *2

    6+ 3p   = 14 + 2p          {subtract 2p form both sides}

6 + 3p - 2p= 14

      6 + p  = 14                {Subtract 6 form both sides}

              p = 14 - 6

              p = 8

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The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

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