PLEASE HELP ME! THANK YOU! :)

1 answer:

4 0

All we need to do is to divide August and September rainfall measurements into two equal halves and then substract 0.6 from one of them and add 0.6 to the other one.

6.2 ÷ 2 = 3.1
3.1 - 0.6 = 2.5
3.1 + 0.6 = 3.7

It's said that in August there was 0.6 inch more rainfall, so the bigger value (3.7 inches) is for August and the smaller value (2.5 inches) is for September.

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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

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