0.44 is equal to 4.4*10^-1. (scientific notation)
Answer:
The probability is ![\frac{56!}{64!}](https://tex.z-dn.net/?f=%20%5Cfrac%7B56%21%7D%7B64%21%7D%20)
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,
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
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!}](https://tex.z-dn.net/?f=%20%5Cfrac%7B8%21%7D%7B64%20%5Cchoose%208%7D%20%3D%20%5Cfrac%7B8%21%7D%7B%5Cfrac%7B64%21%7D%7B8%2156%21%7D%7D%20%3D%20%5Cfrac%7B56%21%7D%7B64%21%7D%20)
The choices are points of a function that lies at coordinates (x,y). Just do trial and error with each choice and see which applies
a) x = 0, y = 0
-5(0)^2 + 20(0) - 17 > 0
-17 > 0 (not true)
b) x = 2, y= 2
-5(2)^2 + 20(2) - 17 > 2
3 > 2 (true)
c) x =5, y=2
-5(5)^2 + 20(5) - 17 > 2
-42 > 2 (not true)
d.) x=1, y=5
-5(1)^2 + 20(1) -17 > 5
-2 > 5 (not true)
So, the answer is B.
I hope you understand my work goodluck :)
i give the answer here
I'm assuming you're looking for an equation so this is what it would look like: (k=kakapo, e=emu)
5k-70=e