Answer:
The probability is 
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

Answer:
76c + 228
Step-by-step explanation:
Use the distributive property
76 × 3 = 228
76 × c = 76c
76c + 228
Answer:
30.25 & 33.25
Step-by-step explanation:
x² + 11x + (5.5)² = 3+ (5.5)²
x² + 11x + 30.25 = 33.25
Answer:
184 in²
Step-by-step explanation:
Given :
Width, w = 6 inches
Length, l = 10 inches
Height, h = 2 inches
To obtain how much wrapping paper is needed ; we take the surface area of the box
Surface area = 2(lw + lh + wh)
Surface area = 2((6*10) + (6*2) + (10*2))
Surface area = 2(60 + 12 + 20)
Surface area = 2(92)
Surface area = 184 in²
The amount of wrapping paper needed = 184 in²
7x^2 - 16x - 8 = 0
using the quadratic formula x = [ -(-16)+/- sqrt((-16^2 - 4*7-8] / 2*7
= (16 +/1 sqrt 480) / 14
= 2.71, -0.42 to nearest hundredth