X*10/100 hope that helps:)
The correct answer is c :)
the point is described by two variables: x and y. it is written like that: (x, y). so the first variable in brackets is the x variable. the other is y. so pick every point a-d, put these values into the equation, solve it and check if the values before and after = are the same. if so - you are correct!
look:
c) (1, 5) so x=1, y=5
equation: y=5^x
putting values: 5=5^1 and this is true!
Answer:
102
Step-by-step explanation:
We have the mean (m) 128.5 and the standard deviation (sd) 8.2, we must calculate the value of z for each one and determine whether or not it is an outlier:
z = (x - m) / sd
In the first case x = 148:
z = (148 - 128.5) /8.2
z = 2.37
In the second case x = 102:
z = (102 - 128.5) /8.2
z = -3.23
In the first case x = 152:
z = (152 - 128.5) /8.2
z = 2.86
The value of this is usually between -3 and 3, therefore when x is 102 it goes outside the range of the value of z, which means that this is the outlier.
To do this you need to work backwards. if you note the number as 'n' you can put
7n = 5600
to find n you simply divide both sides by 7 getting
n = 5600/7
n = 800
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
