The game of roulette uses a wheel containing 38 pockets. Thirty-six pockets are numbered 1,2…36, and the remaining two are marked 0 and 00. The wheel is spun, and a pocket is identified as winner. Assume that the observance of any one pocket is just as likely as any other. Let A be the event that you observe either a 0 or a 00. Find P (A).

step-by-step explan mxfcg'k; uafwgzfgjhklhgation:

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

their is no graph showing

step-by-step explanation: