At a fundraiser, a school group charges $6 for tickets for a "grab bag." You choose one bill at random from a bag that contains 39 $1 bills, 19 $5 bills, 3 $10 bills, 7 $20 bills, and 3 $100 bills. Is it likely that you will win enough to pay for your ticket?

no, the probability is only about 18%

Step-by-step explanation:

there are total of 71 bills of varying denominations in the bag

you need to select a bill that is $10 or more to cover your $6 expense; therefore, P($ ≥ 10) = 3 + 7 + 3 / 71  or 13 /71, about 18%

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

is there supposed to be a picture here?