Of the cartons produced by a company, 4% have a puncture, 8% have a smashed corner , and 1.4% have both a puncture and a smashed corner. Find the probability that a randomly that a randomly selected carton has a carton had a puncture or a smashed corner

yeah and

step-by-step explanation:

you know what it isnt great cheating

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

what i would say the best answer is a. 40.5