Not a single one of your individual tosses has an outcome with the expected average value (namely, 2.5 heads per toss of 5 coins). further, note that the two distributions of 20 outcomes pictured to the right both produce the same average value, namely, 2.5. in the absence of any information about how the average value was obtained,

Step-by-step explanation:

For a coin tossing the event Head or Tail turn up is equally likely

i.e, P(Head turn up) = 0.5 and also P(tail turn up) = 0.5

5 coin tossed then expected value of head is 5*P(Head turn up) = 5*0.5 = 2.5

This is a case of Binomial Distribution... Since mean = np in binomial distribution

here n=5 and p=0.5

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

answer

106 emergencies

step-by-step explanation: