A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 60 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 7000 aspirin tablets actually has a 4% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

the probability to be accepted is 0.302 (30.2%) (many will be rejected)

Step-by-step explanation:

assuming that the rate of 4% applies to the 60 tablets then since each tablet behaves independently, the random variable X= number of tablets with defects out of 60 tablets has a binomial distribution , where:

p(X)=n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of tablets tested = 60

x = number of defective tablets

p= probability to be defective = 0.04

then in order to be accepted x≤0 , then the probability that the batch is accepted Pa is

Pa=P(x≤1) = P(0) + P(1) = (1-p)^n + n*p*(1-p)^(n-1)

replacing values

Pa= (1-p)^n + n*p*(1-p)^(n-1) =  0.96^60 + 60*0.04*0.96^59 = 0.302 (30.2%)

then the probability to be accepted is  0.302 (30.2%) (many will be rejected)

a = 11 : 4

b = 38

c = 1 for every 6

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.