It is known that 10% of the calculators shipped from a particular factory are defective. what is the probability that at least one in a random sample of four calculators is defective

P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)

The brobability of a binomial distribution is given by

where: n = 4

Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439

Delta airlines or untied airlines

The answer is 270 cars.you can do this by making a proportional relationship. for instance, 200 is the whole we are using, so on a “number line” 200 would be over 100%. now, since he sells 35% more, add 35% to 100% to get 135% on the other side of the “number line” on the bottom half, leaving the top half empty.cross multiply 200 by 135 then divide by 100 to get your answer. i hope this , if it did, , if you don’t mind, giving me brainiest! 200/100% } 200 • 135 /100 ? /135%