It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that no more than one in a random sample of four calculators is defective? Multiple Choice 0.2916 0.9477 0.3439 0.6561

The probability that no more than one in a random sample of four calculators is defective is 0.9477.

Step-by-step explanation:

We can model this as a binomial randome variable, because we have a sum of n Bernoulli variables with probability p of success.

In this case, the sample size is n=4 and the probability of an individual defective calculator is p=0.10.

The probability of having k defective calculators in a sample of n calculators is calculated as:

We have to calculate the probability that no more than one out of four calculators is defective (P(x≤1)). This is the same as adding the probability of having no defective calculator in the sample (P(x=0)) and the probabiltity of having one defective in the sample (P(x=1)):

  21.3

step-by-step explanation:

multiplying the given equation by c/sin(25°), we get

  c = 9/sin(25°) ≈ 21.2958 ≈ 21.3

we assume "c" is the length of ab, the side opposite angle c in triangle abc.