Amanufacturer produces a large number of toasters. from past experience, the manufacturer knows that approximately 1% are defective. in a quality control procedure, we randomly select 50 toasters for testing. we want to determine the probability that no more than one of these toasters is defective.

The probability is 0.9106

Step-by-step explanation:

The variable that says the number of defective toasters follows a binomial distribution, where we have n identical and independent events (50 toasters) with a probability p of success (1% are defective) and a probability 1-p of fail (99% are not defective). So the probability that x toasters from the 50 are defective is:

Then, the probability P that no more than one of these toasters is defective is:

P = P(0) + P(1)

So, P(0) and P(1) are calculated as:

Finally, P is equal to:

P = 0.6050 + 0.3056 = 0.9106

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

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