Roduction records indicate that 2.8% of the light bulbs produced in a facility are defective. a random sample of 30 light bulbs was selected. a. use the binomial distribution to determine the probability that fewer than three defective bulbs are found.

Our required probability is 0.947.

Step-by-step explanation:

Since we have given that

Number of light bulbs selected = 30

Probability that the light bulb produced in a facility are defective = 2.8% = 0.028

We need to find the probability that fewer than 3 defective bulbs are found.

We will use "Binomial distribution":

n = 30, p = 0.028

so, P(X>3)=P(X=0)+P(X=1)+P(X=2)

So, it becomes,

and

So, the probability that fewer than three defective bulbs are defective is given by

stop cheating

step-by-step explanation:

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d