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A manufacturing company produces parts that are good (90%), partially defective (2%), or completely defective (8%). These parts

Schach [20]

Answer:

The probability that a part is good given that it passed the inspection machine is P=0.978.

Step-by-step explanation:

As the inspection machine detect and discard any part that is completely defective, only the good and partially defective parts passed this inspection.

Then, if we have:

Probability of being a good part P(G)=0.90

Probability of being a patially defective part P(P)=0.02

Probability of being a completely defective part P(D)=0.08.

Probability of passing the inspection machine = 1-P(D)=1-0.08=0.92

Then, the probability of having a good part, given that it passed the inspection machine is:

$P(G|Pass)=P(G)/P(Pass)=0.90/0.92=0.978$

4 0

3 years ago

Root beer is priced at $2.88 for 6 cans<br><br> what is the cost of one can?

Rainbow [258]

Answer:

0.48

hope this helps

have a good day :)

Step-by-step explanation:

8 0

2 years ago

The expression on the left side of an equation is shown below. 3x+9= []

kozerog [31]

Answer:

C. 3x

Step-by-step explanation:

The first two choices give 1 solution each.

The last choice gives an infinite number of solutions.

Choice C. gives no solution.

3x + 9 = 3x

Subtract 3x from both sides.

9 = 0

Since 9 = 0 is false, equation 3x + 9 = 3x has no solution.

4 0

3 years ago

Read 2 more answers

"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d

dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$