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olasank [31]
4 years ago
13

In a certain manufacturing process, it is known that, on average, 1 in every 100 items is defective.

Mathematics
1 answer:
Rashid [163]4 years ago
5 0

Answer:

A. 0.009899

B. 0.005624

Step-by-step explanation:

Data:

Let the probability that an item is defective = \frac{1}{100}

The probability that the item is not defective = \frac{99}{100}

The probability that the fifth item is defective = \frac{1}{100}* \frac{98}{99}

                                                                            = 0.009899

Probability that one in 5 items is defective  = 0.005624

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Plz help asap i really need it for 30 points
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Answer:

13.04 cm^2

Step-by-step explanation:

Larger Circle diameter is 12.5 cm and the smaller is 3.5 cm

The Area of a circle is A = πr^2 = πd^2/4

Larger Circle divide the diameter by 2 = 6.25 cm

Smaller Circle dive the diameter by 2 = 1.75 cm

Using above formula  A =  122.71846303085 cm^2 &  

Finally we deduct the smaller Area from the larger Area

A = 9.6211275016187 cm^2 - 9.6211275016187 cm^2

122.71846303085 cm^2 - 9.6211275016187 cm^2 =

113.0973355292 cm^2 thus 13.04 is your answer since I used the

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3 years ago
Which line is parallel to the line given below?
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B

Step-by-step explanation:

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AleksandrR [38]

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Need help on knowing where to graph!!<br> y = 4x - 2<br> y = 1/2x +5
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2 years ago
10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.
erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used to solve this question.

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}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

p = \frac{4}{6} = 0.6667

10 normal six sided dice are thrown.

This means that n = 10

Find the probability of obtaining at least 8 failures.

This is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

So

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

P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951

P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867

P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174

Then

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992

0.2992 = 29.92% probability of obtaining at least 8 failures.

8 0
3 years ago
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