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insens350 [35]
3 years ago
7

• A number is 10.678 more than 8.76. What is the number?​

Mathematics
1 answer:
kolbaska11 [484]3 years ago
7 0
10.678 + 8.76 = 19.438

Your answer is 19.438!

Good luck!
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Based on past experience, the main printer in a university computer centre is operating properly 90% of the time. Suppose inspec
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Answer:

a) 38.74% probability that the main printer is operating properly for exactly 9 inspections.

b) Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

c) The expected number of inspections in which the main printer is operating properly is 9.

Step-by-step explanation:

For each inspection, there are only two possible outcomes. Either it is operating correctly, or it is not. The probability of the printer operating correctly for an inspection is independent of any other inspection, 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.

Based on past experience, the main printer in a university computer centre is operating properly 90% of the time.

This means that p = 0.9

Suppose inspections are made at 10 randomly selected times.

This means that n = 10

A) What is the probability that the main printer is operating properly for exactly 9 inspections.

This is P(X = 9). So

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

P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874

38.74% probability that the main printer is operating properly for exactly 9 inspections.

B) What is the probability that the main printer is operating properly for at least 3 inspections?

This is:

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

So

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

P(X = 0) = C_{10,0}.(0.9)^{0}.(0.1)^{10} \approx 0

P(X = 1) = C_{10,1}.(0.9)^{1}.(0.1)^{9} \approx 0

P(X = 2) = C_{10,2}.(0.9)^{2}.(0.1)^{8} \approx 0

Thus:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0 + 0 = 0

Then:

P(X \geq 3) = 1 - P(X < 3) = 1 - 0 = 1

Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

C) What is the expected number of inspections in which the main printer is operating properly?

The expected value for the binomial distribution is given by:

E(X) = np

In this question:

E(X) = 10(0.9) = 9

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What does B equal?

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So you know that he earns $15 per hour so we can write that as 15h, h = # of hours worked

We also know he has to earn AT LEAST $200, so putting all this together we form the inequality(=< means less than or equal to): 15h =< 200

so now we have to solve for h (number of hours worked):

15h =< 200

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So, the painter must work a minimum of 13 and 1/3 hours to earn at least $200.

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