Please help me ahhh im doing 8th grade math with pythagorean theorem

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

yaroslaw [1]

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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16(-4/3)<br> Please help me!!!

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I got -64/3 for this problem.

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Using the segment addition postulate, which is true?

marusya05 [52]

Answer:

D

Step-by-step explanation:

For the segment addition postulate, two segments added together have to equal the third segment. In other words, if we added them together, the sum should be the first letter of the first line and the second letter of the second line.

BC + CD = BD

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The expression should be (2/h)x(k+8).

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A football coach set a goal of keeping opposing teams to an average of less than 200 yard. of total offense each game. After thr

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(701+y)/(3+1)<200

(701+y)/4<200

701+y<800

y<99

So the most yards they can allow in the next game is 98 yards

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