Mathematics, 11.06.2021 21:30 carlybeavers50

Based on past experience, the main printer in a university computer centre is operating properly 90% of the time. Suppose inspections are made at 10 randomly selected times. A) What is the 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? C) What is the expected number of inspections in which the main printer is operating properly?

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Answer from: josephvcarter

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)

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

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

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

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

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

Answer from: Quest

4pounds of live bait and 1 pound of natural bait

Answer from: Quest

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in $\delta bea$ and $\delta dea$ ;

$ba=da$ , as these are radii of our given circle.

$ea=ea$

therefore, $\delta bea\cong \delta dea$ by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

$3x-24=x+10$

$3x-x=24+10$

$2x=34$

$x=\frac{34}{2}$

$x=17$

now let us substitute x=17 in the expression for the length of be.

$\text{length of be}=3x-24$

$\text{length of be}=3\cdot 17-24$

$\text{length of be}=51-24$

$\text{length of be}=27$

therefore, length of be will be 27 units and option c is the correct choice.

In circle a, ∠bae ≅ ∠dae. what is the length of be? 14 units 17 units 27 units 34 units

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