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Readme [11.4K]
2 years ago
9

In the diagram below, points A and B are 3 inches apart. Try to find two points that are 1 inch from point A

Mathematics
1 answer:
____ [38]2 years ago
8 0

Answer:

It is impossible to find two points that are both 1 inch and 1.5 inches from the point A simultaneously as shown in the following proof

From segment addition postulate given a point C that is between the points of a segment AB, we have;

AC + CB = AB

Therefore, where AC = 1 inch and CB = 1.5 inches, we have;

AC + CB = 1 inch + 1.5 inches = 2.5 inches ≠ AB = 3 inches

However it is possible to find two points in which one of the points is 1 inch from the point A while the other point is 1.5 inches from the point B

Step-by-step explanation:

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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm
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(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, P\left(\frac{E}{A}\right)=0.02

Now, from the conditional probability,

P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}

\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)

\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

P\left(\frac{E}{A}\right)=0.40

and \Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)

This is the probability of disregarding the partially defective parts by inspection machine.

P\left(\frac{E}{A}\right)=0.98

and \Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

=P(E\cap C)

= 0.0686 [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)

\Rightarrow P(E)=0.018+0.012+0.0686

\Rightarrow P(E)=0.0986

So, the total probability that the part produced get shipped

=1-P(E)=1-0.0986=0.9014

The probability that the part is good (either defect free or slightly defective)

=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)

=(0.9-0.018)+(0.03-0.012)

=0.9

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}

=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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