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1 day ago

Suppose the coordinate of a is 0, ar = 5, and at =7. What are the possible coordinates of the midpoint of rt ?

Mathematics

1 answer:

Zina [3.9K]1 day ago

4 0

Here, 'a' relates to 0.

There are two scenarios for 'r' and 't'.

Scenario 1.

Both are positioned on the same side to the right of 'a'.

In this case, 'r' would equal 5, and 't' would equal 7.

The midpoint between 'r' and 't' is $\frac{5+7}{2} =6$ .

Scenario 2.

If both are found to the left of 'a'.

Then 'r' would equal -5, while 't' would equal -7.

The midpoint is $\frac{-7-5}{2} =-6$ .

Scenario 3.

If 'r' is right of 'a' and 't' is left of 'a'.

Thus 'r' equals 5 and 't' equals -7.

The midpoint is $\frac{-7+5}{2}=-1$ .

Scenario 4.

If 'r' is left of 'a' while 't' is right of 'a'.

In this case, 'r' corresponds to -5 and 't' corresponds to 7.

The midpoint is $\frac{-5+7}{2}=1$ .

The potential midpoint coordinates for 'rt' are 6, -6, 1, and -1.

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Zina [3914]

Answer:

The chance that there is one or more defective integrated circuits is 33.10%.

Detailed solution:

Each integrated circuit can be either defective or not, which provides only two possible outcomes. As a result, this scenario is well-suited to be analyzed using the binomial probability distribution.

About the binomial distribution

This distribution calculates the likelihood of exactly x successes in n repeated trials where each trial has two possible results.

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

Here, $C_{n,x}$ represents the count of combinations of x items selected from n elements, described by the formula:

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

And $\pi$ is the probability of the event X occurring.

Applying it to the problem

The product contains 40 integrated circuits, so $n = 40$ .

The probability that an individual integrated circuit is defective is 0.01, so $\pi = 0.01$ .

Finding the probability of at least one defective circuit

There are two cases: either at least one integrated circuit is defective (probability $P(X > 0)$ ) or none are defective (probability $P(X = 0)$ ). Since probabilities sum to 1, we want to determine $P(X>0)$ .

$P(X > 0) + P(X = 0) = 1$

$P(X > 0) = 1 - P(X = 0)$

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

$P(X = 0) = C_{40,0}.(0.01)^{0}.(0.99)^{40} = 0.6690$

$P(X > 0) = 1 - P(X = 0) = 1 - 0.6690 = 0.3310$

Hence, the probability of having one or more defective integrated circuits is 33.10%.

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Detailed answer:

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Answer:

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