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damaskus [11]
3 years ago
13

JL is a diameter of circle K. If tangents to circle K are constructed through points L and J, what relationship would exist betw

een the two tangents? Explain.
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
6 0
I. Let t be the line tangent at point J. We know that a tangent line at a  point on a circle, is perpendicular to the diameter comprising that certain point.
So t is perpendicular to JL

let l be the tangent line through L. Then l is perpendicular to JL
 
ii. So t and l are 2 different lines, both perpendicular to line JL.

2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.

Remark. Draw a picture to check the steps
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A service center receives an average of 0.6 customer complaints per hour. Management's goal is to receive fewer than five compla
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Answer:

48.68% probability that managment is unhappy with the number of complaints in the next eight hours.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

A service center receives an average of 0.6 customer complaints per hour.

This means that \mu = 0.6n, in which n is the number of hours.

Eight hours:

This means that n = 8, \mu = 0.6*8 = 4.8

Determine the probability that managment is unhappy with the number of complaints in the next eight hours.

They will be unhappy if they receive five or more complaints.

Either they receive less than five complaints, or they receive at least five. The sum of the probabilities of these events is 1. So

P(X < 5) + P(X \geq 5) = 1

We want P(X \geq 5).

Then

P(X \geq 1) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-4.6}*4.6^{0}}{(0)!} = 0.0101

P(X = 1) = \frac{e^{-4.6}*4.6^{1}}{(1)!} = 0.0462

P(X = 2) = \frac{e^{-4.6}*4.6^{2}}{(2)!} = 0.1063

P(X = 3) = \frac{e^{-4.6}*4.6^{3}}{(3)!} = 0.1631

P(X = 4) = \frac{e^{-4.6}*4.6^{4}}{(4)!} = 0.1875

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0101 + 0.0462 + 0.1063 + 0.1631 + 0.1875 = 0.5132

Finally

P(X \geq 1) = 1 - P(X < 5) = 1 - 0.5132 = 0.4868

48.68% probability that managment is unhappy with the number of complaints in the next eight hours.

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Answer/Step-by-step explanation

1. When adding real numbers with the same sign the sum will have the same sign as the numbers added.

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3. There are a couple of properties of addition:

The additive commutative property tells us that the order in which you add the numbers does not change the sum.

4. And the additive associative property tells us that it also that the order in which we group three or more numbers does not affect the sum.

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