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Vesnalui [34]
3 years ago
13

Use the Distance Formula to find the distance between (5, - 4) and (0, 8).

Mathematics
1 answer:
gogolik [260]3 years ago
8 0

Answer:

13 units

Step-by-step explanation:

Let d be the distance between given points.

\therefore \: d =  \sqrt{ {(5 - 0)}^{2} + ( - 4 - 8)^{2}  }  \\  \hspace{16 pt}=  \sqrt{ {(5)}^{2} + ( - 12)^{2}  }  \\  \hspace{16 pt}=  \sqrt{ 25 + 144 } \\  \hspace{16 pt}=  \sqrt{ 169 }  \\  \therefore \: d =13 \: units \\

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Solve this equation for x. (x+4) (x-2)=0
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Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found th
olchik [2.2K]

Answer: Our required probability is 0.515.

Step-by-step explanation:

Since we have given that

Probability that the respondents voted in favor of Scot Walker P(F) = 56%

Probability that the respondent voted in not favor of Scot Walker = 100-56 = P(F')=44%

Probability of those who did vote in favor had a college degree = 36% = P(C|F)

Probability of those who did vote against had a college degree = 44% = P(C|F')

So, we will use "Conditional Probability":

probability that he voted in favor of Scott Walker given that they had a college degree is given by

P(F|C)=\dfrac{0.56\times 0.36}{0.56\times 0.36+0.44\times 0.43}\\\\P(F|C)=0.515

Hence, our required probability is 0.515.

7 0
3 years ago
A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol
bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution 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.

25% of the incoming calls involve fax messages

This means that p = 0.25

25 incoming calls.

This means that n = 25

a. What is the probability that at most 4 of the calls involve a fax message?

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

In which

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4). Then

P(X \geq 4) = 1 - P(X < 4)

In which

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

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

P(X \leq 4) + P(X > 4) = 1

From a), P(X \leq 4) = 0.214)

Then

P(X > 4) = 1 - 0.214 = 0.786

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0
3 years ago
in comparing the galilean transformations with the lorentz transformations, what multiplicative factor appears in the lorentz tr
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In Lorentz transformations we have velocity based factors that is not present in Galilean ones.

<h3>What is lorentz transformations?</h3>

The relationship between two distinct coordinate frames that are moving relative to one another at a constant speed is known as a Lorentz transformation. Dutch physicist Hendrik Lorentz is credited with coining the name of the transformation. There are two frames of reference: Inertial Frames, which refer to motion that has a constant speed.

<h3>What are Galilean transformations?</h3>

The relationship between two distinct coordinate frames that are moving relative to one another at a constant speed is known as a Lorentz transformation. Dutch scientist Hendrik Lorentz is credited with coining the term transformation. There are two frames of reference: Inertial Frames, which relate to motion that has a constant speed.

Any and all rulers and other self-contained length standards, such as those you might use to set up a measurement frame, can be contracted in length based on velocity. This results in the Lorentz factor at the place in the LT.

Any and all clocks and other physically independent systems, such as those used to record events in a measurement frame, are subject to velocity time dilation. As a result, the LF is temporarily in the LT.

The fact that applying Einstein synchronisation individually for each measurement frame turns out to be natural. Consequently, the position-dependent synchronisation offset between the clocks of one measurement frame and those of the other is represented by the enigmatic second term in the LT for the time.

To learn more about transformations click the following link :-

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