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IgorLugansk [536]
3 years ago
9

Which of the following represents a relation that is NOT a function? A. X 7 -5 10 -7 Y 34 32 40 34 B. X -7 -5 -7 2 Y 34 32 40 34

C. X -7 -5 -1 2 Y 34 32 40 34 D. X -7 -5 7 10 Y 34 32 40 34
Mathematics
1 answer:
Nostrana [21]3 years ago
8 0
The x-value of -7 shows up more than once in relation ...
  B. X -7 -5 -7 2 Y 34 32 40 34

_____
When there is more than one y-value for an x-value, the relation is NOT a function.
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Step-by-step explanation:

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How many hours did they drive in all please explain why
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The Jacobys drove 19 1/2 hours from Monday to Thursday.

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Multiply and Simplify<br> (x-4)(x^2 - 5x - 6)
kirill [66]

Answer:

x^3 -9x^2 +14x +24

Step-by-step explanation:

(x-4)(x^2 - 5x - 6)

Multiply the x by everything in the second term

x * (x^2 - 5x - 6)

x^3 -5x^2 -6x

Multiply the -4 by everything in the second term

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-4x^2 +20x +24

Add everything together

I like to line them up vertically

x^3 -5x^2 -6x

      -4x^2 +20x +24

-------------------------------

x^3 -9x^2 +14x +24


4 0
3 years ago
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The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of si
OverLord2011 [107]

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!

a) The probability that exactly 3 arrivals occur during a particular hour is:

P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\

b) The probability that <em>at least</em> 3 people arrive during a particular hour is:

P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938

c) In this case, t=0.25, so we recalculate the parameter as:

\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

E(x)=\lambda=1.5

3 0
3 years ago
Solve (−7) ⋅ (−4).<br><br> a −28<br> b −11<br> c 28<br> d 11
german
-11 because first you add -4 and -7 because when you subtract negatives you add I hope I helped and have a good day
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