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WARRIOR [948]
4 years ago
6

Help please!!! I’ll mark u as brainliest

Mathematics
1 answer:
sergeinik [125]4 years ago
7 0

Answer:

Answer is below

Step-by-step explanation:

Yes, the graph would still be a function, but a different function than if the values were all different. This would cause two of the values to be the same and have no slope on the line. Before, the function was quadratic, but now it would be quadratic until the two values are equal.

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Simplify the expression
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5x^4 - 3x^3 + 6x - (3x^3 + 11x^2 - 8x) =
5x^4 - 3x^3 + 6x - 3x^3 - 11x^2 + 8x =
5x^4 -6x^3 - 11x^2 + 14x <== yep, I got the same thing 
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4 years ago
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ABC is a straight line.The length of AB is four times the length of BC.AC=75
bearhunter [10]

Answer:

AB = 60

Step-by-step explanation:

Given AC = 75 and AB = 4BC, then

AB + BC = AC, that is

4BC + BC = 75

5BC = 75 ( divide both sides by 5 )

BC = 15

Thus

AB = 4BC = 4 × 15 = 60

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3 years ago
I need the answer to be correct
VladimirAG [237]

Answer:

The GCF is x²

Step-by-step explanation:

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List some words that might be used in a real-world example that can be expressed with an addition problem. (For example: increas
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Answer:

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3 years ago
A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

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 interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
4 years ago
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