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2 years ago

Which of the following is a polynomial?

Mathematics

2 answers:

Zina [86]2 years ago

4 0

Answer:

A, B, C

Step-by-step explanation:

Assuming the choices are:

A. $(x-2)(x^{4}+3)$

B. $x^{2} -1$

C. A+ $x^{4} +16$

D. $\frac{1}{x} +2$

Put simply, a polynomial in this case will be a function that has at least one variable raised to a power greater than 1.

Choice A already has a variable raised to a power greater than 1 ( $x^{4}$ ), so no additional manipulation is needed.

Choice B also has a term that is already raised to a power greater than 1 ( $x^{2}$ ). It is the same for choice C ( $x^{4}$ ).

Choice D does not have an obvious variable raised to a power greater than 1. When the equation is manipulated it becomes $x^{-1} +2$ . Since -1 is less than 1, choice D is not a polynomial.

maxonik [38]2 years ago

3 0

Answer:

Option D

Step-by-step explanation:

$\\ \sf\longmapsto \dfrac{1}{x}+2$

Simplify

$\\ \sf\longmapsto x^{-1}+2$

Degree is less than 1 .It's not a polynomial

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An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

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3 years ago

Plz help I need to pass math!!!!!

serg [7]

You would divide 8 3/4 by 50 to find out how many cups you would get for just one cupcake. So you would do 8 3/4 DIVIDED BY 50= 0.175 which means there will be 0.175 cups of sugar per cupcake.

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3 years ago

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Will the shape shrink or enlarge?

aleksklad [387]

The shape will shrink

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3 years ago

Graph the solution set of: 2x - 4 ≤ 8 and x + 5 ≥ 7

Alex17521 [72]

Answer:

x ≤ 6 and x ≥ 2

Step-by-step explanation:

For now, we will start with doing each problem at a time. Here is your equation:

2x - 4 ≤ 8

First, you want to get the variable by itself. So, you add 4 to both sides. It will look like this:

2x - 4 ≤ 8

+ 4 + 4

The four being added on the left side cancels out, and you add 4 to 8. Now, it should look like this:

2x ≤ 12

Next, you want the x by itself. So, you would divide both sides by 2.

2x ≤ 12

/2 /2

2 divided by 2 cancels out, and 12 divided by 2 equals 6. Now, you have a final answer of:

x ≤ 6

But, you now have to do the other one!

Here is what you start off with:

x + 5 ≥ 7

First, you want the variable side by itself. So, you subtract 5 from both sides.

x + 5 ≥ 7

- 5 -5

Now, you have this:

x ≥ 2

Because the variable is already by itself, you don't need to do any more division and this is you final answer. Now put both answers you got together which equals:

x ≤ 6 and x ≥ 2

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3 years ago

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Shape 1 and shape 2 are plotted on a coordinate plane. Which statement about the shapes is true? A. Shape 1 and shape 2 are not

denpristay [2]

Answer:

i think its D.

A reflection, a rotation, and a translation will prove that shape 2 is congruent to shape 1.

Step-by-step explanation:

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