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chubhunter [2.5K]
2 years ago
7

WILL MARK BRAINLIEST!

Mathematics
2 answers:
Mars2501 [29]2 years ago
7 0

Answer:

it's polynomial

Step-by-step explanation:

hope it helps u

amid [387]2 years ago
5 0

Answer:

multinomial

Step-by-step explanation:

polynomial are equal to multinomial

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a student found the slope of a line that passes through the points (1,14) and (3,4) to be 5. what mistake did she make?
SSSSS [86.1K]
M = (y2-y1)/(x2-x1)
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4 0
3 years ago
A box in the shape of a right rectangular prism has a length of 8.5 inches, a width of 4.5 inches, and a height of 3.75 inches,
jeyben [28]
Just multiply all of them because Volume = Length x Width x Height. So it's 8.5 x 4.5 x 3.75. That equals 143.4375. Cubic inches is just there after you multiply, so you don't have to convert anything. So it's 143.4375 cubic inches.
6 0
3 years ago
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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
Shalika bought a purse for $120. The tax rate is 9%.  What is the total amount Shalika paid? Enter your answer in the box.
ad-work [718]
130.80 is the answer you have to multiply 120 times 1.09
5 0
3 years ago
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Please help.<br> Is algebra.<br> PLEASE HELP NO LINKS OR FILES.<br> I don't want links.
Andru [333]

Answer:

b

Step-by-step explanation:

first, group the terms with the same power together and it would become (-9n⁵-4n⁵)+(n³-9n³)+(11n+6n). then, combine the terms in each group together which would get -13n⁵-8n³+17n as the answer

6 0
3 years ago
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