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lorasvet [3.4K]
3 years ago
10

Please help ILL MARK BRAINLIEST(17 points too)

Mathematics
1 answer:
jonny [76]3 years ago
3 0

Answer:

5, 10, 15, 20, 25, 30

6, 12, 18, 24, 30

Answer: 30

3, 6, 9, 12, 15, 18, 21, 24

24

Answer: 24

14, 28

4, 8, 12, 16, 20, 24, 28, 32

Answer: 28

Hoped it helped :)

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5 0
3 years ago
Which statement is true about polynomial 5s^6t^2 + 6st^9 - 8s^6t^2 - 6t^7 after it has been fully simplified?
Marina86 [1]

Answer:  3t²(2st⁷ - s⁶ - 2t⁵)

<u>Step-by-step explanation:</u>

5s⁶t² + 6st⁹ - 8s⁶t² - 6t⁷     <em>5s⁶t² and - 8s⁶t² are like terms which = -3s⁶t² when combined</em>

=          6st⁹ - 3s⁶t² - 6t⁷      <em>next, factor out the GCF of 3t²</em>

=      3t²(2st⁷ - s⁶ - 2t⁵)

5 0
4 years ago
Read 2 more answers
What's the gcf of 645 and 570
Gnoma [55]

The greatest common factor of 645 and 570 would be <u>35</u>.

4 0
3 years ago
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

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3 years ago
What is 46.2 x 10 negative 2=
Nutka1998 [239]

The Answer is -924.

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