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anyanavicka [17]
3 years ago
10

Step by step to this please

Mathematics
2 answers:
vladimir1956 [14]3 years ago
6 0

Answer:

7

Step-by-step explanation:

1) Substitute the values of t, v, and w into the expression:

(-2)^{2} + (1)(3)

2) Simplify the terms:

(-2)^{2} + (1)(3)\\=4 + 3\\= 7

Drupady [299]3 years ago
6 0

Answer:

7

Step-by-step explanation:

-2^2+3

4+3

7

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K & Christian each have a garden of the same size and shape cake grows flowers one and six her of her garden Christian grows
m_a_m_a [10]

Answer:

Christian will grow more flower as she utilizes all her garden area for the flower growth.

Step-by-step explanation:

Flower growth is difficult process as it involves great care by a gardener. Christian also grows flowers in her three garden which are available to her. She has same size flower which she grows in her garden.

5 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

7 0
3 years ago
You are looking at comparable houses with the same asking prices in neighboring towns. The mill rate in the first town is 20 mil
zalisa [80]
C) is the answer!!!!!!!
3 0
2 years ago
-2/3p - 4 &lt; 8<br><br>help for 15 points.​
wel

Answer:

p > -18

Step-by-step explanation:

-2/3p - 4 < 8

-2/3p < 12

p > 12 x -3/2

p > -18

3 0
2 years ago
Read 2 more answers
Can someone please help me??
babunello [35]
I'll help on 12. You know that 5*5 is 25. therefore what times 5 equals 20? 
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3 years ago
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