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Basile [38]
3 years ago
5

Rebecca's dog runs with an initial speed of 1.5m/s on a waxed floor. It slides to a stop within an acceleration of -.25 m/s/s. H

ow long does it take the dog to stop
Mathematics
1 answer:
yanalaym [24]3 years ago
8 0
The correct  answer would be : 1.75
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On a coordinate plane, a parabola opens down. It has an x-intercept at (negative 5, 0), a vertex at (negative 1, 16), a y-interc
tigry1 [53]

Answer:

I have no idea sorry maybe ome body

Step-by-step explanation:

but first u divide and that's all i know

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3 years ago
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Photo A, B, C are all square photos. The area of Photo C is the same as that of a rectangular photo whose length is the side len
motikmotik

Answer:

A

Step-by-step explanation:

because all off the porf on it was about a

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3 years ago
You draw one card from a standard 52-card deck. In how many ways could the card be face card or a six?
Nady [450]
16/25 because there are 3 face cards which have 4 different suits and 4 different suits for the number 6
8 0
4 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
Whats 1/18 as a decimal can someone please help me
sergij07 [2.7K]

1/18 as a decimal is 0.055555555555556 because you divide one by eighteen and get that number.

Hope this helped!

6 0
3 years ago
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