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pickupchik [31]
3 years ago
12

Need help like now please

Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

-46 -35 -10 y -7 7 8   13 35  x

Step-by-step explanation:

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Please help with this question!!
barxatty [35]

Answer:

  -29/20

Step-by-step explanation:

The cosecant function is the inverse of the sine function. Both sine and cosecant are odd functions, meaning csc(-x) = -csc(x) = -1/sin(x).

  csc(-θ) = -1/sin(θ) = -1/(20/29) = -29/20

6 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
How many numbers of the set {1, 2, 3, 4, . . . , 297, 298, 299, 300} are multiples of 10 but not multiples of 15?
slega [8]

Answer:

I get 20

Step-by-step explanation:

While there probably is some logical way to go about this I just did it in excel

screenshot below

5 0
3 years ago
Read 2 more answers
The sound intensity of a typical vacuum cleaner is 80 dB (decibels). Sound intensity of 160 dB will burst an eardrum. How many t
NemiM [27]

Answer:

Ten thousand times as loud.

5 0
3 years ago
Read 2 more answers
The cosine of an angle is .28. Find the sine of the angle.
musickatia [10]
The ARC cosine of .28 = 73.74 degrees
the sine of 73.74 degrees = .96


5 0
3 years ago
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