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Ksju [112]
3 years ago
8

Solve for x. Enter your answer in the box X=

Mathematics
1 answer:
prohojiy [21]3 years ago
3 0

Answer:

72°

Step-by-step explanation:

\frac{360}{5}  = 72

I know that this is not the correct formula for this type of problem, but since <em>x</em><em> </em>is an acute angle, I found this answer a possibility.

I hope this is correct, and as always, I am joyous to assist anyone at any time.

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52 /? = 20/100<br><br>what is the missing denominator??? ​
mezya [45]

Answer:

1140

Step-by-step explanation:

52/x=20/100

52(20)=100(x)

52(20) = 1040

100(x) = 100

1040+100= 1140

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2 years ago
Last two answer for both or no brainleist
professor190 [17]

Answer:

IM NOT SURE BUT DO U GO TO GEM PREP

Step-by-step explanation:

6 0
3 years ago
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What is the median of the data set?<br><br> 12, 25, 16, 9, 5, 22, 27, 20
spin [16.1K]
The answer is 7. 9 + 5 / 2 equals to 7. 9 and 5 and the two middle numbers.
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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
What is tan K? help pleasee
alina1380 [7]

8/15

Hope this helps!

5 0
3 years ago
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