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sineoko [7]
4 years ago
15

Ayuda!!!! Broo I need help

Mathematics
1 answer:
Leviafan [203]4 years ago
3 0

Answer:

The second one Have a good day :) can I get brainliest please.

Step-by-step explanation:

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These two triangles are similar. Solve for X.
Paul [167]

Answer:

Hello I am not 100% on my answer but I would assume that X is 2.5 and Y is 6.5

Hope This Helps! please correct me if i am wrong

5 0
3 years ago
What is the estimate answer and the exact answer of 17/10 ÷2 4/5
OLEGan [10]
Both 17/10 and 2 4/5 should be rounded up in preparation for this estimation.

17/10 is close to 2 and 2 4/5 is close to 3.  Thus, your estimated answer should be
                     2/3, or approx. 0.6666....

Exact answer:    divide 17/10 by 14/5.  LCD is 10, so convert 14/5 to 28/10.

Now divide 17/10 by 28/10.  Answer:  17/28 = approx.   =  0.607 approx.

These two results are comparable:   0.6666....  and 0.6071 ....
8 0
4 years ago
Can somebody help me
saveliy_v [14]

Answer:

Y=5

You must first divide b (-10) by two and square it to know what you need to make a perfect square (25). You must then add or subtract the amount to make C (-13) into that number (add 38). Don't forget to do it on the other side too! (3 becomes 41) You then factor the left side. ((Y-5)^2=41) The result is your answer. (Y=5)

6 0
3 years ago
PLEASE HELP U WILL GET 15 POINTS A quadrilateral labeled figure A. The sides are labeled 4 centimeters, 2 centimeters, 3 centime
Eva8 [605]
The sides are 2, 1, 1.5, and 2.5 clockwise
3 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
4 years ago
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