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lesya [120]
3 years ago
14

Help with this please

Mathematics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

(6,5)

Step-by-step explanation

you have to use the grid to find the distance:)

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Examine the diagram.
iris [78.8K]

Answer:

5

Step-by-step explanation:

When two lines are crossed by another line (the transversal), the pair of angles on  the opposite and inner side of each of those two lines are called Alternate Interior Angles.  Therefore 5 is the alternate interior angle to 4.

5 0
3 years ago
Daria discovers a pattern in her homework assignment. X shows the missing digit in the pattern. The table shows 12, 84, 156, 228
MrRissso [65]
Your answer would be C 300 your welcome
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
A worker at a computer factory can assemble 8 computers per hour.
enyata [817]
25 hours, 8 computers in 1 hour, take 200 computers and divide by 8 to find hours
7 0
2 years ago
Is 3(a+b) = to 3a+3b​
Tanzania [10]

Answer:

Yes. Distributive property

Step-by-step explanation:

3(a + b)

Opening bracket

= 3(a) + 3(b)

= 3a + 3b

8 0
3 years ago
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