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SVEN [57.7K]
2 years ago
11

Hi so this question is super confusing! I can't even figure it out

Mathematics
1 answer:
frutty [35]2 years ago
6 0

Answer:

i think it's n/2 may be even and 3n+1 is odd. They succumb to 3n if that helps.

Step-by-step explanation:

hope it helps, have a good day/night

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Six hundred seven and four hundred nine one-thousandths. 
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WILL GIVE BRAINLIEST IF YOU GIVE AN EXPLANATION. In the packaging part of a warehouse, boxes of blue cups must be moved from the
Ugo [173]

Answer:

components: ⟨15, 75⟩, magnitude: about 76

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3 ice cream cones cost $8.25. At this rate , how much do 2 ice cream cones cost?
myrzilka [38]

Answer:

5.5

Step-by-step explanation:

8.25/3=2.75

2.75 times 2 is 5.5

4 0
2 years ago
Find the probability​ p(e or​ f) if e and f are mutually​ exclusive, ​p(e)equals 0.25​, and ​p(f)equals 0.51.
Masteriza [31]

The definition of two events being mutually exclusive (or disjoint) only means that it is not possible for the two events to occur together. Given two events, E and F, they are mutually exclusive and also mean independent.

In this case, since events E and F are mutually exclusive, therefore the probability that either E or F will occur will simply be the sum of two events.

P (E or F) = P (E) + P (F)

P (E or F) = 0.25 + 0.51

P (E or F) = 0.76

 

Therefore this means that there is a 76% probability that either E or F will occur.

8 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
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