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WARRIOR [948]
3 years ago
7

Question 1 (1 point)

Mathematics
1 answer:
Leni [432]3 years ago
6 0
<h3>▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪</h3><h3 /><h3><u>Question</u> 1 </h3>

Money spent semi - annually (in 6 months) = $ 150

So, money spent annually (12 months) is equal to :

  • 150 \times 2

  • \$ \: 300
<h3><u>Question</u> 2</h3>

Her budget annually (in 12 months) = $ 900

So, budget monthly is equal to :

  • \dfrac{900}{12}

  • \$ \: 75
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What are you trying to ask can you post a pic with the exact question?
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3 years ago
Xander spends most of his time with his 10 closest friends. he has known 4 of his 10 friends since kindergarten. if he is going
Mademuasel [1]

Probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third not is 1/10

Total friends = 10

Friends from kindergarten = 4

Probability is the chance that a given event will occur. Probability of an event lies within 0 to 1

P(E) = Favourable outcomes / Total outcomes

Probability of getting 1st friend from kindergarten = 4/10

Probability of getting 2nd friend from kindergarten = 3/9

Probability of getting 3rd friend not from kindergarten = 6/8

Since all these probabilities are independent, We can use Multiplicative identity.  Thus,

Required probability is 4/10 * 3/9 * 6/8

= 1/10

Thus, Probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third not is 1/10

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brainly.com/question/7514534

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5 0
2 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

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The answer seems to be D!!
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