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spin [16.1K]
3 years ago
7

PLEASE HELP ILL GIVE BRAINLEST

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
6 0

Answer:

4177

Step-by-step explanation:

add what rhode island is and then do the same for washington. then add both

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While unpacking a box of canned vegetables, you notice that 2 out of 40 cans are dented. Write “2 out of 40” as a fraction, deci
nata0808 [166]
2/40=1/20
Percentage:5%
Decimal-0.05
8 0
3 years ago
Please answer I really need help with these just answer at least one.
aleksandrvk [35]
.50(150)+30=$105
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7 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

7 0
3 years ago
It math bad at it please help:);-)
abruzzese [7]

Answer:

It is -1/7 x (6p-1)

Step-by-step explanation:

Btw download photomath on your phone it helps alot

6 0
3 years ago
Three sets of a sum of a number and four are added to the sum of 7 times the same number and 13
Reika [66]
Let
x--------------> a number

we know that
<span>three sets of a sum of a number and four----------> 3(x+4)
t</span><span>he sum of 7 times the same number and 13--------> 7x+13

therefore

</span>(Three sets of a sum of a number and four) are added (to the sum of 7 times the same number and 13) ----------> [3(x+4)] + [7x+3]

[3(x+4)] + [7x+3]------> [3x+12] + [7x+3]=10x+15

the answer is 10x+15


5 0
3 years ago
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