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mariarad [96]
3 years ago
13

9times 2/3 and than in simplest form

Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0
18/3 then it will be 6 
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I will mark brainliest please help ​
kirill115 [55]

Answer:

15 cups

Step-by-step explanation:

The formula to find cups from quarts C = Q * 4, where c is how many cups, and Q how many quarts. To find how many cups are in a quart, simply multiply the number of quarts by 5. In this case, there are 3 3/4 quarts of coffee, multiply that by 5, and you would get 15 cups of coffee!

5 0
3 years ago
Does anyone know how to do this I need help
baherus [9]

Answer:

NO

Step-by-step explanation:

In order to find the inverse you switch the x and y and solve for y.

x=(3/y )+3

x-3=3/y

y(x-3)=3

y= 3/x-3

no that doesn't equal h(x)

6 0
3 years ago
NEED HELP ASAP <br> WILL GIVE BRAIN
34kurt
Im really dumb i srry!
7 0
3 years ago
Find the midpoint (8, -6) (6,5)​
andrey2020 [161]
I think it’s (7, -0.5)
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
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