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Tom [10]
3 years ago
12

Simplify the expression -6×5-4×5-10+3

Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

-57

Step-by-step explanation:

-6 x 5 - 4 x 5 - 10 + 3

Use PEMDAS

-30 - 20 - 10 + 3

-50 - 10 + 3

-60 + 3

-57

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2x-5y=10 in slope intercept form​
den301095 [7]

Answer:

y= 0.4x + 2

Step-by-step explanation:

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PLEASE HELP WITH MARK BRAINLIEST!<br> What is the value of g?
Sati [7]

Answer:

35

Step-by-step explanation:

this is a right triangle g+55 is a right triangle as shown by the red square in the corner.

90-55=35

so g=35

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I need the answer now in the comments
RUDIKE [14]
Hypotenuse leg would be right
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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
4 years ago
Write the phrase as an expression. Then simplify the expression. The product of 8 and a number y multiplied by 9
Neporo4naja [7]

Answer:

8y x 9  ⇒  72y

Step-by-step explanation:

5 0
4 years ago
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