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HACTEHA [7]
3 years ago
10

What is the answer to numbers 2,3,4,5 and 6

Mathematics
2 answers:
sattari [20]3 years ago
7 0
Lets Start with Number 2
10 1/8 + ( 3 5/8 + 2 7/8)
Lets start with the parenthesis
10 1/8 + ( 3 5/8 + 2 7/8)
                     5 ( 12/8) 
                       6 ( 4/8) 
Now we add  10 1/8 + 6 4/8
                                 16 5/8


Gekata [30.6K]3 years ago
5 0
All u have to do is PEMDAS
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830.76/30.1=27.6 miles
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3 years ago
Which one should I choose
shutvik [7]

Answer:

-32

Step-by-step explanation:

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Solve 2x^2=11x+6 by factoring
Ede4ka [16]

Answer:

(x-6) (2x+1)

Step-by-step explanation:

1) move everything over to the left side, so subtract 11x and 6.

2x^2 -11x - 6 = 0

2)multiply 2(a term) with -6 (c term)

x^2 -11x -12

3) factor and find what multiples to -12 and adds up to -11. in this case its -12 and positive 1

(x-12) (x+1)

4) divide -12 and 1 by the original a term (2)

(x-6) (x+1/2)

5) move the denominator of 2 over to the x.

(x-6) (2x+1)

5 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
PLEASE HELP QUICK
expeople1 [14]

Answer:

The Answer Is A

Step-by-step explanation:

A=12.3

8 0
3 years ago
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