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frez [133]
3 years ago
13

Equivalent ratios to 16/12

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

16/12=8/6=4/3

Step-by-step explanation:

Ok so somethings you can do to get equivalent answers is multiply each of the numbers in the fraction by 2. You can also divide by two if they are even.

Ex: 4/8=2/4=1/2- Dividing

Ex: 6/10=12/20=24/40- Multiplying

MrRa [10]3 years ago
5 0

Answer:

4/3

Step-by-step explanation:

16/12 divided by 4/4 = 4/3

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Simplify the expression
NeTakaya

First, solve the parenthesis

Remember to follow PEMDAS, and the left-> right rule

12/3 = 4

4 x 2 = 8

8 + 3 = 11

11 - 7 = 4

Square the remaining number

4² = 4 x 4 = 16

16 is your answer

hope this helps

7 0
3 years ago
Can someone please help me with this question?
Lina20 [59]

Answer:

402.12

Step-by-step explanation:

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3 years ago
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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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Answer:

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the answer will be 1255.725

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