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3 years ago

Need help on this !!!!!!

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

I think its C because you just have to follow the expressions pattern i hope this helps good luck!!

adelina 88 [10]3 years ago

6 0

The answer is C. This is because there is a continuous pattern throughout the whole table.

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If x+9.8=14.7, what is the value of 8(x-3.7?

yulyashka [42]

First find X by subtracting 9.8 from 14.7
X should equal 4.9
than fill in the equation for x
8(4.9-3.7)
you then subtract in the parenthasis
getting the answer of 1.2
finally multiply 1.2x8
Your final answer is 9.6

3 0

3 years ago

3. [6x - 2y=2.<br> (2 + 6x=y

Hunter-Best [27]

Answer:

3. [6x - 2y=2.

(2 + 6x=y

Step-by-step explanation:

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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

A 12" pizza costs $14. What is the<br> price per square inch?

timama [110]

Answer:

0.857$ (Its really big in answer but This is the average result.

The REAL answer

0.857142857$

4 0

2 years ago

Subtract.<br> (3r²s+ 4rs+2-16r²s²) - (7sr²+10-13r²s²)<br><br> Simplify answer. Do not factor

pentagon [3]

[- 3r²s² - 4sr² + 4rs - 8] is the resultant answer when we subtract the linear equation (7sr²+10-13r²s²) from (3r²s+ 4rs+2-16r²s²).

<h3>What are linear equations?</h3>

A linear equation in one variable is a mathematical expression for an equation with only one variable.
While any two real integers can represent A and B, the ambiguous variable x has just one possible value.
The formula is Ax + B = 0.
9x + 78 = 18 is an example of a linear equation with just one variable.

So, we have:

(3r²s+ 4rs+2-16r²s²) - (7sr²+10-13r²s²)

Now, calculate as follows:

= (3r²s+ 4rs+2-16r²s²) - (7sr²+10-13r²s²)
= 3r²s+ 4rs+2-16r²s² - 7sr²- 10 + 13r²s²
= 13r²s² - 16r²s² - 7sr² + 3r²s + 4rs - 10 + 2
= - 3r²s² - 4sr² + 4rs - 8

Therefore, [- 3r²s² - 4sr² + 4rs - 8] is the resultant answer when we subtract the linear equation (7sr²+10-13r²s²) from (3r²s+ 4rs+2-16r²s²).

Learn more about linear equations:

brainly.com/question/11897796

#SPJ13

7 0

1 year ago