Identify the GCF of 12a^4b^3 + 8a^3b^2. 4a^3b^2 2a^3b 4a^2b 8a^3b^2

In a week, a man received a paycheck for $826.60. The withholding for taxes was $91.40. If he worked 34 hours in that week,

sleet_krkn [62]

Answer:

Step-by-step explanation:

3 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

Find the equation of the parabola with focus (5, 1) and directrix y = -1.

Mumz [18]

Check the picture below.

since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".

keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.

since the parabola is opening upwards, the value for "p" is positive, thus

$\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=5\\
k=0\\
p=1
\end{cases}\implies 4(1)(y-0)=(x-5)^2
\\\\\\
4y=(x-5)^2\implies y=\cfrac{1}{4}(x-5)^2$

8 0

3 years ago

What is the domain of y=4sin(x)?

stealth61 [152]

Answer:

( − ∞ , ∞ ) { x | x ∈ R }

Step-by-step explanation:

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

4 0

3 years ago

The strip below is cut into 9 equal bars. Shade 2/3 of this strip.

Anna35 [415]

Answer:

Shade 6 bars.

Step-by-step explanation:

You are asked to shade 2/3 of the strip. It is given that the strip has 9 equal parts. Multiply the number and fraction together:

$9 * \frac{2}{3} = \frac{9}{1} * \frac{2}{3} = \frac{9 * 2}{1 * 3} = \frac{18}{3} = 6$

You will shade 6 of the 9 bars given.

~

5 0

2 years ago