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Harman [31]
3 years ago
15

What is the zero slope

Mathematics
1 answer:
Mars2501 [29]3 years ago
7 0
It's the first one because all the y's are the same number. That means that when you graph the chart you would get a horizontal line. Horizontal lines on graphs always have a slope of zero....unlike a verticals slope is undefined.
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photoshop1234 [79]
2.75! Good luck hope you do good
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Savannah wants to buy a new tablet that costs $140. She has a coupon expiring this week that is for 15% off any item. Next week,
tankabanditka [31]

Answer:

Its better to buy this week because it is cheaper with using coupon.

Step-by-step explanation:

140 x 0.85 = 119

125 > 119

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What is the value of m?
Gnesinka [82]
M=18

explication: both of the angles are 90 degrees, 90-33=57, 3(18)+3=57
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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
What are the x-intercepts of the equation?<br> <img src="https://tex.z-dn.net/?f=r%28x%29%20%3D%20x-1%2Fx%2B4" id="TexFormula1"
cestrela7 [59]

Answer:

<u>The only x-intercept is x = 1</u>

Step-by-step explanation:

The equation is:

r(x)=\frac{x-1}{x+4}

We can substitute y for r(x), to write in the notation:

r(x)=\frac{x-1}{x+4}\\y=\frac{x-1}{x+4}

To get y-intercept, we put x = 0

and

To get x-intercept, we put y =0

We want to find x-intercepts here, so we substitute 0 into y and solve for x. Shown below:

y=\frac{x-1}{x+4}\\0=\frac{x-1}{x+4}\\0(x+4)=x-1\\0=x-1\\x=1

<u>The only x-intercept is x = 1</u>

3 0
3 years ago
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