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2 years ago

What is the value of x in the figure shown below?

Mathematics

1 answer:

Pie2 years ago

5 0

The value of X is 35 because the angle 35 measures the same as the angle X, hence their values are the same

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Find p(0), p(1) and p(2) for the polynomial P(t) = 2 + t + 2t²-t³

FinnZ [79.3K]

p(0)=2

p(1)=2+1+2-1=4

p(2)=2+2+8-8=4

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2 years ago

on every 3rd day Ivan goes to the gym. On every fifth day Gavin goes to the gym. What day will Ivan and Gavin will see each othe

Marat540 [252]

They will see each other on the fifteenth day

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3 years ago

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Make a table with the domain of {2,3,4,5,6} and draw a graph of the absolute value function y = 2|x-4| + 3.

Jet001 [13]

Answer:

<h2>In the attachment.</h2>

Step-by-step explanation:

$|a|=\left\{\begin{array}{ccc}a&for\ a\geq0\\-a&for\ a$

Put each value of x from the set {2, 3, 4, 5, 6}

to the equation y = 2|x - 4| + 3:

x = 2 → y = 2|2 - 4| + 3 = 2|-2| + 3 = 2(2) + 3 = 4 + 3 = 7 → (2, 7)

x = 3 → y = 2|3 - 4| + 3 = 2|-1| + 3 = 2(1) + 3 = 2 + 3 = 5 → (3, 5)

x = 4 → y = 2|4 - 4| + 3 = 2|0| + 3 = 2(0) + 3 = 0 + 3 = 3 → (4, 3)

x = 5 → y = 2|5 - 4| + 3 = 2|1| + 3 = 2(1) + 3 = 2 + 3 = 5 → (5, 5)

x = 6 → y = 2|6 - 4| + 3 = 2|2| + 3 = 2(2) + 3 = 4 + 3 = 7 → (6, 7)

Mark the points in the coordinates system.

The domain is only five numbers, therefore the graph of this function is only five points.

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2 years ago

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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

Sketch a line with a slope of -2 and another with a slope of -4

WINSTONCH [101]

Answer:

Step-by-step explanation:

Try and become familiar with a program like Desmos. It will do wonderful things for you.

I have graphed

Red: y = -2x + 5 (The 5 is the y intercept. For this question it could be anything.

Blue: y = - 4x + 5 Same note as above.

4 0

2 years ago