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2 years ago

-3 (5 g +1) + 8 g

Mathematics

1 answer:

Komok [63]2 years ago

8 0

Answer:

-7g-3

Step-by-step explanation:

in order to evaluate for this mathematical expression, we need to consider the negative sign carefully and then open the bracket then evaluate.

<u>solutuion</u>

-3 (5 g +1) + 8 g

-15g -3 + 8g

-15g +8g -3

-7g-3

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Fourteen less than the quotient of a number and 5

Ede4ka [16]

Answer:

Step-by-step explanation:

14-(a/5)

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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:

$Var(X^2)= 3-(1)^2 =2$

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3 years ago

Sorry again but i'm having trouble on this one

Lisa [10]

Answer:

m∠PNO = 60°

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Step-by-step explanation:

∡NPO is 87° because it's a vertical angle with the 87° angle

3) ∡PNO is supplementary to the 120° angle (they must add to 180°)

4) m∠O = 180 - (60 + 87) = 33

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2 years ago

What is 55x690-20+438x129=? <br> I will give brainliest.

frosja888 [35]

Answer:

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2 years ago

Read 2 more answers

Intercepts:

Tema [17]

An x-intercept is the point where the function passes the x axis at y=0.
The y-intercept is the point where the function crosses the y axis at x=0

1. It is an x-intercept, so y = 0. The ordered pair would be (-6, 0)
2. It is a y-intercept, so x = 0. The ordered pair would be (0, -2.3)
3. It is a y-intercept, so x=0. The ordered pair would be (0, 3/4)

To find the x-intercept, set y = 0. To find the y-intercept, set x=0

4. y-intercept: y = 3(0) -9. y = -9 The y-intercept is at (0, -9)
x-intercept: 0 = 3x -9. 9 = 3x. x = 3. The x-intercept is at (3, 0)

5. x intercept: 0 = 5x +10. -10 = 5x. x = -2. The x-intercept is at (-2, 0)
y-intercept: y = 5(0) + 10. y = 10. The y-intercept is at (0, 10)

If you look at finding the y-intercepts in the two problems above, you may see there is a pattern forming. The y-intercept is the number that your adding or subtracting that is located after the x (ex. y = 4x - 2 - the y-intercept would be -2)

9. First, find the x and y-intercepts. The y intercept is -3 You’d graph that at (0, -3). 0 = -1/2x - 3. -1/2x = 3. x = -6 so the x-intercept is at (-6, 0). If you only need to graph 2 points, then you can graph just those two points and draw a line between them.

To graph y=-1/2x + 3, start at the y-intercept and use the slope (-1/2) to find other points. Because your slope is -1/2, you’d go down 1 unit and then to the right 2 units. That would be your next point. If you wanted your line to go further up, go up one unit and then to the left 2 units. That would be your next point.

I am not sure what you need to do on 11 and 12

I think you should try 7, 8 and 10 on your own and let me know if you have any questions on them or if you are stuck on anything.

3 0

3 years ago