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3 years ago

How much would it be if 2 1/4 is tripled

Mathematics

1 answer:

Tcecarenko [31]3 years ago

4 0

If you mean added together 3 times then it would be 6 3/4 and just in case this sounds rude I do not want it to be.

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Let F(x,y)=1+sqrt(4-y^2). . (a) Evaluate F(3,1). (b) Find and sketch the domain of F. (c) Find the range of F

charle [14.2K]

The goal of this function is to get a positive value under the square root sign so that the value would not be invalid. Evaluating F(3,1), the value is 1 + <span>sqrt(4-1^2) or equal to 1 + sqrt of 3.</span> In this case, the domain would be x equal to any numbers and the range equal to numbers from -2 ≤ y ≤ 2.

6 0

3 years ago

Read 2 more answers

Gabrielle paints flower pots to sell at the craft fair. She paints 2/5 of a flower pot teal, 1/4 of it yellow, and the rest of i

Stolb23 [73]

13/20
make the common denominator 20
2/5 = 8/20
1/4 = 5/20
8/20 + 5/20 = 13/20

8 0

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

LM = 9, NR = 16, SR = 8. Find the perimeter of triangle LRN.

sweet [91]

<h2>D. 50 units</h2>

Hi, I’m Lena and I will be answering your question to the best of my ability. If you have any further questions after my answer, do not hesitate to ask me! ツ

♡ Let’s look at the main important thing(s) that will lead you in the correct path.

♥ The perimeter is the outside, it is <em>around</em> the figure. When you see those inner triangles, do not pay attention to them! They are just there to make your overthink the question.

♡ Let's solve!

♥ To begin, let's take LM. It already gives you half the segment, so all you need to do is double the number. 9*2=18.

♥ Now, let's take a look at NR. They already give you the segment as a whole, so you do not need to worry about it!

♥ Look at SR. You do the same thing as you did to LM, double the number. 8*2=16.

Now, we add all the numbers to get our final answer.

18+16+16=50.

3 0

3 years ago

Find the graph of the inequality y>-(1/6)x+1.

Leya [2.2K]

Answer:

y > -x/6 + 1

Step-by-step explanation:

Hope this can help

4 0

2 years ago