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2 years ago

Need help with functions

Mathematics

1 answer:

Komok [63]2 years ago

4 0

Answer: ur right

Step-by-step explanation:

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Would you use grams or kilometers as the unit to measure the mass of a strawberry PLEASE ANSWER ALL YHE QUESTIONS!!! Get brainli

skelet666 [1.2K]

Answer:

You'd use grams because the strawberry is so small, and kilograms are emasuring bigger things.

Step-by-step explanation:

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2 years ago

Read 2 more answers

Need the answer to 2. 3. 7. 8. please help! Really confused on how to do this.

Oduvanchick [21]

2. $\overline{SV}||\overline{TU}$ (this is given in the first sentence)

3. $\angle XSV\cong\angle XUT$ and $\angle XVS\cong\angle XTU$ due to ...

7. ... the alternate interior angles theorem (which holds because $\overline{ST}$ and $\overline{TU}$ are parallel)

8. Because the above establishes that $\Delta SXT\cong\Delta UXV$ , which means $\overline{ST}||\overline{UV}$ , which in turn means $VUTS$ is a parallelogram.

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3 years ago

Determine weather the pair of sets is equal ,equivalent,both or neither

spin [16.1K]

Answer:

Equivalent

Step-by-step explanation:

The sets are not equal since computer science is not the same as algebra.

Equal means exactly the same

Equivalent means they have the same number of items in them

If they are equal, they are equivalent

7 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago

I need help with this question ASAP ! ASAP!!!!!! PLEASE!

irina1246 [14]

Answer:

One Triangle = 2.09 in²

Two Triangles = 4.18 in²

Rectangle = 17.48 in²

Total area of whole trapezoid = 21.66 in²

Step-by-step explanation:

Since it was not clarified which region is shaded we will just find the area of each individual part of the shape.

Let's start with the triangles.

1. To find the area of a triangle, the formula is $\frac{1}{2} b*h$ . It is given that the base of one triangle is equal to 1.1 in and the height is equal to 3.8 in., so in the equation, it would look like:

$\frac{1}{2} * 1.1 * 3.8 = 2.09$ in²

2. So now that we know one triangle is equal to 2.09 in², we now know that the other triangle is equal to the same area. To find the total of the two triangles you need to multiply the area by 2:

$2.09 *2=4.18$ in²

Moving on to the rectangle...

1. To find the area of the rectangle we need to use the formula base times height or b x h. It is given that the height is 3.8 in while the length is 4.6 in. So in the equation it would look like:

$4.6 * 3.8 = 17.48$ in²

Now to find the total area of all shapes combined...

1. To do this, we just need to add up all the areas we found, so...

17.48 + 4.18 = 21.66 in²

3 0

2 years ago