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3 years ago

∠2 and ∠6 can be classified as:

Mathematics

2 answers:

aleksklad [387]3 years ago

8 0

Answer:

Corresponding angles

Step-by-step explanation:

Corresponding angles are angles that are on the same side of the transversal ( intersecting lines ) which are congruent ( equal to each other )

Angles 2 and 6 are congruent and are on the same side of the transversal

Lilit [14]3 years ago

6 0

Answer:

A. corresponding angles

Step-by-step explanation:

2 and 6 are on the same side of the transversal and in the corresponding position ( same location) with respect to the parallel lines

This makes the corresponding angles

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Roberto has $200 in spending money. He wants to buy some video games that cost $25.50 each. Write and solve an inequality to fin

Veseljchak [2.6K]

Answer:

f(x) = 25.50x Roberto can buy 7 video games

Step-by-step explanation:

Each game is $25.50 which in math means it has an "x"

So if x=1 then 25.50x says that 1 video game is $25.50

Set up the equation:

200 = 25.50x (divide both sides by 25.50 to isolate the x)

200/25.50 = 7.84

x = 7.84

You can't have 0.84 of a video game, so Roberto can only buy 7 of them.

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3 years ago

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jolli1 [7]

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9 times the sum of 8 and y

Naya [18.7K]

Answer:

9(8+y)

Step-by-step explanation:

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3 years ago

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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

If ab = 8 and a^2+b^2=16, then what is the value of (a+b)^2?

aksik [14]

$ab = 8$ & $a^{2} + b^{2} = 16$ so , $( a+b )^{2} = 32$ .

<u>Step-by-step explanation:</u>

Here we have , ab= 8 & a^2+b^2=16 i.e. $ab = 8$ and $a^{2} + b^{2} = 16$ .

We need to find value of (a+b)^2 i.e. $(a+b)^{2}$ :

It's and identity and we know that $( a+b )^{2} = a^{2} +b^{2} +2ab$

⇒ $( a+b )^{2} = a^{2} +b^{2} +2ab$

⇒ $( a+b )^{2} = (a^{2} +b^{2}) +2(ab)$

⇒ $( a+b )^{2} = (16) +2(8)$

⇒ $( a+b )^{2} = (16) +(16)$

⇒ $( a+b )^{2} = 32$

∴ $ab = 8$ & $a^{2} + b^{2} = 16$ so , $( a+b )^{2} = 32$ .

6 0

3 years ago