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2 years ago

Can someone help me find this arc, please.

Mathematics

1 answer:

alexira [117]2 years ago

6 0

Answer:

Arc AC = 90°

Step-by-step explanation:

Postulate:

The measure of an arc will ALWAYS be equal to the measure of the central angle of the circle.

The angle of arc AC is marked with a square.

That square mark means it is 90°.

So the arc would also measure 90°.

Cheers

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2x+8y=6, -5x-20y=-15 need work because I don't understand

Vlada [557]

Answer:

Infinitely many solutions.

Step-by-step explanation:

Step: Solve 2x + 8y = 6 for x:

2x + 8y = 6

2x + 8y + −8y =6 + −8y (Add -8y to both sides)

2x = −8y + 6

2x/2 = (-8y + 6)/2 (Divide both sides by 2)

x = −4y + 3

Step: Substitute −4y + 3 for x in −5x − 20y = −15:

−5x − 20y = −15

−5 (−4y + 3 ) −20y = −15

−15 = −15 (Simplify both sides of the equation)

−15 + 15 = −15 + 15 (Add 15 to both sides)

0 = 0

4 0

3 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

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3 years ago

If (x)=3x-1 and s(x)=2x+1, which expression is equivalent to r over s (6) ?

katrin [286]

Answer:

$\large\boxed{\dfrac{3(6)-1}{2(6)+1}}$

Step-by-step explanation:

$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\\\\\text{We have}\\\\f(x)=3x-1,\ g(x)=2x+1\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{3x-1}{2x+1}\\\\\left(\dfrac{f}{g}\right)(6)\to\text{put x = 6 to the equation of the function}\ \left(\dfrac{f}{g}\right)(x):\\\\\left(\dfrac{f}{g}\right)(6)=\dfrac{3(6)-1}{2(6)+1}$

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3 years ago

2^_/2^2=2^3. Please help me I really really really really really need this. THX

Ket [755]

The missing number is 6

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2 years ago

Can someone help me find this arc, please.​

Can someone help me find this arc, please.