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I need help with this 14a-72r-c-34d?

Nina [5.8K]

Answer: This can not be simplified anymore. So the answer is 14a-72r-c-34d

Step-by-step explanation: Please give me a brainliest

3 0

3 years ago

Classify each number below as a rational number or an irrational number

klemol [59]

Π - Irrational

Square Root of 16 - Rational

Negative Square Root of 17- Irrational

-48.39 - Rational

-97.97 Repeating - Rational

8 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 :

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

(c) Get the marginal density of Y by integrating with respect to x instead:

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

(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.)

8 0

3 years ago

Nvm i alr know the answer lol

Airida [17]

Answer:

ok lol

Step-by-step explanation:

5 0

3 years ago

A piece if wire 40cm is bent to form a right angled triangle whose hypotenuse is 17cm long find the other two sides of triangle

hjlf

Answer: 150mm

Step-by-step explanation:

Take the two adjecent sides of the triangle to be x and y and X is the angle opposite the side, x.

From the trigonometric identities for a right-angled triangle, it is known that

x=170sin X and;

y=170cos X

Thus the perimeter of the triangle is given by

170sin X + 170cos X + 170 = 400

170(sin X + cos X) = 400 - 170

170(sin X + cos X) = 230

sin X + cos X = 230/170 = 1.3529

Squaring both sides hence,

sin^2(X) + 2sin X.cos X + cos^2(X) = 1.8304

Recall that sin^2(X) + cos^2(X) = 1 and 2sin X.cos X = sin 2X.

Thus, the former expression gives;

1 + sin 2X = 1.8304

sin 2X = 0.8304

2X = arcsin (0.8304)

2X = 56.14° or (180 - 56.14)° = 56.14° or 123.86° for X existing within a triangle.

X = 56.14/2 or 123.86/2

X = 28.07° or 61.93°

Thus the two adjacent sides are ;

x = 170 sin (28.07) = 80mm

y = 170 cos (28.07) = 150mm

Using the value of 61.93° as X would give the same values but interchanged for x and y.

I hope you find this useful.

5 0

3 years ago