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3 years ago

15

The joint F.D.P of a bivariate VA (X, Y) is given

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

2

Step-by-step explanation:

ed

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Question 3 Solve for m. 4 over 3 m equals 4

Lorico [155]

Answer:

$m = 3$

Step-by-step explanation:

$\frac{4}{3} m = 4 \\ divide \: both \: sides \: by \: \frac{4}{3} \\ \\ m = 3$

3 0

3 years ago

Find three consecutive even numbers whose sum is 1674.

Norma-Jean [14]

Answer:

556, 558, and 560

Step-by-step explanation:

1674/3 = 558 (An even number, yay)

So 558 + 558 + 558 = 1674

Subtract two from the first one and add two to the last (still equal)

556 + 558 + 560

adding them equals 1674 so the answer is

556, 558, and 560

7 0

3 years ago

Which of the following best describes how the y values are changing over each interval? x y 1 20 2 40 3 80 4 160 5 320 They are

MAVERICK [17]

<h3>Answer</h3>

They are being multiplied by 2 each time.

4 0

2 years ago

Read 2 more answers

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

3 years ago

Which of the following statement does not describe people with self control<br>

PolarNik [594]

Answer:

None of them.

Step-by-step explanation:

People with self-control are not easily described.

4 0

3 years ago