A symmetrical binomial distribution with large sample size can be approximated by a _________

Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

$f'(x) = k*((x^2)'*ln(1/x) + x^2*(ln(1/x)')) = k*(2x\,ln(1/x)+x^2*(\frac{1}{1/x}*(-\frac{1}{x^2})))\\= k * (2x \, ln(1/x)-x)$

We need to equalize f' to 0

k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
1/x = e^(1/2)
x = 1/e^(1/2) = 1/√e ≅ 0.607

Thus, the value of x that gives the maximum transmission is 1/√e.

7 0

2 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

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

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

1 year ago

use the drop-down menus to describe the key aspects of the function f(x) = –x2 – 2x – 1. the vertex is the . the function is inc

timurjin [86]

Answer:

The vertex of the parabola is the maximum value, i.e.,(-1,0). The function is increasing x<-1. the function is decreasing x>-1. the domain of the function is all real numbers. the range of the function is all real numbers less than or equal to 0.

Step-by-step explanation:

The given function is

$f(x)=-x^2-2x-1$

$f(x)=-[x^2+2x+1]$

$f(x)=-(x+1)^2$ ....(1)

The general vertex form of the parabola is

$f(x)=a(x-h)^2+k$ .....(2)

Where, (h,k) is vertex and a is stretch factor.

On comparing (1) and (2), we get

$a=-1$

$h=-1$

$k=0$

The vertex of the parabola is (-1,0). Since a=-1<1 so it is a downward parabola.

The axis of symmetry is x=-1. So, before -1 the function is increasing and after -1 the function is decreasing.

The vertex of a downward parabola is the point of maxima. So, the rang of the function can not exceed 0.

Therefore the vertex of the parabola is the maximum value, i.e.,(-1,0). The function is increasing x<-1. the function is decreasing x>-1. the domain of the function is all real numbers. the range of the function is all real numbers less than or equal to 0.

8 0

2 years ago

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A symmetrical binomial distribution with large sample size can be approximated by a __________