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4 years ago

2.456times10^11 in standard form

Mathematics

2 answers:

Gelneren [198K]4 years ago

4 0

2.456 x 10^11= 245600000000

professor190 [17]4 years ago

4 0

nothin further can be done 2.456*10 11

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Leno4ka [110]

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3 years ago

si tienes una mesa de 25m de largo por 10 de ancho ¿ en cuanto varíaran cada una de sus dimensiones para que sin alterar su supe

Murrr4er [49]

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3 years ago

A survey was conducted with a large group of teenagers from 6 different states and it asked them for their favourite sport out o

UkoKoshka [18]

Answer:

H0: There is no association between state and sporting preference.

H1: There is an association between state and sporting preference

Step-by-step explanation:

The hypothesis to be tested for is whether the factor 'state' is associated with the factor 'sporting preference'.

The study is therefore about 'association' and whether the distributions of sporting preferences are identical across states. In scenario in this case is the test for association which is the most appropriate test.

Two factors are deemed to not be associated unless there is supporting evidence to suggest otherwise. Since the null hypothesis is the default belief, the correct pair of hypotheses are:

H0: There is no association between state and sporting preference.

H1: There is an association between state and sporting preference

3 0

3 years ago

Let M be the closed surface that consists of the hemisphere

ycow [4]

Since $M$ is closed, you can use the divergence theorem: The flux of $\vec E(x,y,z)$ across $M$ is

$\displaystyle\iint_{\partial M}\vec E\cdot\mathrm d\vec S=\iiint_M(\nabla\cdot\vec E)\,\mathrm dV=54\iiint_M\mathrm dV$

which is 54 times the volume of the hemisphere centered at (0, 0, 0) with radius 1, $\boxed{36\pi}$ .

Judging by the question content, you're supposed to find this value by computing the the integral of $\vec E$ across $M_1$ and $M_2$ .

Across $M_1$ :

Parameterize the hemisphere by

$\vec r(u,v)=(\cos u\sin v,\sin u\sin v,\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $M_1$ to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=(\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v)$

The flux of $\vec E$ across $M_1$ is

$\displaystyle\iint_{M_1}\vec E\cdot\mathrm d\vec S=18\int_0^{\pi/2}\int_0^{2\pi}(\cos u\sin v,\sin u\sin v,\cos v)\cdot\left(\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle18\int_0^{\pi/2}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv=36\pi$

Across $M_2$ :

Parameterize the disk by

$\vec s(u,v)=(u\cos v,u\sin v,0)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal to $M_2$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec u}{\partial v}=(0,0,-u)$

Then the flux across $M_2$ is

$\displaystyle\iint_{M_2}\vec E\cdot\mathrm d\vec S=18\int_0^{2\pi}\int_0^1(u\cos v,u\sin v,0)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv=0$

Then the total flux across $M$ is $36\pi$ , as expected.

4 0

4 years ago