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denis23 [38]
3 years ago
5

Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = x

yezi + xy2z3j − yezk, S is the surface of the box bounded by the coordinate plane and the planes x = 3, y = 6, and z = 1.
Mathematics
1 answer:
kykrilka [37]3 years ago
8 0

\vec F has divergence

\nabla\cdot\vec F=\dfrac{\partial(xye^z)}{\partial x}+\dfrac{\partial(xy^2z^3)}{\partial y}-\dfrac{\partial(ye^z)}{\partial z}=ye^z+2xyz^3-ye^z=2xyz^3

By the divergence theorem, the integral of \vec F across S is equal to the integral of \nabla\cdot\vec F over the interior of S:

\displaystyle\iiint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^6\int_0^32xyz^3\,\mathrm dx\,\mathrm dy\,\mathrm dz=\boxed{\frac{81}2}

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Step-by-step explanation:

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Conference A and conference B each have 10 football teams. The table shows the number of football games that each team in the tw
babymother [125]

Answer with explanation:

1.Mean of a Data set

   =\frac{\text{Sum of all the observation}}{\text{total number of observation}}

=\frac{10+9+11+8+10+12+2+7+8+0+12+10+8+11+3+12+6+7+4+3}{20}\\\\=\frac{153}{20}\\\\=7.65

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3 0
3 years ago
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Help, WILL GIVE BRAINLIEST
Harlamova29_29 [7]

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Step-by-step explanation:

Not 100% sure this is what you needed but this is my try.

6 0
2 years ago
Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
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