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3 years ago

Micahdisu pls help. Look at both pics pls

Mathematics

1 answer:

AlexFokin [52]3 years ago

5 0

The best way to compare the two would be median and Ian generally gets more.

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-2x-7≥41 in algaebra for

dexar [7]

Answer:

They are equal. x= -24

Step-by-step explanation:

-2x - 7 ≥ 41

First you isolate the variable

-2x -7 ≥ 41

+7 +7

-2x≥48

then divide each side by -2

x≥-24

Then go back and plug in

-2(-24)-7=

48-7=41

Therefore

41=41

4 0

3 years ago

Which of the following about the normal distribution is not true?

Lostsunrise [7]

Answer:

Normal distribution is a Continuous distribution.

Step-by-step explanation:

Normal Distribution:

The following information is true about normal distribution.

Theoretically, the mean, median, and mode are the same.

Its parameters are the mean, mu, and standard deviation, sigma.

About two thirds of the observations fall within plus or minus 1 standard deviation from the mean.

But normal distribution is not a discrete distribution. It is a continuous distribution.

The normal distribution is a continuous probability distribution. This has several implications for probability.
The probability that a normal random variable X equals any particular value is 0.

5 0

3 years ago

Which is greater 7/8 or 5/10

Sloan [31]

Find a common denominator 7/8 = 35/40 and 5/10 = 20/40. Also 7/8 is almost a whole and 5/10 is half. 7/8 is greater.

7 0

3 years ago

Read 2 more answers

Divide using a common factor of 6 to find an equivalent fraction for

Savatey [412]

5: six twelfths is one half

6: eight tenths is four fifths

7: four eighths is one half

8: ten tenths is two halves

9: two sixths is one third

10: one quarter, because you divide the top and bottom number by 3

method:

take the first one - four eighths, which is 4/8

the question gives you what to divide them by, 4, so u divide both the top number and bottom number to get 1/2, which is one half :)

7 0

3 years ago

Read 2 more answers

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

3 years ago