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4 years ago

Two angles whose sum is 180?180? are said to be

Mathematics

1 answer:

harina [27]4 years ago

5 0

2 angles whose sum is 180 are supplementary angles

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Rosa has already jarred 12 liters of jam and will jar an additional 1 liter of jam every day. How many days does Rosa need to sp

weqwewe [10]

Answer:

8 days

Step-by-step explanation:

Pretty simple actually if you put your brain to it

every day she adds one liter right?

so to reach 20 liters of jam you would minus 12 from 20 which would be 8 days in order to reach 20.

3 0

3 years ago

The data (1.5.8.5.1) represent a random sample of the number of days absent from school for five students at Monta Vista High. F

lianna [129]

Answer:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

a. mean = 4; standard deviation = 2.68

Step-by-step explanation:

For this case we have the following data given:

1,5,8,5,1

We can find the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

And for the deviation (assuming that the correct approximation is the deviation for a population) we can calculate the deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}$

And replacing we got:

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

a. mean = 4; standard deviation = 2.68

3 0

4 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

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

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$