Given the data set (41, 55, 48, 44)

A spinner has 4 equal-sized sections. To win the game, the pointer must land on a purple section.

sveticcg [70]

Answer:

You didn't include the statements so can you comment them below please.

1 favorable outcome / 4 outcomes in total = 0.25 = 25% to get wanted spot

0.25 purple = 1/4 = 25%

purple = 1/4

If you have any questions you can ask. Hope this helps

Step-by-step explanation:

6 0

4 years ago

A house painter has an annual income of $35,000 and paid $2,500 in income tax. What percent of her income is the income tax?

almond37 [142]

Answer:

The 7.14% of her income is the income tax .

Step-by-step explanation:

Formula

$Percentage = \frac{Part\ value\times 100}{Total\ value}$

As given

A house painter has an annual income of $35,000 and paid $2,500 in income tax.

Here

Part value = $2500

Total value = $35000

Put in the formula

$Percentage = \frac{2500\times 100}{35000}$

$Percentage = \frac{250000}{35000}$

$Percentage = 7.14\ (Approx)$

Therefore the 7.14% of her income is the income tax .

3 0

3 years ago

Someone please help for 15 points!!!

Alenkasestr [34]

Answer:

With what nothing there

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

What is -4/5 * 10/16

Fiesta28 [93]

The answer is -1/2. Hope that helped.

5 0

4 years ago

Read 2 more answers

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