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BlackZzzverrR [31]
3 years ago
10

A number cube with faces labeled from 1 to 6 will be rolled once.

Mathematics
1 answer:
kramer3 years ago
3 0

Answer:

\Omega=\{1,2,3,4,5,6\}

A=\{1,2,3,4\}

Step-by-step explanation:

<u>Sample Space</u>

The sample space of a random experience is a set of all the possible outcomes of that experience. It's usually denoted by the letter \Omega.

We have a number cube with all faces labeled from 1 to 6. That cube is to be rolled once. The visible number shown in the cube is recorded as the outcome. The possible outcomes are listed as the sample space below:

\Omega=\{1,2,3,4,5,6\}

Now we are required to give the outcomes for the event of rolling a number less than 5. Let's call A to such event. The set of possible outcomes for A has all the numbers from 1 to 4 as follows

A=\{1,2,3,4\}

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Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y
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Answer:

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Step-by-step explanation:

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To find total charge on the disk let Q be the total charge and x=r\cos\theta,y=r\sin\theta so that,

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=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA

=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta

=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta

=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta

=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta

=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi

=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi

=256\pi

Hence total charge is 256\pi coulombs per square meter.

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