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Zanzabum
1 year ago
5

a point in space $(x,y,z)$ is randomly selected so that $-1\le x \le 1$,$-1\le y \le 1$,$-1\le z \le 1$. what is the probability

that $x^2 y^2 z^2\le 1$?
Mathematics
1 answer:
Yuliya22 [10]1 year ago
5 0

Probability that x^2+y^2+z^2\leq 1 is 0.523

Given that a point (x, y, z) is in space which is randomly selected as

-1\le x \le 1$,$-1\le y \le 1$,$-1\le z \le 1$

These three equations represents a cube in space.

Length of each side of the cube = 1 - -1 = 2

Now  x^2+y^2+z^2\leq 1  represents a sphere with center (0, 0, 0) and of radius less than or equal to 1.

So we can imagine that this sphere is inside the cube.

So the probability that   x^2+y^2+z^2\leq 1  = Volume of the sphere/Volume of the cube

Volume of sphere with radius 1 = \frac{4}{3} \pi r^3=\frac{4}{3} \pi \times 1 = \frac{4}{3}\times 3.14 = 4.186 cubic unit

Volume of the cube of length 2 = 2^3 = 8 cubic units

So the probability that  x^2+y^2+z^2\leq 1  = 4.186/8 = 0.523

Learn more about probability at brainly.com/question/25870256

#SPJ4

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Answer:

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3 years ago
40.32 is 76.8% of what number?
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Hey there

Lets solve this question together, 

Convert <span>76.8%</span><span> to a decimal.
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<u>Answer:</u>

The correct answer option is c.  $8.05.

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