Classify the following real numbers as rational or irrational:
1 answer:
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Chebyshev's theorem in statistics states that for many probability distributions, no more than 1/k² of measured values will be k standard deviations away from the mean.
Because the area under the probability distribution curve is equal to 1, Chebyshev's theorem means that the shaded area shown in the figure is equal to 1 - 1/k².
When k = 1.75, the shaded area is
1 - 1/1.75² = 0.7635 = 67.35%
Therefore the percent of the area within +/- 1.75 standard deviations from the mean is
67.35/2 = 33.7%, which is at least 33% of the observations.
Answer:
According to the Chebyshev theorem, at least 33% of the observations lie within +/- standard deviations from the mean.
Because the area under the probability distribution curve is equal to 1, Chebyshev's theorem means that the shaded area shown in the figure is equal to 1 - 1/k².
When k = 1.75, the shaded area is
1 - 1/1.75² = 0.7635 = 67.35%
Therefore the percent of the area within +/- 1.75 standard deviations from the mean is
67.35/2 = 33.7%, which is at least 33% of the observations.
Answer:
According to the Chebyshev theorem, at least 33% of the observations lie within +/- standard deviations from the mean.
Answer:
To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?
Step-by-step explanation:
I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.
Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).
Now just form the dot product of those two vectors:
(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)
Now the truth of your claim will be embodied in that dot product being zero:
x^2 + y^2 - (1+z)(1-z) = 0
x^2 + y^2 - (1-z^2) = 0
x^2 + y^2 + z^2 = 1
But that last line is just the definition of points on the surface of a sphere of radius 1, so the claim is proven.
Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.
Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.
