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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.
The test statistic z will be equal to -0.946 and it shows that there is no significant difference in the proportion of rehires between full time and part time.
Given sample sizes of 833 and 386 and result of samples 434 and 189.
Proportion of full time=434/833=0.52
Proportion of part time=189/386=0.49.
Difference in proportion =0.52-0.49
TTF- i∈ rho=0
TTF+i∈ rho≠0.
Mean of difference=0.03
Z=(X-μ)/σ
σ=
=0.0317
σ=0.0317
z=(0-0.03)/0.0317
=-0.03/0.0317
=-0.317
p value will be =0.1736.
Because p value is greater than 0.01 so we will accept the null hypothesis which shows that there is no significant difference in the proportions.
Hence there is no significant difference in the proportion of rehires between full time and part time.
Learn more about z test at brainly.com/question/14453510
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