The answer is C. |-31|=3
The absolute value of -31 should’ve been 31 not 3
The answers that would fill in the blanks are
- 2r
- a circle
- an annulus
- 1/3πr³
- 4/3πr³
<h3>What is the Cavalier's principle?</h3>
This principle states that if two solids are of equal altitude then the sections that the planes would make would have to be parallel and also be at the same distances from their bases which are equal such that the volumes of the solids would be equal.
Now we have to fill in the blanks with the solution.
For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius r and height<u> 2r</u> minus the volume of two cones, each with a radius and height of r. A cross section of the sphere is a <u>circle</u> base of cylinder, is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is an <u>annulus_ </u>.The volume of the cylinder with radius r and height 2r is 2πr³, and the volume of each cone with radius r and height r is 1/3πr³. So the volume of the cylinder minus the two cones is 4/3πr³. Therefore, the volume of the sphere is by Cavalieri's principle
Read more on Cavalieri's principle here
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Answer:
See below.
Step-by-step explanation:
Either 3 packets of 6 pens or
2 packets of 9 pens.
The answer to the question
The coefficient of determination can be found using the following formula:
![r^2=\mleft(\frac{n(\sum ^{}_{}xy)-(\sum ^{}_{}x)(\sum ^{}_{}y)}{\sqrt[]{(n\sum ^{}_{}x^2-(\sum ^{}_{}x)^2)(n\sum ^{}_{}y^2-(\sum ^{}_{}y)^2}^{}}\mright)^2](https://tex.z-dn.net/?f=r%5E2%3D%5Cmleft%28%5Cfrac%7Bn%28%5Csum%20%5E%7B%7D_%7B%7Dxy%29-%28%5Csum%20%5E%7B%7D_%7B%7Dx%29%28%5Csum%20%5E%7B%7D_%7B%7Dy%29%7D%7B%5Csqrt%5B%5D%7B%28n%5Csum%20%5E%7B%7D_%7B%7Dx%5E2-%28%5Csum%20%5E%7B%7D_%7B%7Dx%29%5E2%29%28n%5Csum%20%5E%7B%7D_%7B%7Dy%5E2-%28%5Csum%20%5E%7B%7D_%7B%7Dy%29%5E2%7D%5E%7B%7D%7D%5Cmright%29%5E2)
Using a Statistics calculator or an online tool to work with the data we were given, we get
So the best aproximation of r² is 0.861
