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Vlad1618 [11]
3 years ago
11

Will all slices created by a slicing plane that passes through a sphere be the same size

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
6 0

Answer:

The suggestion proposed considering special cases, so that’s what we’ll do.

One special case we could consider is when x=0. If h=r, then we have a hemisphere. You may have learnt that the surface area of a sphere is 4πr2, so the answer in this case is 2πr2. Likewise, we can consider the most extreme cases: when h=2r, we get an area of 4πr2 (the entire sphere), and when h=0, we get zero area.

So if we do manage to get a general formula, we will be able to check it in these special cases.

Another possible idea when x=0 is to consider what happens when h is very small. In this case, the surface area looks a lot like a circle, but the radius of the circle does not seem particularly easy to work out, so maybe we’ll leave this one for a moment. (There are things we can do to approximate the radius using more advanced techniques, but they are currently beyond us.)

Another thing we could consider is the situation where h is very small. Then the part of the sphere between the planes looks very much like the frustum of a cone, and we know how to find the (surface) area of such a shape – see Cones.

If we recall our answer from that problem, we find the formulae for the surface area

π(R+r)s=π(R+r)hcosθ=π(R2−r2)sinθ.

Here, R is the radius of the base of the frustum, r is the radius of the top, h is the perpendicular height, s is the slant length and θ is the angle the slant makes with the vertical.

Step-by-step explanation:

Studentka2010 [4]3 years ago
6 0

Answer: no

Step-by-step explanation:

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The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir
Kazeer [188]

Answer:

(1) (c) <u>5.30 years</u>.

(2) (b) <u>0.289</u>.

(3) (b) <u>0.80</u>.

(4) (d) <u>0.50</u>.

(5) (a) <u>5.25 years</u>.

Step-by-step explanation:

Let <em>X</em> = age of the children in kindergarten on the first day of school.

The random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 4.8 years and <em>b</em> = 5.8 years.

The probability density function function of <em>X</em> is:

f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a

(1)

The expected value of a Uniform random variable is:

E(X)=\frac{1}{2}(a+b)

Compute the mean of <em>X</em> as follows:

E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3

Thus, the  mean of the distribution is (c) <u>5.30 years</u>.

(2)

The standard deviation of a Uniform random variable is:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289

Thus, the standard deviation of the distribution is (b) <u>0.289</u>.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\

                =\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8

Thus, the probability that a randomly selected child is older than 5 years old is (b) <u>0.80</u>.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

P(5.2

                            =\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) <u>0.50</u>.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of <em>x</em> as follows:

   P (X < x) = 0.45

\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45

        \int\limits^{x}_{4.8} {1}\, dx=0.45

           [x]^{x}_{4.8}=0.45

       x-4.8=0.45\\

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Thus, the age of the child at the 45th percentile is (a) <u>5.25 years</u>.

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Answer:

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x=8/2=4

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