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exis [7]
3 years ago
11

A cone has a radius of 200 m at its base. The cone is 50 m high. How much space is inside the cone?

Mathematics
1 answer:
Sauron [17]3 years ago
7 0

The question "how much space is inside a solid?" is actually asking about the volume of that solid.

The volume of a cone is given by

V = \dfrac{1}{3}\pi hr^2

where h is the height of the cone, and r is its radius.

So, you only need to plug in the values:

V = \dfrac{1}{3}\pi\cdot 50 \cdot 200^2 = \dfrac{2000000\pi}{3}m^3

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Answer:

The linear regression equation that represents  the set of data is:

y=1083.48+23.73x

The predicted number of new cases for 2013 is 1368.

Step-by-step explanation:

The general form of a linear regression is:

y=a+bx

Here,

<em>y</em> = dependent variable

<em>x</em> = independent variable

<em>a</em> = intercept

<em>b</em> = slope

Compute the values of <em>a</em> and <em>b</em> as follows:

\begin{aligned}        a &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} =             \frac{ 6857 \cdot 55 - 15 \cdot 17558}{ 6 \cdot 55 - 15^2} \approx 1083.476 \\ \\b &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2}        = \frac{ 6 \cdot 17558 - 15 \cdot 6857 }{ 6 \cdot 55 - \left( 15 \right)^2} \approx 23.743\end{aligned}

The linear regression equation that represents  the set of data is:

y=1083.48+23.73x

Compute the predicted value of the number of new cases for 2013 (i.e. <em>x</em> = 12) as follows:

y=1083.48+23.73x

  =1083.48+(23.73\times12)\\\\=1083.48+284.76\\\\=1368.24\\\\\approx 1368

Thus, the predicted number of new cases for 2013 is 1368.

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