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kondaur [170]
3 years ago
11

A cone has a height of 11 meters and a radius of 7 meters . What is the volume?

Mathematics
2 answers:
Levart [38]3 years ago
5 0

Answer:

564.512 m³

Step-by-step explanation:

Formula to find the volume of a cone is :

Volume = \frac{1}{3} π r² h

Here,

r = radius = 7 m

h =  height = 11 m

Let us solve now.

Volume = \frac{1}{3} π r² h

Volume = \frac{1}{3} π ×7² × 11

Volume = \frac{1}{3} π × 49 × 11

Volume = \frac{1}{3} π × 539

Volume = \frac{1}{3} × 1693.538

Volume = 564.512 m³

Hope this helps you :-)

Let me know if you have any other questions :-)

baherus [9]3 years ago
3 0

Answer:

V = 564.2 m^3

Step-by-step explanation:

The volume of a cone is given by

V = 1/3 pi r^2 h  where r is the radius and h is the height

V = 1/3 pi (7)^2 (11)

V =539 /3 * pi

Letting pi be approximated by 3.14

V = 564.1533333 m^3

Rounding to the nearest tenth

V = 564.2 m^3

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The ratio of the side lengths of a quadrilateral is 3:2:6:7, and its perimeter is 126 meters. What is the length of the shortest
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since the lengths of all those four sides are in a 3:2:6:7 ratio, and the whole perimeter is 126, what we do is, simply divide the whole by (3+2+6+7) and distribute accordingly.


\bf \stackrel{3\cdot \frac{126}{3+2+6+7}}{3}~~:~~\stackrel{2\cdot \frac{126}{3+2+6+7}}{2}~~:~~\stackrel{6\cdot \frac{126}{3+2+6+7}}{6}~~:~~\stackrel{7\cdot \frac{126}{3+2+6+7}}{7} \\\\\\ 3\cdot \cfrac{126}{18}~~:~~2\cdot \cfrac{126}{18}~~:~~6\cdot \cfrac{126}{18}~~:~~7\cdot \cfrac{126}{18} \\\\\\ 21~~:~~\stackrel{shortest}{14}~~:~~42~~:~~49

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Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer
almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

\to \mu=\frac{\sum xi}{n}

       =22.8

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187

In point b:

\to \mu=\frac{\sum xi}{n}

       =20.6875  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251

In point c:

 \to \mu=\frac{\sum xi}{n}

         =21  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179

In point d:

\to \mu=\frac{\sum xi}{n}

       =20.8375  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\  =\sqrt{0.553166667} \\\\ =0.7438

6 0
3 years ago
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