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zavuch27 [327]
1 year ago
6

What is the surface area of cone below Diameter AB=21 and distance from B to C =36

Mathematics
1 answer:
Vesnalui [34]1 year ago
6 0

Answer:

1533.88261311522

Step-by-step explanation:

<em>Let r be the radius of the (base/circle)</em>

<em>and  L be the slant height of the cone</em>

<em>Formula ………………………………………………………………………………………………………</em>

The surface area of a cone = the (curved/lateral) surface area +  the base

                                             =\pi r^{2}\ \ \ +\ \ \ \pi Lr

=========================================

r=\frac{21}{2} =10.5

L = BC = 36

\text{surface area} =\pi \left( 10.5\right)^{2}  +\pi \times 36\times \left( 10.5\right)

                   =346.360590058275 + 1187.52202305694

                   =1533.88261311522

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a) \bar{x}=4.63 md=4.55 mo= 1.9 b) Sample Standard Deviation≈ 2.58 Coefficient of Variation=55.72% Sample Range=6.9

Step-by-step explanation:

a)

<u>Mean</u>

a)\bar{x}=\frac{1.9+ 2.3+ 5.7+ 5.2+ 1.9+ 8.8+ 3.9+ 7.3}{8} =\frac{37}{8}=4.625=\approx 4.63

For the <u>Median</u>, we have to order the entries. So, ordering it goes:

1.9 1.9 2.3 3.9 5.2 5.7 7.3 8.8

Since we have even entries \frac{\frac{n}{2}+\frac{n}{2} +1}{2}=\frac{4th+5th}{2}=\frac{3.9+5.2}{2}=4.55

mode

The mode for this data 1.9 1.9 2.3 3.9 5.2 5.7 7.3 8.8 is 1.9

b)

<u>Sample Standard Deviation</u>

Here it is the formula to calculate it:

_{x}=\sqrt{\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}}\\ S_{x}=\sqrt{\frac{46.855}{7}}\approx 2.58

<u>Coefficient of Variation</u>

CV is the quocient between sample Standard deviation over Mean and it is used to make comparisons.

CV=\frac{S_x}{\bar{x}}*100%=\frac{2.58}{4.63} \approx 55.72%

<u>Range</u>

The difference between the highest and the lowest value of this sample

8.8-1.9=6.9

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