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katovenus [111]
3 years ago
9

If the basketball shown below has a radius of 6 inches from the center to the outside surface, approximately how many cubic inch

es of air will it hold when fully inflated? A. 904.32 in3 B. 288 in3 C. 1808.64 in3 D. 72 in3
Mathematics
2 answers:
Soloha48 [4]3 years ago
4 0

Answer:

The answer to your question is the letter A. 904.32 in³

Step-by-step explanation:

Data

radius = 6 in

Volume = ?

π = 3.1416

Formula

Volume = 4/3 πr³

-Substitution

Volume = 4/3 (3.1416)(6)³

-Simplification

Volume = 4/3 (3.1416)(216)

Volume = 2714.342 / 3

-Result

Volume = 904.7 in³

The closest answer to my result is A.904.3 in³

gizmo_the_mogwai [7]3 years ago
4 0

Answer:

Step-by-step explanation:

At the part where you said "4/3 (3.1416)(216)" you are suppose to actually multiply 4/3(3.14)(216) Not "3.1416". Because when you multiply "4/3(3.14)(216)" you get 904.32 which is rounded to 904.3.

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For example: To find the standard deviation, you have to add up all the numbers in the data set, then divide by how many numbers there are, and that will get you your answer.

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The Mean is: 9 + 2 + 5 + 4 + 12 + 7 + 8 + 11 + 9 + 3 + 7 + 4 + 12 + 5 + 4 + 10+ 9 + 6 + 9 + 4. Over 20. That equals: 104 over 20 = 7.

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3 years ago
The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?
Serhud [2]

Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi }  =  \sqrt{\frac{a^{2} + b^{2} }{2} }

Squaring Both sides, we get

[\frac{C}{2\pi }]^{2}   =  [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2}  }   =  {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2}  }   =  {{a^{2} + b^{2} }

\frac{C^{2} }{2(\pi )^{2} }  = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} }  -  a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}  = b

∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

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