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Tennis balls with a diameter of 2.5 in. are packaged 3 to a can. The can is a cylinder. Find the volume of the space in the can that is not occupied by tennis balls. Assume that the balls touch the sides, top, and bottom of the can. Round your answer to the nearest hundredth. What is the volume not occupied by balls?

Sagot :

Ryan2idn
a) First we calculate the volume of each ball:

[tex]\boxed{V=\frac{4\pi r^3}{3}}\\ \\ \boxed{V=\frac{4*(3,14)*(1.25)^3}{3}=8,177 \ square \ inch}[/tex]

b) Now we calculate the volume of tree balls:

[tex]\boxed{V_T=3*8,177=24.531 \ square \ inch}[/tex]

c) Now we calculate the volume of cylinder:

c1) Base area:

[tex]\boxed{A_b=\pi r^2}\\ \\ \boxed{A_b=3,14*(1,25)^2=4.906 \ square \ inch}[/tex]

c2) Cylinder height

[tex]\boxed{h=3*(2.5)=7.5 \ inch}[/tex]

c3) Cylinder Volume:

[tex]V_C=A_b*h\\ \\ \boxed{V_C=4.906*7.5=36.795 \ cubic \ inch}[/tex]

d) Finally we calculate the internal space:

[tex]\boxed{\boxed{s=36.795-24.531= 12,264\ cubic \ inch}}[/tex]
Lilithidn
[tex]d=2.5 \ in \\ \\ r=\frac{d}{2}=\frac{2.5}{2}=1.25 \ in \\ \\ \pi=3.14 \\ \\Volume \ of \ a \ Cylinder : \\ \\V = \pi r^2 h[/tex]

[tex]h=3\cdot d = 3*2.5 = 7.5 \ in \\ \\V_{c} = 3.14\cdot (1.25)^2\cdot 7.5 = 23.55 * 1.5625=36.80 \ in^3[/tex]


[tex]The \ volume \ of \ tree \ balls : \\ \\ V_{3b}= 3 \cdot \frac{4}{3} \pi r^3 = 4\pi r^3 \\ \\V_{3b}=4\cdot 3.14\cdot (1.25)^3= 12.56\cdot 1.9531 \approx 24.53 \ in^3 \\ \\V_{c}-V_{3b}=36.80-24.53 =12.27 \ in^3 \\ \\ Answer : \ The \ volume \ not \ occupied \ by \ balls \ it \ 12.27 \ in^3[/tex]



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