You are starting a Shaved Ice business and need to decide on a cup.  Your choices are between a cone and a cylindrical shaped cup.  The cone shaped cup has a diameter of 3.25 inches and is 3.75 inches tall.  The cylindrical cup has a diameter of 4 inches and height of 4.75 inches.  The spherical scoop you have chosen has a diameter of 2.75 inches.  Which cup should you choose if you want two scoops of ice to best fit in the cup?

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You are starting a Shaved Ice business and need to decide on a cup.  Your choices are between a cone and a cylindrical shaped cup.  The cone shaped cup has a diameter of 3.25 inches and is 3.75 inches tall.  The cylindrical cup has a diameter of 4 inches and height of 4.75 inches.  The spherical scoop you have chosen has a diameter of 2.75 inches.  Which cup should you choose if you want two scoops of ice to best fit in the cup?

Mathematics
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step #1 we have to compute the volume V of the spherical scoop: \[\begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {1.375^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {1.375^3} = ...? \hfill \\ \end{gathered} \]
since if the diameter is 2.75 inches, then the radius is: 2.75/2=1.375 inches

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Other answers:

okay
I got 2.374625
I got this: \[\Large \begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {1.375^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {1.375^3} = \frac{{12.56}}{3} \times 2.600 = 10.884inche{s^3} \hfill \\ \hfill \\ \end{gathered} \]
\[\large \begin{gathered} V = \frac{{4 \times 3.14}}{3}{r^3} = \frac{{12.56}}{3} \times {1.375^3} = \hfill \\ \hfill \\ = \frac{{12.56}}{3} \times {1.375^3} = \frac{{12.56}}{3} \times 2.600 = 10.884inche{s^3} \hfill \\ \hfill \\ \end{gathered} \]
now I have to do the same for the cylindrical cup
yes!
the volume V of the cylindrical cup is: \[\large \begin{gathered} V = 3.14 \times R \times R \times H = \hfill \\ \hfill \\ = 3.14 \times 2 \times 2 \times 4.75 = ...inche{s^3} \hfill \\ \end{gathered} \] since the radius is 4/2=2 inches
I have used this approximation: pi = 3.14
what do you get?
59.66
correct!
sorry if Im going slow on the questions I writing the answers and work down
ok! no worries! :)
I have to calculate the scoops
we can see that the cylindrical cup is too large since two scoops of ice occupy 10.884*2= 21.768 inches^3
okay
thank yu for helping me :D
:)
now we have to compute the volume of the cone shaped cup
okay
would is be 59.6 and 10.88
here is that volume: \[\begin{gathered} V = \frac{1}{3}3.14 \times R \times R \times H = \frac{{3.14 \times 1.625 \times 1.625 \times 3.75}}{3} = \hfill \\ \hfill \\ = \frac{{3.14 \times 1.625 \times 1.625 \times 3.75}}{3} = \frac{{3.14 \times 2.64 \times 3.75}}{3} = ...inche{s^3} \hfill \\ \end{gathered} \]
since the radius is: 3.25/2= 1.625 inches
okay
what do you get?
1.625
that is the radius, whereas the volume is: \[\begin{gathered} V = \frac{1}{3}3.14 \times R \times R \times H = \frac{{3.14 \times 1.625 \times 1.625 \times 3.75}}{3} = \hfill \\ \hfill \\ = \frac{{3.14 \times 1.625 \times 1.625 \times 3.75}}{3} = \frac{{3.14 \times 2.64 \times 3.75}}{3} = \hfill \\ \hfill \\ = 10.364inche{s^3} \hfill \\ \end{gathered} \]
so the cone shaped cup is too small, since it can not contain 2 scoops of ice, therefore we have to choose the cylindrical cup

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