Question

a sno-cone at the carnival has the shape of a hemisphere on top of an inverted cone. what is the volume of the sno-cone if its radius is 5.7 cm and the height of the conical portion is 9.2 cm? volume (to the nearest tenth) blank select an answer question help: message instructor submit question

Explanation:

Step1: Recall formulas for cone and hemisphere volumes

The volume of a cone is $V_{cone}=\frac{1}{3}\pi r^{2}h$, and the volume of a hemisphere is $V_{hemisphere}=\frac{2}{3}\pi r^{3}$. The total volume of the sno - cone is the sum of the volume of the cone and the volume of the hemisphere.

Step2: Calculate the volume of the conical portion

Given that $r = 5.7\space cm$ and $h=9.2\space cm$. Substitute these values into the cone volume formula:
\[

$$\begin{align*} V_{cone}&=\frac{1}{3}\times\pi\times(5.7)^{2}\times9.2\\ &=\frac{1}{3}\times\pi\times32.49\times9.2\\ &=\frac{1}{3}\times\pi\times298.908\\ & = 99.636\pi \end{align*}$$

\]

Step3: Calculate the volume of the hemispherical portion

Given $r = 5.7\space cm$. Substitute into the hemisphere volume formula:
\[

$$\begin{align*} V_{hemisphere}&=\frac{2}{3}\times\pi\times(5.7)^{3}\\ &=\frac{2}{3}\times\pi\times185.193\\ &=123.462\pi \end{align*}$$

\]

Step4: Calculate the total volume

The total volume $V = V_{cone}+V_{hemisphere}=99.636\pi + 123.462\pi=223.098\pi$
Now, substitute $\pi\approx3.14159$:
\[

$$\begin{align*} V&=223.098\times3.14159\\ &\approx223.098\times3.1416\\ &\approx700.9 \end{align*}$$

\]

Answer:

The volume of the sno - cone is approximately $\boldsymbol{700.9}\space cm^{3}$ (to the nearest tenth).