Question
you buy a sphere-shaped piece of choclate that has a creamy center. the chocolate piece measures 1 inch in diameter. houw much creamy filling can fit inside the chocolate piece? calculate your answer in terms of pi and round to the nearest tenth
___ inches^3
___ inches^3
Answers
GPT-4o mini
To find the volume of the creamy filling inside the sphere-shaped chocolate piece, we first need to calculate the volume of the sphere. The formula for the volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
The diameter of the chocolate piece is given as 1 inch, so the radius \( r \) is:
\[
r = \frac{\text{diameter}}{2} = \frac{1 \text{ inch}}{2} = 0.5 \text{ inches}
\]
Now, we can substitute \( r \) into the volume formula:
\[
V = \frac{4}{3} \pi (0.5)^3
\]
Calculating \( (0.5)^3 \):
\[
(0.5)^3 = 0.125
\]
Now substitute this value back into the volume equation:
\[
V = \frac{4}{3} \pi (0.125)
\]
Now multiply:
\[
V = \frac{4 \cdot 0.125}{3} \pi = \frac{0.5}{3} \pi
\]
Calculating \( \frac{0.5}{3} \):
\[
\frac{0.5}{3} \approx 0.1667
\]
Therefore, the volume becomes:
\[
V \approx 0.1667 \pi \text{ inches}^3
\]
Now, rounding \( 0.1667 \) to the nearest tenth:
\[
0.1667 \approx 0.2
\]
Thus, the volume of the creamy filling that can fit inside the chocolate piece is:
\[
\frac{1}{6} \pi \text{ inches}^3 \quad \text{(exact form)} \text{ or } 0.2 \text{ inches}^3 \quad \text{(rounded)}
\]
Final answer in terms of \(\pi\):
\[
\frac{2}{12} \pi \text{ or } 0.2 \text{ inches}^3 \text{ (to the nearest tenth)}
\]
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
The diameter of the chocolate piece is given as 1 inch, so the radius \( r \) is:
\[
r = \frac{\text{diameter}}{2} = \frac{1 \text{ inch}}{2} = 0.5 \text{ inches}
\]
Now, we can substitute \( r \) into the volume formula:
\[
V = \frac{4}{3} \pi (0.5)^3
\]
Calculating \( (0.5)^3 \):
\[
(0.5)^3 = 0.125
\]
Now substitute this value back into the volume equation:
\[
V = \frac{4}{3} \pi (0.125)
\]
Now multiply:
\[
V = \frac{4 \cdot 0.125}{3} \pi = \frac{0.5}{3} \pi
\]
Calculating \( \frac{0.5}{3} \):
\[
\frac{0.5}{3} \approx 0.1667
\]
Therefore, the volume becomes:
\[
V \approx 0.1667 \pi \text{ inches}^3
\]
Now, rounding \( 0.1667 \) to the nearest tenth:
\[
0.1667 \approx 0.2
\]
Thus, the volume of the creamy filling that can fit inside the chocolate piece is:
\[
\frac{1}{6} \pi \text{ inches}^3 \quad \text{(exact form)} \text{ or } 0.2 \text{ inches}^3 \quad \text{(rounded)}
\]
Final answer in terms of \(\pi\):
\[
\frac{2}{12} \pi \text{ or } 0.2 \text{ inches}^3 \text{ (to the nearest tenth)}
\]