1. **State the problem:** Albert's flour fills a cylindrical jar with diameter 4 inches and height 6 inches. We want to find which containers can hold the same volume of flour. 2. **Calculate the volume of Albert's jar:** The volume $V$ of a cylinder is given by the formula: $$V = \pi r^2 h$$ where $r$ is the radius and $h$ is the height. Since the diameter is 4 inches, the radius is: $$r = \frac{4}{2} = 2$$ So the volume is: $$V = \pi \times 2^2 \times 6 = \pi \times 4 \times 6 = 24\pi$$ cubic inches. 3. **Check each container:** - **Cylinder with radius 2 inches and height 6 inches:** Volume is: $$V = \pi \times 2^2 \times 6 = 24\pi$$ cubic inches. This matches Albert's jar volume exactly. - **Square prism with side length 4 inches and height 6 inches:** Volume is: $$V = \text{side}^2 \times h = 4^2 \times 6 = 16 \times 6 = 96$$ cubic inches. Since $24\pi \approx 75.4$, $96 > 75.4$, so this container can hold all the flour. - **Cone with diameter 4 inches and height 6 inches:** Radius $r = \frac{4}{2} = 2$ inches. Volume of a cone: $$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 2^2 \times 6 = \frac{1}{3} \pi \times 4 \times 6 = 8\pi$$ cubic inches. Since $8\pi \approx 25.1$, this is less than $24\pi$, so it cannot hold all the flour. - **Sphere with diameter 4 inches:** Radius $r = 2$ inches. Volume of a sphere: $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 2^3 = \frac{4}{3} \pi \times 8 = \frac{32}{3} \pi \approx 33.51$$ cubic inches. Since $33.51 < 75.4$, it cannot hold all the flour. 4. **Final answer:** The containers that can hold all the flour are: - Cylinder with radius 2 inches and height 6 inches. - Square prism with side length 4 inches and height 6 inches.