1. **Stating the problem:** We have a graduated cylinder with water. When 8 marbles are dropped in, the water level is 10 milliliters. When 11 marbles are dropped in, the water level is 12 milliliters. We need to find: - d. The volume of one marble. - e. The initial volume of water before any marbles were added. - f. The water height after 14 marbles are dropped in. - g. Whether the relationship between water volume and number of marbles is linear, and the meaning of the slope if it is. 2. **Formula and rules:** The volume of water plus marbles is a linear function of the number of marbles: $$ V = V_0 + n \times v_m $$ where $V$ is the total volume after $n$ marbles, $V_0$ is the initial water volume, $v_m$ is the volume of one marble. 3. **Using the given data:** - For $n=8$, $V=10$ ml - For $n=11$, $V=12$ ml Set up equations: $$ 10 = V_0 + 8 v_m $$ $$ 12 = V_0 + 11 v_m $$ 4. **Find $v_m$ by subtracting the first equation from the second:** $$ 12 - 10 = (V_0 + 11 v_m) - (V_0 + 8 v_m) $$ $$ 2 = 3 v_m $$ $$ v_m = \frac{2}{3} $$ ml 5. **Find $V_0$ by substituting $v_m$ back into one equation:** $$ 10 = V_0 + 8 \times \frac{2}{3} $$ $$ 10 = V_0 + \frac{16}{3} $$ $$ V_0 = 10 - \frac{16}{3} = \frac{30}{3} - \frac{16}{3} = \frac{14}{3} $$ ml 6. **Find the water height after 14 marbles:** $$ V = V_0 + 14 v_m = \frac{14}{3} + 14 \times \frac{2}{3} = \frac{14}{3} + \frac{28}{3} = \frac{42}{3} = 14 $$ ml 7. **Is the relationship linear?** Yes, the volume increases by a constant amount per marble, so the relationship is linear. **Meaning of the slope:** The slope $v_m = \frac{2}{3}$ ml means each marble displaces $\frac{2}{3}$ milliliters of water, increasing the water level by that amount. **Final answers:** - d. Volume of one marble: $\frac{2}{3}$ ml - e. Initial water volume: $\frac{14}{3}$ ml - f. Water height after 14 marbles: 14 ml - g. The relationship is linear; slope represents volume per marble.