An unknown metal has been found and the following experimental results have been tabulated in the table below. The table contains the grams of the unknown metal and the volume in milliliters of water displacement. Find a linear model that expresses volume as a function of the mass. \begin{tabular}{|r|r|} \hline grams & Volume in ml \\ \hline 19 & 374.4 \\ \hline 20.5 & 404 \\ \hline 22 & 420.8 \\ \hline 23.5 & 444.9 \\ \hline 25 & 478.2 \\ \hline 26.5 & 501.7 \\ \hline 28 & 530.1 \\ \hline \end{tabular} A) Write the linear regression equation for the data in the chart. Volume $=$ $x+$ decimal places where $x$ is the grams of the unknown metal. Round your answers to 3 B) If the mass of an unknown metal is 17 , using your un-rounded regression equation find its predicted volume. Round your answer to 1 decimal place. $m L$

Step 1 ：We are given a table of data that shows the relationship between the mass of an unknown metal in grams and the volume of water displaced in milliliters. We are asked to find a linear model that expresses volume as a function of mass.

Step 2 ：The linear regression model is of the form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.

Step 3 ：Using the given data, we can calculate the slope and y-intercept of the best fit line. The slope (\(m\)) is approximately 19.048 and the y-intercept (\(c\)) is approximately 0.000.

Step 4 ：Therefore, the linear regression equation for the data in the chart is \(\text{Volume} = 19.048 \times \text{grams} + 0.000\).

Step 5 ：We can use this equation to predict the volume for a given mass. If the mass of an unknown metal is 17 grams, we substitute 17 for \(x\) in the equation to get the predicted volume.

Step 6 ：The predicted volume is approximately 339.2 milliliters.

Step 7 ：Final Answer: If the mass of an unknown metal is 17 grams, its predicted volume is \(\boxed{339.2}\) mL.