The chart to the right shows a country's annual egg production. Model the data in the chart with a linear function, using the points $(0,51.7)$ and $(4,60.2)$. Let $x=0$ represent 1994, $x=1$ represent 1995, and so on, and let $y$ represent the egg production (in billions). Predict egg production in 2000. How does the result compare to the actual data given in the table, 69.7 ? \begin{tabular}{|c|c|} \hline Year & $\begin{array}{c}\text { Egg production } \\ \text { (in billions) }\end{array}$ \\ \hline 1994 & 51.7 \\ \hline 1995 & 52.5 \\ \hline 1996 & 54.4 \\ \hline 1997 & 57.1 \\ \hline 1998 & 60.2 \\ \hline 1999 & 63.8 \\ \hline 2000 & 69.7 \\ \hline \end{tabular} The linear model for the data is $\square$. (Type an equation using $\mathrm{x}$ as the variable. Type your answer in slope-intercept form. Use integers or decimals for any numbers in the equation. Round to the nearest thousandth as needed.)

Step 1 ：First, we need to find the slope of the line that passes through the points (0,51.7) and (4,60.2). The slope of a line is given by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

Step 2 ：Using the given points, we find that \(m = \frac{60.2 - 51.7}{4 - 0} = 2.125\).

Step 3 ：Next, we use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), to find the equation of the line. Using the point (0,51.7) and the slope we just found, we get the equation \(y = 2.125x + 51.7\).

Step 4 ：We can use this equation to predict the egg production in 2000, which corresponds to \(x = 6\). Substituting \(x = 6\) into the equation, we get \(y = 2.125(6) + 51.7 = 64.45\).

Step 5 ：Finally, we compare this prediction to the actual data given in the table. The actual egg production in 2000 was 69.7 billion, which is greater than our prediction of 64.45 billion.

Step 6 ：\(\boxed{\text{The linear model for the data is } y = 2.125x + 51.7. \text{ The model predicts an egg production of 64.45 billion in 2000, which is less than the actual production of 69.7 billion.}}\)