Annual high temperatures in a certain location have been tracked for several years. Let $X$ represent the year and $Y$ the high temperature. Based on the data shown below, calculate the regression line (each value to two decimal places).
\[
y=
\]
$x+$
\begin{tabular}{|c|c|}
\hline$x$ & $y$ \\
\hline 4 & 8.9 \\
\hline 5 & 9.3 \\
\hline 6 & 7.1 \\
\hline 7 & 6.4 \\
\hline 8 & 8.8 \\
\hline 9 & 8.5 \\
\hline
\end{tabular}

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Step 1 ：The problem is asking for the regression line based on the given data. The regression line is a tool used to predict the value of one variable (Y) based on the value of another variable (X). It is represented by the equation \(y = mx + b\), where m is the slope of the line and b is the y-intercept.

Step 2 ：To calculate the regression line, we need to find the values of m and b. The formula for m (slope) is given by: \(m = \frac{NΣXY - (ΣX)(ΣY)}{NΣX^2 - (ΣX)^2}\) and the formula for b (y-intercept) is given by: \(b = \frac{ΣY - m(ΣX)}{N}\).

Step 3 ：Where: N is the number of observations, ΣXY is the sum of the product of X and Y, ΣX is the sum of X, ΣY is the sum of Y, ΣX^2 is the sum of the squares of X.

Step 4 ：We can calculate these values using the given data. X = [4 5 6 7 8 9], Y = [8.9 9.3 7.1 6.4 8.8 8.5], N = 6, sum_X = 39, sum_Y = 49.0, sum_XY = 316.4, sum_X2 = 271.

Step 5 ：Using these values, we can calculate m and b. m = -0.12 and b = 8.95 (rounded to two decimal places).

Step 6 ：This means the regression line is \(y = -0.12x + 8.95\). This line can be used to predict the high temperature (Y) based on the year (X).

Step 7 ：\(\boxed{y = -0.12x + 8.95}\) is the final answer.