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The relationship of average temperature and the average wind speed in given in the question.

The formula used:$Residual=y鈭�\stackrel{铃�}{y}$

The general equation of the least square regression line is:

$\stackrel{铃�}{y}=b_0+b_{1}x$

As a result, the estimate of the constant $b_0$ is given in the computer output's row "Intercept" and column "Estimate" as:

$b_0=11.897762$

In the computer output, the estimate of the slope $b_1$ is given in the row "Avg temp" and the column "Estimate" as:

$b_1=鈭�0.041077$

As a result, when we plug the values into the general equation, we get:

$$\begin{gathered} \stackrel{铃�}{y}=b_0+b_{1}x \\ 鈬�\stackrel{铃�}{y}=11.897762鈭�0.041077x \end{gathered}$$

As a result, the average wind speed at $42$degrees Fahrenheit is:

$$\begin{gathered} \stackrel{铃�}{y}=11.897762鈭�0.041077x \\ =11.897762鈭�0.041077\left(42\right) \\ =10.172528 \end{gathered}$$

Thus the residual will be calculated as:.

$$\begin{gathered} Residual=y鈭�\stackrel{铃�}{y} \\ =2.2鈭�10.172528 \\ =鈭�7.972528 \end{gathered}$$

This means that while using the regression line to make a prediction, we underestimated the average wind speed on the day with an average temperature of $40F$by$7.972528$ mph.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{铃�}{y}=11.897762鈭�0.041077x$

The slope is the coefficient of $x$ in the least-squares regression equation, and it reflects the average rise or decrease of y per unit of $x$ as we all know. Thus,

$b_0=鈭�10.041077$

As a result, the average wind speed reduces by$0.041077$ miles per hour per degree Fahrenheit.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{铃�}{y}=11.897762鈭�0.041077x$

The standard error of the estimate s is calculated as follows in the computer output following "root mean square error":

$s=3.655950$

The standard error of the estimations, as we all know, is the average error of forecasts, and thus the average difference between actual and predicted values. As a result, using the equation of the least square regression line, the predicted average wind speed differed by $3.655950$ mph on average from the actual average wind speed.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{铃�}{y}=11.897762鈭�0.041077x$

The coefficient of determination is given as follows in the computer output after "RSquare":

$$\begin{gathered} r^2=0.0478742 \\ =4.7874\% \end{gathered}$$

The coefficient of determination, as we all know, represents the proportion of variance in the answers $y$ variable that can be explained by a least square regression model using the explanatory $x$ variable. As a result, we can state that the least square regression line employing average temperature as an explanatory variable explains $4.7874$ percent of the variation in average wind speed.