The given data can be represented on a scatter chart as shown below.

It can be seen that there is a negative correlation between quantity and price.

The line of best fit is obtained by minimizing the squared errors of the data values i.e. the square of the difference between the predicted and actual values of the dependent variable with respect to the independent variable.

$\begin{array}{ccccccc}x& y& x-\overline{x}& (x-\overline{x}{)}^2& (y-\overline{y})& (x-\overline{x}{)}^2& (x-\overline{x})(y-\overline{y})\\ 10& 120& -20& 400& 39& 1521& -780\\ 20& 105& -10& 100& 24& 576& -240\\ 30& 85& 0& 0& 4& 16& 0\\ 40& 55& 10& 100& -26& 676& -260\\ 50& 40& 20& 400& -41& 1681& -820\\ & & & \sum =1000& & \sum =4470& \sum =-2100\end{array}$

$\overline{x}=\frac{\sum x}{n}=\frac{150}{5}=30$
$\overline{y}=\frac{\sum y}{n}=\frac{405}{5}=81$

$r=\frac{{\sum }_{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sqrt{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}\sqrt{{\sum }_{i=1}^n(y_i-\overline{y}{)}^2}}=\frac{-2100}{\sqrt{1000\times 4470}}=-\frac{2100}{2114.24}=-0.99$

$r=\frac{{\sum }_{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sqrt{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}\sqrt{{\sum }_{i=1}^n(y_i-\overline{y}{)}^2}}$

A negative correlation indicates that one quantity decreases as the other increases. So, if the slope of the line of best fit is negative, the correlation is also negative.  Hence, the given statement is true.

$\begin{array}{ccccc}X& Y& X-\overline{X}& Y-\overline{Y}& (X-\overline{X})(Y-\overline{Y})\\ 4& 5& -4& -6& 24\\ 6& 9& -2& -2& 4\\ 8& 12& 0& 1& 0\\ 10& 14& 2& 3& 6\\ 12& 15& 4& 4& 16\\ & & & & \sum =50\end{array}$

$\overline{X}=\frac{\sum X}{N}=\frac{40}{5}=8$
$\overline{Y}=\frac{\sum Y}{N}=\frac{55}{5}=11$

$Cov(X,Y)=\frac{(X-\overline{X})(Y-\overline{Y})}{N}=\frac{50}{5}=10$

$Let{A}_x=150000\&h_x=50000Let{A}_y=25\&h_y=5$ 

$\begin{array}{ccccccc}{X}_i& Y_i& U_i=\frac{X_i-A_x}{h_x}& V_i=\frac{Y_i-A_y}{h_y}& U_i^2& V_i^2& U_i{V}_i\\ 50000& 40& -2& 3& 4& 9& -6\\ 100000& 30& -1& 1& 1& 1& -1\\ 150000& 25& 0& 0& 0& 0& 0\\ 200000& 15& 1& -2& 1& 4& -2\\ 250000& 10& 2& -3& 4& 9& -6\\ & & \sum =0& \sum =-1& \sum =10& \sum =23& \sum =-15\end{array}$

$r=\frac{N\sum \left(U_i{V}_i\right)–(\sum U_i)(\sum V_i)}{\sqrt{N\sum U_i^2–(\sum U_i{)}^2}\sqrt{N\sum V_i^2–(\sum V_i{)}^2}}=\frac{-75}{7.07\times 10.68}=-0.993$

Variables in a correlation are called co-variables.

Correlation shows the degree to which variations in one variable explain the variation of the second variable. It does not show that there is a causal relationship between two variables.