Question Number 122142 by A8;15: last updated on 14/Nov/20
Answered by ebi last updated on 14/Nov/20
$${y}={mx}+{c} \\ $$$${m}=\frac{{N}\Sigma\left({xy}\right)−\Sigma{x}\Sigma{y}}{{N}\Sigma{x}^{\mathrm{2}} −\left(\Sigma{x}\right)^{\mathrm{2}} } \\ $$$${c}=\frac{\Sigma{y}−{m}\Sigma{x}}{{N}} \\ $$$$\begin{vmatrix}{{x}}\\{−\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{1}}\\{\Sigma{x}=\mathrm{0}}\end{vmatrix}\begin{vmatrix}{{y}}\\{\mathrm{3}}\\{\mathrm{2}}\\{\mathrm{4}}\\{\Sigma{y}=\mathrm{9}}\end{vmatrix}\begin{vmatrix}{{x}^{\mathrm{2}} }\\{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{1}}\\{\Sigma{x}^{\mathrm{2}} =\mathrm{2}}\end{vmatrix}\begin{vmatrix}{{xy}}\\{−\mathrm{3}}\\{\mathrm{0}}\\{\mathrm{4}}\\{\Sigma{xy}=\mathrm{1}}\end{vmatrix} \\ $$$${N}=\mathrm{3}\:\left({no}.\:{of}\:{data}\right) \\ $$$$ \\ $$$$\therefore\:{m}=\frac{\mathrm{3}\left(\mathrm{1}\right)−\left(\mathrm{0}\right)\left(\mathrm{9}\right)}{\mathrm{3}\left(\mathrm{2}\right)−\left(\mathrm{0}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}=\frac{\mathrm{9}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}\right)}{\mathrm{3}}=\mathrm{3} \\ $$$$ \\ $$$${a}+{b}={m}+{c}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{3}=\frac{\mathrm{7}}{\mathrm{2}}\:\left({B}\right) \\ $$$$ \\ $$$$ \\ $$