An experiment has shown the relationship between Voltage $(V)$ and Current $($A$).$ The following are the results:

$\backslash$table$[[$Voltage $($V$)($x$),$Current $($I$)($y$)],[11,15\mathrm{.}2],[13,17\mathrm{.}7],[15,19\mathrm{.}3],[17,21\mathrm{.}5],[19,23\mathrm{.}9],[21,25\mathrm{.}4]]$

The least$-$square model $y=bx+a.$

where

$a=\frac{{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i^2{\displaystyle \underset{i=1}{\overset{n}{}}}{y}_i-{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{y}_i}{n{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i^2-({\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{)}^2},b=\frac{n{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{y}_i-{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{\displaystyle \underset{i=1}{\overset{n}{}}}{y}_i}{n{\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i^2-({\displaystyle \underset{i=1}{\overset{n}{}}}{x}_i{)}^2}$

This can be replaced by current, $I=(\frac{1}{R})V+a,$ where $(\frac{1}{R})=b$

Round the answer in $4$ decimals.