What conclusion can you draw from the following observations on a resistor made of alloy manganin?

$I (A)$	$V (V)$	$I (A)$	$V (V)$
$0.2$	$3.94$	$3.0$	$59.2$
$0.4$	$7.87$	$4.0$	$78.8$
$0.6$	$11.8$	$5.0$	$98.6$
$0.8$	$15.7$	$6.0$	$118.5$
$1.0$	$19.7$	$7.0$	$138.2$
$2.0$	$39.4$	$8.0$	$158.0$

(N/A) To determine the nature of the conductor,we calculate the ratio $R = V/I$ for several data points:
For $I = 0.2 \, A, V = 3.94 \, V \implies R = 3.94 / 0.2 = 19.7 \, \Omega$.
For $I = 1.0 \, A, V = 19.7 \, V \implies R = 19.7 / 1.0 = 19.7 \, \Omega$.
For $I = 8.0 \, A, V = 158.0 \, V \implies R = 158.0 / 8.0 = 19.7 \, \Omega$.
Since the ratio $V/I$ remains constant at $19.7 \, \Omega$ for all given values,the resistor obeys Ohm's law.
Therefore,we conclude that manganin is an ohmic conductor with a constant resistance of $19.7 \, \Omega$.