Answer
(a) $I = 11.6~A$
(b) $R = 10.3~\Omega$
Work Step by Step
(a) We can find the average power:
$P = \frac{1000\times 10^3~W~h}{(30)(24~h)} = 1390~W$
We can find the average current:
$P = IV$
$I = \frac{P}{V}$
$I = \frac{1390~W}{120~V}$
$I = 11.6~A$
(b) We can find the average resistance:
$R = \frac{V}{I}$
$R = \frac{120~V}{11.6~A}$
$R = 10.3~\Omega$
