5V + 2.0V = 5.5V$.
The power supplied by the 4.5V battery is $P = IV$, where $I$ is the current and $V$ is the voltage.
The current supplied by the batteries is given as $13.3 mA = 13.3 \times 10^{-3} A$.
The power supplied by the 4.5V battery is $P = (13.3 \times
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