## College Physics (4th Edition)

The percentage change in the RBC is a decrease of $4.0~\%$
$RBC_0 = 5.0\times 10^6~\mu L^{-1}$ $RBC_f = 4.8\times 10^6~\mu L^{-1}$ Let $x$ be the decrease in RBC. We can find $x$: $R_0~(1-x) = R_f$ $R_0~x = R_0-R_f$ $x = \frac{R_0-R_f}{R_0}$ $x = \frac{5.0\times 10^6~\mu L^{-1}- 4.8\times 10^6~\mu L^{-1}}{ 5.0\times 10^6~\mu L^{-1}}$ $x = 0.040 = \frac{4.0}{100} = 4.0\%$ The percentage change in the RBC is a decrease of $4.0~\%$.