Answer
$7200\frac{L}{day},7630\frac{Kg}{day}$
Work Step by Step
We can find the volume of the blood pumped in one day as
$\frac{\Delta V}{\Delta t}=(5.00 \frac{L}{min})(60\frac{min}{h})(24\frac{h}{day})=7200\frac{L}{day}$
Now we can find the mass of the blood pumped in one day as
$\frac{\Delta m}{\Delta t}=\rho(\frac{\Delta V}{\Delta t})$
We plug in the known values to obtain:
$\frac{\Delta m}{\Delta t}=(1060\frac{Kg}{m^3})(7200L)(\frac{m^3}{10^3L})=7630\frac{Kg}{day}$
