Explanation:
Recall that
[tex]1\:\text{micron} = 1\:\mu\text{m} = 10^{-6}\:\text{m}[/tex] so the speed of the bacterium is
[tex]v = \dfrac{6.82×10^{-6}\:\text{m}}{3.5\:\text{s}} = 1.9×10^{-6}\:\text{m/s}[/tex]
Next, we convert this speed to km/hr. Recall that
[tex]1\:\text{km} = 1000\:\text{m}[/tex]
[tex]1\:\text{hr} = 60\:\text{minutes} = 3600\:\text{s}[/tex]
Therefore,
[tex]1.9×10^{-6}\:\dfrac{\text{m}}{\text{s}}×\left(\dfrac{1\:\text{km}}{1000\:\text{m}}\right)×\left(\dfrac{3600\:\text{s}}{1\:\text{hr}}\right)[/tex]
[tex]= 6.8×10^{-6}\:\text{km/hr}[/tex]
