The length of an $textit{E. Coli}$ bacterium is $5cdot 10^{-6} text{m}$. Table 1.4 shows that $1 text{km}=1000 text{m}$. To calculate the length in kilometers, we use a conversion factor, with the unit we want in the numerator, and the unit we want cancelled in the denominator. Therefore,

$$
begin{align*}
(5cdot 10^{-6} text{m})left(dfrac{1 text{km}}{1000 text{m}}right)=boxed{5cdot 10^{-9} text{km}}\
end{align*}
$$

The length of an $textit{E. Coli}$ bacterium in kilometers is

                                 $boxed{5cdot 10^{-9} text{km}}$

$textbf{(a)}$      One kilodollar is 1000 dollars. Using the same principle as in the previous exercise to convert dollars to kilodollars, we have that the price of the minivan in kilodollars is

$$
begin{align*}
(33 200 text{dollars})left(dfrac{1 text{kilodollar}}{1000 text{dollars}}right)=boxed{33.2 text{kilodollars}}\
end{align*}
$$

$textbf{(b)}$      One megadollar is 1 000 000 dollars, or, more elegantly, $10^{6} text{dollars}$. The price of the minivan in megadollars is

$$
begin{align*}
(33 200 text{dollars}left(dfrac{1 text{megadollar}}{1 000 000 text{dollars}}right)=boxed{0.0332 text{megadollars}}\
end{align*}
$$

$textbf{(a)}$      The price in kilodollars is

                    $boxed{33.2 text{kilodollars}}$

$textbf{(b)}$      The price in megadollars is

                $boxed{0.0332 text{megadollars}}$

If a honeybee flaps its wings 200 times per second, then the time required for one wingbeat is

$$
begin{align*}
dfrac{1 text{s}}{200}=5cdot 10^{-3} text{s}\
end{align*}
$$

One millisecond is one thousandth of a second, $1 text{ms}=10^{-3} text{s}$, so the time required for one wingbeat expressed in milliseconds is

$$
begin{align*}
(5cdot 10^{-3} text{s})left(dfrac{1 text{ms}}{10^{-3} text{s}}right)=boxed{quad 5 text{ms}quad}\
end{align*}
$$

The time required for one wingbeat is

                        $boxed{quad 5 text{ms}quad}$