All Solutions
Page 21: Practice Problems
$$
begin{align*}
L&=(0.012 text{km})left(dfrac{1 text{m}}{10^{-3} text{km}}right)=12 text{m}\
W&=(1050 text{cm})left(dfrac{1 text{m}}{100 text{cm}}right)=10.5 text{m}\
H&=(495 text{cm})left(dfrac{1 text{m}}{100 text{cm}}right)=4.95 text{m}\
end{align*}
$$
The volume is simply length $times$ width $times$ height,
$$
begin{align*}
V&=Ltimes Wtimes H\
&=12 text{m}times 10.5 text{m}times 4.95 text{m}\
&=boxed{623.7 text{m}^{3}}\
end{align*}
$$
$$
begin{align*}
(1.2 text{L})left(dfrac{1 text{mL}}{10^{-3} text{L}}right)=boxed{1 200 text{mL}}\
end{align*}
$$
$boxed{1 200 text{mL}}$
1.2 mathrm{L}=(1.2 L)left( frac{1 mL}{0.001 L} right)=1200 mL
$$
$$
begin{align*}
1 text{carat}&=(0.2 text{g})left(dfrac{1 text{kg}}{1000 text{g}}right)=2cdot 10^{-4} text{kg}\
end{align*}
$$
so, $530.2 text{carats}$ is equal to
$$
begin{align*}
(530.2 text{carats})left(dfrac{2cdot 10^{-4} text{kg}}{1 text{carat}}right)=0.10604 text{kg}\
end{align*}
$$
Finally, given that $1 text{kg}=2.21 text{lb}$, the weight of the Star of Africa (rounded to two decimals) in pounds is
$$
begin{align*}
(0.10604 text{kg})left(dfrac{2.21 text{lb}}{1 text{kg}}right)=boxed{0.23 text{lb}}\
end{align*}
$$
$boxed{0.23 text{lb}}$