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(a) Suppose that a person has an average heart rate of 72.0 beats $/ \mathrm{min}$. How many beats does he or she have in $2.0 y$ ? (b) In $2.00 y$ ? (c) In $2.000 y$ ?
Solution (a) To calculate the number of beats she has in 2.0 years, we need to multiply 72.0 beats/minute by 2.0 years and use conversion factors to cancel the units of time:
$$
\frac{72.0 \text { beats }}{1 \mathrm{~min}} \times \frac{60.0 \mathrm{~min}}{1.00 \mathrm{~h}} \times \frac{24.0 \mathrm{~h}}{1.00 \mathrm{~d}} \times \frac{365.25 \mathrm{~d}}{1.00 \mathrm{y}} \times 2.0 \mathrm{y}=7.5738 \times 10^7 \text { beats }
$$
Since there are only 2 significant figures in 2.0 years, we must report the answer with 2 significant figures: $7.6 \times 10^7 \text { beats. }$
(b) Since we now have 3 significant figures in 2.00 years, we now report the answer with 3 significant figures: $7.57 \times 10^7$ beats.
(c) Even though we now have 4 significant figures in 2.000 years, the 72.0 beats/minute only has 3 significant figures, so we must report the answer with 3 significant figures: $7.57 \times 10^7$ beats.
Solution To calculate the heart rate, we need to divide the number of beats by the time and convert to beats per minute.
$$
\frac{\text { beats }}{\text { minute }}=\frac{40 \text { beats }}{30.0 \mathrm{~s}} \times \frac{60.0 \mathrm{~s}}{1.00 \mathrm{~min}}=80 \text { beats } / \mathrm{min}
$$
To calculate the uncertainty, we use the method of adding percents.
$$
\% \text { unc }=\frac{1 \text { beat }}{40 \text { beats }} \times 100 \%+\frac{0.5 \mathrm{~s}}{30.0 \mathrm{~s}} \times 100 \%=2.5 \%+1.7 \%=4.2 \%=4 \%
$$
Then calculating the uncertainty in beats per minute:
$\delta A=\frac{\% \text { unc }}{100 \%} \times A=\frac{4.2 \%}{100 \%} \times 80$ beats $/ \mathrm{min}=3.3$ beats $/ \mathrm{min}=3$ beats $/ \mathrm{min}$
Notice that while doing calculations, we keep one EXTRA digit, and round to the correct number of significant figures only at the end.
So, the heart rate is $80 \pm 3$ beats $/ \mathrm{min}$.
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