Problem 1-15: Solving for the number of heartbeats of a person with the correct significant figure

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PROBLEM:

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?

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SOLUTION:

Part a

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

Convert 72 beats/min to beats/year.

\begin{align*}
72 \ \text{beats/min} & = \left(\frac{72.0\:\text{beats}}{1\:\bcancel{\text{min}}}\right)\left(\frac{60.0\:\bcancel{\text{min}}}{1.00\:\bcancel{\text{hr}}}\right)\left(\frac{24.0\:\bcancel{\text{hr}}}{1.00\:\bcancel{\text{day}}}\right)\left(\frac{365.25\:\bcancel{\text{days}}}{1\:\text{year}}\right)\\
& =3.7869\times 10^7\:\text{beats/year}
\end{align*}

In 2 years, the number of beats of an average person is

\begin{align*}
& = 3.7869 \times 10^{7} \ \text{beats/year} \times 2 \ \text{years} \\
& = 7.5738 \times 10^7 \ \text{beats} 
\end{align*}

For part a, the answer is limited to 2.0 which is 2 significant figures. The answer is 7.6 \times 10^{7} \ \text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part b

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part c

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is  7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

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