Let's solve the problem step by step: ### **Given:** - Number of beats, \( N = 24 \pm 1 \) - Time interval, \( t = 20.0 \pm 0.5 \) s We want the heart rate in **beats per minute** and its uncertainty. --- ## **Step 1: Calculate the Heart Rate** Heart rate in beats per second is: \[ \text{Heart Rate} = \frac{N}{t} \] To convert to beats per minute (bpm): \[ \text{Heart Rate (bpm)} = \frac{N}{t} \times 60 \] Plug in the values: \[ \text{Heart Rate (bpm)} = \frac{24}{20.0} \times 60 = 1.2 \times 60 = 72~\text{bpm} \] --- ## **Step 2: Calculate the Uncertainty** Let: - \( \Delta N = 1 \) - \( \Delta t = 0.5 \) The relative uncertainty in a quotient is: \[ \left(\frac{\Delta R}{R}\right) = \sqrt{ \left(\frac{\Delta N}{N}\right)^2 + \left(\frac{\Delta t}{t}\right)^2 } \] Where: - \( R = \frac{N}{t} \) - \( \Delta R \) is the absolute uncertainty in \( R \) Calculate each term: - \( \frac{\Delta N}{N} = \frac{1}{24} \approx 0.0417 \) - \( \frac{\Delta t}{t} = \frac{0.5}{20.0} = 0.025 \) So, \[ \left(\frac{\Delta R}{R}\right) = \sqrt{ (0.0417)^2 + (0.025)^2 } \approx \sqrt{0.00174 + 0.000625} \approx \sqrt{0.002365} \approx 0.0486 \] The heart rate in bpm is \( 72 \), so the uncertainty is: \[ \Delta \text{(bpm)} = 0.0486 \times 72 \approx 3.5 \] --- ## **Step 3: Final Answer** \[ \boxed{72 \pm 4~\text{beats per minute}} \] (Rounded uncertainty to 1 significant figure, and heart rate to match.) --- ### **Final Answer:** **72 ± 4 beats per minute**

Certainly! Here's a concise four-line theoretical explanation: The heart rate is calculated by dividing the number of beats by the time interval, then converting to beats per minute by multiplying by 60. The uncertainties in measurement are propagated using the fractional uncertainties of both the number of beats and the time. Since the measurements involve division, the relative uncertainties add in quadrature to find the overall uncertainty in the heart rate. This method provides an estimated value along with its associated measurement uncertainty.