Questions: Sarah and lygen are working on a physics lab together. They each take measurements of three different timed experiments. Sarah has an analog stopwatch and records experimental times of 7 1/9, 8 5/7, and 8 2/3 seconds. lygen has a digital stopwatch and records experimental times of 7.1, 8.6, and 8.3 seconds. Determine the average time for the experiments, to the nearest tenth.

Solution Steps

To find the average time for the experiments, we need to first convert Sarah's recorded times from mixed fractions to decimal form. Then, we will calculate the average of all the times recorded by both Sarah and Lygen. Finally, we will round the result to the nearest tenth.

Sarah's recorded times are given as mixed fractions: \(7 \frac{1}{9}\), \(8 \frac{5}{7}\), and \(8 \frac{2}{3}\). We convert these to decimal form:

• \(7 \frac{1}{9} = \frac{64}{9} \approx 7.1111\)
• \(8 \frac{5}{7} = \frac{61}{7} \approx 8.7143\)
• \(8 \frac{2}{3} = \frac{26}{3} \approx 8.6667\)

Lygen's times are already in decimal form: \(7.1\), \(8.6\), and \(8.3\). We combine all the times:

• Sarah's times: \(7.1111\), \(8.7143\), \(8.6667\)
• Lygen's times: \(7.1\), \(8.6\), \(8.3\)

To find the average time, we sum all the times and divide by the number of measurements: \[ \text{Average} = \frac{7.1111 + 8.7143 + 8.6667 + 7.1 + 8.6 + 8.3}{6} \approx 8.0820 \]

We round the average time to the nearest tenth: \[ 8.0820 \approx 8.1 \]

Final Answer