Answer :
To solve the problem and fill in the table with calculated averages for each set of times, we can follow these steps:
### Data Overview
We have the times taken for a washer to travel distances of 0.25 meters and 0.50 meters for different numbers of washers. The datasets are organized by the number of washers and trials for each distance.
### Calculating Averages
For each set of trials, the average time can be computed by summing all the recorded times for that set and then dividing by the number of trials.
Let's go through the detailed steps for each number of washers:
#### 1 Washer
##### Times to travel 0.25 meters
- Trial 1: 2.24 seconds
- Trial 2: 2.21 seconds
- Trial 3: 2.23 seconds
Average = [tex]\(\frac{2.24 + 2.21 + 2.23}{3} = \frac{6.68}{3} \approx 2.227\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 3.16 seconds
- Trial 2: 3.08 seconds
- Trial 3: 3.15 seconds
Average = [tex]\(\frac{3.16 + 3.08 + 3.15}{3} = \frac{9.39}{3} \approx 3.13\)[/tex] seconds
#### 2 Washers
##### Times to travel 0.25 meters
- Trial 1: 1.94 seconds
- Trial 2: 1.95 seconds
- Trial 3: 1.87 seconds
Average = [tex]\(\frac{1.94 + 1.95 + 1.87}{3} = \frac{5.76}{3} \approx 1.92\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 2.69 seconds
- Trial 2: 2.72 seconds
- Trial 3: 2.42 seconds
Average = [tex]\(\frac{2.69 + 2.72 + 2.42}{3} = \frac{7.83}{3} \approx 2.61\)[/tex] seconds
#### 3 Washers
##### Times to travel 0.25 meters
- Trial 1: 1.33 seconds
- Trial 2: 1.35 seconds
- Trial 3: 1.34 seconds
Average = [tex]\(\frac{1.33 + 1.35 + 1.34}{3} = \frac{4.02}{3} \approx 1.34\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 1.88 seconds
- Trial 2: 1.93 seconds
- Trial 3: 1.67 seconds
Average = [tex]\(\frac{1.88 + 1.93 + 1.67}{3} = \frac{5.48}{3} \approx 1.83\)[/tex] seconds
#### 4 Washers
For 4 washers, there are no recorded times for either 0.25 meters or 0.50 meters. Thus, the averages cannot be calculated and remain None.
### Final Table
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Number of Washers} & \text{Trial} & \multicolumn{2}{|c|}{\text{Time to travel 0.25 m } (t_1(s))} & \multicolumn{2}{|c|}{\text{Time to travel 0.50 m } (t_2(s))} \\ \hline \multirow{3}{}{\text{1 washer, mass} = 4.9 \text{g}} & \text{Trial \#1} & 2.24 & \multirow{3}{}{2.23} & 3.16 & \multirow{3}{*}{3.13} \\ \hline & \text{Trial \#2} & 2.21 & & 3.08 & \\ \hline & \text{Trial \#3} & 2.23 & & 3.15 & \\ \hline \multirow{3}{}{\text{2 washers, mass} = 9.8 \text{g}} & \text{Trial \#1} & 1.94 & \multirow{3}{}{1.92} & 2.69 & \multirow{3}{*}{2.61} \\ \hline & \text{Trial \#2} & 1.95 & & 2.72 & \\ \hline & \text{Trial \#3} & 1.87 & & 2.42 & \\ \hline \multirow{3}{}{\text{3 washers, mass} = 14.7 \text{g}} & \text{Trial \#1} & 1.33 & \multirow{3}{}{1.34} & 1.88 & \multirow{3}{*}{1.83} \\ \hline & \text{Trial \#2} & 1.35 & & 1.93 & \\ \hline & \text{Trial \#3} & 1.34 & & 1.67 & \\ \hline \text{4 washers, mass} = 19.6 \text{g} & \text{Trial \#1} & & \text{Average} & & \text{Average} \\ \hline & \text{Trial \#2} & & & & \\ \hline & \text{Trial \#3} & & & & \\ \hline \end{array} \][/tex]
In summary, the averages for the time taken to travel 0.25 meters and 0.50 meters have been calculated for each set of washers available in the given trials. The missing data for 4 washers was acknowledged, and thus no average could be computed for those instances.
### Data Overview
We have the times taken for a washer to travel distances of 0.25 meters and 0.50 meters for different numbers of washers. The datasets are organized by the number of washers and trials for each distance.
### Calculating Averages
For each set of trials, the average time can be computed by summing all the recorded times for that set and then dividing by the number of trials.
Let's go through the detailed steps for each number of washers:
#### 1 Washer
##### Times to travel 0.25 meters
- Trial 1: 2.24 seconds
- Trial 2: 2.21 seconds
- Trial 3: 2.23 seconds
Average = [tex]\(\frac{2.24 + 2.21 + 2.23}{3} = \frac{6.68}{3} \approx 2.227\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 3.16 seconds
- Trial 2: 3.08 seconds
- Trial 3: 3.15 seconds
Average = [tex]\(\frac{3.16 + 3.08 + 3.15}{3} = \frac{9.39}{3} \approx 3.13\)[/tex] seconds
#### 2 Washers
##### Times to travel 0.25 meters
- Trial 1: 1.94 seconds
- Trial 2: 1.95 seconds
- Trial 3: 1.87 seconds
Average = [tex]\(\frac{1.94 + 1.95 + 1.87}{3} = \frac{5.76}{3} \approx 1.92\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 2.69 seconds
- Trial 2: 2.72 seconds
- Trial 3: 2.42 seconds
Average = [tex]\(\frac{2.69 + 2.72 + 2.42}{3} = \frac{7.83}{3} \approx 2.61\)[/tex] seconds
#### 3 Washers
##### Times to travel 0.25 meters
- Trial 1: 1.33 seconds
- Trial 2: 1.35 seconds
- Trial 3: 1.34 seconds
Average = [tex]\(\frac{1.33 + 1.35 + 1.34}{3} = \frac{4.02}{3} \approx 1.34\)[/tex] seconds
##### Times to travel 0.50 meters
- Trial 1: 1.88 seconds
- Trial 2: 1.93 seconds
- Trial 3: 1.67 seconds
Average = [tex]\(\frac{1.88 + 1.93 + 1.67}{3} = \frac{5.48}{3} \approx 1.83\)[/tex] seconds
#### 4 Washers
For 4 washers, there are no recorded times for either 0.25 meters or 0.50 meters. Thus, the averages cannot be calculated and remain None.
### Final Table
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Number of Washers} & \text{Trial} & \multicolumn{2}{|c|}{\text{Time to travel 0.25 m } (t_1(s))} & \multicolumn{2}{|c|}{\text{Time to travel 0.50 m } (t_2(s))} \\ \hline \multirow{3}{}{\text{1 washer, mass} = 4.9 \text{g}} & \text{Trial \#1} & 2.24 & \multirow{3}{}{2.23} & 3.16 & \multirow{3}{*}{3.13} \\ \hline & \text{Trial \#2} & 2.21 & & 3.08 & \\ \hline & \text{Trial \#3} & 2.23 & & 3.15 & \\ \hline \multirow{3}{}{\text{2 washers, mass} = 9.8 \text{g}} & \text{Trial \#1} & 1.94 & \multirow{3}{}{1.92} & 2.69 & \multirow{3}{*}{2.61} \\ \hline & \text{Trial \#2} & 1.95 & & 2.72 & \\ \hline & \text{Trial \#3} & 1.87 & & 2.42 & \\ \hline \multirow{3}{}{\text{3 washers, mass} = 14.7 \text{g}} & \text{Trial \#1} & 1.33 & \multirow{3}{}{1.34} & 1.88 & \multirow{3}{*}{1.83} \\ \hline & \text{Trial \#2} & 1.35 & & 1.93 & \\ \hline & \text{Trial \#3} & 1.34 & & 1.67 & \\ \hline \text{4 washers, mass} = 19.6 \text{g} & \text{Trial \#1} & & \text{Average} & & \text{Average} \\ \hline & \text{Trial \#2} & & & & \\ \hline & \text{Trial \#3} & & & & \\ \hline \end{array} \][/tex]
In summary, the averages for the time taken to travel 0.25 meters and 0.50 meters have been calculated for each set of washers available in the given trials. The missing data for 4 washers was acknowledged, and thus no average could be computed for those instances.
