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Sagot :
### Step-by-Step Solution:
#### Part (a): Use the histogram to complete the table.
First, let's review the provided information in the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time } (t \text{ minutes}) & \text{Frequency} \\ \hline 0 < t < 30 & 45 \\ \hline 30 < t < 40 & 35 \\ \hline 40 \leqslant t < 60 & \\ \hline 60 \leqslant t < 80 & 20 \\ \hline 80 \leqslant t < 100 & 10 \\ \hline \end{array} \][/tex]
Given the frequencies for certain intervals, we also need to determine the total number of patients observed. Based on the histogram, the total number of patients is known to be 110.
Now, we can sum up the known frequencies:
- [tex]\(0 < t < 30\)[/tex]: 45 patients
- [tex]\(30 < t < 40\)[/tex]: 35 patients
- [tex]\(60 \leqslant t < 80\)[/tex]: 20 patients
- [tex]\(80 \leqslant t < 100\)[/tex]: 10 patients
Adding these frequencies together:
[tex]\[ 45 + 35 + 20 + 10 = 110 \][/tex]
We find that:
[tex]\[ \text{Total number of patients} = 110 \][/tex]
Since we have already reached the total number of patients accounted for (110), it means there are no patients in the missing interval:
[tex]\[ 40 \leqslant t < 60 \][/tex]
Therefore, the completed table is:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time } (t \text{ minutes}) & \text{Frequency} \\ \hline 0 < t < 30 & 45 \\ \hline 30 < t < 40 & 35 \\ \hline 40 \leqslant t < 60 & 0 \\ \hline 60 \leqslant t < 80 & 20 \\ \hline 80 \leqslant t < 100 & 10 \\ \hline \end{array} \][/tex]
So for part (a), the missing frequency is [tex]\(0\)[/tex].
#### Part (b): Use the table to complete the histogram.
Now, we use the completed table to fill in the histogram. The frequencies for each interval should be:
- [tex]\(0 < t < 30\)[/tex]: 45 patients
- [tex]\(30 < t < 40\)[/tex]: 35 patients
- [tex]\(40 \leqslant t < 60\)[/tex]: 0 patients (this interval will have no bar in the histogram)
- [tex]\(60 \leqslant t < 80\)[/tex]: 20 patients
- [tex]\(80 \leqslant t < 100\)[/tex]: 10 patients
To represent this information in a histogram:
- Draw bars for each interval on the horizontal axis corresponding to the ranges of waiting times.
- The heights of the bars should correspond to the frequencies given above.
- The bar for [tex]\(0 < t < 30\)[/tex] should be 45 units high.
- The bar for [tex]\(30 < t < 40\)[/tex] should be 35 units high.
- There should be no bar (height = 0) for the [tex]\(40 \leqslant t < 60\)[/tex] interval.
- The bar for [tex]\(60 \leqslant t < 80\)[/tex] should be 20 units high.
- The bar for [tex]\(80 \leqslant t < 100\)[/tex] should be 10 units high.
This visualization will complete both the table and the histogram.
#### Part (a): Use the histogram to complete the table.
First, let's review the provided information in the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time } (t \text{ minutes}) & \text{Frequency} \\ \hline 0 < t < 30 & 45 \\ \hline 30 < t < 40 & 35 \\ \hline 40 \leqslant t < 60 & \\ \hline 60 \leqslant t < 80 & 20 \\ \hline 80 \leqslant t < 100 & 10 \\ \hline \end{array} \][/tex]
Given the frequencies for certain intervals, we also need to determine the total number of patients observed. Based on the histogram, the total number of patients is known to be 110.
Now, we can sum up the known frequencies:
- [tex]\(0 < t < 30\)[/tex]: 45 patients
- [tex]\(30 < t < 40\)[/tex]: 35 patients
- [tex]\(60 \leqslant t < 80\)[/tex]: 20 patients
- [tex]\(80 \leqslant t < 100\)[/tex]: 10 patients
Adding these frequencies together:
[tex]\[ 45 + 35 + 20 + 10 = 110 \][/tex]
We find that:
[tex]\[ \text{Total number of patients} = 110 \][/tex]
Since we have already reached the total number of patients accounted for (110), it means there are no patients in the missing interval:
[tex]\[ 40 \leqslant t < 60 \][/tex]
Therefore, the completed table is:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time } (t \text{ minutes}) & \text{Frequency} \\ \hline 0 < t < 30 & 45 \\ \hline 30 < t < 40 & 35 \\ \hline 40 \leqslant t < 60 & 0 \\ \hline 60 \leqslant t < 80 & 20 \\ \hline 80 \leqslant t < 100 & 10 \\ \hline \end{array} \][/tex]
So for part (a), the missing frequency is [tex]\(0\)[/tex].
#### Part (b): Use the table to complete the histogram.
Now, we use the completed table to fill in the histogram. The frequencies for each interval should be:
- [tex]\(0 < t < 30\)[/tex]: 45 patients
- [tex]\(30 < t < 40\)[/tex]: 35 patients
- [tex]\(40 \leqslant t < 60\)[/tex]: 0 patients (this interval will have no bar in the histogram)
- [tex]\(60 \leqslant t < 80\)[/tex]: 20 patients
- [tex]\(80 \leqslant t < 100\)[/tex]: 10 patients
To represent this information in a histogram:
- Draw bars for each interval on the horizontal axis corresponding to the ranges of waiting times.
- The heights of the bars should correspond to the frequencies given above.
- The bar for [tex]\(0 < t < 30\)[/tex] should be 45 units high.
- The bar for [tex]\(30 < t < 40\)[/tex] should be 35 units high.
- There should be no bar (height = 0) for the [tex]\(40 \leqslant t < 60\)[/tex] interval.
- The bar for [tex]\(60 \leqslant t < 80\)[/tex] should be 20 units high.
- The bar for [tex]\(80 \leqslant t < 100\)[/tex] should be 10 units high.
This visualization will complete both the table and the histogram.
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