Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
Answer:
0, 4
Step-by-step explanation:
The best way to approach this problem is to look at the graph of the given function. Replace values of x from 1 to 24 to indicate the numbers of hours in a day. As seen on the graph, there is only one point where the port is at high tide. That would be at 1:00 am.
Looking at the graph, it would be safe for the boats to be in the port when the graph levels off at around 10 to 24. That's from 10 am to before 12 midnight. Then, they would have to stay away between 12 midnight to before 10 am.