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:
The average rate of change is -3.
Step-by-step explanation:
We are given the function:
And we want to find the average rate of change from <em>x </em>= 0 to <em>x </em>= 3.
In other words, we will compute the function at the two endpoints, and then find the slope of the line that crosses the two points.
For our first endpoint at <em>x</em> = 0, our function evaluates to:
So, our first point is (0, 9).
For our second endpoint at <em>x</em> = 3, our function evaluates to :
So, our second point is (3, 0).
Then by the slope formula, our average rate of change will be:
Answer:
Don't put it on the news because it's a high risk personnel?
Answer:
Step-by-step explanation:
The missing coefficient of 
is .
Answer:
It’s B maybe not 100%
sure
Step-by-step explanation:
lol tell me if I’m correct
