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:
39
Step-by-step explanation:
1. Subtract the numbers ... 3(3×3+4)
2. Multiply the numbers ... 3(9+4)
3. Add the numbers ... 3×13
4. Multiply the numbers ... 39
Answer:
-3
Step-by-step explanation:
The equation is in the slope intercept form
y= mx +b where m is the slope and b is the y intercept
y = -3x +0
The slope is -3 and the y intercept is 0
Answer:
the soulution is (5,10)
Step-by-step explanation:
Answer: 5
Step-by-step explanation:
5 multiples by itself 4 times is 625.
5 x 5 x 5 x 5
