Jen has 46 more than Jay, “Jen has 46 more than Jay.” You answered your own question.
The answer is 73 because 8 is more than 5 so it rounds up to be 73
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
<span>When adding up fractions, the idea is to bring them to a common denominator. In our case, the common denominator is 12. So we must amplify each fraction in order to bring its denominator to 12. 1/6 becomes 1*2/6*2 = 2/12. 2/3 becomes 2*4/3*4. 1/4 becomes 1*3/4*3. When we sum them up we get 2/12 + 8/12 + 3/13 or 13/12, which is (12+1)/12 or 12/12 + 1/12 or 1+ 1/12. So the answer is C.</span>
