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:
see below
Step-by-step explanation:
butterfly = x minutes
breast = y minutes
total is 50 minutes
breast stroke = butterfly +20
Part A
x+y = 50
y = x+20
Part B
Substitute the second equation into the first
x+ (x+20) = 50
2x+20 = 50
2x+20-20 = 50-20
2x = 30
2x/2 = 30/2
x = 15
y = x+20
y = 20+15
y = 35 minutes
35 minutes on the breast stroke
Part C
45 minutes on the butterfly is not reasonable
We only have a total of 50 minutes
50-45 = 5
But he spend more time on the breast stroke ( 20 minutes more), but if he spend 45 minutes on the butterfly, he would have to spend less on the breast stroke.
Answer:
7 and 12
Step-by-step explanation:
X-intercept would be (-5,0)
and the y-intercept would be (0,-5)
