Answer:
Class interval 10-19 20-29 30-39 40-49 50-59
cumulative frequency 10 24 41 48 50
cumulative relative frequency 0.2 0.48 0.82 0.96 1
Step-by-step explanation:
1.
We are given the frequency of each class interval and we have to find the respective cumulative frequency and cumulative relative frequency.
Cumulative frequency
10
10+14=24
14+17=41
41+7=48
48+2=50
sum of frequencies is 50 so the relative frequency is f/50.
Relative frequency
10/50=0.2
14/50=0.28
17/50=0.34
7/50=0.14
2/50=0.04
Cumulative relative frequency
0.2
0.2+0.28=0.48
0.48+0.34=0.82
0.82+0.14=0.96
0.96+0.04=1
The cumulative relative frequency is calculated using relative frequency.
Relative frequency is calculated by dividing the respective frequency to the sum of frequency.
The cumulative frequency is calculated by adding the frequency of respective class to the sum of frequencies of previous classes.
The cumulative relative frequency is calculated by adding the relative frequency of respective class to the sum of relative frequencies of previous classes.
Answer:
B
Explanation:
Slope = rise/run = -2/2 = -1
Answer:
<em>The elephants' moat should be 3.2 meters deep.</em>
Step-by-step explanation:
Suppose, the depth of the elephants' moat is 
meter.
Width of the lions' moat is 5 meter and width of the elephants' moat is 4 meter.
Given that, the lion’s moat is 4 meter deep.
So, <u>according to the ratio of width and depth</u>, the equation will be.......
Thus, the elephants' moat should be 3.2 meters deep.
Answer:
1) pounds
2) grams
3) tons
Step-by-step explanation:
The suitable measurements for the scenarios are:
1) A piano weighs 5.5 × 10² units
550 Ib (pounds)
2) A grain of salt weighs 1.02 × 
units
0.00102 grams
3) A tractor trailer weighs 8.8 × 
units
88 ton
10²g = 1kg
10^-3g = miligrams
10^1 g = decagram
