Supposing we begin with n=1, the first few terms of this arith sequence are
5, 1, -3, -7, and so on. 4 is the common difference.
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: 15º and 50º
Explanation:
The angles in a triangle add up to 180.
180 - 115 = 65
3x + 10x = 13x
13x = 65
x = 5
3x = 15
10x = 50
Answer: alternate interior angles
Step-by-step explanation:
Got it Right on the test. :)
