Answer:
Step-by-step explanation:
As you can see from the graph I attached you, the possible solutions in the interval from 0 to 2π are approximately:
So, it's useful to solve the equation too, in order to verify the result:
Taking the inverse sine of both sides:
Using this result we can conclude the solutions in the interval from 0 to 2π are approximately:
Steps to solve:
k + 10 = 27
~Subtract 10 to both sides
k = 17
Best of Luck!
3(4+3r)
= 12 + 9r
Hope it helped :)
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:
THANKS FOR THE POINTS
Step-by-step explanation:
