Answer:
The relative frequency is found by dividing the class frequencies by the total number of observations
Step-by-step explanation:
Relative frequency measures how often a value appears relative to the sum of the total values.
An example of how relative frequency is calculated
Here are the scores and frequency of students in a maths test
Scores (classes) Frequency Relative frequency
0 - 20 10 10 / 50 = 0.2
21 - 40 15 15 / 50 = 0.3
41 - 60 10 10 / 50 = 0.2
61 - 80 5 5 / 50 = 0.1
81 - 100 <u> 10</u> 10 / 50 = <u>0.2</u>
50 1
From the above example, it can be seen that :
- two or more classes can have the same relative frequency
- The relative frequency is found by dividing the class frequencies by the total number of observations.
- The sum of the relative frequencies must be equal to one
- The sum of the frequencies and not the relative frequencies is equal to the number of observations.
By order of operation
5+1x10
5+10
15
By normal solving
5+1x10
6x10
60
Answer:
<h2>
29</h2>
Step-by-step explanation:
Answer:
Kindly check explanation
Step-by-step explanation:
The hypothesis :
H0 : μ1 = μ2
H1 : μ1 > μ2
Given :
x1 = 21.1 ; n1 = 53 ; s1 = 1.1
x2 = 20.7 ; n2 = 46 ; s2 = 1.2
The test statistic :
(x1 - x2) / √[(s1²/n1 + s2²/n2)]
(21.1 - 20.7) / √[(1.1²/53 + 1.2²/46)]
0.4 / 0.2326682
Test statistic = 1.719
The degree of freedom using the conservative method :
Comparing :
Degree of freedom = n - 1
Degree of freedom 1 = 53 - 1 = 52
Degree of freedom 2 = 46 - 1 = 45
Smaller degree of freedom is chosen ;
The Pvalue from Test statistic, using df = 45
Pvalue = 0.0462
Pvalue < α ; Hence, there is significant evidence to conclude that average age of Gorka student is higher than Yaphoa.
Answer:
Can not be determined
Step-by-step explanation:
regularly, you would +20 to both sides, but without a full =, you cant solve. If the equation is 6x-20=0, x would equal 3 2/6.
Hope this helps!
