For the table 0.2 would be (1/5) 0.5 would be (1/2). And 0.7 would be (7/10).
For number 2 I would say the denominator changed each time
Answer:
m=17/40 
Step-by-step explanation:
So you 5/7m - 1/7 = 9/56 
You Simplify both sides Then
well I Don´t know I hope this helped u, I tried ^^¨
To find this, first find the factor or rate of which the numbers are moving. To do so do as follows.
subtract 1 from 3
3-1=2
So each number is having 2 added to it.
Now add two to 7 and the numbers afterwards till you get the 12th term
7+2=9
1+3+5+7+9
9+2=11
1+3+5+7+9+11
11+2=13
1+3+5+7+9+11+13
13+2=15
1+3+5+7+9+11+13+15
15+2=17
1+3+5+7+9+11+13+15+17
17+2=19
1+3+5+7+9+11+13+15+17+19
19+2=21
1+3+5+7+9+11+13+15+17+19+21
21+2=23
1+3+5+7+9+11+13+15+17+19+21+23
So 23 is the 12th term
Answer:
A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.
B) slope: (5.6-1.8)/(2-0) = 1.9
y = 1.9x + 1.8
(Line of best fit)
x = 13,
y = 1.9(13) + 1.8 = 26.5
Predicted no. of games won after 13 months of practice is 26.5
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.
