<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u> </u>√15 units
Step-by-step explanation:
Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .
As we know ,
Distance(D) = √(x^1-x^2) +(y^1-y^2)
Now,
D= √(x^1-x^2) +(y^1-y^2)
= √(6-1) +(1+9)
= √5+10
= √15 units
: Therefore the distance between (6,1) and (1,-9) is √15 units.
1/3= 5/15
2/5= 6/15
5/15 + 6/15 = 11/15
15/15 - 11/15 = 4/15
Part C= 4/15
Answer:
4
Step-by-step explanation:
Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:
CW = UCL - LCL
Where,
CW= Class width
UCL= Upper class limit
LCL= Lower class limit
From the table above:
For class 1, CW = 64 - 60 = 4
For class 2, CW = 69 - 65 = 4
For class 3, CW = 74 - 70 = 4
For class 4, CW = 79 - 75 = 4
For class 5, CW = 84 - 80 = 4
Therefore, the class width of the grouped data = 4
5.75 is the answer.
If you subtract 43.9 FROM 49.65, that's what you get.
