Answer:
EG = 19
Step-by-step explanation:
* Lets explain how to solve the problem
- If a line bisects another line that means the point of intersection
divides the second line into two equal parts
∵ EF bisects CD at G
∴ CG = GD
∵ CG = 5x - 1
∵ GD = 7x - 13
∴ 7x - 13 = 5x - 1
* Lets solve the equation
∵ 7x - 13 = 5x - 1
- Subtract 5x from both sides and add 13 to both sides
∴ 7x - 5x = 13 - 1
∴ 2x = 12
- Divide both sides by 2
∴ x = 6
- Point G divides EF into two parts EG and GF
∴ EF = EG + GF
∵ EF = 6x - 4
- Substitute the value of x to find EF
∵ x = 6
∴ EF = 6(6) - 4 = 36 - 4 = 32
∴ EF = 32
∵ GF = 13
- Substitute the values of EF and GF in the equation of EF
∴ 32 = EG + 13
- Subtract 13 from both sides
∴ 19 = EG
* EG = 19
the answer would be 3 I believe
<span>Just list all elements that belong both to A and to B
A ∩ B = {6,12}</span>
There is no hard and fast rule to select the class width. It largely depends on our application.However, one thing that should be kept in mind is that the number of classes should neither to be too low nor too high. So keeping this thing in mind, the class width is select.
The range of the data is = Maximum- Minimum = 96 - 11 = 85
10 classes will be most suited for this data.
The class width for each data can be calculated as:
Class Width = Range / Number of Classes = 85/10 = 8.5
Class width is always rounded to nearest next integer. So the class width will be 9 in this case.
So, the best value of class width or interval width for the given data will be 9.