Answer:
Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:
Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:
Class 66 - 68:
When the computation is completed, the frequency distribution will be:
Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:
<em>See attachment for result of 1-VarStats</em>
Answer:
Option C is true - The number of students who read 4 books or fewer
Step-by-step explanation:
Here, to find the mean and median, we do not have proper information or specific values. We just have the ranges, so both the mean and median cannot be determined. So, options A and B are not correct.
We can see two bins covering the 9 or more books, so option D cannot be true either.
Therefore, only option C is left and that is true.- the number of students who read 4 books or fewer.
Answer:
rise over run
Step-by-step explanation:
how much it goes up over how far it goes to the side
Answer:
(x + 4) and (x + 1)
Step-by-step explanation:
x² + 5x + 4
Consider the factors of the constant term (+ 4) which sum to give the coefficient of the x- term (+ 5)
The factors are 4 and 1 , since
4 × 1 = 4 and 4 + 1 = 5 , then
x² + 5x + 4 = (x + 4)(x + 1) ← in factored form
Part A: 3x + 15 = 5x - 25
Part B: x = 20
