Answer:
It is not normally distributed as it has it main concentration in only one side.
Step-by-step explanation:
So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).
So, let us begin the groupings into their different classes, shall we?
Data given:
0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.
(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.
(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31
(3). 0.4 - 0.6 : there are 2 values that falls into this category.
(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.
Class interval frequency.
0.00 - 0.20. 15.
0.20 - 0.40. 2.
0.4 - 0.6. 2.
Answer:
$.75
Step-by-step explanation:
9 ÷ 12 = 0.75
.75 × 12 = 9
The surface area will increase by a factor of 49/16
Answer:
He needs 5x+56 ft
Step-by-step explanation:
To find how much of fencing he needs , we find the perimeter of the given figure
All sides are equal in a square
To find perimeter of the square we add all the sides
4 sides we have for the square
one side is x, so perimeter of square = x+x+x+x= 4x
Now we find perimeter of rectangle
Opposite sides of rectangle are equal
Here for rectangle we consider only three sides
because fourth side is common for rectangle and square
So perimeter of the rectangle (with 3 sides) = 28 +x+ 28 = 56+x
Total fencing = perimeter of square + perimeter of rectangle
4x + 56 + x= 5x+56
Answer:
100
Step-by-step explanation:
Since the second 4 is two decimal points away from the first 4, the answer should be 100 (two zeros)