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.
Well cutting off the small 4x2 you get 8 plus’s the 7x2 is 14 so just add those together you should get 22in ^2
Answer:
9 packages of chocolate bars
Step-by-step explanation:
Let he bought c packages of chocolate bars and t packages of toffee bars,
Since, he bought 1 fewer package of chocolate bars than toffee bars.
⇒ c = t - 1 -----(1)
Also, he handed out out
of the chocolate bars and
of the toffee bars,
If he handed out the same number of each kind of candy bar.

( By cross multiplication )
( Division property of equality )
From equation (1),





Hence, he bought 9 packages of chocolate bars.
Answer:
see below
Step-by-step explanation:
f(x) = –(x + 6)(x + 2)
The function is increasing until it reaches the vertex, so it will increase until x=-4. The function will decrease after the vertex, so after x = -4
increasing: -∞ < x < -4
decreasing : -4 < x < ∞