<span>E = {x|x is a perfect square between 1 and 9} = {1,4
</span><span>F = {x|x is an even number greater than or equal to 2 and less than 9}
</span><span>D = {x|x is a whole number}
</span>
answer
D ∩ (E ∩ F) = 4
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.
Here is your answer
The angle opposite of the larger side is greater and vice versa.
Here
smallest side= AC (8cm)< BC (9cm) <AB (10cm)
So,
angles opposite to these sides are
B <A <C
HOPE IT IS USEFUL
Seven add five is 12 . A clock has four quarters , the first quarter is fifteen minutes.
Therefore, 5 1/5 hours after 7:00 = 12:15
178 * 18 + (98-2)
3204 + 96
3300