Answer:
Correct option is
C
36.25
Modal class =30−40
So we have, l=30,f0=12,f1=32,f2=20 and h=10
⇒ Mode=l+2f1−f0f2f1−f0×h
=30+2×32−12−2032−12×10
=30+6.25
=36.25
∴ Mode =36.25
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:
x-(x times 33%) or x times (100-33)%
Step-by-step explanation:
8x(5x2+3)=8x(15+3)=8x13=104
Answer:
(31)2 is the answer to (62)4
Step-by-step explanation: