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.
7*8-4(6-2)+18
56-4(4)+18
56-16+18
40+18
58
6^2-(8*3)+2^2(7*3)
36-24+4(21)
12+84
96
Answer:
51 + 29
Step-by-step explanation:
Answer:
Machine's useful number of years = 9 years
Step-by-step explanation:
Using the straight line method, depreciation is calculated as the difference between the cost of the equipment minus the salvage value, all divided by the number of useful years.
Yearly Depreciation
= (Cost - Salvage value) ÷ (Number of useful years)
Yearly depreciation = P20,000
Cost = P200,000
Salvage Value = P20,000
Number of useful years = n
20000 = (200000 - 20000) ÷ n
20000 = (180000/n)
n = (180000/20000) = 9 years
Hope this Helps!!!
The absolute value of zero is zero