Answer:
Mean x¯¯¯ 2.2
Median x˜ 1
Range = 9
Q1 --> 0
Q2 --> 1
Q3 --> 3
Interquartile range: 3
Mean Absolute Deviation (MAD): 2.08
Standard deviation : 2.71
Step-by-step explanation:
1. Number of absent students:
0,1,1,0,3,2,5,1,9,0
Find the:
Mean
0, 0, 0, 1, 1, 1, 2, 3, 5, 9/10
= 22/10
= 2.2
Median
0, 0, 0, 1), 1, 1, (2, 3, 5, 9
= 1 + 1/2
= 2/2
= 1
Lower quartile
Upper quartile
Range
= 9 - 0
= 9
Interquartile range = 3
Mean absolute deviation = 2.08
Standard deviation = 2.71
Step-by-step explanation:
This problem bothers on the density of substances
Given data
Mass of substance m= 16kg
Volume of substance v = 2L
the equation that relates the volume. V. to the mass, m is the expression for density
ρ= mass/volume
Substituting our given data
ρ= 16/2= 8kg/L
Using the density calculated above we can get the volume liters, of 128kg of liquid
8= 128/v
v=128/8
v= 16L
The volume is 16L
The density of a substance is expressed as the mass per unit volume
4+|k+1|<15
-4 -4 15-4=11
|k+1|<11 |k+1|=k+1
k+1<11
-1 -1
k<10
Answer:
ALL of the above
Step-by-step explanation:
I'm very positive this is right
on second thought...
