Answer: A (x,y) is your original. Your answer after would be A' (-x,-y) you would change the sign. If its negative it becomes positive, and if it'd positive it becomes negative.
Answer:
2
Step-by-step explanation:
Given the data : 29, 2, 28, 30, 26, 31
Outlier ;
Lower :Q1 - (1.5 * IQR)
Upper : Q3 + (1.5 * IQR)
Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range
Using calculator :
Q1 = 26
Q3 = 30
IQR = (Q3 - Q1) = 30 - 26 = 4
Lower : 26 - (1.5 * 4) = 20
Upper : 30 + (1.5 * 4) = 36
Hence, the number in the given data which falls outside the range is 2
2000 divided by 5 is 400. 400 time 3 is 1200. You could also do 2000*.6 (because .6 = 3/5)
Data Set 1: (2, 2, 3, 4, 4, 5) Data Set 2: (5, 5, 10, 15, 15, 20) The difference between the interquartile ranges of the data se
vlabodo [156]
I R for data set 1 = 4-2 = 2
I R for data set2 = 15-5 = 10
Required difference is 10-2 = 8
Its 8.
