You have :
--------------
DE arc = ( pi ) ( AD ) ( 2.36 radians / 2 pi radians ) = ( 2/3 ) ( AB ) ( 2.36 radians / 2 )
DE arc = ( 2/3 ( AB ) ( 1.18 radians )
BC arc = ( pi ) ( AB ) ( 1.18 radians / 2 pi radians )
BC arc = ( AB ) ( 0.59 radians )
BC arc / DE arc = ( AB ) ( 0.59 radians ) / ( 2/3 ) ( AB ) ( 1.18 radians )
BC arc / DE arc = ( AB ) ( 0.59 rad ) / ( 2/3 ) ( AB ) ( 1.18 rad )
BC arc / DE arc = ( 3/2 ) ( .59 rad / 1.18 rad ) = 3/4 <-------
Remmber
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)
conver to improper
4 and 1/5=20/5+1/5=21/5
2 and 1/3=6/3+1/3=7/3
(21/5)/(7/3)=(21/5)(3/7)=63/35=9/5=1 and 4/5
Given inequality: 2y−x ≤ −6
Option-1 : (-3,0)
2×0 - (-3) = 0 + 3 = 3 > -6
Not satisfied
Option-2 : (6,1)
2×1 - 6 = 2 - 6 = -4 > -6
Not satisfied
Option-3 : (1, -4)
2×(-4) - 1 = -8 - 1 = -9 < -6
Satisfied.
Thus, (1, -4) is a solution.
Option-4 : (0, -3)
2×(-3) - 0 = -6 - 0 = -6 = -6
Satisfied.
Thus, (0, -3) is a solution.
Option-5 : (2, -2)
2×(-2) - 2 = -4 - 2 = -6 = -6
Satisfied.
Thus, (2, -2) is a solution.
Solutions are: (1, -4), (0, -3) , (2, -2)
Answer:
The median weight for shelter A is greater than that for shelter B.
The median weight for shelter B is greater than that for shelter A.
The data for shelter A are a symmetric data set.
The data for shelter B are a symmetric data set.
The interquartile range of shelter A is greater than the interquartile range of shelter B.
