m square +84 is the answer
Yes it is possible. Consider the following scenarios
Scenario A:
Min = 5
Q1 = 10
Median = 12
Q3 = 18
Max = 22
The IQR is equal to the difference of Q3 and Q1
IQR = Q3-Q1 = 18-10 = 8
The range is the difference of the min and max
Range = Max - Min = 22 - 5 = 17
So in summary for scenario A, we have
IQR = 8
Range = 17
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Now consider another scenario, call it scenario B, where
Min = 100
Q1 = 102
Median = 105
Q3 = 110
Max = 117
I claim that the IQR and Range for scenario B is going to be the same as in Scenario A. Let's find out
IQR = Q3 - Q1 = 110 - 102 = 8
Range = Max - Min = 117 - 100 = 17
So
IQR = 8
Range = 17
which is identical to scenario A.
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Scenario B has completely different data than scenario A, yet the IQR and Range are equal to scenario A's counterparts. This shows that it is possible to have 2 completely sets of data yet have the same IQR and range.
The wrap up here, and the answer to the question, is "yes it is possible" with the explanation given above.
Since the figures are similar, we can establish a rule of three as follows.
We know that the area of the smaller figure is
, and its volume is
. We also know that the area of the larger figure is
; since we don't now its volume, lets represent it with
:
We can conclude that the volume of the larger figure is
; therefore, the correct answer is
a.
-5 + x/4 < -7
X/4 < -7 + 5
X<-2 x 4
X = -8
