Answer:
x = -0.8
<CAD = -8
Step-by-step explanation:
10x + 4 = 5x - 4 (Clean up)
10x = -8 (Divide each side by 10)
x = -0.8
<CAD = 5x - 4
<CAD = 5(-0.8) - 4
<CAD = -4 - 4
<CAD = -8
Answer:
C. I used to not get this but now I do. hope I helped
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.
There are105 different choices.
For 1 appetizer, there are 15 different combinations of entrees and desserts. 1 dessert has 5 entrees you can mix, and since there are 3 desserts, there are 15 choices. It goes the same for every appetizer until there is 7x15. 7Ax15choices equals 105 options.
The customer has 105 meal choices.
