On the RS graph, given a point E, if r decreases and q decreases, the new point will be _______ point E.
to the right below 

to the right above 
to the right parallel to 

all of the above are possible 
none of the above 
On the RS graph, given a point E, if r decreases and q INCREASES, the new point will be _______ point E.
to the right above 

to the right below 
to the right parallel to 

all of the above are possible 
How can we logically explain that the area of partial insurance is below the full insurance line?
Full insurance requires q < I_{H}  I_{S} for a given value of q. To turn this into partial insurance, we need a q (partial) lower than q to satisfy q(partial) < I_{H}  I_{S} So for every point on the full insurance curve we lower q which would lower the intersection point between I_{H}' and I_{S}' at every level of I_{H}' and all these points are below the full insurance line. 
Full insurance requires q = I_{H}  I_{S} for a given value of q. To turn this into partial insurance, we need a q (partial) HIGHER than q to satisfy q(partial) < I_{H}  I_{S} So for every point on the full insurance curve we lower q which would lower the intersection point between I_{H}' and I_{S}' at every level of I_{H}' and all these points are below the full insurance line. 
Full insurance requires q = I_{H}  I_{S} for a given value of q. To turn this into partial insurance, we need a q (partial) LOWER than q to satisfy q(partial) > I_{H}  I_{S} So for every point on the full insurance curve we lower q which would lower the intersection point between I_{H}' and I_{S}' at every level of I_{H}' and all these points are below the full insurance line. 
Full insurance requires q = I_{H}  I_{S} for a given value of q. To turn this into partial insurance, we need a q (partial) LOWER than q to satisfy q(partial) < I_{H}  I_{S} So for every point on the full insurance curve we INCREASE q which would lower the intersection point between I_{H}' and I_{S}' at every level of I_{H}' and all these points are below the full insurance line. 
Full insurance requires q = I_{H}  I_{S} for a given value of q. To turn this into partial insurance, we need a q (partial) LOWER than q to satisfy q(partial) < I_{H}  I_{S} So for every point on the full insurance curve we lower q which would lower the intersection point between I_{H}' and I_{S}' at every level of I_{H}' and all these points are below the full insurance line. 
Given a point E in an IHIS space, if r decreases and q decreases where is the new point (E') relative to point E? a. above left of point E b. above right of point E c. below left of point E d. below right of point E e. could either be above right, parallel right, or below right of point E
Answer : The point E' will lie to the c. below left of point E as given that both r decreases and q decreases at this new point (E') relative to point E which is only possible to the left down point to E as shown in the diagram below:
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