The violent crime rate (number of violent crimes per 1000 residents)in Meadowbrook is 60% higher now than it was four years ago. The corresponding increase for Parkdale is only 10%. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.
The argument above is flawed because it fails to take into account
a) Changes in population density of both Parkdale and Meadowbrook over the past four yrs.
b) how the rate of population growth in Meadowbrook over the past four yrs compares to the corresponding rate for Parkdale
c) the ratio of violent to non-violent crimes committed during the past four yrs in Meadowbrook and Parkdale
d) the violent crime rates in Meadowbrook and Parkdale four years ago
e) how Meadowbrook's expenditures for crime prevention over the past four yrs compare to Parkdale's expenditures
[spoiler]OA:D.
Why is A incorrect?[/spoiler]
The argument above is flawed because it fails to take into account
a) Changes in population density of both Parkdale and Meadowbrook over the past four yrs.
b) how the rate of population growth in Meadowbrook over the past four yrs compares to the corresponding rate for Parkdale
c) the ratio of violent to non-violent crimes committed during the past four yrs in Meadowbrook and Parkdale
d) the violent crime rates in Meadowbrook and Parkdale four years ago
e) how Meadowbrook's expenditures for crime prevention over the past four yrs compare to Parkdale's expenditures
[spoiler]OA:D.
Why is A incorrect?[/spoiler]
so D is right
