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# The violent crime rate (number of violent crimes per 1,000

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The violent crime rate (number of violent crimes per 1,000 [#permalink]

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10 Sep 2006, 23:08
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The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. 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 the population density of both Parkdale and Meadowbrook over the past four years

B:how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale

C:the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale

D:the violent crime rates in Meadowbrook and Parkdale four years ago

E:how Meadowbrookâ€™s expenditures for crime prevention over the past four years compare to Parkdaleâ€™s expenditures

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10 Sep 2006, 23:15
I think this is (D)
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10 Sep 2006, 23:55
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One more for D.

A:changes in the population density of both Parkdale and Meadowbrook over the past four years - violent crime rate = number of violent crimes per 1,000 residents. So, the violent crime rate already takes the number of people into account .

B:how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale - rate of population growth is irrelevant.

C:the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale - non violent crimes??? Out of scope

D:the violent crime rates in Meadowbrook and Parkdale four years ago - YES. Suppose M had a violent crime rate of 10 and P had a crime rate of 20. With a 60% rise M has a rate of 16. With a 10% rise P has a rate of 22. M > P : shows that argument is flawed.

E:how Meadowbrookâ€™s expenditures for crime prevention over the past four years compare to Parkdaleâ€™s expenditures - Irrelevant.

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Re: CR- violent crime rate (weaken) [#permalink]

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10 Sep 2006, 23:58
jerrywu wrote:
The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. 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 the population density of both Parkdale and Meadowbrook over the past four years

B:how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale

C:the ratio of violent to nonviolent crimes committed during the past four years 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 years compare to Parkdale?s expenditures

Between A and B
I ll take B
The violent crime rate (number of violent crimes per 1,000 residents)
in Meadowbrook is 60 percent higher now than it was four years ago and what if the population declined significantly over four year period?
The corresponding increase for Parkdale is only 10 percent.
If population for example doubled?[/i]
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11 Sep 2006, 00:11
ago
kripalkavi wrote:
One more for D.

A:changes in the population density of both Parkdale and Meadowbrook over the past four years - violent crime rate = number of violent crimes per 1,000 residents. So, the violent crime rate already takes the number of people into account .

B:how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale - rate of population growth is irrelevant.

C:the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale - non violent crimes??? Out of scope

D:the violent crime rates in Meadowbrook and Parkdale four years ago - YES. Suppose M had a violent crime rate of 10 and P had a crime rate of 20. With a 60% rise M has a rate of 16. With a 10% rise P has a rate of 22. M > P : shows that argument is flawed.

E:how Meadowbrook?s expenditures for crime prevention over the past four years compare to Parkdale?s expenditures - Irrelevant.

OK lets suppose that violent crime rates of both cities four years ago.But stop man,It was 4 years ago. And we are talking about present days
During four year period rates could change many times!
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11 Sep 2006, 01:41
Yurik79 wrote:
ago
kripalkavi wrote:
One more for D.

A:changes in the population density of both Parkdale and Meadowbrook over the past four years - violent crime rate = number of violent crimes per 1,000 residents. So, the violent crime rate already takes the number of people into account .

B:how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale - rate of population growth is irrelevant.

C:the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale - non violent crimes??? Out of scope

D:the violent crime rates in Meadowbrook and Parkdale four years ago - YES. Suppose M had a violent crime rate of 10 and P had a crime rate of 20. With a 60% rise M has a rate of 16. With a 10% rise P has a rate of 22. M > P : shows that argument is flawed.

E:how Meadowbrook?s expenditures for crime prevention over the past four years compare to Parkdale?s expenditures - Irrelevant.

OK lets suppose that violent crime rates of both cities four years ago.But stop man,It was 4 years ago. And we are talking about present days
During four year period rates could change many times!

conclusion is: residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale

don't you think that D directly weakens it?
The rate of population growth is quite irrelevant becase the crime rate mentioned is in terms of 1000 people.

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11 Sep 2006, 01:49
D for sure.

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11 Sep 2006, 01:51
D for sure.

The likelihood of something is a percentage. so population number is irrelevant. but 60% and 10% are compared with their own 4 years ago. so what are the current rate? whose current rate is higher? you gotta know those rates for years ago.

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11 Sep 2006, 01:58
straight D, this is pure math

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11 Sep 2006, 01:58
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# The violent crime rate (number of violent crimes per 1,000

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