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请教t-4-q20

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楼主
发表于 2007-10-15 09:00:00 | 只看该作者

请教t-4-q20

T-4-Q20 天山-7-22

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.

文章涉及两个城市MP本身4年前和现在犯罪率的比较,结论是现在MP犯罪受害者多。

横向比较背景一致

The argument above is flawed because it fails to take into account

 

  1. Changes in the population density of both Meadowbrook and Parkdale over the past four years.

  2. How the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale

  3. The ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale

  4. The violent crime rates in Meadowbrook and Parkdale four years ago

  5. How Meadowbrooks’ expenditures for crime prevention over the past four years compare to Parkdale’s expenditures.

我怎么觉得应该选B呢?应该和人口的增长率来确定现在犯罪的百分比吧,请指教

沙发
 楼主| 发表于 2007-10-15 22:29:00 | 只看该作者
ding
板凳
 楼主| 发表于 2007-10-16 20:45:00 | 只看该作者
ding
地板
发表于 2007-10-16 20:54:00 | 只看该作者
题目说:M的犯罪率上升了60%,P的犯罪率上升了10%,结论是M的犯罪率比P的犯罪率高.
B选项说:M和P在这4年里的人口增长率------这是无关项,因为人口增长率与犯罪率无必然联系.
D选项说:4年前M和P的犯罪率------正确,因为知道了基数和变化量后,才能比较谁的犯罪率高.
5#
发表于 2009-7-18 15:24:00 | 只看该作者
6#
发表于 2009-8-4 09:37:00 | 只看该作者
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