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A new study has found that it may be possible to train people to be more intelligent,increasing the brainpower they had at birth. Until now,it has been widely assumed that the kind of mental ability that allows us to solve new problems without having any relevant previous experience—what psychologists call fluid intelligence—is innate and cannot be taught(though people can raise their grades on tests of it by practicing). But in the new study,researchers describe a method for improving this skill,along with experiments to prove it works. The key, researchers found, was carefully structured training in working memory—the kind that allows memorization of a telephone number just long enough to dial it. This type of memory is closely related to fluid intelligence,so the researchers reasoned that improving it might lead to improvements in fluid intelligence. First they measured fluid intelligence of volunteers using standard tests. Then they trained each in a complicated memory task—the child’s card game,in which they had to recall a card they saw and heard. During the course, they needed to ignore irrelevant items,monitor ongoing performance,manage two tasks at the same time and connect related items to one another in space and time. The four groups experienced a half-hour of training daily for 8,12,17 and 19 days,respectively. To make sure they were not just improving their test-taking skills,the researchers compared them with control groups that took the tests without the training. The results, published Monday in The Proceedings of the National Academy of Sciences,were striking. Improvement in the trained groups was a lot greater. Moreover,the longer they trained, the higher their scores were. All performers,from the weakest to the strongest,showed significant improvement. “Our results show you can increase your intelligence with proper training.”said Dr Jaeggi, a co-author of the paper.“No one knows how long the gains will last after training stops,”he added,“and the experiment’s design did not allow the researchers to determine whether more training would continue to produce further gains.” 67.The researchers thought the key to improving the intelligence was ______________. A.memorizing telephone numbers B.improving working memory C.training in concentration D.recalling a card 68.The following aspects of the training help increase intelligence EXCEPT___________. A.ignoring irrelevant items B.monitoring ongoing performance C.managing two tasks at the same time D.using previous experience 69.When the experiment was conducted,the researchers______________. A.trained the four groups for the same period of time B.only made comparisons between the four groups C.compared the four groups with control groups D.trained the four groups together 70.By writing the article,the writer intends to ______________. A.inform the readers of a new study B.call on people to be trained to increase intelligence C.prove one’s born brainpower can be improved D.tell people the improved intelligence will last forever
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【单选题】( )即可处理的电力电缆(设备)缺陷,由运行维护(检查)人员与管理或检修部门的有关技术人员具体研究处理方案,以便随时安排检修处理。
A.
计划停电
B.
非计划停电
C.
无须停电
【判断题】停电检修时必须把各方面电源完全断开,与停电设备有关的变压器和电压互感器还必须从低压侧断开。
A.
正确
B.
错误
【简答题】某工程钻孔 3号土样3-1粉质黏土和3-2淤泥质黏土的压缩试验数据列于下表,试计算压缩系数a 并对其压缩性进行评价: 垂直压力( KPa ) 0 50 100 200 300 400 孔隙比 土样 3-1 0.866 0.799 0.770 0.736 0.721 0.714 土样 3-1 1.085 0.960 0.890 0.803 0.748 0.707
【多选题】糖皮质激素的生理作用包括
A.
促进蛋白质分解
B.
使淋巴细胞减少
C.
升高血糖
D.
使胃酸和胃蛋白酶增加
E.
刺激二型肺泡细胞产生二软脂酰软磷脂
【简答题】动物园里有大象9头,金丝猴4只。金丝猴的数量是大象的几分之几?
【简答题】设inta,b=5;,则执行表达式a=b-2==3后a的值是______。
【单选题】某正常固结土层厚 3.5m ,平均自重应力 p cz =100kPa ,压缩试验数据如表所示,建筑物平均附加应力 p 0 =200kPa ,试计算该土层的最终沉降量 =( )cm 压力 p (kPa) 0 59 100 200 300 400 孔隙比 e 0.984 0.900 0.828 0.752 0.710 0.680
A.
226
B.
22.6
C.
242
D.
24.2
【单选题】检修部分若分为几个在电气上不相连接的部分[如分段母线以隔离开关(刀闸)或断路器(开关)隔开分成几段];则各段应分别验电接地短路。降压变电所全部停电时,应将各个可能来电侧的部分接地短路,其余部分不必每段都装设接地线或合上接地刀闸。接地线、接地刀闸与检修设备之间不得连有()。
A.
隔离开关
B.
断路器(开关)或熔断器
C.
断路器(开关)和隔离开关
D.
断路器(开关)或隔离开关
【多选题】糖皮质激素的生理作用包括
A.
升高血糖
B.
促进蛋白质合成
C.
促进脂肪分解
D.
轻度的潴钠排钾
E.
降低血钙
【单选题】如图,已知△ACD ∽ △ABC,∠1=∠B,下列各式正确的是(  )
A.
AD AB = AC AB = CD BC
B.
AD AC = AC AB = CD BC
C.
AD CD = AB AC = CD BC
D.
AD AB = AB AC = CD BC
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