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【单选题】
听力原文: A potato farmer was sent to prison just at the time when he should have been digging the ground for planting the new crop of potatoes. He knew that his wife would not be strong enough to do the digging by herself, but that she could manage to do the planting and he knew that he did not have any friends or neighbors who would be willing to do the digging for them. So he wrote a letter to his wife which said, 'Please do not dig the potato field. I hid the money and the gun there.' Ten days later he got a letter from his wife. It said, 'I think somebody read your letter before it got out of the prison. Someone arrived here two days ago and dug up the whole potato field. What shall I do now?' The prisoner wrote back at once, 'Plant the potatoes, of course.' (30)
A.
Because his wife was ill in bed.
B.
Because the farm work needed him badly.
C.
Because the weather was so bad that his wife could not dig up the land.
D.
Because his wife didn't know how to do any farm work.
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【单选题】超星期刊的网址是?
A.
qikan.chaoxing.com
B.
www.qikan.com
C.
www.ssqikan.com
D.
ssqikan.com
【多选题】超星期刊的网址是什么
A.
www.chaoxing.com
B.
qikan.chaoxing.com
C.
www.faxian.com
D.
www.yz.chaoxing.com
【判断题】效用理论中,序数效用论与基数效用论的差异之一在于前者假定效用具有可加性
A.
正确
B.
错误
【单选题】以下为顺序表的查找算法,分析算法,请在空白处填上正确的语句 int LocateElem_Sq ( SqList L, ElemType X) //在 L 中查找第一值等于 X 的元素。若找到则返回该元素位序号;否则返回 0 { ________; while ( ( i <=L.length ) && ( L.elem[i-1] !=X) ) i++; if( ________ ) ...
A.
i=0, i
B.
i=1, i<=L.length< /div>
C.
i=1, i
D.
i=0, i<=L.length< /div>
【判断题】交流电桥只需复阻抗模的平衡就可以。( )
A.
正确
B.
错误
【多选题】现代管理思想阶段主要的流派有( )
A.
科学管理理论学派
B.
行为管理理论学派
C.
权变理论学派
D.
社会系统学派
E.
管理科学学派
【单选题】超星期刊的网址是()
A.
http://qikan.chaoxing.com
B.
http://www.qikan. com/
C.
http://qikan.superlib.com
D.
http://qikan.chaoxin.com/
【简答题】利用课程设计中中国药典的链接,查阅阿司匹林的质量标准,总结药品标准的主要内容包括哪几个部分。
【单选题】超星期刊的网址是?
A.
qikan.chaoxing.com
B.
www.cxqk.com
C.
www.cnki.net
D.
www.ldzy.com
【多选题】伦理问题,主要包括:创业者与原雇主之间的伦理,创业者与其他利益关系者涉及的伦理问题,比如等。
A.
法律法规
B.
人事伦理
C.
利益冲突
D.
客户欺诈
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【单选题】以下为顺序表的查找算法,分析算法,请在空白处填上正确的语句 int LocateElem_Sq ( SqList L, ElemType X) //在 L 中查找第一值等于 X 的元素。若找到则返回该元素位序号;否则返回 0 { ________; while ( ( i <=L.length ) && ( L.elem[i-1] !=X) ) i++; if( ________ ) ...
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