farmer has the ability to grow either corn or cotton or some
combination of the two. Given no other information, it follows that
the farmer’s opportunity cost of a bushel of corn multiplied by
his opportunity cost of a bushel of cotton ( )
A: is
equal to 0.
B: is
between 0 and 1.
C: is
equal to 1.
D: is
greater than 1.
举一反三
- Individuals Lin<br/>and Li<br/>both produce good X. Lin has a<br/>comparative advantage in the production of good X if Lin ( ) A: has a lower opportunity cost of producing good X than<br/>has Li. B: has a lower opportunity cost of producing good X than<br/>of producing good C: can produce more units of X in a given time period<br/>than can Li. D: can produce X using newer technology than can Li.
- 执行下面的语句后,输出结果是( )。[br][/br]s = 0[br][/br]for a in range(1, 5) :[br][/br]for b in range(1, a) :[br][/br]s += 1[br][/br]print(s) A: 0 B: 1 C: 5 D: 6
- The<br/>spinning process of combing pure cotton is followed by cotton<br/>assorting, cleaning and carding,(),<br/>()<br/>drawing 1~2<br/>passes, roving, () and finishing, to form the the desired cotton<br/>yarns or threads. A: Preparation for combing B: Combing C: Roving D: Spinning
- The<br/>American belief in equality of opportunity means that _____. A: all<br/>Americans are rich B: Americans<br/>believe that everyone should be equal C: everyone<br/>should have an equal chance to succeed D: all<br/>Americans are equally born
- 下面代码的输出结果是[br][/br]s=0[br][/br]while(s[=1):<br] print('计数:',s)[br][/br] s=s+1 A: 计数:1 B: 出错 C: 计数:0 D: 计数:0<br>计数:1
内容
- 0
以下能正确计算1*2*3*……*10的程序段是( )。 A: do<br>{<br>i=1; s=1;<br>s*=i;<br>i++;<br>} while(i<=10); B: do<br>{<br>i=1; s=0;<br>s*=i;<br>i++;<br>} while(i<=10); C: i=1; s=1;<br>do<br>{<br>s*=i;<br>i++;<br>} while(i<=10); D: i=1; s=0;<br>do<br>{<br>s*=i;<br>i++;<br>} while(i<=10);
- 1
下列Python语句后的显示结果是()。[br][/br]x=2[br][/br]y=2.0[br][/br]if(x==y):[br][/br] print("Equal")[br][/br]else:[br][/br] print("Not Equal") A: Equal B: Not Equal C: 编译错误 D: 运行时错误
- 2
求下面矩阵的 Cholesky 分解 （다음 행렬의 Cholesky factorization을 구하시오）. \begin{bmatrix}<br/>1\ \,\, 3\ \,\, 7\\ <br/>3\ 10\ 26\\ <br/>7\ 26\ 75\\<br/>\end{bmatrix} A: \(U=\begin{bmatrix}<br/>1\ 3\ 7\\ <br/>0\ 1\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) B: \(U=\begin{bmatrix}<br/>1\ 2\ 7\\ <br/>0\ 3\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) C: \(U=\begin{bmatrix}<br/>1\ 3\ 7\\ <br/>0\ 2\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) D: \(U=\begin{bmatrix}<br/>1\ 3\ 1\\ <br/>0\ 1\ 5\\ <br/>0\ 0\ 7\\<br/>\end{bmatrix}\) E: \(U=\begin{bmatrix}<br/>1\ 2\ 7\\ <br/>0\ 3\ 1\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\)
- 3
The sum of frequencies for all classes will always equal 1. (<br/>)
- 4
若a = 1，下列各式的结果是什么? 1. ! a | a 2. ~ a | a 3. a ^ a 4. a >> 2 [br][/br] [br][/br] 解： 1． _____ 2．______ 3．_______ 4．_______