美国学术能力评估考试(SAT)

填空题If 6w is an integer and 21/2 < 5w < 31/2, what is the value of w?____

题目
填空题
If 6w is an integer and 21/2 < 5w < 31/2, what is the value of w?____
参考答案和解析
正确答案: 2/3 or 0.666
解析:
因为根据题干2/2 < 5w < 3/2,则5/2 < 5w < 7/2,5/10 < w <7/10,又因为6w是正数,所以w一定是1/6的倍数,所以w的值为4/6 = 0.666.
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相似问题和答案

第1题:

单击命令按钮时,下列程序的执行结果为 Private Sub Command1_Click() Dim a As Integer,b As Integer,c As Integer a=2:b=3:C=4 Print P2(c,b,A)End Sub Private Function P1(x As Integer,y As Integer,z As Integer) P1=2 * X + y + 3 * z End Function Private Function P2(x As Integer,y As Integer,z As Integer) P2=P1(z,x,y) + X End Function

A.21

B.19

C.17

D.34


正确答案:A
解析:对于多个过程或函数依次调用和处理与简单过程调用处理一样,本题先调用函数P2,而函数P2又调用函数P1,结果返回的顺序是从P1到P2,P2计算后输出到调用的地方。

第2题:

单击一次窗体之后,下列程序代码的执行结果为______。 Private Sub Command1_ Click() Dim a As Integer, b As Integer, c As Integer a = 1: b = 2: c = 4 Print Value(a, b,C)End Sub Function Num(x As Integer, y As Integer, z As Integer) Num = x * x + y * y + z * z End Function Function Value(x As Integer, y As Integer, z As Integer) Value = Num(x, y, z) + 2 * x End Function

A.21

B.23

C.19

D.35


正确答案:B

第3题:

单击命令按钮时,下列程序的执行结果为

Private Sub Command1_Click()

Dim a As Integer, b As Integer, c As Integer

a=2: b=3: c=4

Print P2(c, b, A.

End Sub

Private Function P1(x As Integer, y As Integer, z As Integer)

P1=2 * x + y + 3 * z

End Function

Private Function P2(x As Integer, y As Integer, z As Integer)

P2=P1(z, x, y) + x

End Function ( )。

A.21

B.19

C.17

D.34


正确答案:A

第4题:

Q4 :

int iVal1 = 0, iVal2 = 0;

int * ipVal;

ipVal = &iVal1; What is the Value of ipVal = *ipVal =

ipVal = (int*)iVal1; What is the Value of ipVal = *ipVal =

iVal2 = (int)&iVal1; What is the Value of ipVal = *ipVal =


正确答案:
 

第5题:

以下程序运行后,单击命令按钮,窗体显示的结果是 ______。 Private Function p1(x As Integer,y As Integel,z As Integer) p1=2*x+y+3*z End Function Private Function p2(x As Integer,y As Integer,z As Integer) p2=p1(2,y,x)+x End Function Private SubCommandl_Click() Dim a As Integer Dim b As Integer Dim c As Integer a=2:b=3:c=4 Print p2(c,b,A)

End Sub

A.19

B.21

C.22

D.23


正确答案:D
解析:当程序执行到Print p2(c,b,a)语句时,首先调用函数过程p2,将实参c、b、a,的值4、3、2按顺序传递给p2的形参x、y、z,使形参x、y、z的值分别是4、3、2。当执行到函数过程p2中的语句p2=p1(z,y,x)+x时,又去调用函数过程p1,并将实参x、y、x的值2、3、4按顺序传递给形参x、y、z,使函数过程p1的形参x、y、z的值分别是2、3、4。于是执行p1中的语句p1=2*x+y+3*x=2*2+3+3*4=4+3+12=19。即函数过程p2中p1(2,y,x)的值为19,因此p2=p1(2,y,x)+x=19+4=23。该值返回给Command1_Click事件过程中的p2(c,b,a),所以窗体显示的值是23。

第6题:

单击命令按钮时,下列程序代码的执行结果为 ( ) Function FirProc(x As Integer, y As Integer, z As Integer) FirProc=2*x+y+3*z End Function Function SecProc(x As Integer, y As Integer, z As Integer) SecProc=FirProc(z, x, y)+x End Function Private Sub Commandl Click() Dim a As Integer, b As Integer, c As Integer a=2 :b=3 :c=4 Print SecProc(c, b,A)End Sub

A.21

B.19

C.17

D.34


正确答案:A
解析:执行语句Print SecProc (c,b,a)时,调用SecProc函数,此时将实参c, b,a的值对应传递给形参x,y,z。得SecProc =FirProc(a,c,b)+c,此时又需要调用Fir- Proc函数将a,c,b的值传递给对应形参x,y, z。在FirProc函数中执行语句FirProc=2*x +y+3*z即执行语句FirProe=2*a+c+3 *b其结果值为2*2+4+3*3即17。故 FirProe(a,c,b)的返回值为17。再与c相加即得SecProc函数的返回值结果21。选项A正确。

第7题:

单击一次命令按钮,下列程序代码的执行结果为( )。 Private Sub Commandl_Click() Dima As Integer,b As Integer,CASInteger a=2:b=3:c=4 PrintP2(c,b,A) End Sub PrivateFunctionPl(X As Integer,y As Intege;z As Integer) P1=2*x+y+3*Z End Function Priva

A.21

B.19

C.17

D.34


正确答案:A

第8题:

下列程序运行后,单击命令按钮,窗体显示的结果为( )。 Private Function pl(x As Integer,y As Integer,z As Integer) pl=2*x+y+3*z End Function Private Function p2(X As Integer,y As Integer,z As Integer) p2=p1(z,y,x)+x End Function Private Sub Commandl_Click()

A.23

B.19

C.21

D.22


正确答案:A

第9题:

Q2 : Declare Function Multiply_Add Lib “test” (ByRef A as integer ,VeRef B as integer , ByVal c as integer ) as integer

Q2.1: Explain what is Declare Function used for ?

Ans :

Q2.2 : Explain the difference between ByRef and ByVal ?

Ans :


正确答案:
 

第10题:

下列程序的执行结果为 Private Sub Comrnandl_Click( ) Dim p As Integer, q As Integer p=12:q=20 Call Value(p, q) Print p; q End Sub Private Sub Value(ByVal m As Integer, ByVal n As Integer) m=m * 2: n=n - 5 Print m; n End Sub

A.20 12 20 15

B.12 20 12 25

C.24 15 12 20

D.24 12 12 15


正确答案:C
解析:被调过程Value的两个参数m和n前面都有关键字“ByVal”修饰,即在主调过程调用此过程时,实参与形参之间是以传值方式传递信息的,而当实参与形参以传值方式相结合时,形参的改变并没有影响到实参,单单就这一点来说,我们可知,当程序代码执行主调过程中的PrimP;q语句时,p和q的值应不变,还是12,20,这样就能够排除选项A)和选项B)了。当主调过程调用被调过程时,把实参p,q的值分别传给形参m和n,这就是说此时形参m值为12,n值为20,执行被调过程中的m=m*2:n=n-5语句后,m值为24,n值为15,所以Printm;n后,程序输出的结果为24和15。最后程序输出的结果如选项C)所示。