单选题A bag contains three green marbles, four blue marbles, and two orange marbles. If a marble is picked at random, what is the probability that an orange marble will NOT be picked?A 1/4B 1/3C 4/11D 1/2E 9/7

题目
单选题
A bag contains three green marbles, four blue marbles, and two orange marbles. If a marble is picked at random, what is the probability that an orange marble will NOT be picked?
A

1/4

B

1/3

C

4/11

D

1/2

E

9/7


相似考题
更多“单选题A bag contains three green marbles, four blue marbles, and two orange marbles. If a marble is picked at random, what is the probability that an orange marble will NOT be picked?A 1/4B 1/3C 4/11D 1/2E 9/7”相关问题
  • 第1题:

    对于这样的一个枚举类型:

    enum Color:byte

    {

    Red,

    Green,

    Blue,

    Orange

    }


    正确答案:
    答:string[] ss=Enum.GetNames(typeof(Color));
         byte[] bb=Enum.GetValues(typeof(Color));

  • 第2题:

    7岁的儿童用药剂量为成人的

    A、1/5~1/4

    B、1/4~1/3

    C、1/3~2/5

    D、2/5~1/2

    E、1/2~3/5


    参考答案:D

  • 第3题:

    以下对枚举类型名的定义中正确的是( )。

    A.enum a={one,two,three};

    B.enum a{one=9,two=-1.three=200};

    C.enum a={"one","two","four"};

    D.enum a{"nine","two","three"};


    正确答案:B
    解析:枚举类型的定义类似于结构体和共用体,所以选项A和C不正确。大括号中的枚举元素应该使用合法标识符来定义,而不是字符串常量,所以选项D不正确。故应该选择B。

  • 第4题:

    在列表:c=["black","red","green","yellow","orange","blue"]中,设定变量j=3,那么运行代码turtle.pencolor(c[j])后,画笔的颜色将变成黄色。


    正确答案:正确

  • 第5题:

    Which of the following is the correct arrangement of the data pairs in a T568B wire pattern?()

    • A、Orange & orange/white on pins 1 & 2
    • B、Blue & blue/white on pins 4 & 5
    • C、Orange & orange/white on pins 4 & 5
    • D、Brown & brown/white on pins 1 & 2
    • E、Green & green/white on pins 3 & 6

    正确答案:A,E

  • 第6题:

    问答题
    If there are only red, blue, and green marbles in a jar, what is the ratio of red to blue marbles?  (1) The ratio of red to green marbles is 2:3.  (2) The ratio of green to blue marbles is 6:5.

    正确答案: C
    解析:
    题目中含三个未知量,(1)、(2)条件相结合可以计算红色和蓝色珠子的比例,两个条件无法单独决定,故本题选C项。

  • 第7题:

    单选题
    A jar contains 10 blue, 8 green, and 6 red marbles. Every time a marble is removed from the jar, it is not replaced. What is the probability, to the nearest hundredth, that the second marble chosen is green if the first marble chosen is green?
    A

    0.28

    B

    0.29

    C

    0.30

    D

    0.31

    E

    0. 32


    正确答案: C
    解析:
    Because the first marble drawn is green that means there are a total of 23 marbles remaining. Seven of these are green. P (second green) = 7/23= 0.30

  • 第8题:

    单选题
    A student is instructed to arrange four cards in a row on a table. She has six cards to choose from, each of which has a different color: black, red, blue, green, yellow, and brown. If the student follows these instructions but otherwise chooses her cards randomly, what is the probability that her arrangement will be blue, red, yellow, and green, in that order?
    A

    1/90

    B

    1/ 180

    C

    1/360

    D

    1/540

    E

    1/720


    正确答案: B
    解析:
    从6张不同颜色的卡片中选出其中四张排列为1列,第一张卡片有6中选择,第二张卡片有5中选择,以此类推,第三张和第三张卡片分别有4种和3种选择,所以将4张卡片排在一起的可能性为6 x 5 x 4 x 3 = 360种。所以概率为1/360。

  • 第9题:

    单选题
    Which of the following is the correct pin out for T568B?()
    A

    Orange/white, orange, green/white, blue, blue/white, green, brown/white, brown

    B

    Blue, blue/white, orange/white, orange, brown/white, brown, green/white, green

    C

    Green/white, green, orange/white, blue, blue/white, orange, brown/white, brown

    D

    Orange/white, orange, brown/white, brown, green/white, green, blue, blue/white


    正确答案: C
    解析: 暂无解析

  • 第10题:

    多选题
    Which of the following is the correct arrangement of the data pairs in a T568B wire pattern?()
    A

    Orange & orange/white on pins 1 & 2

    B

    Blue & blue/white on pins 4 & 5

    C

    Orange & orange/white on pins 4 & 5

    D

    Brown & brown/white on pins 1 & 2

    E

    Green & green/white on pins 3 & 6


    正确答案: A,D
    解析: 暂无解析

  • 第11题:

    单选题
    Each of seven identical cards is numbered on one of its sides with a different integer from 1 to 7. The cards are turned over and then shuffled so that they are not in a predictable order. If two cards are selected at random without replacement, what is the probability that their sum is at least l1?
    A

    4/9

    B

    4/21

    C

    49/81

    D

    7/10

    E

    4/7


    正确答案: A
    解析:
    The number of all possible sums is 7×6=42. Since 8, {(7, 6), (7, 5), (7, 4), (6, 7), (6, 5), (5, 7), (5, 6), (4, 7)}, of the 42 possible sums are 11 or more, the probability of picking tow cards whose sum will be at least 11 is 8/42=4/21.

  • 第12题:

    单选题
    What is the purpose of the final sentence of the passage?
    A

    To express pride in the cultural treasures of Athens.

    B

    To refute the argument that more people can see the Elgin Marbles at the British Museum than at the Acropolis.

    C

    To comment on the relative number of tourists at two of Europe’s most famous tourist attractions.

    D

    To express concern that the large number of tourists on the Acropolis will damage the Elgin Marbles, should they be returned.

    E

    To provide an appropriate end to a rousing debate.


    正确答案: D
    解析:
    推断题。根据第二段“They further argue that more people can see the treasures if they are proudly displayed in a major museum”可知,文章最后一句是为了强调Elgin Marbles送返希腊的原因,故B项最合适。

  • 第13题:

    以下对枚举类型名的定义中正确的是( )

    A.enum a={one,two,three};

    B.enum a{one=9,two=-1,three=200};

    C.enum a={"one","two","four"};

    D.enum a{"nine","two","three"};


    正确答案:B

  • 第14题:

    阅读以下说明和 Java 代码,填补代码中的空缺,将解答填入答题纸的对应栏内。 【说明】 设计 RGB 方式表示颜色的调色板,进行绘图。其类图如图 6-1 所示。该程序的 Java代码附后。图6-1 类图

    【Java 代码】 //颜色类 class MyColor { private int red ,green, blue; public MyColor( ) { red = o; green = 0; blue = 0; } public MyColor(int red ,int green ,int blue) { this.red = red; this.green = green; this.blue = blue; } //其他方法略 public String toString( ) { return "Red: " + red + "\tGreen: " + green + "\tBlue " + blue; } } //调色板类 class Palette { public int number; / /颜色数 private (1)palette; //颜色表 public Palette( ) { number = 256; palette = new MyColor[number); } public Palette(MyColor[] palette ,int number) { (2)= number; (3)= palette; } //其他方法略 public String toString( ) { String str = ""; for (int i = 0; i < number; i++) { str +=i+ " : " + palette[i] + "\n"; } return str; } //绘图类 class Drawing { public (4) int COLORNUMBER = 16; public static void main(String[] args) { Palette palette; int red ,green ,blue; MyColor[] color = new MyColor[COLORNUMBER]; for (int i = 0; i < COLORNUMBER; i++) { red = (int) (Math.random( ) * 256); green = (int) (Math.random( ) * 256); blue = (int) (Math.random( ) * 256); color [i] = (5) (red ,green ,blue); } palette = new Palette(color ,COLORNUMBER); System.out.println(palette); } }


    正确答案:(1) MyColor[]
    (2) this.number
    (3) this.palette
    (4) static final
    (5) new MyColor

  • 第15题:

    You are preparing to take another vessel in tow.Which signal indicates,Hawser is made fast ______.

    A.Firing of a green star signal

    B.Firing of a red star signal

    C.An orange smoke signal

    D.Three white star rockets at 1-minute intervals


    正确答案:A

  • 第16题:

    晚幼粒细胞核凹陷程度为()

    • A、<1/4
    • B、<1/3
    • C、<1/2
    • D、>1/2
    • E、>1/3

    正确答案:C

  • 第17题:

    Which of the following is the correct pin out for T568B?()

    • A、Orange/white, orange, green/white, blue, blue/white, green, brown/white, brown
    • B、Blue, blue/white, orange/white, orange, brown/white, brown, green/white, green
    • C、Green/white, green, orange/white, blue, blue/white, orange, brown/white, brown
    • D、Orange/white, orange, brown/white, brown, green/white, green, blue, blue/white

    正确答案:A

  • 第18题:

    单选题
    A jar contains 54 marbles each of which is blue, green, or white. The probability of selecting a blue marble at random from the jar is 1/3, and the probability of selecting a green marble at random is 4/9. How many white marbles does the jar contain?
    A

    6

    B

    8

    C

    9

    D

    12

    E

    18


    正确答案: A
    解析:
    Since the probability of selecting a blue marble at random from the jar is 1/3 and the probability of selecting a green marble at random from the jar is 4, the probability of selecting a white marble is 1-(1/3 + 4/9) = 1- (7/9) = 2/9. If x marbles of the 54 marbles in the jar are white, then x/54 = 2/9, 9x = 108, x = 108/9 = 12.

  • 第19题:

    单选题
    A jar contains 4 red, 1 green, and 3 yellow marbles. If 2 marbles are drawn from the jar without replacement, what is the probability that both will be yellow?
    A

    3/8

    B

    3/28

    C

    1/4

    D

    3/56

    E

    5/56


    正确答案: B
    解析:
    There are eight marbles in the jar. The probability of choosing the first yellow marble is 3/8. The probability that your second marble will also be yellow is 2/7. The probability that both will be yellow is 3/8×2/7=6/56=3/28.

  • 第20题:

    判断题
    在列表:c=["black","red","green","yellow","orange","blue"]中,设定变量j=3,那么运行代码turtle.pencolor(c[j])后,画笔的颜色将变成黄色。
    A

    B


    正确答案:
    解析: 暂无解析

  • 第21题:

    单选题
    A bag contains three green marbles, four blue marbles, and two orange marbles. If a marble is picked at random, what is the probability that an orange marble will NOT be picked?
    A

    1/4

    B

    1/3

    C

    4/11

    D

    1/2

    E

    9/7


    正确答案: A
    解析:
    The total number of marbles in the bag is 3+4+2 or 9. Of these 9 marbles, 3+4 or 7 marbles are not orange. Hence, the probability that an orange marble will NOT be picked is 7/9.

  • 第22题:

    单选题
    Three fair coins are tossed at the same time. What is the probability that all three coins will come up heads or all will come up tails?
    A

    1/8

    B

    1/6

    C

    1/4

    D

    1/3

    E

    3/8


    正确答案: E
    解析:
    If three fair coins are tossed at the same time, the probability that all three will come up heads is (1/2)×(1/2)×(1/2) or 1/8. probability that all three coins will come up tails is also (1/2) × (1/2) × (1/2) or 1/8. Hence, the probability that all three coins will come up heads or all will come up tails is (1/8)+(1/8) or 1/4.

  • 第23题:

    单选题
    Three coins are tossed at the same time. What is the probability that exactly two heads are face up?
    A

    1/8

    B

    1/4

    C

    3/8

    D

    1/2

    E

    5/8


    正确答案: B
    解析:
    First, list all the possible outcomes when three coins are tossed in the air. When a coin is tossed, there are two outcomes, Heads or Tails. When three coins are tossed, there are 2×2×2= 8 possible outcomes:
    {HHH, HHT, HTH, THH, HTI', THT, TTH, TIT} Three of these outcomes have exactly two heads. Therefore, P (exactly two heads) =3/8.

  • 第24题:

    问答题
    Read the passage carefully to find the answers for Questions 1 to 5. Answer each question in a maximum of 10 words. Remember to write the answers on the Answer Sheet. Questions 1 to 5 are based on the following passage.Children’s Thinking  One of the most eminent of psychologists, Clark Hull, claimed that the essence of reasoning lies in the putting together of two “behavior segments” never actually performed before, in some novel way, so as to reach a goal.  Two followers of Clark Hull, Howard and Tracey Kendler, devised a test for children that was explicitly based on Clark Hull’s principles. The children were given the task of learning to operate a machine so as to get a toy. In order to succeed they had to go through a two-stage sequence. The children were trained on each stage separately. The stages consisted merely of pressing the correct one of two buttons to get a marble and of inserting the marble into a small hole to release the toy.  The Kendlers found that the children could learn the separate bits readily enough. Given the task of getting a marble by pressing the button they could got the marble; given the task of getting a toy when a marble was handed to them, they could use the marble to get the toy. (All they had to do was put it in a hole.) However, they did not for the most part “integrate”, to use the Kendlers’ terminology. They did not press the button to get the marble and then proceed without further help to use the marble to get the toy. Therefore, the Kendlers concluded that they were incapable of deductive reasoning.  The mystery at first appears to deepen when we learn, from another psychologist, Michael Cole, and his colleagues, that adults in an African culture apparently cannot do the Kendlers’ task either. It lessens, on the other hand, when we learn that a task was devised which was strictly analogous to the Kendlers’ one but much easier for the African males to handle.  Instead of the button-pressing machine, Cole used a locked box and two differently colored match-boxes, one of which contained a key that would open the box. Notice that there are still two behavior segments— “open the right match-box to get the key” and “use the key to open the box”—so the task seems formally to be the same. But psychologically it is quite different. Now the subject is dealing not with a strange machine but with familiar meaningful objects; and it is clear to him what he is meant to do. It then turns out that the difficulty of “integration” is greatly reduced.  Recent work by Simon Hewson is of great interest here for it shows that, for young children, too, the difficulty lies not in the inferential processes which the task demands, but in certain perplexing features of the apparatus and the procedure. When these are changed in ways which do not at all affect the inferential nature of the problem, five-year-old children solve the problem as well as college students did in the Kendlers’ own experiments.  Hewson made two crucial changes. First, he replaced the button-pressing mechanism in the side panels by drawers in these panels which the child could open and shut. This took away the mystery from the first stage of training. Then he helped the child to understand that there was no “magic” about the specific marble which, during the second stage of training, the experimenter handed to him so that he could pop it in the hole and get the reward.  A child understands nothing, after all, about how a marble put into a hole can open a little door. How is he to know that any other marble of similar size will do just as well? Yet he must assume that if he is to solve the problem. Hewson made the functional equivalence of different marbles clear by playing a “swapping game” with the children.  The two modifications together produced a jump in success rates from 30 percent to 90 percent for five-year-olds and from 35 percent to 72.5 percent for four-year-olds. For three-year-olds, for reasons that are still in need of clarification, no improvement— rather a slight drop in performance—resulted from the change.  We may conclude, then, that children experience very real difficulty when faced with the Kendler apparatus; but this difficulty cannot be taken as proof that they are incapable of deductive reasoning.  Questions:1.Howard and Tracey Kendler trained their subjects _______ in the two stages of their experiment.  2.What did the Kendlers conclude?  3.What objects did Cole use to do his experiment?  4.Who used a machine to measure deductive reasoning that replaced button-pressing with drawer   opening?  5.Hewson’s modifications resulted in a higher success rate for _______ children.

    正确答案:
    1.separately 题干中subjects指的是孩子们。由第三段The children were trained on each stage separately可知。
    2.Children were incapable of deductive reasoning. 由the Kendlers concluded that they were incapable of deductive reasoning可知。
    3.A locked box and two differently coloured match-boxes 由Cole used a locked box and two differently colored match-boxes可知他用的实验器材。
    4.Simon Hewson 由第八段Hewson made two crucial changes. First, he replaced the button-pressing mechanism in the side panels by drawers in these panels which the child could open and shut可知,Hewson做了两个重要的变化,第一个就是用机器代替了按扭装置。
    5.four-year-old and five-year-old 由文章倒数第二段The two modifications together produced a jump in …ercent to 72.5 percent for four-year-olds可知答案。
    解析: 暂无解析