12.6 对概率规则的补充

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  概率规则的补充规则

  

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  Freddy wants to be a part of the school newspaper staff.  He is hoping the meetings are either on Monday or Wednesday since he has soccer practice on all the other days.  What is the probability that the meetings will not be scheduled on Tuesday, Thursday, Friday, Saturday, or Sunday?
  ::佛莱迪想成为学校报纸工作人员的一部分。 他希望会议是在星期一或星期三,因为他在所有其他日子都参加足球练习。 星期二、星期四、星期五、星期六或星期天,会议可能不会安排在星期二、星期五、星期六或星期天。

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  In this concept, you will learn about the Complementary Rule and how to calculate the probability of a complementary event occurring.
  ::在此概念中,您将了解补充规则以及如何计算发生补充事件的概率。

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  Complement Rule for Probability
  ::概率规则的补充规则

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  When one of two disjoint events must occur, the two events are said to be complementary .  Since one or the other event must occur, the sum of the probabilities of the two complementary events adds up to 1, or 100 percent of the outcomes of the events.
  ::当两起脱节事件之一必须发生时,这两起事件据说是相辅相成的。 由于其中一起或另一起事件必须发生,两个互补事件概率的总和相当于事件结果的1%或100%。

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  For example, the probability of the spinner below landing on either red or blue is 1.  The arrow will land on either red or blue 100 percent of the time.
  ::例如,在红色或蓝色下方着陆的旋转器的概率为 1。 箭头将全部降落在红色或蓝色上。 100%的时间, 箭头将降落在红色或蓝色上 。

  

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  $$\begin{array}{rl}P(\text{red or blue})& =P(\text{red})+P(\text{blue})\\ & =\frac{1}{2}+\frac{1}{2}=1\end{array}$$

  
  ::P(红或蓝色)=P(红)+P(蓝色)=12+12=1

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  Here are some situations which involve complementary events.
  ::以下是一些涉及补充活动的情况。

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  • Flipping a coin heads or flipping a coin tails.
    ::抛硬币头或抛硬币尾巴
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  • Turning on a light switch on or turning a light switch off.
    ::打开灯开关或关掉灯开关。
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  • Locking a door or unlocking a door.
    ::锁上门或锁上门。

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  Although some complementary events are 50-50 events, such as flipping a coin, not all are.
  ::虽然有些补充活动是50-50事件,例如抛硬币,但并非全都是。

  

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  For example, for the spinner shown:
  ::例如,对于显示的旋转器:

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  $$\begin{array}{rl}P(\text{blue or yellow})& =P(\text{blue})+P(\text{yellow})\\ & =\frac{3}{4}+\frac{1}{4}\\ & =1\end{array}$$

  
  ::P(蓝色或黄色)=P(蓝色)+P(黄)=34+14=1

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  The events $\begin{array}{r}B(\text{blue})\end{array}$ and $\begin{array}{r}Y(\text{yellow})\end{array}$ are complementary because their probabilities add up to 1. But the two complements are not equal in size.
  ::B(蓝)和Y(黄)事件是相辅相成的,因为它们的概率可达1,但两者的大小不相等。

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  Note that some disjoint events are NOT complementary events. Here, $\begin{array}{r}R(\text{red})\end{array}$ and $\begin{array}{r}B(\text{blue})\end{array}$ are disjoint events. However, their probabilities do NOT add up to 1 or 100 percent:
  ::请注意,有些脱节事件不是补充性事件。 这里, R( red) 和 B( B) 是脱节事件。 但是, 它们的概率不等于1%或100% :

  

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  $$\begin{array}{rl}P(\text{red or blue})& =P(\text{red})+P(\text{blue})\\ & =\frac{1}{4}+\frac{1}{4}\\ & =\frac{1}{2}\end{array}$$

  
  ::P(红或蓝色)=P(红)+P(蓝色)=14+14=12

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  Since the sum of any two complements is 1, if you know the probability of one complement , you can find the probability of the other.
  ::由于任何两个补充的总数是1, 如果你知道一个补充的概率, 你可以找到另一个的概率。

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  For events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ , suppose the probability of $\begin{array}{r}B\end{array}$ is 0.4. That means:
  ::对于事件A和B,假设B的概率是0.4。

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  $$\begin{array}{rl}P(A)+P(B)& =1\\ P(A)+0.4& =1\end{array}$$

  
  ::P(A)+P(B)=1P(A)+0.4=1

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  Therefore, the probability of $\begin{array}{r}P(A)\end{array}$ is 0.6, because:
  ::因此,P(A)的概率为0.6,因为:

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  $$\begin{array}{rl}P(A)+P(B)& =1\\ 0.6+0.4& =1\end{array}$$

  
  ::P(A)+P(B)=10.6+0.0.4=1

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  The Complement Rule states that for any two complements, $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ , the value of $\begin{array}{r}P(A)=1-P(B)\end{array}$ .  In other words, subtract the complement you know from 1 to find the unknown complement.
  ::补充规则规定,对于任何两个补充,A和B,P(A)=1-P(B)的值为P(A)=1-P(B)。 换句话说,将所知道的补充从1减去,以找到未知的补充。

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  $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are complements. $\begin{array}{r}P(B)=0.3\end{array}$ . Find $\begin{array}{r}P(A)\end{array}$ .
  ::A和B是补充。 P(B)=0.3. 查找P(A)。

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  To figure this out, subtract the complement you know, 0.3, from 1 to find $\begin{array}{r}P(B)\end{array}$
  ::要找出答案, 请减去您知道的 0. 3, 从 1 到 找到 P( B)

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  $$\begin{array}{rl}P(B)& =1-P(A)\\ & =1-0.3\\ & =0.7\end{array}$$

  
  ::P(B)=1-P(A)=1-0.3=0.7

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  What is the probability that the arrow will land on red, green, or yellow?
  ::箭头落下红、绿、黄的概率是多少?

  

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  The events are disjoint so the probability of one of them occurring is the sum of their individual probabilities.
  ::这些事件是脱节的,所以其中一种事件的发生概率是其个别概率的总和。

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  $$\begin{array}{rl}P(\text{red or blue or green})& =P(\text{red})+P(\text{blue})+P(\text{green})\\ & =\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\\ & =\frac{3}{4}\end{array}$$

  
  ::P(红色或蓝色或绿色)=P(红色)+P(蓝色)+P(绿色)=14+14+14=34

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  The probability of the complementary event of the spinner landing on yellow is:
  ::旋转器在黄色上着陆的补充事件的概率是:

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  $$\begin{array}{rl}P(\text{yellow})& =1-P(\text{red or blue or green})\\ & =1-\frac{3}{4}\\ & =\frac{1}{4}\end{array}$$

  
  ::P(黄)=1-P(红或蓝或绿)=1-34=14

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  The probability of the Mets winning tonight’s game is 0.6. Predict how likely it is for the Mets to lose tonight’s game.
  ::大都会队赢得今晚比赛的概率是0.6。 预测大都会队输掉今晚比赛的可能性。

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  Winning the game and losing the game are complementary events. 
  ::赢得比赛和输掉比赛是相辅相成的事件。

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  First, substitute the value for the known probability into the formula for complementary events.
  ::首先,将已知概率的值替换为互补事件的公式。

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  $$\begin{array}{rl}P(\text{lose})& =1-P(\text{win})\\ & =1-0.6\end{array}$$

  
  ::P( 关闭) = 1 - P( win) = 1 - 0.6

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  Next, subtract the known probability from 1.
  ::下一步,从 1 中减去已知的概率。

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  $$\begin{array}{rl}P(\text{lose})& =1-P(\text{win})\\ & =1-0.6\\ & =0.4\end{array}$$

  
  ::P( 关闭) = 1 - P( win) = 1 - 0. 6= 0. 4

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  Then, state the answer as the probability for the complementary event.
  ::然后,指出答案作为补充事件的概率。

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  The answer is the probability that the Mets will lose tonight's game is 0.4.  In other words, there is a 40% chance that the Mets will lose the game tonight.
  ::答案是大都会队今晚输球的概率是0.4。换句话说,大都会队今晚输球的概率是40%。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were given a problem about Freddy's school newspaper club.
  ::之前,你得到一个问题 关于弗雷迪的学校 报纸俱乐部。

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  If his soccer practice occurs on Tuesdays, Thursdays, Fridays, Saturdays, and Sundays, what is the probability that the newspaper club meetings will not fall on a soccer practice day?
  ::如果他的足球练习发生在星期二、星期四、星期五、星期六和星期天,

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  First, substitute the value for the known probability into the formula for complementary events.  We know that Freddy has soccer practice on 5 out of 7 days:

  $$\begin{array}{rl}P(\text{non-soccer practice days})& =1-P(\text{soccer practice days})\\ & =1-\frac{5}{7}\end{array}$$

  Next, subtract the known probability from 1:

  $$\begin{array}{rl}P(\text{non-soccer practice days})& =1-P(\text{soccer practice days})\\ & =1-\frac{5}{7}\\ & =\frac{2}{7}\end{array}$$

  Then, state the answer as the probability for the complementary event:
  ::首先,将已知概率的值替换为互补活动的公式。 我们知道弗雷迪在7天中的5天有足球练习:P(非足球练习日)=1-P(足球练习日)=1-57Next,减去1:P(非足球练习日)=1-P(足球练习日)=1-57=27, 回答是补充活动的概率:

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  The answer is the probability that the newspaper club will decide to meet on a day when soccer practice is not taking place is   $\begin{array}{r}\frac{2}{7}\end{array}$ , or  29%. 
  ::答案是,报纸俱乐部在足球练习不进行之日决定开会的概率为27人,占29%。

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  Example 2
  ::例2

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  Y and Z are complements. If the probability of Y occurring is 14%, what is the probability of Z occurring?
  ::Y和Z是补充。如果Y的概率是14%,Z的概率是多少?

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  First, substitute the value for the known probability into the formula for complementary events.  In this case, since, the known probability is reported in percentage, we will subtract from 100 instead of 1:

  $$\begin{array}{rl}P(\text{Z})& =100percent-P(\text{Y})\\ & =100percent-14percent\end{array}$$

   Next, subtract the known probability from 100 percent:

  $$\begin{array}{rl}P(\text{Z})& =100percent-P(\text{Y})\\ & =100percent-14percent\\ & =86percent\end{array}$$

  Then, state the answer as the probability for the complementary event:
  ::首先,将已知概率的值替换为互补事件的公式中。 在这种情况下,由于已知概率以百分比报告,我们将从100(而不是1:P(Z)=100%-P(Y)=100%-14%)中减去:P(Z)=100%-P(Y)=100%-100%-P(Y)=100%-14%=86%)中减去已知概率。

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  The answer is the probability that Z will occur is 86%.  
  ::答案是Z的概率是86%。

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  Example 3
  ::例3

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  A and C are complements. If C is .67, find A.
  ::A和C是补充的,如果C是.67,请找到A

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  First, substitute the value for the known probability into the formula for complementary events:

  $$\begin{array}{rl}P(\text{A})& =100percent-P(\text{C})\\ & =100percent-67percent\end{array}$$

  Next, subtract the known probability from 100 percent:

  $$\begin{array}{rl}P(\text{A})& =100percent-P(\text{C})\\ & =100percent-67percent\\ & =33percent\end{array}$$

  Then, state the answer as the probability for the complementary event:
  ::首先,将已知概率值替换为补充事件的公式:P(A) =100% - P(C) = 100% - 67% 下一步,从100% 中减去已知概率:P(A) = 100% - P(C) = 100% - 67% = 33%。 回答是补充事件的概率 :

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  The answer is the probability that  A will occur is 33%.  
  ::答案是A的发生概率是33%。

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  Example 4
  ::例4

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  If the Yankees have a 45% chance of winning tonight, what is the probability that they won't win?
  ::如果洋基队今晚有45%的胜选机会 他们赢不了的几率是多少?

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  First, substitute the value for the known probability into the formula for complementary events.  In this case, since, the known probability is reported in percentage, we will subtract from 100 instead of 1:

  $$\begin{array}{rl}P(\text{won't win})& =100percent-P(\text{win})\\ & =100percent-45percent\end{array}$$

  Next, subtract the known probability from 100 percent:

  $$\begin{array}{rl}P(\text{won't win})& =100percent-P(\text{win})\\ & =100percent-45percent\\ & =55percent\end{array}$$

  Then, state the answer as the probability for the complementary event:
  ::首先, 将已知概率的值替换为互补事件的公式 。 在此情况下, 由于已知概率以百分比报告, 我们将从 100 中减去 1: P( 不会赢) = 100% - P( win) = 100% - 45% 下一步, 从100% 中减去已知概率 : P( 不会赢) = 100% - P( win) = 100% - 45% = 55% , 说明补充事件的概率 :

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  The answer is the probability that the Yankees won't win is 55%.  
  ::答案是扬基队赢不了的概率是55%

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  Example 5
  ::例5

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  D and E are complements. If the probabiity of D is 0.02, what is E?
  ::如果D的正直程度是0.02,那么E是什么?

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  First, substitute the value for the known probability into the formula for complementary events:

  $$\begin{array}{rl}P(\text{E})& =1-P(\text{D})\\ & =1-0.2\end{array}$$

  Next, subtract the known probability from 1:

  $$\begin{array}{rl}P(\text{E})& =1-P(\text{D})\\ & =1-0.2\\ & =0.8\end{array}$$

  Then, state the answer as the probability for the complementary event:
  ::首先,将已知概率值替换为补充事件公式中的已知概率值:P(E)=1-P(D)=1-0.2Next,减去1:P(E)=1-P(D)=1-P(D)=1-0.2=0.8的已知概率,然后,将回答作为补充事件的概率:

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  The answer is the probability that E will occur is 0.8, or 80%.  
  ::答案是E的概率是0.8, 或80%。

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  Review
  ::回顾

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  Find the complement.
  ::寻找补充。

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  1. $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are complements. $\begin{array}{r}P(B)=0.15\end{array}$ . Find $\begin{array}{r}P(A)\end{array}$ .
     ::A和B是补充。P(B)=0.15。查找P(A)。
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  2. $\begin{array}{r}C\end{array}$ and $\begin{array}{r}D\end{array}$ are complements. $\begin{array}{r}P(C)=0.8\end{array}$ . Find $\begin{array}{r}P(D)\end{array}$ .
     ::C和D是补充。 P(C)=0.8。 查找 P(D) 。
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  3. $\begin{array}{r}G\end{array}$ and $\begin{array}{r}H\end{array}$ are complements. $\begin{array}{r}P(H)=49\mathrm{\%}\end{array}$ . Find $\begin{array}{r}P(G)\end{array}$ .
     ::G和H是补充。 P(H)=49%。 查找 P(G) 。
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  4. $\begin{array}{r}T\end{array}$ and $\begin{array}{r}S\end{array}$ are complements. $\begin{array}{r}P(T)=\frac{3}{8}\end{array}$ . Find $\begin{array}{r}P(S)\end{array}$ .
     ::T和S是补充。 P(T)=38。 查找 P(S) 。
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  5. $\begin{array}{r}L\end{array}$ and $\begin{array}{r}K\end{array}$ are complements. $\begin{array}{r}P(K)=0.07\end{array}$ . Find $\begin{array}{r}P(L)\end{array}$ .
     ::L和K是补充。 P( K) =0.07. 查找 P( L) 。
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  6. $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are complements. $\begin{array}{r}P(B)=0.125\end{array}$ . Find $\begin{array}{r}P(A)\end{array}$ .
     ::A和B是补充。 P(B)=0.125. Find P(A).
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  7. $\begin{array}{r}N\end{array}$ and $\begin{array}{r}M\end{array}$ are complements. $\begin{array}{r}P(N)=96.1\mathrm{\%}\end{array}$ . Find $\begin{array}{r}P(M)\end{array}$ .
     ::N和M是补充。 P(N)=96.1%。 查找 P(M) 。
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  8. $\begin{array}{r}Q\end{array}$ and $\begin{array}{r}Z\end{array}$ are complements. $\begin{array}{r}P(Q)=\frac{1}{5}\end{array}$ . Find $\begin{array}{r}P(Z)\end{array}$ .
     ::Q和Z是补充。 P( Q)=15。 查找 P( Z) 。

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  Write complementary or not complementary .
  ::写作补充或不补充。

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  9. Percentage of votes that 2 candidates get in a 2-candidate election
     ::2名候选人在2名候选人选举中获得的选票百分比
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  10. Percentage of votes that 3 candidates get in a 3-candidate election
      ::3名候选人在3轮选举中获得的选票百分比
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  11. Winning a game or losing a game
      ::赢一场游戏或输一场游戏
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  12. Choosing an odd or even integer from the set of positive integers
      ::从正整数组中选择奇数甚至整数
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  13. Choosing a number between 1 and 5
      ::选择 1 和 5 之间的数字
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  14. Choosing a color of paint
      ::选择油漆颜色

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。

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  Resources
  ::资源