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(Passage 2, Lines 1-2, Para. 5)
Scientists have reason to think that a man can put up with far more radiation than 0.1 rem without being damaged;
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更多“（Passage 2, Lines 1-2, Para. 5…”相关的问题
(Passage 3, Lines 4-5, Para. 2)
... we find that we have to deal with things like climate, soil, plants, and such-like factors common to all biological situations;
Which of the following terms is defined in the passage？
A．pheromones (line 2)
B．colony integrity (lines 11-12)
C．mass migrations (line 14)
D．private messages (lines 18-19)
S3.(Lines 5-6, Para. 2, Passage 3)
What easier way is there for a nurse, a policeman, a barber, or a waiter to lose professional identity (身份) than to step out of uniform？
S2.(Lines 1-2, Para. 6, Passage 2)
But for many, the fact that poor people are able to support themselves almost as well without government aid as they did with it is in itself a huge victory.
(Passage 2, Lines 1-2, Para. 5)
While both groups did better than chance would predict, nearly half the participants in each group made the wrong choice two or more times.
Which of the following terms is defined in the passage？
A．resonator (line 2)
B．solo (line 7)
C．left-hand technique (line 25)
D．fingering patterns (lines 25-26)
A、#1-初始参数 N=1000 #重复次数 n=c(22,40,50,64) #样本数 y1=y2=y3=c() #空向量，用于保存频数、频率和理论频率 #2-计算 #随机抽取n个数并进行判断，是则返回TRUE，否则返回FALSE fun0305=function(n){ t1=round(runif(n,1,365),digits=0) #在1-365中随机抽取n个数，并四舍五入 if(any(diff(sort(t1))==0)) return(T) else return(F) #进行判断 } #重复1000次，并返回结果 for(i in 1:length(n)){ y=sapply(rep(n[i],N),fun0305) y1[i]=sum(y) #计算频数 } y2=y1/N #计算频率 y3=1-choose(365,n)*factorial(n)/365^n #计算理论值 z=rbind(y1,y2,y3) colnames(z)=paste("r=",n,sep="") #列名 rownames(z)=c("频数","频率","理论频率") #行名 z
B、n=c(100,1000,5000,10000) fun0306=function(x){ sum(abs(runif(x,0,60)-runif(x,0,60))＜=20) } y="sapply(n,fun0306)" rbind(y,y＞ C、# 程序ch010304.R n=c(100,1000,5000) p0=0.9;p1=0.45;p2=0.55;p3=0.6 pb=1-(1-p1)*(1-p2)*(1-p3) #计算能够回答出结果的次数 fun0307=function(n,p){ sum(runif(n)＜=p) } #进行计算 sapply(n,fun0307,p="p0)"＞ D、#参数 sigma=1 u=c(-2,0,2) #计算和绘图 x=seq(-6,6,0.1) t1=t2=list() for(i in 1:3){ t1[[i]]=dnorm(x,u[i],sigma) t2[[i]]=pnorm(x,u[i],sigma) } par(mar=c(2,2,2,1)) plot(x,t1[[1]],xlim=c(-6,6),type="l",lty=2,col=2) lines(x,t1[[2]]) lines(x,t1[[3]],col=4,lty=3) text(u,0.35,paste("u=",u,sep=""),col=c(2,1,4)) plot(x,t2[[1]],xlim=c(-6,6),type="l",lty=2,col=2) lines(x,t2[[2]]) lines(x,t2[[3]],col=4,lty=3) text(u,0.5,paste("u=",u,sep=""),col=c(2,1,4))
(Passage 4, Lines 2-3, Para. 1)
Even a skilled writer probably could not describe all the features that make one face different from another.
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