Resolution typically is(71)2 in dots per inch(dpi), and is stated as the number of columns and rows of dots.A．computedB．stackC．measuredD．searched
Resolution typically is(71)2 in dots per inch(dpi), and is stated as the number of columns and rows of dots.
A．computed
B．stack
C．measured
D．searched
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● Routing protocols use different techniques for assigning（71）to individual networks．Further, each routing protocol forms a metric aggregation in a different way. Most routing protocols can use multiple paths if the paths have an equal（72）.Some routing protocols can even use multiple paths when paths have an unequal cost.In either case,load（73）can improve overall allocation of network bandwidth. When multiple paths are used,there are several ways to distribute the packets. The two most common mechanisms are per-packet load balancing and per-destination load balancing.Per-packet load balancing distributes the（74）across the possible routes in a manner proportional to the route metrics,Per-destination load balancing distributes packets across the possible routes based on（75）.（71）A. calls B. metrics C. links D. destinations （72）A. user B. distance C. entity D. cost （73）A. bracketing B. balancing C. downloading D. transmitting （74）A. destinations B. resources C. packets D. sources （75）A. destinations B. resources C. packets D. sources
Two sources of variation in the length of hex bolts exist. Production variation and measurement error. The QC manager knows that the standard deviation from these sources is 0.24" and 0.43" , respectively. What is the total standard deviation.A . .10 inch.B . .19 inch.C . .34 inch.D . .49 inch.E . .67 inch.
183 Two sources of variation in the length of hex bolts exist. Production variation and measurement error. The QC manager knows that the standard deviation from these sources is 0.24" and 0.43" , respectively. What is the total standard deviation.A. .10 inch.B. .19 inch.C. .34 inch.D. .49 inch.E. .67 inch
65 Two sources of variation in the length of hex bolts exist. Production variation and measurement error. The QC manager knows that the standard deviation from these sources is 0.24" and 0.43" , respectively. What is the total standard deviation.A. .10 inch.B. .19 inch.C. .34 inch.D. .49 inch.E. .67 inch.
Basically, a(71)is a standalone process or a set of integrated processes that runs on a router or server to control the flow of(72)application traffic(73)through it. Typically, firewalls are(74)on the entry point to a public network such as the Internet. They could be considered traffic cops. The firewall' s(75)is to ensure that all communication between an organization's network and the Internet conform. to the organization's security policies.A．routerB．clientC．firewallD．switch
阅读下面程序import javax.swing.JOptionPane；public class Comparison{public static void main(String args[]){String firstNumber, //用户输入第1个数据变量second Number, //用户输入第2个数据变量result； //输出结果变量int number1， //用于比较的第1个数number2； //用于比较的第2个数//用户输入第1个数据的字符串firstNumber=JOptionPane. showInputDialog(“输入第1个整数：”)；//用户输入第2个数据的字符串secondNumber=JOptionPane.showlnputDialog(“输入第2个整数：”)；//将字符串转换为整数类型number1=Integer.parseInt(firstNumber)；number2=Integer.parseInt(secondNumber)；//初始化结果变量___________________;//比较两个数据if(number1＝number2)result += number1 +" =" + number2;if(number1 != number2 )result += numberl +" !=" + number2;if( number1 ＜ number2 )result = result + "\n" + numberl +" ＜" + number2;if(number1 ＞ number2 )result = result + "\n" + number1 + "＞" + number2;if( number1 ＜= number2 )result = result + "\n" + number1 +" ＜=" + number2;if (number1 ＞= number2 )result = result + "\n" + number1 + "＞=" + number2;//显示结果JOptionPane.showMessageDialog(null,result," 比较结果",JOptionPane. INFORMATION_MESSAGE);System.exit(0);}}为使程序能正确运行并得到合理的输出结果，初始化结果变量语句(下划线处)应是A．result = ""B．result = nullC．result = number1D．result = number2
阅读下面程序import javax.swing．JOptionPane；public class Comparison {public static void main(String args[]) {String firstNumber， //用户输入第1个数据变量secondNumber， //用户输入第2个数据变量result； //输出结果变量int number1， //用于比较的第1个数number2； //用于比较的第2个数//用户输入第1个数据的字符串firstNuinber=JOptionPane.showInputDialog("输入第1个整数:")；//用户输入第2个数据的字符串secondNumber=JOptionPane.showInputDialog("输入第2个整数:")；//将字符串转换为整数类型number1=Integer.parseInt(firstNumber)；number2=Integer.parseInt(secondNumber)；//初始化结果变量______//比较两个数据if(number1==number2)result+=number1+"=="+number2；if(number1! =number2) result+=number1+"!="+number2；if(number1＜number2) result=result+"\n"+number1+"＜"+number2；if(number1＞number2) result=result+"\n"+number1+"＞"+number2；if(number1＜=number2) result=result+"\n"+number1+"＜="+number2；if(number1＞=number2) result=result+"\n"+number1+"＞="+number2：//显示结果JOptionPane.showMessageDialog(null,result."比较结果"，JOptionPane.INFORMATION_MESSAGE)；System．exit(0)；}}为使程序能正确运行并得到合理的输出结果，初始化结果变量语句(下画线处)应是A．result=" "；B．result=nullC．result=number1D．result=number2
用递归函数求解斐波那契序列，在划线处填上正确语句。 int fib（number){ if （__________ || number == 2) return 1; else: return fib（number - 1) + fib（number - 2);A．number == 1B．number == 0C．number == 3D．number != 1
23、假定 number=4，下面哪个语句不会出错？A．print(format(number, "2d"), format(number ** 1.5, "4d"))B．print(format(number, "2d"), format(number ** 1.5, "4.2d"))C．print(format(number, "2d"), format(number ** 1.5, "4.2f"))D．print(format(number, "2f"), format(number ** 1.5, "4.2d"))