When packets are routed on the Internet, the route for a particular packet is not pre- determined. Instead, the next hop is decided by each router along the way.
How does that increase the fault tolerance of routing on the Internet?

Data is processed, stored, and transmitted as a series of 1s and 0s. each 1 or 0 is called a(n) . a series of eight 0s and 1s, called a(n) , represents one character—a letter, number, or punctuation mark. data becomes when it is presented in a format that people can understand and use. digital signals are represented by two different , such as +5 volts and +.2 volts. digital data can also take the form of light and dark spots etched onto the surface of a cd or the positive and negative orientation of particles on the surface of a hard disk. data consists of numbers that might be used in arithmetic operations. it can be represented digitally using the number system. data is composed of letters, symbols, and numerals that are not used in arithmetic operations. computers represent this type of data using , ebcdic, unicode, or utf-8. data is quantified using terms such as or kibibyte (1024 bytes), and prefixes such as or mebi (1,048,576), and giga or (1,073,741,824). to reduce file size and transmission times, digital data can be compressed. compression provides the means to compress data and reconstitute it into its original state. compression throws away some of the original data during the compression process. compressed files usually have at the end of the file name and are represented with special icons. these files have to be or unzipped before you can view the data they contain.

Your company has 1,500 desktop computers running windows 7. you want to upgrade them to windows 10. which type of microsoft license would be best suited in this situation?

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }