Click here to see the source code for the kitty class: kitty. java the array mycats is an 8 element array of type kitty that has already been declared and initialized. write the expression(s) in the box below that prints the name of the fourth kitty (in the cell with index 3) to the console:

Problems: 1. using textbooks, reference books, and internet as your source of research, draw the following microprocessor microarchitectures i. intel 8086 ii. motorola 68000 i atmel atmega32 iv. mips single cycle v. arm cortex-m3 write an hdl module for a hexadecimal seven-segment display decoder. the input is 4-bit binary representing a hex number (0-f), and the output is 8-bit seven segment display bits (a-h). thus, the decoder must handle the digits 10 - 15 to display a-f respectively, in addition to 0-9 numbers. 2. design a 4-bit left and right rotator (both outputs). first sketch schematic diagrams of your design. then implement your design using hdl coding. 3. 4. design a modified priority encoder that receives an 8-bit input, a7: 0 and produces a 3-bit output, y2o. y indicates the most significant bit of the input that is true. y should be 0 if none of the inputs are true. give a simplified boolean equation, sketch a schematic, and write an hdl code. 5.write an 8: 1 multiplexer module called mux8 with selection inputs s, data input d, and data output y. data input (d) and data output (v) are 32-bit wide

The most common battery cable terminal is a that provides a large surface contact area with the ability to tighten the terminal onto the battery post using a nut and bolt.

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; }