'E14-22 (LO 5) (Preferred Dividends) Matt Schmidt Company's ledger sho.docx

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The document discusses how to allocate total declared dividends of $366,000 among preferred and common stock under different conditions. If the preferred stock is cumulative and fully participating, it receives all unpaid prior year dividends plus a portion of the current year amount. If noncumulative and nonparticipating, preferred stock receives no dividends. If noncumulative but participating above a 10% common dividend, preferred stock receives dividends only after common stock reaches the 10% threshold.

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// C Program How might I modify this code to handle any number of commands piped together? such as ls | wc or ls | wc -l | ps -elf etc.. #include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <string.h> #include <sys/types.h> #include <sys/wait.h> // keeps track of cmd parameters typedef struct param { char *param; struct param *next; } pt; // parses individual commands typedef struct cmd_list { char *cmd; int param_count; pt *param_list; char * input_file_name; char *output_file_name; struct cmd_list *next; } cl; // keeps track of main commands that are piped typedef struct list { cl *head; cl *tail; int count; } pt; void execommands(clt *cmds) { clt *cmd = cmds; for (int i = 1; i < cmds->count-1; i++) { int pipes[2] = {-1, -1}; pid_t pid = -1; clt *temp = NULL; pipe(pipes); pid = fork(); switch (pid) { case -1: perror("fork failed"); break; case 0: { char **rhp_argv = NULL; char *rhp = NULL; pt *temp = NULL; int i = 1; int flag_count = 0; char flag_string[100] = ""; temp = cmd->param_list; if (dup2(pipes[STDIN_FILENO], STDIN_FILENO) < 0) { perror("child process failed dup2"); _exit(EXIT_FAILURE); } close(pipes[STDIN_FILENO]); close(pipes[STDOUT_FILENO]); rhp = cmd->cmd; if(0 != cmd->param_count) { rhp_argv = (char **) calloc((cmd->param_count), sizeof(char *)); rhp_argv[0] = rhp; while(temp != NULL) { if (flag_count > 0 && temp->param[0] == '-') memmove(temp->param, temp->param + 1, strlen(temp->param)); } strcat(flag_string, temp->param) flag_count++; temp = temp->next } rhp_argv[i] = strdup(flag_string); i++; } else { rhp_argv = (char **) calloc(2, sizeof(char *)); rhp_argv[0] = rhp; } execvp(rhp_argv[0], rhp_argv); perror("child cannot exec program"); _exit(EXIT_FAILURE); if(cmd->param_count != 0) { for(int celm = 0; celm < flag_count; i++) { free(rhp_argv[i]); rhp_argv[i] = NULL; } } free(rhp_argv); rhp_argv = NULL; free(temp); } break; default: int status; waitpid(pid, &status, 0); break; } // end switch } // end for } .

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(1) In the tubular flowers of foxgloves, wild-type coloration is red, while a mutation called white produces white flowers. Another mutation, called peloria, causes the flowers at the apex of the stem to be huge. Yet another mutation, called dwarf, affects stem length. You cross a white-flowered plant (otherwise phenotypically wild type) to a plant that is dwarf and peloria but has wild-type red flower color. All the F1 plants are tall with white, normal-sized flowers. You cross an F1 plant back to the dwarf and peloria parent, and you see the 543 progeny shown in the chart to the right. (Only mutant traits are noted.) a. White, tall, and normal sized flowers/are the dominant alleles because these are what were expressed in the F1 generation, meaning that these hybrid offspring are exhibiting the dominant trait. b. What were the genotypes of the parents in the original cross? How do you know? c. Draw a map showing the linkage relationships of these three loci. Why is the calculated distance value for the two outside genes smaller than the value you get when you sum the two shorter distances? How can you correct this mistake? Make sure to show all your work. d. Is there interference? If so, calculate the coefficient of coincidence and the interference value (show your work!), and state the interpretation of the interference value. .

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(shady) conditions will most likely have a.fewer chloroplasts than sun-loving plant leaves b.more chloroplasts than sun-loving plant leaves c.the same number of chloroplasts as sun-loving plant leaves d.no chloroplasts 2- A guard cell is a.part of the leaf's protection against animals b.a protective part of the seed coat c. part of a structure controlling the stomata d.none of the above 3- A leaf that is not divided into leaflets is termed aa.simple leaf c.compound leaf b.palmate leaf d.pinnate leaf 4- The primary function of a leaf is to a.engage in photosynthesis and produce carbohydrates b.catch water from rain c.protect the plant from drying out by providing shade d.produce seeds for the next generation of plants 5- An internode is the a.space between two successive nodes on the same stem b.dormant meristematic tissue at the base of a leaf c.vascular tissue that connects a lateral node to a leaf d.structure that supports the leaf node 6- The attachment site of a leaf to a stem is called the a.petiole c.node b.axillary bud d.blade 7- Given the diagram above, the letter O represents a ( n ) a.node c.internode b.axillary bud d.shoot 8- The vein of a leaf is a.support tissue c.epidermal tissue b.vascular tissue d.both a and b .

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(i) Werck the icon to vere the trpssictons) More info Requirements 1. Joumake thin trangackons 2. Cwcinte the balince in Retaned Eamingen en Decutber 31,2027 Asturhe the balerio on dantary 1, 2027 ans f 5 Dod atid red liccene les tho yoar was t44s 000 3. Fiepare the stocktoldes' equity secton of the hadince shent as of December 31, 202 t flefe was to prefored alock nsepid pror to the 2007 tiansptions. Jan. 1. Lssued 70,000 stares of $10 par vatue common stock for a total of $1 , 050 , 000 . Dec. 31: Paid the cash dividends Dec 31. Distributed the stock dividend Requirement 2, Cakulate the balance in Retained Earnings on December 31, 2027. Assume the bakance on danuary 1,2027 was $5 , 000 and net income for tho yoar was $445 , 000 Complote the table below to calculate the balance in Rotained Earnings on December 31, 2027. Bevicre the iournal entyes from fequicencent i. .

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#include <iostream> #include <vector> #include "Date.h" #include "Person.h" #include "DateException.h" #include "PersonExceptions.h" int main() { std::string firstName; std::string lastName; int birthMonth; int birthDay; int birthYear; float height; float weight; char answer = 'y'; std::vector<CIST2362::Person> personDatabase; // vector of Person Objects // input person objects // You must catch exceptions properly while (toupper(answer) == 'Y') { std::cin >> firstName; std::cin >> lastName; std::cin >> birthMonth; std::cin >> birthDay; std::cin >> birthYear; std::cin >> weight; std::cin >> height; // place person objects into vector // ENTER CODE HERE std::cin >> answer; } // output the person objects // ENTER CODE HERE return 0; } (Date.cpp) #include "Date.h" #include "DateException.h" #include <sstream> namespace CIST2362 { Date::Date() { // ENTER CODE HERE } Date::Date(int m, int d, int y) { // ENTER CODE HERE } void Date::setNames() { names[0] = "January"; names[1] = "Febraury"; names[2] = "March"; names[3] = "April"; names[4] = "May"; names[5] = "June"; names[6] = "July"; names[7] = "August"; names[8] = "September"; names[9] = "October"; names[10] = "November"; names[11] = "December"; } void Date::setMonth(int m) { // ENTER CODE HERE } void Date::setDay(int d) { // ENTER CODE HERE } void Date::setYear(int y) { // ENTER CODE HERE } std::string Date::getDateShort() { // ENTER CODE HERE return "M/D/Y"; } std::string Date::getDateLong() { // ENTER CODE HERE return "Month Day, Year"; } } // CIST2362 (Date.h) #ifndef LESSON_5_PROGRAMMING_ASSIGNMENT_DATE_H #define LESSON_5_PROGRAMMING_ASSIGNMENT_DATE_H #include <iostream> #include <string> namespace CIST2362 { // Constants const int NUM_MONTHS = 12; class Date { private: int month; int day; int year; // An array of strings to hold // the names of the months std::string names[NUM_MONTHS]; // Private member function to assign // the month names to the names array void setNames(); public: // Constructors Date(); Date(int, int, int); // Mutators void setMonth(int m); void setDay(int d); void setYear(int y); // Functions to print the date std::string getDateShort(); std::string getDateLong(); }; } // CIST2362 #endif //LESSON_5_PROGRAMMING_ASSIGNMENT_DATE_H (DateException.cpp) #include "DateException.h" namespace CIST2362 { // ENTER CODE HERE } // CIST2362 (DateException.h) #ifndef LESSON_5_PROGRAMMING_ASSIGNMENT_DATEEXCEPTION_H #define LESSON_5_PROGRAMMING_ASSIGNMENT_DATEEXCEPTION_H #include <exception> namespace CIST2362 { // Exception classes class InvalidDay: public std::exception { public: // ENTER CODE HERE }; class InvalidMonth: public std::exception { public: // ENTER CODE HERE }; class InvalidYear: public std::exception { public: // ENTER CODE HERE }; } // CIST2362 #endif //LESSON_5_PROGRAMMING_ASSIGNMENT_DATEEXCEPTION_H (Person.cpp) #include <sstream> #include <iomanip> #include "Person.h" #include ".

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