Think about the definition of epidemiology. How many of the following.pdf

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Think about the definition of epidemiology. How many of the following refer to the rapplication of epidemiology? Assist in carrying out public health programs Act as a resource person Communication public health information Sound methods of scientific investigation Measure the frequency of a disease relative to an at-risk population Describe the frequency of a disease according to race/ethnicity Draw conclusions about causality based on the totality of evidence Provide information useful in health planning and decision making for establishing health programs with appropriate priorities O a. 5 d. 2 Solution Answer is a.5 Epidemiology is the branch of medicine which concentrates on the incidence of a disease and its spread and possible control of diseases and collecting all other factors related to it. If you come to application part, like studying history of the health population, estimating individual risks, causes of health and disease, identifying syndromes etc. Question 2 Answer is a. A non-living intermediary such as a fomite, food, or water that conveys the infectious agent from its reservoir to a susceptible host-definition of carrier. c.An infected person that contains, spreads or harbors an infectious organisms-definition of vehicle. Carrier will harbors the infectious organism by shows no symptoms of the disease and transmit to other hosts. Example cooking or eating vessels and vehicle is the nonliving intermediary (fomite, food, or water) which is capable of conveying the infectious agent from its reservoir to a susceptible host.

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Problem 1: Create Node class (or use what you have done in Lab4) • Add methods to get and set, Data and Link • Here the data is an object. (java object) your Linked List can hold any java class. a) Create MyLinkedList2 class and write methods to • Methods to insert nodes at the beginning, at the end and at a specific position. insertFirst(Node x), insertLast(Node x) , insertLast(Node x, int pos) • Methods to delete node from the beginning and from the end. delFirst(), delLast() • Method to search the Linked List for a specific object. searchList(x Object) • Method to Sort the Linked List in ascendant or descendant order according to the object value(s). for example sorted by name, by age, by gpa. sortAsc(),sortDesc(). • Method to display all nodes objects (printing an object is related to its toString() method). • Method to return the largest and smallest object in the list – maxList(), minList() according to some specific object members. . • Method to return the number of nodes – listSize() • Method to merge two linked Lists into one. MyLinkedList2 list3= mergeList(MyLinkedList2 list1, MyLinkedList2 list2) b) Create Main class : MyLinkedList2Demo • Create a linked List LL1 and insert some objects to it. Use any a student java class. • Fill LL1 by adding students from an input file. • Display all the objects (students) in the List LL1 • Search for a particular object (Student by name or by Id or both) in LL1 and print true or false • Sort LL1 and display it ( using asc and desc on multiple members) • Perform all the delete methods on LL1. • Perform all the insert methods on LL1. • Call and run all the implemented methods in a sequential order. NB: Organize a nice output for all the questions above. Problem 2: Same as Pb1 using double linked lists. Create Nod3e and MyLinkedList3 classes. -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------- The input file : id fname lname age gpa year college cuCourses totalCourses 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201112345 Amna chamisi 17 2.1 2014 ENG 31 32 201412345 Amna chamisi 17 2.1 2014 COE 5 29 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 Solution /** * Node.java * @author * */ public class Node { public Object getObj() { return obj; } public Node setObj(Object obj) { this.obj = obj; return this; } public Node getNext() { return next; } public Node setNext(Node next) { this.next = next; return this; } private Object obj; private Node next; } import java.util.Collections; import java.util.Comparator; import java.util.LinkedList; /** * MyLinkedList2.java * @author * */ public class MyLinkedList2 { public void insertFirst(Node x) { if (head == null) { head = x; } else { x.setNext(head); } } public void insertLast(Node x) { if (head == null) { head = x; r.

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One of your friends has several brothers and sisters, each quite different in appearance, interests, and abilities, even though all have the same parents. Explain in detail how these siblings reflect crossing over and independent assortment. Solution Meiosis process increases the genetic diversity in a species. In prophase I, crossing over takes place and in metaphase I and anaphase I, independent assortment occurs. When homologous chromosomes are paired together, during meiosis, the chromosomes make contact with the other pair of chromosomes. This point of contact is responsible for the exchange of genetic information between the chromosomes and is called the chiasmata. Although the same genetic information is contained by the two homologous chromosomes, the assortment of the chromosomes determine which information is present in each of the four gametes produced. For a particular phenotype, an allele can determine which characteristic will be expressed by the organism. For example, Chromosome 1 contains an allele for brown eyes while chromosome 2 contains an allele for black eyes. Thus, crossing over and independent assortment are responsible for different appearence, abilities and interest..

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In Java: Write a program that reads a text file that contains a grade (for instance, 87) on each line. Calculate and print the average of the grades. Solution import java.io.BufferedReader; import java.io.FileReader; // importing filereader import java.io.IOException; // importing file io exceptions public class grade { public static void main(String[] args) { BufferedReader br = null; double count=0; int noOfgrades=0; try { String sLine; br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer reader while ((sLine = br.readLine()) != null) { // rading line by line noOfgrades=noOfgrades+1; // getting number of grades int num=Integer.parseInt(sLine); // convert string into integer count=count+num; // adding each grade to count } System.out.println(\"count:\\t\"+count); System.out.println(\"noOfgrades:\\t\"+noOfgrades); System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades)); } catch (IOException e) { e.printStackTrace(); // catch exception } finally { try { if (br != null)br.close(); // closing buffer } catch (IOException ex) { ex.printStackTrace();// printing exceptions } } } } /********** out file****** fileName:sample.txt file contains \" 23 34 56 7 88 89 52 34 65 \" */ /************output************* count: 448.0 noOfgrades: 9 Average of the grades: 49.77 */.

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For Programming Embedded Systems Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)? A) Line 28 B) Line 36 C) Line 25 D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101 VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks) /*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103 switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0 break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0; state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1; tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct 6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1] , state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks [1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137 139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1 Solution c) Line 25 is correct Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished tasks.

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For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why don’t we draw T-h or P-t diagram? Solution For analysis of thermodynamics cycle we draw various diagrams, In all these diagram T-s,P-h,T- h and P-T diagram are some of them. Every diagram give some specific informations about system working like T-s diagram give information about Heat transfer, P-h diagram use in air conditioning because air is mixture so pressure affect more and enthalphy is measure of energy. like these two diagram we use P-t diagram use in pure substance to see the transformation at triple point. and T-h diagram use for visualise how energy change with temprature..

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Based on your reading of the GCU introduction and the textbooks, what is the main difference between the modern western worldview and the postmodern worldview? Which parts, if any, do you identify with? Explain. Cite references from your reading to support your answer. Solution Postmodernism is a response against modernism. According to everyone has a point of view through which they see the world. Premodern (the worldview before modernism)- says that natural events are caused by supernatural causes. God, angels, demons, and the like are a reality to these folks..

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Answer using Giardias, Shiga-toxin producing E.coli 1.Causative Agnet of Disease : 2.Type of Agent ( eg. Virus,bacterium) : 3.Further Classification ( eg.Gram type,viral serotype) : 4.How is disease diagnosed? 5.How is the disease treated? 6.How can the disease be prevented? 7.Distinguishing characteristics of organism (what makes the organism unique such as virulence factors that are required for disease, unique attributes that aid or prohibit the identificaiton of the organism ): Solution 1). Giardiasis is an infectious disease caused by a parasite called Giardia lamblia. It is spread through eating contaminated food, contact with the infected people, contaminated water and through pets. This organism is generally found thought the worlds, particularly in the developing countries..

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Abraham Shine recently opaned his own accounting fiem on April 1, which he operates as a corporation. The name of the new entity is Abraham Shine CPA Shine xparienced the following events during the organizing phase of the new business and ies Srst month of operations in 2018 (Cick the icon to view the transactions) Read the teaments Requirement 1. Analyze the effects of the events on the accounšing equation of Abraham Shine, CPA Anal ze the events chronolo caly one transaction at a rme egning with the transaction on the 6th caklate the balance in each account after analyzing the efect of the transaction on the account g equation Aher calculating the ending balance of each account on the 30m caloulate total assets and totall Babilities and equity (Complete only the nacessary answer boxes for your transaction ines Do not enter any zeros for your transaction lines 1 Car down all balances to the al line including zero balance accounts entering \"Q for any zero balances Enter decrease in ar account with 4 minus sign or pare theses Abbreviations used Ap Accounts Payable A Accounts Receivable. Com Common Contr “Conttuted. Div . Dividends. Exp . Expense. Fum . Fumitu, Liab. Liablies Rev-Revenue Sup . Supplies, Uni-Uilties) ASSETS LIAB+ EQUITY Contr. Retained Earnings ServiceRent Exp Cap Cash + AR + Ofice + Furn AP + Com Die UtI Sup Stock Rev Exp Enter any inumber in the edit fhelds and then click Check Answer 13-taring Clear Al Check Answer Solution Answer Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-30 -4000 + + + = + - 4000 + - - TOTAL 81810 0 200 7500 0 80000 4000 15500 1600 390 Income Service Revenue 15500 Expenses Rent Expense 1600 Utilities expense 390 Total expense 1990 Net Income 13510 Opening Bal 0 Add: Net Income 13510 Less: Dividend paid 4000 9510 Closing balance 9510 A. Owner\'s Equity Common Stock 80000 Retained Earnings 9510 89510 B. Liabilities Accounts payable 0 Total Owners\' Equity and Liabilities 89510 C. Assets Cash 81810 Accounts Receivables 0 Office Supplies 200 Furniture 7500 Total Assets 89510 Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-3.

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29. This figure shows the principal coronary blood vessels. Which one is the left coronary artery (LCA)? B. 8 C. 5 D. 3 E. 1 will cause a more severe myocardial infarction (MI) 30. Obstruction of the than the obstruction of any of the others A. left marginal vein B. left coronary artery (LCA) C. posterior interventricular vein D. anterior interventricular branch E. circumflex branch Solution 29. D. 3 30. B. left coronary artery 31. A. cardiac muscle fibers have striations 32. C. Sinatrial (SA) node. 33. B. . The tendinous cords. 34. e. They have about the same endurance as skeletal muscle fibers. 35. SA node, AV node, AV bundle, Purkinje fibers, cardiocyte in LV 36. A. Na+ inflow 37. E. slow CA2+ channels.

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14. Suppose I have collected the names of people in several separate files based on their starting letter. The names in the first file begin with A through G, the names in the second file begin with H through N, the names in the third file beginning with O through T, and the names in the fourth file contain names beginning with U through Z. None of the files are sorted. instead, the names that appear in each file occur in any order. Describe in detail how I could take advantage of multithreading to create one sorted list of all the names. Make sure you include details such as what your Thread class would do and what your main class that uses the Threads would do. (3 points) Solution We can slice strings with negative indexes as well. This IDLE transcript should clarify the rules for slicing where both indexes are specified: >>> word = \"PYTHONISCOOL\" >>> print(word[2:7]) THONI >>> print(word[6:6]) >>> print(word[4:2]) >>> print(word[5:11]) NISCOO >>> print(word[8:12]) COOL >>> >>> print(word[-7:-2]) NISCO >>> print(word[-9:8]) HONIS >>> print(word[-3:-7]) >>> Notice if you attempt to slice a string where the starting index refers to a position that occurs at or after the ending index, the slice is the empty string, containing no characters. A string can also be sliced using only one index. If only one index is given, Python must know if it’s the start or end of the slice. It assumes that the omitted index must refer to the beginning or end of the string, accordingly. These examples should clarify slicing using only one index: >>> print(word[8:]) COOL >>> print(word[-6:]) ISCOOL >>> print(word[:6]) PYTHON >>> print(word[:-4]) PYTHONIS.

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