3. According to Conger (1990), what is the dark side of leadership a.pdf

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3. According to Conger (1990), what is the dark side of leadership and what are the forces that give rise to this side of leadership? Solution Dark side of leadership is the leadership that focuses upon the personal gains and becomes out of touch with reality. Under dark side of leadership, leaders put in their personal agenda ahead and use impression management to show it as an organizational objective. Forces that give rise to this side of leadership are as following..

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The output should now look like this: Row 1 sum: 30 Row 2 sum: 48 Row 3 sum: 55 Count: 12 Grand total: 133 Mean: 11.08 Even numbers: 7 Odd numbers: 5 How to add even and odd numbers counts to existing code public class Problem { public static void main(String [] args) { int iVal; int iRow; int iRowSum; int iToken; int totalSum = 0; int totalEle = 0; iRowSum = 0; iRow = 1; try { Scanner ifsInput = new Scanner(new File(\"c:\\home\\student\\lastname/9_1_Input.txt\")); while(ifsInput.hasNextLine()) { while(ifsInput.hasNextInt()) { for(iToken = 0; iToken <= 3; iToken++) { iVal = ifsInput.nextInt(); iRowSum = iRowSum + iVal; // increasing count of total element totalEle++; } // adding into totalSum totalSum = totalSum + iRowSum; System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum); iRowSum = 0; iRow++; } } // added new code to print grand total and mean value System.out.println(\"Count: \"+totalEle); System.out.println(\"Grand total: \"+totalSum); System.out.println(\"Mean: \"+String.format(\".2f\", (double)/totalSum/totalEle); } catch (FileNotFoundException sMsg) { System.out.println(\"The file cannot be found or opened.\"); } } } Solution Input999.txt( Save this file under D Drive Then the path of the file pointing to it is D:\\\\Input999.txt) 11 22 33 44 22 33 44 55 33 44 55 66 ___________________________ Problem.java import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class Problem { public static void main(String [] args) { //Declaring variables int iVal; int iRow; int iRowSum; int iToken; double mean=0.0; int evenCount=0,oddCount=0; int totalSum = 0; int totalEle = 0; iRowSum = 0; iRow = 1; try { Scanner ifsInput = new Scanner(new File(\"D:\\\\Input999.txt\")); while(ifsInput.hasNextLine()) { while(ifsInput.hasNextInt()) { for(iToken = 0; iToken <= 3; iToken++) { iVal = ifsInput.nextInt(); //Calculating the even count if(iVal%2==0) evenCount++; //Calculating the odd count else oddCount++; //Calculating the sum of elements in each row iRowSum = iRowSum + iVal; // increasing count of total element totalEle++; } // adding into totalSum totalSum = totalSum + iRowSum; //Displaying the sum of elements in each row System.out.println(\"Row \" + iRow + \" sum: \" + iRowSum); iRowSum = 0; iRow++; } } //Displaying the number of elements System.out.println(\"Count: \"+totalEle); //Displaying the sum of all elements System.out.println(\"Grand total: \"+totalSum); //Calculating the mean mean=(double)totalSum/totalEle; //Displaying the mean System.out.printf(\"Mean: %.2f\ \",mean); //Displaying the number of even nos System.out.println(\"Even Numbers :\"+evenCount); //Displaying the number of odd nos System.out.println(\"Odd Numbers :\"+oddCount); } catch (FileNotFoundException sMsg) { System.out.println(\"The file cannot be found or opened.\"); } } } _________________________________ Output: Row 1 sum: 110 Row 2 sum: 154 Row 3 sum: 198 Count: 12 Grand total: 462 Mean: 38.50 Even Numbers :6 Odd Numbers :6 ______.

The output should now look like this Row 1 sum 30 Row 2 sum.pdf

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The \"end problem\" in eukaryotic DNA replication a. is solved by telomerase b. refers to loss of chromosome ends during replication c. is solved by an RNA dependent DNA polymerase d. A and B e. all of the above Solution Answer : The correct answer is E ( All of the above) Explanation : Telomerases are the ribonucleoprotein complexes containing a small RNA that serves as a template for addition of new six nucleotide repeats. Or Telomerase, the enzyme that synthesizes the telomere ends of linear chromosomes, is a specialized reverse transcriptase that contains an internal RNA template. Telomeres undergo cycles of shortening of the lagging strands due to the inability to complete synthesis. Most normal cells lose telomerase activity with increaseing age of the organism and repeated generations of the cell division. In other words, after certain number of divisions normal cells reaches the Hayflicks limit beyond which cells will no longer devide instead they die. In germ-line cells, which contain telomerase activity, telomere lengths are maintained; for example, telomerase is highly expressed only in cells that need to divide regularly especially in male sperm cells. Whereas somatic cells lack telomerase activity due to which they die after certain number of generations. In cancer cells/tumor cells, end replication problems observed in normal cells will be taken care by the telomerase enzyme which makes the cells to bypass the Hayflick limit and become immortal, thus avoiding cell death Many cancer cells are considered \'immortal\' because telomerase activity supports them to live longer than somatic cells. 90% of the cancer cells retain the ability to express telomerase and acquires the ability to proloferate indefinitely..

The end problem in eukaryotic DNA replicationa. is solved by t.pdf

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Suppose that one obtained the following DNA sequence data for the four fly families and a non- fly insect (the outgroup) from a gene that is not related to morphology. We will call the DNA sequence positions 1 through 10, for reference (in fly family A, position 1 is a T, while position 10 is an A). Which of the characters (nucleotide positions) are variable among the fly families (not counting the outgroup? Which of the characters (nucleotide positions) are phylogenetically informative for cladistics? (Phylogenetically informative traits are those for which there is variation in the group of interest, and for which there is evidence of synapomorphies.) Solution Purines are 2 ringed structure= A,G whereas, pyramidines aresingle ringed structure = C,T Generally A = T (A double bonds with T) and G=C (G triple H bonds with C). Mutation causes a change in theDNA sequence. Two types 1. Transversions: Purine instead of a pyramidine or vice versa (Less frequent so more important for cladistics) A/G to C/T or vice versa. 2 ringed structure to 1 ringed structure See this is seen in all fly families at 7th(ABCD:AGGC) , 8th (ABCD:CTAC) and 9th (ABCD:GCTG)positions 2. Transitions: Purine instead of a puirne and pyramidine instead of a pyramidine (Easy and frequent) The Nucleotide positions are Bolded in the table are variable among the fly families. Among all Fly family C is most variable. (Take A, B, C, D and leave Out group). Phylogenetically informative traits are those for which there is variation in the group of interest and for which there is evidence of synapomorphies. All this is to know the members of a group which share a common history and to know their closely related species. Synapomorphy means any characteristic which is present in an ancestral species and that is shared exclusively (either in more or less modified form) by its evolutionary descendants. Mostly there is a much variation in the 7th, 8th and 9th position of the nucleotide in all fly families sequence so this will be more phylogenetically informative for cladistics and the 5th sequence is the most stable one. From the sequence fly family C is much more related to the out group sequence than all familiesFLY FAMILYDNA Sequence alignedpositions order12345678910ATGCGTTACGABTGCATTGTCACTCCGTAGATTDGCAGTTCCGAOU T GROUPGCAGTTGATT.

Suppose that one obtained the following DNA sequence data for the fou.pdf

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Set up a JavaFX GUI-based program that shows a 10 times 10 grid of labels that forms a multiplication table, with the labels displaying the multiplication problems, rather than the answers. Provide a text input field and a button with an event handler that reads an integer value from the text input and changes the CSS styling in some obvious way for all problems in the table with the given answer. When a new answer is entered and the button is clicked, change all the labels back to the original style, then change the labels showing problems for the new answer to the new style. Use CSS, not setters, for all the styling. The main point of this lab is to learn JavaFX, so the grading will include a heavy emphasis on the neatness and general appearance of your GUI. Solution Code: //multiTable.java // import required packages import java.util.List; import javafx.application.Application; import javafx.event.Event; import javafx.event.EventHandler; import javafx.geometry.Insets; import javafx.geometry.Pos; import javafx.scene.Node; import javafx.scene.Scene; import javafx.scene.control.Button; import javafx.scene.control.Label; import javafx.scene.control.TextField; import javafx.scene.input.MouseEvent; import javafx.scene.layout.BorderPane; import javafx.scene.layout.GridPane; import javafx.scene.layout.HBox; import javafx.stage.Stage; // class for revrse multiplication table public class multiTable extends Application { // over ride nethod for setup the grid @Override public void start(Stage primaryStage) { BorderPane pane1 = new BorderPane(); pane1.setTop(getHbox1()); HBox h_box = new HBox(15); h_box.setPadding(new Insets(15, 15, 15, 15)); h_box.setAlignment(Pos.TOP_CENTER); h_box.getStyleClass().add(\"hbox2\"); Label labl = new Label(\"Enter Answer: \"); h_box.getChildren().add(labl); TextField textfield1 = new TextField(); h_box.getChildren().add(textfield1); GridPane grid_pane1 = setUpGrid(); gridmaker griddp = new gridmaker(grid_pane1); Button Answer = new Button(\"Find problems\"); Answer.addEventHandler(MouseEvent.MOUSE_CLICKED, new EventHandler() { @Override public void handle(Event arg0) { List fact = factors(textfield1); for (int[] RCx1 : fact) { Node node = griddp.get_chldren()[RCx1[0]][RCx1[1]]; node.setStyle(\"-fx-background-color: green\"); } } }); h_box.getChildren().add(Answer); pane1.setCenter(h_box); pane1.setBottom(grid_pane1); Scene scene = new Scene(pane1, 550, 650); scene.getStylesheets().add(getClass().getResource(\"application.css\").toExternalForm()); primaryStage.setTitle(\"lab 6\"); primaryStage.setScene(scene); primaryStage.show(); } // method for grid layout private HBox getHbox1() { HBox hbox2 = new HBox(15); hbox2.setPadding(new Insets(15, 15, 15, 15)); hbox2.setAlignment(Pos.TOP_CENTER); hbox2.getStyleClass().add(\"hbox1\"); Label labl = new Label(\"Multiplication Table\"); hbox2.getChildren().add(labl); return hbox2; } // method to setup a grid public GridPane setUpGrid() { GridPane pane1 = new GridPane(); Label[][].

Set up a JavaFX GUI-based program that shows a 10 times 10 grid of la.pdf

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Sample of DNA isolated from bacterium X contains 17.5% of adenine bases. What percentages of guanine, thymine, and cytosine bases would you expect to find in the same sample of DNA? Match the following types of transport across a biological membrane (designated I, II, or III) with the statements designated A-D: Simple diffusion (SD) Facilitated diffusion (FD) Active transport (AT) Write down your answers in the blank space next to each statement below. Note, that more than one answer may apply each statement: Process is energy dependent: ___ Process needs transmembrane transporter/channel: ___ Process occurs down the concentration gradient: ___ Processes can establish a concentration gradient: ___ Solution Answer 2. Given that adenine bases in the isolated DNA of the bacterium is 17.5%. Since Adenine is known to pair with thymine. So, the percentage of thymine will also be 17.5%. Adenine and thymine will make 35 % of the total nitrogen bases of the DNA. Rest of the DNA will be made from Guanine and Cytosine. Guanine and Cytosine are known to pair with each other. So, their content in the DNA will also be same. 100-35% will give us 65 %. So Guanine will be 32.5% and Cytosine will be 32.5% Answer 3. Process occur down the concentration gradient - Simple Diffusion(SD) Process need transmembrane transporter/Channel- Facilitated Diffusion (FD) Process is energy dependent- Active transport (AT) Process can establish a concentration gradient- Facilitated Diffusion (FD) and Active transport (AT).

Sample of DNA isolated from bacterium X contains 17.5 of adenine.pdf

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Practice using layouts Anderson, Franceschi import javax..pdf

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Please, I need a correct answer and clear explanation. open image and zoom for a clear view. Thanks, Using the West Jet Financial statements, answer the following questions R1 2-4 p.106-109) b. Why do you think West Jet does not report a cost of goods sold c. Explain what “Advanced ticket sales” in the current liabilities section represents e. How much did West Jet use in 2013 to acquire new aircraft? Which statement did you find this information on? RI2-4 (Determination of items from a Canadian company\'s financial statements) Calgary-based WestJet Airlines Ltd. provides services throughout Canada and to some international destinations. Excerpts from its 2013 financial statements are in Exhibits 2-19A to 2- 19C. Exhibit 2-19A WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF EARNINGS Consolidated Statement of Earnings For the years ended December 31 (Stated in thousands of Canadian dollars, except per share amounts) Note 2013 2012 Revenue: Guest 3,337,569 3,133,492 Other 324,628 293,917 3,662,197 3,427,409 Operating expenses: Aircraft fuel 1,039,448 992,787 Airport operations 459,465 424,911 Flight operations and navigational charges 410,052 376,050 Sales and distribution 356,988 333,106 Marketing, general and administration 222,567 202,398 Depreciation and amortization 200,840 185,401 Inflight 176,907 162,633 Aircraft leasing 175,646 173,412 Maintenance 169,197 154,406 Employee profit share 51,577 46,585 3,262,687 3,051,689 Earnings from operations 399,510 375,720 Non-operating income (expense): Finance income 15 17,848 18,391 Finance cost 15 (43,447) (48,900) Gain on foreign exchange 1,136 1,061 Gain (loss) on disposal of property and equipment (2,962) 469 Loss on fuel derivatives — (6,512) (27,425) (35,491) Earnings before income tax 372,085 340,229 Income tax expense (recovery): Current 154,964 66,230 Deferred (51,601) 31,607 11 103,363 97,837 Net earnings 268,722 242,392 Exhibit 2-19B WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF FINANCIAL POSITION Consolidated Statement of Financial Position At December 31 (Stated in thousands of Canadian dollars) Note 2013 2012 Assets Current assets: Cash and cash equivalents 5 1,256,005 1,408,199 Restricted cash 6 58,106 51,623 Accounts receivable 19 42,164 37,576 Prepaid expenses, deposits and other 19 133,263 101,802 Inventory 19 36,722 35,595 1,526,260 1,634,795 Non-current assets: Property and equipment 7 2,487,734 1,985,599 Intangible assets 8 58,691 50,808 Other assets 19 70,778 75,413 Total assets 4,143,463 3,746,615 Liabilities and Shareholders\' Equity Current liabilities: Accounts payable and accrued liabilities 19 543,167 460,003 Advance ticket sales 19 551,022 480,947 Non-refundable guest credits 19 46,975 47,859 Current portion of maintenance provisions 9 76,105 34,135 Current portion of long-term debt 10 189,191 164,909 1,406,460 1,187,853 Non-current liabilities: Maintenance provisions 9 142,411 145,656 Long-term debt 10 689,204 574,139 Other liabilities 19 8,834 9,914 .

Please, I need a correct answer and clear explanation. open image an.pdf

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Patient 2 Patient #2: Jan Johnson Next, Dr. Gupta sees Jan Johnson. She is a 12-year-old female. She is in the 7th grade and has been on the school swim team for the past four years. She is a good student and is generally a happy child. Lately, Jan\'s mother has been worried about her daughter. Jan has been very irritable the past few months and at first, her mother thought it was related to the onset of puberty. In addition to mood changes, Mrs. Johnson has also noted a few other changes in Jan she is often too tired to attend swimming practice and has complained of feeling too weak to swim and of having headaches. Despite her decreased physical activity, Jan has been eating almost non-stop but has lost about 5 lbs, she is also drinking more than usual and has been vomiting frequently. Jan saw her primary care physician, and after running some tests, has sent Jan to follow up with Dr. Gupta. Jan\'s blood work and patient information from his chart are listed below. Solution 1.Jan is easily fatigued, eating too much (polyphagia) drinking too much water (polydipsia), sudden weight loss and nausea are the symptoms. All these symptoms relate to the metabolic disorder called Diabetes. 2.jan is having high blood glucose level than what is considered to be normal. This indicates the insulin receptors have become resistant to insulin which is type 2 diabetes or adequate amounts of insulin is not being secreted by the pancreas which is type 1 diabetes. If untreated, diabetes can be life threatening in the long term. It leads to diabetic retinopathy, kidney parenchymal changes, coronary artet disease, diabetic foot etc. 3. I must collect information about her diet, frequency of urination, recent injuries, family history of diabetes, any blurred vision,..

Patient 2 Patient #2 Jan Johnson Next, Dr. Gupta sees Jan Johnson. .pdf

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Non-math and physics question, but engineering orientated. Identify and discuss what you know are the significant milestones in materials used for commercial aircraft since the end of the Second World War? Include the manufacturing, maintenance and reliability aspects if relevant. Limit your answer to three principal types of material. Provide essay answers with a maximum of 500 words. Use current APA format for all references. (minimum of 3 references) Thanks! Solution On December 17, 1903, Orville and Wilbur Wright capped four years of research and design efforts with a 120-foot, 12-second flight at Kitty Hawk, North Carolina - the first powered flight in a heavier-than-air machine. Prior to that, people had flown only in balloons and gliders. The first person to fly as a passenger was Leon Delagrange, who rode with French pilot Henri Farman from a meadow outside of Paris in 1908. Charles Furnas became the first American airplane passenger when he flew with Orville Wright at Kitty Hawk later that year. Aircraft Innovations For the airlines to attract passengers away from the railroads, they needed both larger and faster airplanes. They also needed safer airplanes. Accidents, such as the one in 1931 that killed Notre Dame Football Coach Knute Rockne along with six others, kept people from flying Aircraft manufacturers responded to the challenge. There were so many improvements to aircraft in the 1930s that many believe it was the most innovative period in aviation history. Air-cooled engines replaced water-cooled engines, reducing weight and making larger and faster planes possible. Cockpit instruments also improved, with better altimeters, airspeed indicators, rate-of- climb indicators, compasses, and the introduction of artificial horizon, which showed pilots the attitude of the aircraft relative to the ground - important for flying in reduced visibility. designers already knew that as an aircraft approaches the speed of sound (Mach 1), in the transonic region, shock waves begin forming, causing a large increase in drag. Wings, already thin, had to become thinner and finer. Fineness is a measure of how thin the wing is compared to its front-to-back chord. A small, highly loaded wing has less drag and so some early types used this type, including the Bell X-1 rocket plane and the Lockheed F-104 Starfighter. But these craft had high takeoff speeds, the Starfighter causing significant pilot deaths during takeoff, and small wings fell out of use. An approach pioneered by German designers during the war was to sweep the wing at an angle, delaying the buildup of shock waves. But this made the wing structure longer and more flexible, making the aircraft more likely to suffer from bending or aeroelasticity and even causing a reversal in the action of the flight controls. Stall behaviour of the swept wing was also poorly understood and could be extremely sharp. Other problems included divergent oscillations which could build up lethal forces. In re.

Non-math and physics question, but engineering orientated.Identify.pdf

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Name three properties of the waveform that can be changed with the m.pdf

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Inventory Valuation: FIFO, LIFO, and Average The company reported the following inventory data for the year: Practice 9-7 Units Cost per Unit 300 $1750 Beginning inventory Purchases: 900 1,200 400 18.00 18.25 March 23. Units remaining at year-end. . . . . . .. Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b)LIFOin- ventory valuation, and (c) average cost inventory valuation. The company uses a periodic inventory system Practice 9-8 Inventory Valuation: Complications with a Perpetual System Refer to Practice 9-7. Assume that the sales occurred as follows: Units Sold 100 600 1,300 July 15 November 1 Total Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b) LIFO inventory valuation, and (c) average cost inventory valuation, The company uses a perpetual inventory system Solution Answer Units sold = 2400 – 400 = 2000 FIFO Cost of Goods available for sale Cost of Goods Sold Endging Inventory Units Cost/unit COG for sale Units sold Cost/unit COGS Units Cost/unit Ending inventory Beginning Inventory 300 17.5 5250 300 17.5 5250 0 17.5 0 Purchases: Mar-23 900 18 16200 900 18 16200 0 18 0 Sep-16 1200 18.25 21900 800 18.25 14600 400 18.25 7300 TOTAL 2400 43350 2000 36050 400 7300 LIFO Cost of Goods available for sale Cost of Goods Sold Endging Inventory Units Cost/unit COG for sale Units sold Cost/unit COGS Units Cost/unit Ending inventory Beginning Inventory 300 17.5 5250 0 17.5 0 300 17.5 5250 Purchases: Mar-23 900 18 16200 800 18 14400 100 18 1800 Sep-16 1200 18.25 21900 1200 18.25 21900 0 18.25 0 TOTAL 2400 43350 2000 36300 400 7050 ---Average Cost Valuation Cost of opening units + Purchases = $43350 Opening units + Purchased units = 2400 Average cost = 43350 / 2400 = $18.0625 per unit Cost of Goods Sold = 2000 units sold x $18.0625 = $36125 Value of ending Inventory = 400 units x $18.0625 = $7225 FIFO Opening + Purchases Cost of Goods Sold Closing Inventory Opening units 300 17.50 5250 300 17.50 5250.00 Jan 16 Sale 100 17.50 1750 200 17.50 3500.00 Mar 23 Purchase 900 18.00 16200 200 17.50 3500.00 900 18.00 16200.00 July 15 Sale 200 17.50 3500 500 18.00 9000.00 400 18.00 7200 Sept 16 Purchase 1200 18.25 21900 500 18.00 9000.00 1200 18.25 21900.00 Nov 1 Sale 500 18.00 9000 400 18.25 7300.00 800 18.25 14600 TOTAL 2400 43350 2000 36050 400 7300.00 Units Value Cost of Goods Sold 2000 36050 Closing Inventory 400 7300 LIFO Opening + Purchases Cost of Goods Sold Closing Inventory Opening units 300 17.50 5250 300 17.50 5250.00 Jan 16 Sale 100 17.50 1750 200 17.50 3500.00 Mar 23 Purchase 900 18.00 16200 200 17.50 3500.00 900 18.00 16200.00 July 15 Sale 600 18.00 10800 200 17.50 3500.00 0 0.00 0 300 18.00 5400.00 Sept 16 Purchase 1200 18.25 21900 200 17.50 3500.00 300 18.00 5400.00 1200 18.25 21900.00 Nov 1 Sale 1200 18.25 21900 200 17.50 3500.00 100 18.00 1800 200 18.00 3600.00 TOTAL 2400 43350.00 2000 36250.00 400 7100 Units Value Cost of Goods Sold 2000 36250 .

Inventory Valuation FIFO, LIFO, and Average The company reported the.pdf

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In December 2009, a 45-year-old female presented to the emergency department (ED) 2 days following abrupt onset of sore throat, nonproductive cough, chills, and mild fever. A chest radiograph was performed, which was normal. She was diagnosed with bronchitis and asked to follow up with her primary care physician, who subsequently started her on levofloxacin and albuterol. Four days later she presented again to the ED with worsening cough, dispread, fever (38.3 degree C; 101 degree F), and generalized lethargy. Additionally, she reported new symptoms including a global headache, dizziness, myalgias, and arthralgias. She had no abdominal pain, but reported nausea and anorexia. Her chest radiograph showed diffuse reticulonodular opacities throughout the left lung, which were not present on her visit 4 days previously. The patient was admitted for further evaluation. Questioning revealed the following: she had a history of diabetes and hypertension. she smoked an average of a pack of cigarettes daily, and she had received the seasonal influenza vaccine. Her husband was recently ill with cough, but no order symptoms. On day 2 of hospitalization the patient\'s respiratory rate increased from 22 to 46 and her oxygen saturation dropped while on oxygen administered by nasal canola. She was transferred to an intensive care unit, where her respiratory status quickly deteriorated, necessitating emergency intubation. A new chest radiograph showed bilateral involvement, and she was begun on vancomycin, anreonam, and azithromycin. Blood cultures drawn on her admission were negative, and an expectorated sputum sample taken at the same time was not processed due to poor specimen quality. A PCR test performed on a nasopharyngeal swab was positive for a viral agent, revealing the etiology of her infection (Fig. 10.1). What is the agent causing her infection? What are the key virulence factors of this agent? How does this virus change over time? What made this virus unique in 2009? Solution Answer= 1] The agent causing her infection is influenza virus Human influenza virus can evolve rapidly by 2 mechanisms 1] antigenic shift = genetic reassortment between a human & non human virus in non human host 2] antigenic drift = accumulation of mutations that facilitate evasion of host immune response New influenza viruses can be constantly emeges from environment 2] influenza constantly change by 2 different ways antigenic drift = These are small changes in genes of influenza viruses that happen continuously over time as the virus replicates Antigenic shift = is an abrupt major changes resulting in hemagglutinin & neuraminidase proteins in influenza viruses that infect humans Such a shift occured in 2009 when H1N1 virus with a new combination of genes emerged to infect people & quickly spread causing a pandemic when shift happens most people has no protection against H1N1.

In December 2009, a 45-year-old female presented to the emergency dep.pdf

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In a longitudinal study of attitude toward statistics, a doctoral student surveyed 10 doctoral students at four times during their program: (a) before taking any statistics courses; (b) after having taken one course; (c) after all required statistics courses; and (d) one semester after completing their preliminary examination. In each instance, alternate forms of the Attitude Towards Statistics instrument were used. The data were: 1. Analyze the data using an alpha of .05 in SPSS (Free trial of SPSS here: http://www- 01.ibm.com/software/analytics/spss/products/statistics/downloads.html) 2. Writing the result in APA format. Solution SPSS out put GLM v1 v2 v3 v4 /WSFACTOR=factor1 4 Polynomial /METHOD=SSTYPE(3) /EMMEANS=TABLES(factor1) COMPARE ADJ(BONFERRONI) /PRINT=DESCRIPTIVE /CRITERIA=ALPHA(.05) /WSDESIGN=factor1. Descriptive Statistics Mean Std. Deviation N v1 40.50 11.008 10 v2 42.40 11.806 10 v3 43.00 11.126 10 v4 36.80 10.390 10 Multivariate Testsa Effect Value F Hypothesis df Error df Sig. factor1 Pillai\'s Trace .768 7.725b 3.000 7.000 .013 Wilks\' Lambda .232 7.725b 3.000 7.000 .013 Hotelling\'s Trace 3.311 7.725b 3.000 7.000 .013 Roy\'s Largest Root 3.311 7.725b 3.000 7.000 .013 a. Design: Intercept Within Subjects Design: factor1 b. Exact statistic Mauchly\'s Test of Sphericitya Measure: MEASURE_1 Within Subjects Effect Mauchly\'s W Approx. Chi-Square df Sig. Epsilonb Greenhouse-Geisser Huynh-Feldt Lower-bound factor1 .509 5.209 5 .394 .688 .896 .333 Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. a. Design: Intercept Within Subjects Design: factor1 b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table. Tests of Within-Subjects Effects Measure: MEASURE_1 Source Type III Sum of Squares df Mean Square F Sig. factor1 Sphericity Assumed 234.275 3 78.092 15.916 .000 Greenhouse-Geisser 234.275 2.064 113.520 15.916 .000 Huynh-Feldt 234.275 2.687 87.190 15.916 .000 Lower-bound 234.275 1.000 234.275 15.916 .003 Error(factor1) Sphericity Assumed 132.475 27 4.906 Greenhouse-Geisser 132.475 18.574 7.132 Huynh-Feldt 132.475 24.182 5.478 Lower-bound 132.475 9.000 14.719 Tests of Within-Subjects Contrasts Measure: MEASURE_1 Source factor1 Type III Sum of Squares df Mean Square F Sig. factor1 Linear 55.125 1 55.125 20.066 .002 Quadratic 164.025 1 164.025 18.401 .002 Cubic 15.125 1 15.125 4.946 .053 Error(factor1) Linear 24.725 9 2.747 Quadratic 80.225 9 8.914 Cubic 27.525 9 3.058 Tests of Between-Subjects Effects Measure: MEASURE_1 Transformed Variable: Average Source Type III Sum of Squares df Mean Square F Sig. Intercept 66178.225 1 66178.225 138.576 .000 Error 4298.025 9 477.558 Estimated Marginal Means factor1 Estimates Measure: MEASURE_1 factor1 Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound 1 40.500 3.481 32.626 48..

In a longitudinal study of attitude toward statistics, a doctoral st.pdf

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If events A and D are mutually exclusive, are A and D independent?(yes or no). Why? If random variables X and Y has the same mean and variance, are the cumulative distribution functions of X and Y the same? (yes or no). Solution 6) No, this is not condition for independence. For example heads and tails are mutually exclusive outcomes, but P(H | T) = 0 which is not same as P(H) = 0.5 7) No, they need not be same..

If events A and D are mutually exclusive, are A and D independent(ye.pdf

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Identify and discuss the characteristics of lifeSolutionThe 7 .pdf

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For the code below complete the preOrder() method so that it perform.pdf

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Here is the Person Class Provided: Use the following files: Person.java Here is the tester file: InsruanceTester.java I have done three of the methods i just need the last one public double monthly premium method completed. Here is my code: public class Person { private String name; private String gender; private int age; /** * Consructs a Person object * @param name the name of the person * @param gender the gender of the person either * m for male or f for female * @param age the age of the person */ public Person(String name, String gender, int age) { this.name = name; this.gender = gender; this.age = age; } /** * gets the age of this Person * @return the age of this Person */ public int getAge() { return age; } /** * gets the gender of this Person * @return the gender of this Person */ public String getGender() { return gender; } /** * gets the name of this Person * @return the name of this Person */ public String getName() { return name; } /** * Increases the age of this Person by 1 year */ public void birthday() { age = age + 1; } } /** * Models an Insurance client */ public class Insurance { private Person client; /** * Constructs an Insurance object with the given Person * @param p the Person for this Insurance */ public Insurance(Person p) { client = p; } public int clientAge(){ return client.getAge(); } public String clientGender(){ return client.getGender(); } public void incrementAge(){ client.birthday(); } } Solution //Person.java public class Person { private String name; private String gender; private int age; /** * Consructs a Person object * @param name the name of the person * @param gender the gender of the person either * m for male or f for female * @param age the age of the person */ public Person(String name, String gender, int age) { this.name = name; this.gender = gender; this.age = age; } /** * gets the age of this Person * @return the age of this Person */ public int getAge() { return age; } /** * gets the gender of this Person * @return the gender of this Person */ public String getGender() { return gender; } /** * gets the name of this Person * @return the name of this Person */ public String getName() { return name; } /** * Increases the age of this Person by 1 year */ public void birthday() { age = age + 1; } } -------------------------------------------------------------------------------------------------------------- /** * Models an Insurance client */ public class Insurance { private Person client; /** * Constructs an Insurance object with the given Person * @param p the Person for this Insurance */ public Insurance(Person p) { client = p; } public int clientAge(){ return client.getAge(); } public String clientGender(){ return client.getGender(); } public void incrementAge(){ client.birthday(); } /** * The method monthlyPremium that sets the insurance * value based on gender and age * */ public double monthlyPremium() { double insurance=0; //Checking if age of client object is <16 //return -1 if(client.getAg.

Here is the Person Class ProvidedUse the following filesPerson.pdf

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God bless... It is often useful to assert that something exists and is unique. For instance, There exists a unique course at UW with the course number CSE 311. (a) 4 Points] Let the domain of discourse be all vases. Let B be a constant representing \"the painted vase in Shayan\'s office.\" Translate \"If there is a vase that is painted, then it is the painted vase in Shayan\'s office.\" into first-order logic. Don\'t forget to define predicates Hint: Remember that a \"constant\', is a particular object in the domain. For example, \"0\" and .1\" are constants in the domain of natural numbers. (b) [4 Points] Let the domain of discourse be all university courses. Now, use an idea similar to part (a) to translate the sentence about the CSE 311 course number into first-order logic. Don\'t forget to define predicates. Solution Ans- We prove only the results for the inverse image of a union and the image of an intersection; the proof of the remaining two results is similar. If x f 1 S jJ Yj , then there exists y S jJ Yj such that f(x) = y. Then y Yj for some j J and x f 1 (Yj ), so x S jJ f 1 (Yj ). It follows that f 1 [ jJ Yj [ jJ f 1 (Yj ). 10 1. Sets and Functions Conversely, if x S jJ f 1 (Yj ), then x f 1 (Yj ) for some j J, so f(x) Yj and f(x) S jJ Yj , meaning that x f 1 S jJ Yj . It follows that [ jJ f 1 (Yj ) f 1 [ jJ Yj , which proves that the sets are equal. If y f T iI Xi , then there exists x T iI Xi such that f(x) = y. Then x Xi and y f(Xi) for every i I, meaning that y T iI f (Xi). It follows that f \\ iI Xi ! \\ iI f (Xi). The only case in which we don’t always have equality is for the image of an intersection, and we may get strict inclusion here if f is not one-to-one. Example 1.25. Define f : R R by f(x) = x 2 . Let A = (1, 0) and B = (0, 1). Then A B = and f(A B) = , but f(A) = f(B) = (0, 1), so f(A) f(B) = (0, 1) 6= f(A B). Next, we generalize the Cartesian product of finitely many sets to the product of possibly infinitely many sets. Definition 1.26. Let C = {Xi : i I} be an indexed collection of sets Xi . The Cartesian product of C is the set of functions that assign to each index i I an element xi Xi . That is, Y iI Xi = ( f : I [ iI Xi : f(i) Xi for every i I ) . For example, if I = {1, 2, . . . , n}, then f defines an ordered n-tuple of elements (x1, x2, . . . , xn) with xi = f(i) Xi , so this definition is equivalent to our previous one. If Xi = X for every i I, then Q iI Xi is simply the set of functions from I to X, and we also write it as XI = {f : I X} . We can think of this set as the set of ordered I-tuples of elements of X. Example 1.27. A sequence of real numbers (x1, x2, x3, . . . , xn, . . .) R N is a function f : N R. We study sequences and their convergence properties in Chapter 3. Example 1.28. Let 2 = {0, 1} be a set with two elements. Then a subset A I can be identified with its characteristic function A : I 2 by: i A if and only if A(i) = 1. Thus, A 7 A is a one-to-one map from P(I) onto 2 I . Before giving.

God bless... It is often useful to assert that something exists and .pdf

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Define PKI (Public Key Infrastructure) and list and discuss the types of protection that it offers. Give an example of where PKI is utilized in daily activity,(industry). Solution Answer:- PKI (Public Key Infrastructure) : Public Key Infrastructure (PKI) is a popular encryption and authentication approach used by both small businesses and large enterprises. What Is Public Key Infrastructure (PKI) : The PKI environment is made up of five components: 1) Certification Authority (CA) -- serves as the root of trust that authenticates the identity of individuals, computers and other entities in the network. 2) Registration Authority (RA) : -- is certified by a root CA to issue certificates for uses permitted by the CA. In a Microsoft PKI environment, the RA is normally called a subordinate CA. 3) Certificate Database : -- saves certificate requests issued and revoked certificates from the RA or CA. 4) Certificate Store :-- saves issued certificates and pending or rejected certificate requests from the local computer. 5) Key Archival Server :-- saves encrypted private keys in a certificate database for disaster recovery purposes in case the Certificate Database is lost. 6) PKI is a very effective method for implementing multi-factor authentication. Some companies, such as Unisys, require that devices that are attached to the corporate network must be able to use PKI for the encrypted and authenticated exchange of information. 7) In cryptography, a PKI is an arrangement that binds public keys with respective identities of entities (like persons and organizations). 8) A public key infrastructure (PKI) is a system for the creation, storage, and distribution of digital certificates which are used to verify that a particular public key belongs to a certain entity. Types Protection: 1) Encryption and/or sender authentication of e-mail messages . 2) Encryption and/or authentication of documents . 3) Authentication of users to applications (e.g., smart card logon, client authentication with SSL). There\'s experimental usage for digitally signed HTTP authentication in the Enigform and mod_openpgp projects . 4) Bootstrapping secure communication protocols such as Internet key exchange (IKE) and SSL. In both of these, initial set-up of a secure channel security association uses asymmetric key , public key methods, whereas actual communication uses faster symmetric key, secret key methods. 5) Mobile signatures are electronic signatures that are created using a mobile device and rely on signature or certification services in a location independent telecommunication environment..

Define PKI (Public Key Infrastructure) and list and discuss the type.pdf

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Consider the graph given above. a. What is the order of the graph? b. What is the degree of vertex SS? c. What is the degree of vertex RR? d. How many components does the graph have? Solution a) we need to find the order of the given graph. we know that the order of the raph is the number of vertices It has total 5 vertice that is Q, R , S , T , U so we can say that order of the graph is 5 b) we need to find the degree of vertex S The degree of a graph vertex of a graph is the number of graph edges which touch . vertex S touches 3 edges so we can say that degree of vertex S is 3 c) we need to find the degree of vertex R vertex R touches only one edge so we can say that degree of vertex R is 1 d) given graph does not form any loop. so we can say that it has only 1 component..

Consider the graph given above.a. What is the order of the graph.pdf

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