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(1) c. Two-tailed test (2)d. Ho: Men and women are the same in their political understanding. (3)b. Ho: Solution (1) c. Two-tailed test (2)d. Ho: Men and women are the same in their political understanding. (3)b. Ho:.
(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
anwarsadath111
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer; when deprotonated the compounds become more soluble in the aqueous layer then the non- aqueous and \"jump\" into the aqueous layer. Solution We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer; when deprotonated the compounds become more soluble in the aqueous layer then the non- aqueous and \"jump\" into the aqueous layer..
We use a base (in this case sodium bicarbonate) during the separatio.pdf
We use a base (in this case sodium bicarbonate) during the separatio.pdf
anwarsadath111
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted from the calculation. Solution While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted from the calculation..
While computing the dilluted earnings per share Dividend paid on Pre.pdf
While computing the dilluted earnings per share Dividend paid on Pre.pdf
anwarsadath111
There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protein · Post-translation: After protein synthesis. Solution There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protei.
There are several ways in which the gene regulation in eukaryotes di.pdf
There are several ways in which the gene regulation in eukaryotes di.pdf
anwarsadath111
The water must be boiled to remove traces of dissolved carbon dioxide (CO2). CO2 dissolves in water to form carbonic acid (H2CO3): CO2 + H2O => H2CO3 H2CO3 is a weak acid and partially ionizes to give H+ ions: H2CO3 => H+ + HCO3- HCO3- => H+ + CO32- The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate according to the reaction: 2 S2O32- + H+ => HSO3- + S Solution The water must be boiled to remove traces of dissolved carbon dioxide (CO2). CO2 dissolves in water to form carbonic acid (H2CO3): CO2 + H2O => H2CO3 H2CO3 is a weak acid and partially ionizes to give H+ ions: H2CO3 => H+ + HCO3- HCO3- => H+ + CO32- The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate according to the reaction: 2 S2O32- + H+ => HSO3- + S.
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
anwarsadath111
adsorb onto silica gel Solution adsorb onto silica gel.
adsorb onto silica gel Solu.pdf
adsorb onto silica gel Solu.pdf
anwarsadath111
Solution : Plants show different types of symptoms when there is any nutritional vacancy in plants.Nitrogen is one of the most common nutrient deficiency in plants.Nitrogen deficiency leads to decrease in growth.Plants are mainly short.Some leaves not able to produce chlorophyll which leads to green color to them.So it give rise to yellow color in leaves..
Solution Plants show different types of symptoms when there is an.pdf
Solution Plants show different types of symptoms when there is an.pdf
anwarsadath111
SF4 Solution SF4.
SF4SolutionSF4.pdf
SF4SolutionSF4.pdf
anwarsadath111
Recommandé
(1) c. Two-tailed test (2)d. Ho: Men and women are the same in their political understanding. (3)b. Ho: Solution (1) c. Two-tailed test (2)d. Ho: Men and women are the same in their political understanding. (3)b. Ho:.
(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
anwarsadath111
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer; when deprotonated the compounds become more soluble in the aqueous layer then the non- aqueous and \"jump\" into the aqueous layer. Solution We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer; when deprotonated the compounds become more soluble in the aqueous layer then the non- aqueous and \"jump\" into the aqueous layer..
We use a base (in this case sodium bicarbonate) during the separatio.pdf
We use a base (in this case sodium bicarbonate) during the separatio.pdf
anwarsadath111
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted from the calculation. Solution While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted from the calculation..
While computing the dilluted earnings per share Dividend paid on Pre.pdf
While computing the dilluted earnings per share Dividend paid on Pre.pdf
anwarsadath111
There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protein · Post-translation: After protein synthesis. Solution There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protei.
There are several ways in which the gene regulation in eukaryotes di.pdf
There are several ways in which the gene regulation in eukaryotes di.pdf
anwarsadath111
The water must be boiled to remove traces of dissolved carbon dioxide (CO2). CO2 dissolves in water to form carbonic acid (H2CO3): CO2 + H2O => H2CO3 H2CO3 is a weak acid and partially ionizes to give H+ ions: H2CO3 => H+ + HCO3- HCO3- => H+ + CO32- The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate according to the reaction: 2 S2O32- + H+ => HSO3- + S Solution The water must be boiled to remove traces of dissolved carbon dioxide (CO2). CO2 dissolves in water to form carbonic acid (H2CO3): CO2 + H2O => H2CO3 H2CO3 is a weak acid and partially ionizes to give H+ ions: H2CO3 => H+ + HCO3- HCO3- => H+ + CO32- The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate according to the reaction: 2 S2O32- + H+ => HSO3- + S.
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
anwarsadath111
adsorb onto silica gel Solution adsorb onto silica gel.
adsorb onto silica gel Solu.pdf
adsorb onto silica gel Solu.pdf
anwarsadath111
Solution : Plants show different types of symptoms when there is any nutritional vacancy in plants.Nitrogen is one of the most common nutrient deficiency in plants.Nitrogen deficiency leads to decrease in growth.Plants are mainly short.Some leaves not able to produce chlorophyll which leads to green color to them.So it give rise to yellow color in leaves..
Solution Plants show different types of symptoms when there is an.pdf
Solution Plants show different types of symptoms when there is an.pdf
anwarsadath111
SF4 Solution SF4.
SF4SolutionSF4.pdf
SF4SolutionSF4.pdf
anwarsadath111
Quick ratio = (Cash + accounts receivable)/Accounts payable For 2008 the quick ratio = (140 + 780)/1120 = 920/1120 = .82 A .82 Solution Quick ratio = (Cash + accounts receivable)/Accounts payable For 2008 the quick ratio = (140 + 780)/1120 = 920/1120 = .82 A .82.
Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf
Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf
anwarsadath111
Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). The infection is strep throat and it is caused by streptococcal bacteria (group B). These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Apart from antibiotics, the major future intervention to treat strep throat is to implement vaccination to boost innate immune levels in the infant because antibiotics such as macrolides have potential inhibitor effects on the growth of the useful gutfolra in infants & toddlers. Probiotics, gicing more Vitamin D & C and natural remedies are major future interventions to avoid strep throat in children. Immunity stimulation is specific future intervention to combat strep throat in infants as vaccination promotes immune factors to reduce the bacteria that cause inflammation of the blood vessels near throat. The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial cell type\" of body cell provides an indication that the sputum specimen represents material from an active infection. Other cells are present in the sputum during the active infection are polymorphonuclear leukocytes (PMNs) Ques-2) What is the significance of facial swelling? List the possible causes of facial swelling? 1). Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). 2). The bacteria causes inflammation of the blood vessels supplying glomeruli, which impairs the glomerular filtration. Decreased urine filtration causes edema (swelling of body organs). Ques-3) What are the possible causes of dark urine? Dark urine is mainly due to \"presence of RBC\" in the blood. Urine was positive for 3+ protein and for3+ blood by dipstick. The examination of urinary sediment revealed numerous RBCs, many of which were dismorphic. Red blood cell casts were also present. Solution Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibioti.
Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf
Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf
anwarsadath111
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination. Solution Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination..
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
anwarsadath111
Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Return on capital(ROC) 24.15% Return on asset(ROA) 20.07% Return on Equity(ROE) 41.30% Efficiency Ratios Profit margin=Net Profit/Sales Operating Profit margin=Operating Profit/Sales Asset turnover=Sales/Average Total Assets Inventory turnover(cogs)=Cost of goods sold/Average total inventory Days in inventory=365/inventory turnover(cogs) Inventory turnover is also calculated from sales: Inventory turnover (Sales)=Sales/average total inventory Days sales in inventory=365/inventory turnover (sales) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Operating Income 6946 E Sales 30274 F Total assets 25707 25509 25608 G Cost of goods sold 15383 H Total inventory 3518 3706 3612 I=C/E Profit margin 0.16974962 J=D/E Operating Profit margin 0.229437801 K=E/F Asset turnover 1.182208685 L=G/H Inventory turnover(cogs) 4.258859358 M=365/L Days in inventory 85.70369889 N=E/H Inventory turnover (Sales) 8.381506091 P=365/N Days sales in inventory 43.54825923 Profit margin 16.97% Operating Profit margin 22.94% Asset turnover 1.18220868 Inventory turnover(cogs) 4.25885936 Days in inventory 85.7036989 Days Inventory turnover (Sales) 8.38150609 Days sales in inventory 43.5482592 Days A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Solution Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 124.
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
anwarsadath111
package employeeType.employee; public class Employee { private String firstName; private String lastName; private char middleInitial; private boolean fulltime; private char gender; private int employeeNum; public Employee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime) { super(); this.firstName = firstName; this.lastName = lastName; this.middleInitial = middleInitial; this.fulltime = fulltime; this.gender = gender; this.employeeNum = employeeNum; } public int getEmployeeNumber() { return this.employeeNum; } public void setEmployeeNumber(int empNum) { this.employeeNum = empNum; } public String getFirstName() { return firstName; } String getLastName() { return lastName; } char getMiddleInitial() { return middleInitial; } public char getGender() { return gender; } public void setFirstName(String fn) { this.firstName = fn; } public void setLastName(String ln) { this.lastName = ln; } public void setMiddleI(char m) { this.middleInitial = m; } public void setGender(char g) { this.gender = g; } public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Employee) { Employee employee = (Employee) obj; if (employee.getEmployeeNumber() == this.getEmployeeNumber()) return true; else return false; } else return false; } public String toString() { // TODO Auto-generated method stub return getEmployeeNumber() + \"\ \" + getFirstName() + \" \" + getLastName() + \"\ Gender:\" + getGender() + \"\ Status:\" + ((fulltime) ? \"Full Time\" : \"Part Time\"); } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class HourlyEmployee extends Employee { private double wage; private double hoursWorked; public HourlyEmployee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime, double wage) { super(firstName, lastName, middleInitial, gender, employeeNum, fulltime); // TODO Auto-generated constructor stub this.wage = wage; this.hoursWorked = 0.0d; } public void increaseHours(double hours) { if (hours > 0) this.hoursWorked += hours; else System.out.println(\"Error! Invalid hours\"); } public String toString() { // TODO Auto-generated method stub return super.toString() + \"\ Wage: \" + wage + \"\ Hours Worked: \" + hoursWorked; } public double calculateWeeklyPay() { double pay; if (hoursWorked > 40) { pay = (40 * wage) + (2 * (hoursWorked - 40) * wage); } else { pay = wage * hoursWorked; } return pay; } public void annualRaise() { DecimalFormat decimalFormat = new DecimalFormat(\"#.##\"); wage = Double.valueOf(decimalFormat.format(wage * 0.05)); } public double holidayBonus() { return 40 * wage; } public void resetWeek() { this.hoursWorked = 0; } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class SalaryEmployee extends Employee { double salary; public SalaryEmployee(String firstName, String lastName, char middleInitia.
package employeeType.employee;public class Employee { private .pdf
package employeeType.employee;public class Employee { private .pdf
anwarsadath111
P = 2/12 option A Solution P = 2/12 option A.
P = 212option ASolutionP = 212option A.pdf
P = 212option ASolutionP = 212option A.pdf
anwarsadath111
OSI MODEL: It has 7 layers INTERNET MODEL: It has5 Layers DOD MODEL: It has 4 Layers Solution OSI MODEL: It has 7 layers INTERNET MODEL: It has5 Layers DOD MODEL: It has 4 Layers.
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
anwarsadath111
Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains pyrimidine ring which is linked to thiazole ring Solution Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains pyrimidine ring which is linked to thiazole ring.
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
anwarsadath111
NH4NO3 = N2O + 2 H2O Moles of NH4NO3 = 1/2 x moles of H2O = 1/2 x 0.625 = 0.3125 mol Solution NH4NO3 = N2O + 2 H2O Moles of NH4NO3 = 1/2 x moles of H2O = 1/2 x 0.625 = 0.3125 mol.
NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf
NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf
anwarsadath111
No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416 Solution No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416.
No. of shares outstanding = Total assets x weight of equity price .pdf
No. of shares outstanding = Total assets x weight of equity price .pdf
anwarsadath111
molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M Solution molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M.
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
anwarsadath111
import java.util.Scanner; public class Main { public static int gridArray1[][] = null; public static int gridArray2[][] = null; public static boolean playerTurn = true; public static int count = 0; public static void setGridDimensions(int x_max, int y_max) { Main.gridArray1 = new int[x_max][y_max]; Main.gridArray2 = new int[x_max][y_max]; for (int i = 0; i < x_max; i++) { for (int j = 0; j < y_max; j++) { gridArray1[i][j] = -1; gridArray2[i][j] = -1; } } } static void placeShip(int starting_x, int starting_y, int length, int direction) { count++; int ships[] = new int[10]; int battleshipArrayPositions1[] = new int[4]; int cruiserArrayPositions1[] = new int[3]; int submarineArrayPositions1[] = new int[3]; int destroyerArrayPositions1[] = new int[2]; int carrierArrayPositions2[] = new int[5]; int battleshipArrayPositions2[] = new int[4]; int cruiserArrayPositions2[] = new int[3]; int submarineArrayPositions2[] = new int[3]; int destroyerArrayPositions2[] = new int[2]; if (playerTurn) { if (direction == 0) { for (int i = starting_x; i < starting_x + length; i++) { Main.gridArray1[i][starting_y] = 0; } } else { for (int i = starting_y; i < starting_y + length; i++) { Main.gridArray1[starting_x][i] = 0; } } } else { if (direction == 0) { for (int i = starting_x; i < starting_x + length; i++) { Main.gridArray2[i][starting_y] = 0; } } else { for (int i = starting_y; i < starting_y + length; i++) { Main.gridArray2[starting_x][i] = 0; } } } } boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) { return false; } static int shoot(int x, int y) { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (Main.gridArray2[i][j] == 0) { return 0; } } } } else { } return 0; } boolean hasBeenAttempted(int x, int y) { return false; } static void displayGrid(boolean showShips) { int count = 0; System.out.print(\" 0 1 2 3 4 5 6 7 8 9\"); if (showShips) { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray1[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray1[i][j] == 0) { System.out.print(\"@ \"); } else if (Main.gridArray1[i][j] == 1) { System.out.print(\"+ \"); } else { System.out.print(\"X \"); } } } } else { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray2[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray2[i][j] == 0) { System.out.print(\"@ \"); } else if (Main.gridArray2[i][j] == 1) { System.out.print(\"+ \"); } else { System.out.print(\"X \"); } } } } } else { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray1[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray1[i][j] == 0) { System.out.print(\"- \"); } el.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;public class Main { public static in.pdf
anwarsadath111
hi Q is nearly done,just give me dia of shaft Solution hi Q is nearly done,just give me dia of shaft.
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
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O2 is simply oxygen gas. Gases are usually made up of two of the same elements. (Cl2 = chlorine gas, N2=nitrogen gas) Solution O2 is simply oxygen gas. Gases are usually made up of two of the same elements. (Cl2 = chlorine gas, N2=nitrogen gas).
O2 is simply oxygen gas. Gases are usually made .pdf
O2 is simply oxygen gas. Gases are usually made .pdf
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CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISPR/Cpf1 systems activity has three stages: 1. Adaptation: Cas1 and Cas2 proteins facilitate the adaptation of small fragments of DNA into the CRISPR array. 2. Formation of crRNAs: processing of pre-cr-RNAs producing of mature crRNAs to guide the Cas protein. 3. Interference: the Cpf1 is bound to a crRNA to form a binary complex to identify and cleave a target DNA sequence. Solution CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISP.
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
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C. Router(config-if)#ip access-group 110 in is the right answer. ip-access-group is a command in interface configuration mode which is used to place an access list on an interface. Solution C. Router(config-if)#ip access-group 110 in is the right answer. ip-access-group is a command in interface configuration mode which is used to place an access list on an interface..
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
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C is false Solution C is false.
C is falseSolutionC is false.pdf
C is falseSolutionC is false.pdf
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B). 1/2 Solution B). 1/2.
B). 12SolutionB). 12.pdf
B). 12SolutionB). 12.pdf
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Answer: Note: LinkedList.cpp is written and driver program main.cpp is also given. //ListInterface.h #ifndef _LIST_INTERFACE #define _LIST_INTERFACE template class ListInterface { public: virtual bool isEmpty() const = 0; virtual int getLength() const = 0; virtual bool insert(int newPosition, const ItemType& newEntry) = 0; virtual bool remove(int position) = 0; virtual void clear() = 0; virtual ItemType getEntry(int position) const = 0; virtual void replace(int position, const ItemType& newEntry) = 0; }; // end ListInterface #endif //Node.h #ifndef NODE_ #define NODE_ template class Node { private: ItemType item; // A data item Node* next; // Pointer to next node public: Node(); Node(const ItemType& anItem); Node(const ItemType& anItem, Node* nextNodePtr); void setItem(const ItemType& anItem); void setNext(Node* nextNodePtr); ItemType getItem() const ; Node* getNext() const ; }; // end Node #endif //Node.cpp #include \"Node.h\" template Node::Node() : next(nullptr) { } // end default constructor template Node::Node(const ItemType& anItem) : item(anItem), next(nullptr) { } // end constructor template Node::Node(const ItemType& anItem, Node* nextNodePtr) : item(anItem), next(nextNodePtr) { } // end constructor template void Node::setItem(const ItemType& anItem) { item = anItem; } // end setItem template void Node::setNext(Node* nextNodePtr) { next = nextNodePtr; } // end setNext template ItemType Node::getItem() const { return item; } // end getItem template Node* Node::getNext() const { return next; } // end getNext //PrecondViolatedExcept.h #ifndef PRECOND_VIOLATED_EXCEPT_ #define PRECOND_VIOLATED_EXCEPT_ #include #include class PrecondViolatedExcept : public std::logic_error { public: PrecondViolatedExcept(const std::string& message = \"\"); }; // end PrecondViolatedExcept #endif //PrecondViolatedExcept.cpp #include \"PrecondViolatedExcept.h\" PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message) : std::logic_error(\"Precondition Violated Exception: \" + message) { } // end constructor //LinkedList.h #ifndef LINKED_LIST_ #define LINKED_LIST_ #include \"ListInterface.h\" #include \"Node.h\" #include #include \"PrecondViolatedExcept.h\" template class LinkedList : public ListInterface { private: Node* headPtr; // Pointer to first node in the chain; // (contains the first entry in the list) int itemCount; // Current count of list items // Locates a specified node in this linked list. Node* getNodeAt(int position) const; public: LinkedList(); LinkedList(const LinkedList& aList); virtual ~LinkedList(); bool isEmpty() const; int getLength() const; bool insert(int newPosition, const ItemType& newEntry); bool remove(int position); void clear(); ItemType getEntry(int position) const; void replace(int position, const ItemType& newEntry); }; // end LinkedList #endif //LinkedList.cpp #include \"LinkedList.h\" #include #include #include #include //Default constructor template LinkedList::LinkedList() : headPtr(nullptr), itemCount(0) { .
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
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Answer: The nucleus is separated from the cytoplasm by a phospholipid bilayer known as nuclear envelope. Transport of macromolecules across the nuclear membrane takes place through specialized nuclear pores. Proteins found in the nucleus are synthesized in the cytoplasm and imported into the nucleus through nuclear pore complexes. Such proteins contain a nuclear- localization signal (NLS) that directs their selective transport into the nucleus.4 proteins are required for nuclear localization- importin , importin , NTF-2 & Ran. Answer-A: It has been previously said that nuclear localization signal(NLS) is located within the amino acid sequence of a protein which needs to be localized in the nucleus. Importins actually identify this signal within the protein and interacts with it and facilitates the transfer of nuclear protein from cytosol to nucleus. If the NLS is not present and has already been deleted, the importins won\'t be able to identify the protein and transport them into the nucleus. In this present case, deletion mutation of the protein has been performed to identify the part of the amino acid sequence acting as NLS. In addition leptomycin B has also been added which prevents nuclear export and is not relevant with respect to nuclear localization and NLS. NLS is located in the amino acid range- 240-357. It is so because it can be observed from the available results that localization into the nucleus takes place for deletion mutants having amino acids - 1-576, 115- 576, 240-576 , 240- 357 & 115-357. Conclusions: 115-576 - NLS is not located in amino acid sequence 1-115 240-576 - NLS is not located within the amino acid sequence - 1-239 240-357 - NLS is not located in amino acid sequence 1-240 & 358-576 115-357 - NLS is not located within 1-115 & 358-576 Localization takes place in all the above cases and the deletion mutants are always found in the nucleus and they may or may not be found in the cytoplasm. From assessment of these conclusions, it can only be inferred that NLS is located in the range 240-357. Answer B: There are some “shuttling” proteins that contain a nuclear-export signal (NES) that stimulates their export from the nucleus to the cytoplasm through nuclear pores, in addition to an NLS that results in their reuptake into the nucleus. Export of proteins from the nucleus is mediated by proteins known as Exportins which first forms a complex with Ran·GTP and then binds the NES in a cargo protein. Once it crosses the nuclear pore, the Ran GAP associated with the NPC cytoplasmic filaments stimulates conversion of Ran·GTP to Ran·GDP. The accompanying conformational change in Ran leads to dissociation of the complex. The NES-containing cargo protein is released into the cytosol, while exportin 1 and Ran·GDP are transported back into the nucleus through NPCs. In this case, NES is located in the amino acid region 115-240. It is speculated to be so because: The deletion mutant bearing amino acids 115-576 possess the NES since i.
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
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Quick ratio = (Cash + accounts receivable)/Accounts payable For 2008 the quick ratio = (140 + 780)/1120 = 920/1120 = .82 A .82 Solution Quick ratio = (Cash + accounts receivable)/Accounts payable For 2008 the quick ratio = (140 + 780)/1120 = 920/1120 = .82 A .82.
Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf
Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf
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Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). The infection is strep throat and it is caused by streptococcal bacteria (group B). These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Apart from antibiotics, the major future intervention to treat strep throat is to implement vaccination to boost innate immune levels in the infant because antibiotics such as macrolides have potential inhibitor effects on the growth of the useful gutfolra in infants & toddlers. Probiotics, gicing more Vitamin D & C and natural remedies are major future interventions to avoid strep throat in children. Immunity stimulation is specific future intervention to combat strep throat in infants as vaccination promotes immune factors to reduce the bacteria that cause inflammation of the blood vessels near throat. The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial cell type\" of body cell provides an indication that the sputum specimen represents material from an active infection. Other cells are present in the sputum during the active infection are polymorphonuclear leukocytes (PMNs) Ques-2) What is the significance of facial swelling? List the possible causes of facial swelling? 1). Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). 2). The bacteria causes inflammation of the blood vessels supplying glomeruli, which impairs the glomerular filtration. Decreased urine filtration causes edema (swelling of body organs). Ques-3) What are the possible causes of dark urine? Dark urine is mainly due to \"presence of RBC\" in the blood. Urine was positive for 3+ protein and for3+ blood by dipstick. The examination of urinary sediment revealed numerous RBCs, many of which were dismorphic. Red blood cell casts were also present. Solution Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibioti.
Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf
Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf
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Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination. Solution Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination..
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
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Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Return on capital(ROC) 24.15% Return on asset(ROA) 20.07% Return on Equity(ROE) 41.30% Efficiency Ratios Profit margin=Net Profit/Sales Operating Profit margin=Operating Profit/Sales Asset turnover=Sales/Average Total Assets Inventory turnover(cogs)=Cost of goods sold/Average total inventory Days in inventory=365/inventory turnover(cogs) Inventory turnover is also calculated from sales: Inventory turnover (Sales)=Sales/average total inventory Days sales in inventory=365/inventory turnover (sales) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Operating Income 6946 E Sales 30274 F Total assets 25707 25509 25608 G Cost of goods sold 15383 H Total inventory 3518 3706 3612 I=C/E Profit margin 0.16974962 J=D/E Operating Profit margin 0.229437801 K=E/F Asset turnover 1.182208685 L=G/H Inventory turnover(cogs) 4.258859358 M=365/L Days in inventory 85.70369889 N=E/H Inventory turnover (Sales) 8.381506091 P=365/N Days sales in inventory 43.54825923 Profit margin 16.97% Operating Profit margin 22.94% Asset turnover 1.18220868 Inventory turnover(cogs) 4.25885936 Days in inventory 85.7036989 Days Inventory turnover (Sales) 8.38150609 Days sales in inventory 43.5482592 Days A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Solution Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 124.
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
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package employeeType.employee; public class Employee { private String firstName; private String lastName; private char middleInitial; private boolean fulltime; private char gender; private int employeeNum; public Employee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime) { super(); this.firstName = firstName; this.lastName = lastName; this.middleInitial = middleInitial; this.fulltime = fulltime; this.gender = gender; this.employeeNum = employeeNum; } public int getEmployeeNumber() { return this.employeeNum; } public void setEmployeeNumber(int empNum) { this.employeeNum = empNum; } public String getFirstName() { return firstName; } String getLastName() { return lastName; } char getMiddleInitial() { return middleInitial; } public char getGender() { return gender; } public void setFirstName(String fn) { this.firstName = fn; } public void setLastName(String ln) { this.lastName = ln; } public void setMiddleI(char m) { this.middleInitial = m; } public void setGender(char g) { this.gender = g; } public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Employee) { Employee employee = (Employee) obj; if (employee.getEmployeeNumber() == this.getEmployeeNumber()) return true; else return false; } else return false; } public String toString() { // TODO Auto-generated method stub return getEmployeeNumber() + \"\ \" + getFirstName() + \" \" + getLastName() + \"\ Gender:\" + getGender() + \"\ Status:\" + ((fulltime) ? \"Full Time\" : \"Part Time\"); } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class HourlyEmployee extends Employee { private double wage; private double hoursWorked; public HourlyEmployee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime, double wage) { super(firstName, lastName, middleInitial, gender, employeeNum, fulltime); // TODO Auto-generated constructor stub this.wage = wage; this.hoursWorked = 0.0d; } public void increaseHours(double hours) { if (hours > 0) this.hoursWorked += hours; else System.out.println(\"Error! Invalid hours\"); } public String toString() { // TODO Auto-generated method stub return super.toString() + \"\ Wage: \" + wage + \"\ Hours Worked: \" + hoursWorked; } public double calculateWeeklyPay() { double pay; if (hoursWorked > 40) { pay = (40 * wage) + (2 * (hoursWorked - 40) * wage); } else { pay = wage * hoursWorked; } return pay; } public void annualRaise() { DecimalFormat decimalFormat = new DecimalFormat(\"#.##\"); wage = Double.valueOf(decimalFormat.format(wage * 0.05)); } public double holidayBonus() { return 40 * wage; } public void resetWeek() { this.hoursWorked = 0; } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class SalaryEmployee extends Employee { double salary; public SalaryEmployee(String firstName, String lastName, char middleInitia.
package employeeType.employee;public class Employee { private .pdf
package employeeType.employee;public class Employee { private .pdf
anwarsadath111
P = 2/12 option A Solution P = 2/12 option A.
P = 212option ASolutionP = 212option A.pdf
P = 212option ASolutionP = 212option A.pdf
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OSI MODEL: It has 7 layers INTERNET MODEL: It has5 Layers DOD MODEL: It has 4 Layers Solution OSI MODEL: It has 7 layers INTERNET MODEL: It has5 Layers DOD MODEL: It has 4 Layers.
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
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Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains pyrimidine ring which is linked to thiazole ring Solution Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains pyrimidine ring which is linked to thiazole ring.
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
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NH4NO3 = N2O + 2 H2O Moles of NH4NO3 = 1/2 x moles of H2O = 1/2 x 0.625 = 0.3125 mol Solution NH4NO3 = N2O + 2 H2O Moles of NH4NO3 = 1/2 x moles of H2O = 1/2 x 0.625 = 0.3125 mol.
NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf
NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf
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No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416 Solution No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416.
No. of shares outstanding = Total assets x weight of equity price .pdf
No. of shares outstanding = Total assets x weight of equity price .pdf
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molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M Solution molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M.
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
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import java.util.Scanner; public class Main { public static int gridArray1[][] = null; public static int gridArray2[][] = null; public static boolean playerTurn = true; public static int count = 0; public static void setGridDimensions(int x_max, int y_max) { Main.gridArray1 = new int[x_max][y_max]; Main.gridArray2 = new int[x_max][y_max]; for (int i = 0; i < x_max; i++) { for (int j = 0; j < y_max; j++) { gridArray1[i][j] = -1; gridArray2[i][j] = -1; } } } static void placeShip(int starting_x, int starting_y, int length, int direction) { count++; int ships[] = new int[10]; int battleshipArrayPositions1[] = new int[4]; int cruiserArrayPositions1[] = new int[3]; int submarineArrayPositions1[] = new int[3]; int destroyerArrayPositions1[] = new int[2]; int carrierArrayPositions2[] = new int[5]; int battleshipArrayPositions2[] = new int[4]; int cruiserArrayPositions2[] = new int[3]; int submarineArrayPositions2[] = new int[3]; int destroyerArrayPositions2[] = new int[2]; if (playerTurn) { if (direction == 0) { for (int i = starting_x; i < starting_x + length; i++) { Main.gridArray1[i][starting_y] = 0; } } else { for (int i = starting_y; i < starting_y + length; i++) { Main.gridArray1[starting_x][i] = 0; } } } else { if (direction == 0) { for (int i = starting_x; i < starting_x + length; i++) { Main.gridArray2[i][starting_y] = 0; } } else { for (int i = starting_y; i < starting_y + length; i++) { Main.gridArray2[starting_x][i] = 0; } } } } boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) { return false; } static int shoot(int x, int y) { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (Main.gridArray2[i][j] == 0) { return 0; } } } } else { } return 0; } boolean hasBeenAttempted(int x, int y) { return false; } static void displayGrid(boolean showShips) { int count = 0; System.out.print(\" 0 1 2 3 4 5 6 7 8 9\"); if (showShips) { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray1[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray1[i][j] == 0) { System.out.print(\"@ \"); } else if (Main.gridArray1[i][j] == 1) { System.out.print(\"+ \"); } else { System.out.print(\"X \"); } } } } else { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray2[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray2[i][j] == 0) { System.out.print(\"@ \"); } else if (Main.gridArray2[i][j] == 1) { System.out.print(\"+ \"); } else { System.out.print(\"X \"); } } } } } else { if (playerTurn) { for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (j % 10 == 0) { System.out.print(\"\ \" + count + \" \"); count++; } if (Main.gridArray1[i][j] == -1) { System.out.print(\"- \"); } else if (Main.gridArray1[i][j] == 0) { System.out.print(\"- \"); } el.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;public class Main { public static in.pdf
anwarsadath111
hi Q is nearly done,just give me dia of shaft Solution hi Q is nearly done,just give me dia of shaft.
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
anwarsadath111
O2 is simply oxygen gas. Gases are usually made up of two of the same elements. (Cl2 = chlorine gas, N2=nitrogen gas) Solution O2 is simply oxygen gas. Gases are usually made up of two of the same elements. (Cl2 = chlorine gas, N2=nitrogen gas).
O2 is simply oxygen gas. Gases are usually made .pdf
O2 is simply oxygen gas. Gases are usually made .pdf
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CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISPR/Cpf1 systems activity has three stages: 1. Adaptation: Cas1 and Cas2 proteins facilitate the adaptation of small fragments of DNA into the CRISPR array. 2. Formation of crRNAs: processing of pre-cr-RNAs producing of mature crRNAs to guide the Cas protein. 3. Interference: the Cpf1 is bound to a crRNA to form a binary complex to identify and cleave a target DNA sequence. Solution CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISP.
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
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C. Router(config-if)#ip access-group 110 in is the right answer. ip-access-group is a command in interface configuration mode which is used to place an access list on an interface. Solution C. Router(config-if)#ip access-group 110 in is the right answer. ip-access-group is a command in interface configuration mode which is used to place an access list on an interface..
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
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C is false Solution C is false.
C is falseSolutionC is false.pdf
C is falseSolutionC is false.pdf
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B). 1/2 Solution B). 1/2.
B). 12SolutionB). 12.pdf
B). 12SolutionB). 12.pdf
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Answer: Note: LinkedList.cpp is written and driver program main.cpp is also given. //ListInterface.h #ifndef _LIST_INTERFACE #define _LIST_INTERFACE template class ListInterface { public: virtual bool isEmpty() const = 0; virtual int getLength() const = 0; virtual bool insert(int newPosition, const ItemType& newEntry) = 0; virtual bool remove(int position) = 0; virtual void clear() = 0; virtual ItemType getEntry(int position) const = 0; virtual void replace(int position, const ItemType& newEntry) = 0; }; // end ListInterface #endif //Node.h #ifndef NODE_ #define NODE_ template class Node { private: ItemType item; // A data item Node* next; // Pointer to next node public: Node(); Node(const ItemType& anItem); Node(const ItemType& anItem, Node* nextNodePtr); void setItem(const ItemType& anItem); void setNext(Node* nextNodePtr); ItemType getItem() const ; Node* getNext() const ; }; // end Node #endif //Node.cpp #include \"Node.h\" template Node::Node() : next(nullptr) { } // end default constructor template Node::Node(const ItemType& anItem) : item(anItem), next(nullptr) { } // end constructor template Node::Node(const ItemType& anItem, Node* nextNodePtr) : item(anItem), next(nextNodePtr) { } // end constructor template void Node::setItem(const ItemType& anItem) { item = anItem; } // end setItem template void Node::setNext(Node* nextNodePtr) { next = nextNodePtr; } // end setNext template ItemType Node::getItem() const { return item; } // end getItem template Node* Node::getNext() const { return next; } // end getNext //PrecondViolatedExcept.h #ifndef PRECOND_VIOLATED_EXCEPT_ #define PRECOND_VIOLATED_EXCEPT_ #include #include class PrecondViolatedExcept : public std::logic_error { public: PrecondViolatedExcept(const std::string& message = \"\"); }; // end PrecondViolatedExcept #endif //PrecondViolatedExcept.cpp #include \"PrecondViolatedExcept.h\" PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message) : std::logic_error(\"Precondition Violated Exception: \" + message) { } // end constructor //LinkedList.h #ifndef LINKED_LIST_ #define LINKED_LIST_ #include \"ListInterface.h\" #include \"Node.h\" #include #include \"PrecondViolatedExcept.h\" template class LinkedList : public ListInterface { private: Node* headPtr; // Pointer to first node in the chain; // (contains the first entry in the list) int itemCount; // Current count of list items // Locates a specified node in this linked list. Node* getNodeAt(int position) const; public: LinkedList(); LinkedList(const LinkedList& aList); virtual ~LinkedList(); bool isEmpty() const; int getLength() const; bool insert(int newPosition, const ItemType& newEntry); bool remove(int position); void clear(); ItemType getEntry(int position) const; void replace(int position, const ItemType& newEntry); }; // end LinkedList #endif //LinkedList.cpp #include \"LinkedList.h\" #include #include #include #include //Default constructor template LinkedList::LinkedList() : headPtr(nullptr), itemCount(0) { .
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
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Answer: The nucleus is separated from the cytoplasm by a phospholipid bilayer known as nuclear envelope. Transport of macromolecules across the nuclear membrane takes place through specialized nuclear pores. Proteins found in the nucleus are synthesized in the cytoplasm and imported into the nucleus through nuclear pore complexes. Such proteins contain a nuclear- localization signal (NLS) that directs their selective transport into the nucleus.4 proteins are required for nuclear localization- importin , importin , NTF-2 & Ran. Answer-A: It has been previously said that nuclear localization signal(NLS) is located within the amino acid sequence of a protein which needs to be localized in the nucleus. Importins actually identify this signal within the protein and interacts with it and facilitates the transfer of nuclear protein from cytosol to nucleus. If the NLS is not present and has already been deleted, the importins won\'t be able to identify the protein and transport them into the nucleus. In this present case, deletion mutation of the protein has been performed to identify the part of the amino acid sequence acting as NLS. In addition leptomycin B has also been added which prevents nuclear export and is not relevant with respect to nuclear localization and NLS. NLS is located in the amino acid range- 240-357. It is so because it can be observed from the available results that localization into the nucleus takes place for deletion mutants having amino acids - 1-576, 115- 576, 240-576 , 240- 357 & 115-357. Conclusions: 115-576 - NLS is not located in amino acid sequence 1-115 240-576 - NLS is not located within the amino acid sequence - 1-239 240-357 - NLS is not located in amino acid sequence 1-240 & 358-576 115-357 - NLS is not located within 1-115 & 358-576 Localization takes place in all the above cases and the deletion mutants are always found in the nucleus and they may or may not be found in the cytoplasm. From assessment of these conclusions, it can only be inferred that NLS is located in the range 240-357. Answer B: There are some “shuttling” proteins that contain a nuclear-export signal (NES) that stimulates their export from the nucleus to the cytoplasm through nuclear pores, in addition to an NLS that results in their reuptake into the nucleus. Export of proteins from the nucleus is mediated by proteins known as Exportins which first forms a complex with Ran·GTP and then binds the NES in a cargo protein. Once it crosses the nuclear pore, the Ran GAP associated with the NPC cytoplasmic filaments stimulates conversion of Ran·GTP to Ran·GDP. The accompanying conformational change in Ran leads to dissociation of the complex. The NES-containing cargo protein is released into the cytosol, while exportin 1 and Ran·GDP are transported back into the nucleus through NPCs. In this case, NES is located in the amino acid region 115-240. It is speculated to be so because: The deletion mutant bearing amino acids 115-576 possess the NES since i.
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
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AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
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