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“According to new research in the Journal of Research in Personality, extroverts prefer the ocean, and introverts go for the mountains.” This is the general theme uncovered by a study that was performed by a team of researchers at the University of Virginia.There are areas where mountains and ocean cohabitate in an incredible adventure and view feast. Apparently travel habits are influenced by our personalities.In general, extroverts are described as outgoing individuals who tend to socialize, suggesting they would gravitate towards places with more recreational activities and people-watching. Introverts, on the other hand, are defined as quieter people who don\'t need a lot of outside stimulus. \"We argue that beaches are typically noisier, with more people to watch, talk to, and hang out with than mountains,” said the study. “In contrast, mountains offer many secluded places, which facilitate isolation.” People preferred the ocean over mountains when they wanted to socialize with others, but they preferred the mountains and the ocean equally when they wanted to decompress alone.Mountains are not normally a place for socialization. All that open space just screams quiet and peacefulness. However, decompressing on a busy beach can be difficult. Maybe the participants of the study were thinking about a more peaceful beach scenario. Solution “According to new research in the Journal of Research in Personality, extroverts prefer the ocean, and introverts go for the mountains.” This is the general theme uncovered by a study that was performed by a team of researchers at the University of Virginia.There are areas where mountains and ocean cohabitate in an incredible adventure and view feast. Apparently travel habits are influenced by our personalities.In general, extroverts are described as outgoing individuals who tend to socialize, suggesting they would gravitate towards places with more recreational activities and people-watching. Introverts, on the other hand, are defined as quieter people who don\'t need a lot of outside stimulus. \"We argue that beaches are typically noisier, with more people to watch, talk to, and hang out with than mountains,” said the study. “In contrast, mountains offer many secluded places, which facilitate isolation.” People preferred the ocean over mountains when they wanted to socialize with others, but they preferred the mountains and the ocean equally when they wanted to decompress alone.Mountains are not normally a place for socialization. All that open space just screams quiet and peacefulness. However, decompressing on a busy beach can be difficult. Maybe the participants of the study were thinking about a more peaceful beach scenario..
“According to new research in the Journal of Research in Personality.pdf
“According to new research in the Journal of Research in Personality.pdf
sharnapiyush773
True as each officer has fixed paroleeles ratio = no of parolles/no of officer Solution True as each officer has fixed paroleeles ratio = no of parolles/no of officer.
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
sharnapiyush773
to upper: Solution to upper:.
to upperSolutionto upper.pdf
to upperSolutionto upper.pdf
sharnapiyush773
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base). Solution there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base)..
there are several theory on defining acid and base, Lewis acids and .pdf
there are several theory on defining acid and base, Lewis acids and .pdf
sharnapiyush773
The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics Solution The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics.
The Statement is False. As there are many applications of hyperbo;ic.pdf
The Statement is False. As there are many applications of hyperbo;ic.pdf
sharnapiyush773
The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following items (among others) for each annual and quarterly report: • They have reviewed the report • Based on their knowledge, the financial information included in the report is fairly presented • Based on their knowledge, the report does not contain any untrue statement of material fact or omit a material fact that would make the financial statements misleading • They acknowledge their responsibility for establishing and maintaining internal controls over financial reporting and other disclosures • They have evaluated the effectiveness of these controls, presented their conclusion as to effectiveness and disclosed any material changes in the company’s controls. SOX required companies to provide audit committees with the resources and authority to engage independent counsel and advisers to help them carry out their duties. SOX also required audit committees to establish procedures for receiving whistleblower complaints regarding accounting, auditing and internal control irregularities and to provide for the confidential and anonymous treatment of employee concerns regarding such matters. In addition, SOX enhanced the external auditor’s required communications with the audit committee to include the following: • A discussion of all critical accounting policies and practices used by the company • All alternative accounting treatments that have been discussed with management, the ramifications of the use of alternative disclosures and accounting treatments and the accounting treatment preferred by the audit firm • Other material written communications between the auditor and management. Conclusion: I think the provisions of Sarbanes oxely Act are adequate, We should also focus on the practicality of ever changing corporote world . A recent report showed that chances of reinstatement of financial statements are 46% higher for the companies who do follow sarbanes- oxley act. See this provision of Sarbanes oxely act. Officers and directors are prohibited from trading during pension “blackout” periods: The Act prohibits corporate officers and directors from trading company securities during a pension fund “blackout” period . It was never there . It was aptly introduced by Sarbanes-oxely Act. Solution The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following it.
The Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
The Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
sharnapiyush773
The main motivation behind software reuse is to avoid wastage of time and money in doing the same tasks repeatedly: The five artifacts are as follows: Portability: It means a product should have easier way of modification so that it can run on a whole new system without incorporating any new component into it or further increasing its cost. Suppose a product or a software which is developed and used at one company it may not usefull for the company now, it may give other charitable companies and only because of portability it is possible. Reusability: it itself a artifact of reuse because reusability means using parts of one product to facilitate the implementation of other product with different or similar features. COTS: it means commercial off the shelf, Actually these are the readily made parts of a software which can be incorporated into other software part to give a specific functionality. COMPONENTS: In reuse, the components which we are using they should be feasible to conform to architectural model of component model which is to build and integrated. Checklists, Requirement specifications and intermediate results of product design pattern are other reusable artifacts. Solution The main motivation behind software reuse is to avoid wastage of time and money in doing the same tasks repeatedly: The five artifacts are as follows: Portability: It means a product should have easier way of modification so that it can run on a whole new system without incorporating any new component into it or further increasing its cost. Suppose a product or a software which is developed and used at one company it may not usefull for the company now, it may give other charitable companies and only because of portability it is possible. Reusability: it itself a artifact of reuse because reusability means using parts of one product to facilitate the implementation of other product with different or similar features. COTS: it means commercial off the shelf, Actually these are the readily made parts of a software which can be incorporated into other software part to give a specific functionality. COMPONENTS: In reuse, the components which we are using they should be feasible to conform to architectural model of component model which is to build and integrated. Checklists, Requirement specifications and intermediate results of product design pattern are other reusable artifacts..
The main motivation behind software reuse is to avoid wastage of tim.pdf
The main motivation behind software reuse is to avoid wastage of tim.pdf
sharnapiyush773
The expression evaluation is compiler dependent, and may vary. A general rule is still followed by many compilers, the rule of precedence and associativity, which is as follows: Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefix ––sizeof & * + – ~ ! Unary Right to left typecasts Unary Right to left * / % Multiplicative Left to right + – Additive Left to right << >> Bitwise shift Left to right < > <= >= Relational Left to right == != Equality Left to right & Bitwise-AND Left to right ^ Bitwise-exclusive-OR Left to right | Bitwise-inclusive-OR Left to right && Logical-AND Left to right || Logical-OR Left to right ? : Conditional-expression Right to left = *= /= %= += –= <<= >>=&= ^= |= Simple and compound assignment2 Right to left , Sequential evaluation Left to right As per the above table the expression (5+11)/(16-100) will be evaluated as 5+11 = 16 then 16-100 = -84 and then 16 / -84 = -0.190476 but if you look at the output in a program it will show it as 0 since it will take the int value. to see the float value you can use this expression (5.0+11.0)/(16.0-100.0) i.e. just add .0 (decimal 0 to convert it to a float). Q2. 65-78+90*4 90*4=360 is evaluated first due to high precedance now, expression becomes 65-78+360 and + and - have the same precedance and the expression is evaluated from left to right 65-78 = -13 now -13+360 = 347 Q3. 10%2/91*(4-12) first braces (4-12) = -8 10 % 2 / 91 * -8 now all operators here have the same precedence, and they will be evaluated from left to right so, 10%2 = 0 now, the exp. is 0 / 91 * -8 0/91 = 0 now, the exp. is 0 * -8 = 0 2.) the backslash (\\) is used as escape sequence in C/C++, escape sequences when used in strings represent a different character, to print the given string we need to print the \\ character itself which is represented by a double backslash i.e. \\\\. printf(\"The backslash (\\\\) is used in a folder path.\ Don\'t use more than two together.\"); Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefix ––sizeof & * + – ~ ! Unary Right to left typecasts Unary Right to left * / % Multiplicative Left to right + – Additive Left to right << >> Bitwise shift Left to right < > <= >= Relational Left to right == != Equality Left to right & Bitwise-AND Left to right ^ Bitwise-exclusive-OR Left to right | Bitwise-inclusive-OR Left to right && Logical-AND Left to right || Logical-OR Left to right ? : Conditional-expression Right to left = *= /= %= += –= <<= >>=&= ^= |= Simple and compound assignment2 Right to left , Sequential evaluation Left to right Solution The expression evaluation is compiler dependent, and may vary. A general rule is still followed by many compilers, the rule of precedence and associativity, which is as follows: Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefi.
The expression evaluation is compiler dependent, and may vary. A g.pdf
The expression evaluation is compiler dependent, and may vary. A g.pdf
sharnapiyush773
Recommended
“According to new research in the Journal of Research in Personality, extroverts prefer the ocean, and introverts go for the mountains.” This is the general theme uncovered by a study that was performed by a team of researchers at the University of Virginia.There are areas where mountains and ocean cohabitate in an incredible adventure and view feast. Apparently travel habits are influenced by our personalities.In general, extroverts are described as outgoing individuals who tend to socialize, suggesting they would gravitate towards places with more recreational activities and people-watching. Introverts, on the other hand, are defined as quieter people who don\'t need a lot of outside stimulus. \"We argue that beaches are typically noisier, with more people to watch, talk to, and hang out with than mountains,” said the study. “In contrast, mountains offer many secluded places, which facilitate isolation.” People preferred the ocean over mountains when they wanted to socialize with others, but they preferred the mountains and the ocean equally when they wanted to decompress alone.Mountains are not normally a place for socialization. All that open space just screams quiet and peacefulness. However, decompressing on a busy beach can be difficult. Maybe the participants of the study were thinking about a more peaceful beach scenario. Solution “According to new research in the Journal of Research in Personality, extroverts prefer the ocean, and introverts go for the mountains.” This is the general theme uncovered by a study that was performed by a team of researchers at the University of Virginia.There are areas where mountains and ocean cohabitate in an incredible adventure and view feast. Apparently travel habits are influenced by our personalities.In general, extroverts are described as outgoing individuals who tend to socialize, suggesting they would gravitate towards places with more recreational activities and people-watching. Introverts, on the other hand, are defined as quieter people who don\'t need a lot of outside stimulus. \"We argue that beaches are typically noisier, with more people to watch, talk to, and hang out with than mountains,” said the study. “In contrast, mountains offer many secluded places, which facilitate isolation.” People preferred the ocean over mountains when they wanted to socialize with others, but they preferred the mountains and the ocean equally when they wanted to decompress alone.Mountains are not normally a place for socialization. All that open space just screams quiet and peacefulness. However, decompressing on a busy beach can be difficult. Maybe the participants of the study were thinking about a more peaceful beach scenario..
“According to new research in the Journal of Research in Personality.pdf
“According to new research in the Journal of Research in Personality.pdf
sharnapiyush773
True as each officer has fixed paroleeles ratio = no of parolles/no of officer Solution True as each officer has fixed paroleeles ratio = no of parolles/no of officer.
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
sharnapiyush773
to upper: Solution to upper:.
to upperSolutionto upper.pdf
to upperSolutionto upper.pdf
sharnapiyush773
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base). Solution there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base)..
there are several theory on defining acid and base, Lewis acids and .pdf
there are several theory on defining acid and base, Lewis acids and .pdf
sharnapiyush773
The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics Solution The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics.
The Statement is False. As there are many applications of hyperbo;ic.pdf
The Statement is False. As there are many applications of hyperbo;ic.pdf
sharnapiyush773
The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following items (among others) for each annual and quarterly report: • They have reviewed the report • Based on their knowledge, the financial information included in the report is fairly presented • Based on their knowledge, the report does not contain any untrue statement of material fact or omit a material fact that would make the financial statements misleading • They acknowledge their responsibility for establishing and maintaining internal controls over financial reporting and other disclosures • They have evaluated the effectiveness of these controls, presented their conclusion as to effectiveness and disclosed any material changes in the company’s controls. SOX required companies to provide audit committees with the resources and authority to engage independent counsel and advisers to help them carry out their duties. SOX also required audit committees to establish procedures for receiving whistleblower complaints regarding accounting, auditing and internal control irregularities and to provide for the confidential and anonymous treatment of employee concerns regarding such matters. In addition, SOX enhanced the external auditor’s required communications with the audit committee to include the following: • A discussion of all critical accounting policies and practices used by the company • All alternative accounting treatments that have been discussed with management, the ramifications of the use of alternative disclosures and accounting treatments and the accounting treatment preferred by the audit firm • Other material written communications between the auditor and management. Conclusion: I think the provisions of Sarbanes oxely Act are adequate, We should also focus on the practicality of ever changing corporote world . A recent report showed that chances of reinstatement of financial statements are 46% higher for the companies who do follow sarbanes- oxley act. See this provision of Sarbanes oxely act. Officers and directors are prohibited from trading during pension “blackout” periods: The Act prohibits corporate officers and directors from trading company securities during a pension fund “blackout” period . It was never there . It was aptly introduced by Sarbanes-oxely Act. Solution The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following it.
The Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
The Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
sharnapiyush773
The main motivation behind software reuse is to avoid wastage of time and money in doing the same tasks repeatedly: The five artifacts are as follows: Portability: It means a product should have easier way of modification so that it can run on a whole new system without incorporating any new component into it or further increasing its cost. Suppose a product or a software which is developed and used at one company it may not usefull for the company now, it may give other charitable companies and only because of portability it is possible. Reusability: it itself a artifact of reuse because reusability means using parts of one product to facilitate the implementation of other product with different or similar features. COTS: it means commercial off the shelf, Actually these are the readily made parts of a software which can be incorporated into other software part to give a specific functionality. COMPONENTS: In reuse, the components which we are using they should be feasible to conform to architectural model of component model which is to build and integrated. Checklists, Requirement specifications and intermediate results of product design pattern are other reusable artifacts. Solution The main motivation behind software reuse is to avoid wastage of time and money in doing the same tasks repeatedly: The five artifacts are as follows: Portability: It means a product should have easier way of modification so that it can run on a whole new system without incorporating any new component into it or further increasing its cost. Suppose a product or a software which is developed and used at one company it may not usefull for the company now, it may give other charitable companies and only because of portability it is possible. Reusability: it itself a artifact of reuse because reusability means using parts of one product to facilitate the implementation of other product with different or similar features. COTS: it means commercial off the shelf, Actually these are the readily made parts of a software which can be incorporated into other software part to give a specific functionality. COMPONENTS: In reuse, the components which we are using they should be feasible to conform to architectural model of component model which is to build and integrated. Checklists, Requirement specifications and intermediate results of product design pattern are other reusable artifacts..
The main motivation behind software reuse is to avoid wastage of tim.pdf
The main motivation behind software reuse is to avoid wastage of tim.pdf
sharnapiyush773
The expression evaluation is compiler dependent, and may vary. A general rule is still followed by many compilers, the rule of precedence and associativity, which is as follows: Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefix ––sizeof & * + – ~ ! Unary Right to left typecasts Unary Right to left * / % Multiplicative Left to right + – Additive Left to right << >> Bitwise shift Left to right < > <= >= Relational Left to right == != Equality Left to right & Bitwise-AND Left to right ^ Bitwise-exclusive-OR Left to right | Bitwise-inclusive-OR Left to right && Logical-AND Left to right || Logical-OR Left to right ? : Conditional-expression Right to left = *= /= %= += –= <<= >>=&= ^= |= Simple and compound assignment2 Right to left , Sequential evaluation Left to right As per the above table the expression (5+11)/(16-100) will be evaluated as 5+11 = 16 then 16-100 = -84 and then 16 / -84 = -0.190476 but if you look at the output in a program it will show it as 0 since it will take the int value. to see the float value you can use this expression (5.0+11.0)/(16.0-100.0) i.e. just add .0 (decimal 0 to convert it to a float). Q2. 65-78+90*4 90*4=360 is evaluated first due to high precedance now, expression becomes 65-78+360 and + and - have the same precedance and the expression is evaluated from left to right 65-78 = -13 now -13+360 = 347 Q3. 10%2/91*(4-12) first braces (4-12) = -8 10 % 2 / 91 * -8 now all operators here have the same precedence, and they will be evaluated from left to right so, 10%2 = 0 now, the exp. is 0 / 91 * -8 0/91 = 0 now, the exp. is 0 * -8 = 0 2.) the backslash (\\) is used as escape sequence in C/C++, escape sequences when used in strings represent a different character, to print the given string we need to print the \\ character itself which is represented by a double backslash i.e. \\\\. printf(\"The backslash (\\\\) is used in a folder path.\ Don\'t use more than two together.\"); Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefix ––sizeof & * + – ~ ! Unary Right to left typecasts Unary Right to left * / % Multiplicative Left to right + – Additive Left to right << >> Bitwise shift Left to right < > <= >= Relational Left to right == != Equality Left to right & Bitwise-AND Left to right ^ Bitwise-exclusive-OR Left to right | Bitwise-inclusive-OR Left to right && Logical-AND Left to right || Logical-OR Left to right ? : Conditional-expression Right to left = *= /= %= += –= <<= >>=&= ^= |= Simple and compound assignment2 Right to left , Sequential evaluation Left to right Solution The expression evaluation is compiler dependent, and may vary. A general rule is still followed by many compilers, the rule of precedence and associativity, which is as follows: Symbol1 Type of Operation Associativity [ ] ( ) . –> postfix ++ and postfix –– Expression Left to right prefix ++ and prefi.
The expression evaluation is compiler dependent, and may vary. A g.pdf
The expression evaluation is compiler dependent, and may vary. A g.pdf
sharnapiyush773
public class Storm { //Attributes private String stormName; private int stormYear; private String stormStart; private String stormEnd; private int stormMag; /** * Constructor * * @param stormName * @param stormYear * @param stormStart * @param stormEnd * @param stormMag */ public Storm(String stormName, int stormYear, String stormStart, String stormEnd, int stormMag) { this.stormName = stormName; this.stormYear = stormYear; this.stormStart = stormStart; this.stormEnd = stormEnd; this.stormMag = stormMag; } /** * @return the stormName */ public String getStormName() { return stormName; } /** * @param stormName the stormName to set */ public void setStormName(String stormName) { this.stormName = stormName; } /** * @return the stormYear */ public int getStormYear() { return stormYear; } /** * @param stormYear the stormYear to set */ public void setStormYear(int stormYear) { this.stormYear = stormYear; } /** * @return the stormStart */ public String getStormStart() { return stormStart; } /** * @param stormStart the stormStart to set */ public void setStormStart(String stormStart) { this.stormStart = stormStart; } /** * @return the stormEnd */ public String getStormEnd() { return stormEnd; } /** * @param stormEnd the stormEnd to set */ public void setStormEnd(String stormEnd) { this.stormEnd = stormEnd; } /** * @return the stormMag */ public int getStormMag() { return stormMag; } /** * @param stormMag the stormMag to set */ public void setStormMag(int stormMag) { this.stormMag = stormMag; } @Override public String toString() { return \"\ \" + getStormYear() + \": \" + getStormName() + \" \" + ((getStormMag() == -1) ? \"(no info)\" : ((getStormMag() == 0) ? \"(tropical storm)\" : \"(hurricane level \" + getStormMag() + \")\")) + \": \" + ((getStormStart().equals(\"\")) ? \"(no start)\" : getStormStart().substring(0, 2) + \"/\" + getStormStart().substring(2)) + \" - \" + ((getStormEnd().equals(\"\")) ? \"(no end)\" : getStormEnd().substring(0, 2) + \"/\" + getStormEnd().substring(2)); } } import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class Database { private static final int MAX_SIZE = 50; //Attributes private Storm[] stormArr; private int count; /** * Constructor * Accepts a file and attempts to read it * Fills the storm array with the data */ public Database(File fileName) { //Initialize array this.stormArr = new Storm[MAX_SIZE]; this.count = 0; //Scanner to read from the file Scanner in = null; try { in = new Scanner(fileName); //Read data from the file while(in.hasNextLine()) { //Year of storm/ Name of storm/ mmdd storm started/ mmdd storm ended/ magnitude of storm String line = in.nextLine(); String[] data = line.replaceAll(\"/\", \"/ \").split(\"/\"); if(data.length < 5) System.out.println(\"Database entry not in the correct format: \" + line); //Add data to array this.stormArr[this.count] = new Storm(data[1].trim(), Integer.parseInt(data[0].trim()), data[2].trim(), data[3].trim(), (data[4].trim()..
public class Storm { Attributes private String stormName;.pdf
public class Storm { Attributes private String stormName;.pdf
sharnapiyush773
picture is missing Solution picture is missing.
picture is missingSolutionpicture is missing.pdf
picture is missingSolutionpicture is missing.pdf
sharnapiyush773
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 ) { setFirstName(firstName); setLastName(lastName); setMiddleInitial(middleInitial); setFulltime(fulltime); setGender(gender); setEmployeeNum(employeeNum); } public String getFirstName() { return firstName; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getLastName() { return lastName; } public void setLastName(String lastName) { this.lastName = lastName; } public char getMiddleInitial() { return middleInitial; } public void setMiddleInitial(char middleInitial) { this.middleInitial = middleInitial; } public boolean isFulltime() { return fulltime; } public void setFulltime(boolean fulltime) { this.fulltime = fulltime; } public char getGender() { return gender; } public void setGender(char gender) { if(gender==\'M\' || gender==\'F\') { this.gender = gender; } else { this.gender=\'F\'; } } public int getEmployeeNum() { return employeeNum; } public void setEmployeeNum(int employeeNum) { while(true) { if(employeeNum>=10000 && employeeNum<=99999) { this.employeeNum = employeeNum; } else { new IllegalArgumentException(\"Invalid Employee Num.should be between 10000 and 99999(inclusive) \"); continue; } } } Override(Have to keep at the rate symbol before the override where ever it is in this project) public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Employee other = (Employee) obj; if (employeeNum != other.employeeNum) return false; if (firstName == null) { if (other.firstName != null) return false; } else if (!firstName.equals(other.firstName)) return false; if (fulltime != other.fulltime) return false; if (gender != other.gender) return false; if (lastName == null) { if (other.lastName != null) return false; } else if (!lastName.equals(other.lastName)) return false; if (middleInitial != other.middleInitial) return false; return true; } Override public String toString() { System.out.println(getEmployeeNum()); System.out.println(getFirstName()+\", \"+getLastName()+\" \"+getMiddleInitial()); System.out.println(\"Gender: \"+getGender()); if(isFulltime()) { System.out.println(\"Status: Full Time\"); } else { System.out.println(\"Status: Part Time\"); } return \" \"; } } _____________________________________________________________________ package employeeType.subTypes; 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, ful.
package employeeType.employee;public class Employee { private.pdf
package employeeType.employee;public class Employee { private.pdf
sharnapiyush773
OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.In the Open Systems Interconnection (OSI) communications model, the Network layer knows the address of the neighboring nodes in the network, packages output with the correct network address information, selects routes and quality of service, and recognizes and forwards to the Transport layer incoming messages for local host domains.It is the third layer, situated in detween transport layerand data-link layer.The network layer is the third level of the Open Systems Interconnection Model (OSI Model) and the layer that provides data routing paths for network communication. Data is transferred in the form of packets via logical network paths in an ordered format controlled by the network layer. The main aim of this layer is to deliver packets from source to destination across multiple links (networks). If two computers (system) are connected on the same link then there is no need for a network layer. It routes the signal through different channels to the other end and acts as a network controller.It also divides the outgoing messages into packets and to assemble incoming packets into messages for higher levels.The Network Layer is responsible for end-to-end (source to destination) packet delivery including routing through intermediate hosts, whereas the Data Link Layer is responsible for node-to-node (hop-to-hop) frame delivery on the same link. Internet Protocol, or IP, is the method that governs how computers share data across the Internet. When one computer sends data, such as an email or a web form, its message gets parsed into small packets that contain the sending computer\'s Internet address, the receiving computer\'s address, and part of the message. Internet Protocol serves several basic functions. The Internet Protocol (IP) is the method or protocol by which data is sent from one computer to another on the Internet. Each computer (known as a host) on the Internet has at least one IP address that uniquely identifies it from all other computers on the Internet. .IP packet headers contain addresses that identify the sending computer and the receiving computer. Routers use this information to guide each packet across communication networks and connect the sending and receiving computers.TCP/IP stands for Transmission Control Protocol / Internet Protocol. It is the communication protocol used for Internet and similar networks such as Internet and Extranet. It controls and manages the data transmission over the Internet. It also defines a mechanism through which every computer on the Internet is identified separately. Every computer on the Internet must have this protocol. Solution OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.I.
OSI (Open Systems Interconnection) is reference model for how applic.pdf
OSI (Open Systems Interconnection) is reference model for how applic.pdf
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Note: Modified code code: #include #include #include #include #include using namespace std; #pragma warning(disable: 4996) typedef enum { male = 0, female = 1 } gender; // enumeration type gender struct dog { char name[30]; gender genderValue; char breed[30]; int age; float weight; }; int count = 0; // the amount of dogs currently stored in the list (initialized at 0) struct dog list[30]; // initialize list of dogs // forward declaration of functions void flush(); void branching(char); void helper(char); int add(char*, char*, char*, int, float, struct dog*); // 30 points char* search(char*, int, struct dog*); // 10 points void display(); void save(char* fileName); void load(char* fileName); // 10 points int main() { load(\"Dog_List.txt\"); // load list of dogs from file (if it exists) char ch = \'i\'; printf(\"Assignment 5: Array of Structs and Enum Types\ \ \"); printf(\"Dog Adoption Center\ \ \"); do { printf(\"Please enter your selection:\ \"); printf(\"\\ta: add a new dog to the list\ \"); printf(\"\\ts: search for a dog on the list\ \"); printf(\"\\td: display list of dogs\ \"); printf(\"\\tq: quit and save your list\ \"); ch = tolower(getchar()); flush(); branching(ch); } while (ch != \'q\'); save(\"Dog_List.txt\"); // save list of dogs to file (overwrite if it exists) return 0; } // consume leftover \'\ \' characters void flush() { int c; do c = getchar(); while (c != \'\ \' && c != EOF); } // branch to different tasks void branching(char c) { switch (c) { case \'a\': case \'s\': helper(c); break; case \'d\': display(); break; case \'q\': break; default: printf(\"Invalid input!\ \"); } } // The helper function is used to determine how much information is needed and which function to send that information to. // It uses values that are returned from some functions to produce the correct ouput. // There is no implementation needed here, but you should study this function and know how it works. // It is always helpful to understand how the code works before implementing new features. // Do not change anything in this function or you risk failing the automated test cases. void helper(char c) { char input[100]; if (c == \'a\') { printf(\"\ Please enter the dog\'s information in the following format:\ \"); printf(\"\\tname:gender:breed:age:weight\ \"); fgets(input, sizeof(input), stdin); // discard \'\ \' chars attached to input input[strlen(input) - 1] = \'\\0\'; char* name = strtok(input, \":\"); // strtok used to parse string char* genderValueString = strtok(NULL, \":\"); char* breed = strtok(NULL, \":\"); int age = atoi(strtok(NULL, \":\")); // atoi used to convert string to int float weight = atof(strtok(NULL, \":\")); // atof used to convert string to float int result = add(name, genderValueString, breed, age, weight, list); if (result == 0) printf(\"\ That dog is already on the list\ \ \"); else printf(\"\ Dog added to list successfully\ \ \"); } else // c = \'s\' { printf(\"\ Please enter the dog\'s information in the following forma.
Note Modified codecode#includeiostream #include stdio.h.pdf
Note Modified codecode#includeiostream #include stdio.h.pdf
sharnapiyush773
Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about. Solution Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about..
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
sharnapiyush773
n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million Solution n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million.
n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf
n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf
sharnapiyush773
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if he felt that the application was wrongly rejected. The situation should have been handled before Vitek hacking the system. The reason for this is because he affected the larger populace and environment by his act. The sewage spill was significant. It polluted over 500 metres of open drain in a residential area and flowed into a tidal canal. Cleaning up the spill and its effects took days and required the deployment of considerable resources. \"Marine life died, the creek water turned black and the stench was unbearable for residents,\" said Janelle Bryant, investigations manager for the Australian Environmental Protection Agency. Solution Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if.
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
sharnapiyush773
It lacks the origin of replication which is most important for the initiation of replication Solution It lacks the origin of replication which is most important for the initiation of replication.
It lacks the origin of replication which is most important for the i.pdf
It lacks the origin of replication which is most important for the i.pdf
sharnapiyush773
In developmental biology, an embryo is divided into two hemispheres: the animal pole and the vegetal pole within a blastula. The animal pole consists of small cells that divide rapidly, in contrast with the vegetal pole below it. In some cases, the animal pole is thought to differentiate into the later embryo itself, forming the three primary germ layers and participating in gastrulation. The vegetal pole contains large yolky cells that divide very slowly, in contrast with the animal pole above it. In some cases, the vegetal pole is thought to differentiate into the extraembryonic membranes that protect and nourish the developing embryo, such as the placentain mammals and the chorion in birds. The development of the animal-vegetal axis occurs prior to fertilisation. Sperm entry can occur anywhere in the animal hemisphere. The point of sperm entry defines the dorso-ventral axis - cells opposite the region of sperm entry will eventually form the dorsal portion of the body a. Males release so many sperm that the egg is covered by them. b. The egg has a plasma membrane, a vitelline envelope, and a jelly coat. c. Acrosome enzymes digest away the zona pellucida around the egg as it extrudes a filament that attaches to a receptor on the vitelline jelly layer envelope. d. This interaction between filament and receptor is a lock-and-key reaction that is species- specific. e. The egg plasma membrane and the sperm nuclear membrane fuse, allowing the nucleus to enter. f. Fusion takes place and the zygote begins development. g. As soon as the plasma membranes of sperm and egg fuse, the plasma membrane and the vitelline envelope undergo changes that prevent entrance of any other sperm. h. The vitelline envelope now becomes the fertilization envelope. Early Developmental Stages 1. Development includes events and processes that occur as a single cell becomes a complex organism. 2. All chordate embryos go through same early developmental stages: zygote, morula, blastula, early and late gastrula. 3. The presence of yolk, dense nutrient material, affects how the embryonic cells complete the first three stages. 4. Following fertilization, a zygote undergoes cleavage, cell division without growth. a. DNA replication and mitosis occur repeatedly; the cells get smaller each division. b. As deuterostomes, lancelets have a radial and indeterminate pattern of cleavage. 1) In radial cleavage, any plane passing through will divide the embryo into symmetrical halves. 2) In indeterminate cleavage, cells have not differentiated; if separated, each one develops a complete organism. 5. Because the lancelet has little yolk, the cell divisions are equal in the resulting morula. 6. A cavity called the blastocoel develops forming the hollow ball called the blastula. 7. Gastrulation is invagination of some cells of the blastocyst into blastocoel to form three primary germ layers. a. The outer layer of cells becomes the ectoderm. b. The inner layer of cells becomes the endoderm. c.
In developmental biology, an embryo is divided into two hemispheres.pdf
In developmental biology, an embryo is divided into two hemispheres.pdf
sharnapiyush773
Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water. Solution Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water..
Bryophytes- Development of primitive vasculature for water transport.pdf
Bryophytes- Development of primitive vasculature for water transport.pdf
sharnapiyush773
given Solution given.
givenSolutiongiven.pdf
givenSolutiongiven.pdf
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Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame. Solution Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame..
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
sharnapiyush773
content analysis refers to an analytical process for measuring the semantic content for communication Solution content analysis refers to an analytical process for measuring the semantic content for communication.
content analysis refers to an analytical process for measuring the s.pdf
content analysis refers to an analytical process for measuring the s.pdf
sharnapiyush773
Because acetone is soluble in water.Since both are miscible They cannot be separated and hence are not used for liquid-liquid extraction Solution Because acetone is soluble in water.Since both are miscible They cannot be separated and hence are not used for liquid-liquid extraction.
Because acetone is soluble in water.Since both are miscible They c.pdf
Because acetone is soluble in water.Since both are miscible They c.pdf
sharnapiyush773
Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect. Solution Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect..
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
sharnapiyush773
Answer: Note: Entire skeleton of the code is provided. The below code is implemented as per the declarations provided. Program code: #include #include #include typedef uint64_t weatherlog_t; unsigned int add(unsigned int, unsigned int); unsigned int sub(unsigned int, unsigned int); unsigned int mul(unsigned int, unsigned int); void print_half_nybbles(unsigned int); unsigned int reverse_half_nybbles(unsigned int); int has_odd(unsigned int); unsigned int make_odd(unsigned int); int is_negative(int); weatherlog_t pack_log_entry(unsigned int, unsigned int, unsigned int, unsigned int, int, int, unsigned int, unsigned int); unsigned int get_year(weatherlog_t entry); unsigned int get_month(weatherlog_t entry); unsigned int get_day(weatherlog_t entry); unsigned int get_zip(weatherlog_t entry); unsigned int get_high(weatherlog_t entry); unsigned int get_low(weatherlog_t entry); unsigned int get_precip(weatherlog_t entry); unsigned int get_wind(weatherlog_t entry); int main(int argc, char **argv) { unsigned int i, j; int x, y; unsigned int year, month, day, zip, high_temp, low_temp, precip, avg_wind_speed; weatherlog_t log_entry; printf(\"Enter an integer: \"); scanf(\"%u\", &i); printf(\"Enter another integer: \"); scanf(\"%u\", &j); printf(\"One more integer, please: \"); scanf(\"%d\", &x); printf(\"Please enter a positive integer: \"); scanf(\"%d\", &y); printf(\"i + j = %u\ \", add(i,j)); printf(\"i - j = %u\ \", sub(i,j)); printf(\"i * j = %u\ \", mul(i,j)); if (is_negative(x)) printf(\"%d is negative\ \", x); else printf(\"%d is non-negative\ \", x); if (has_odd(y)) { printf(\"%x has an odd number of bits in its binary representation\ \", y); } else { printf(\"%x has an even number of bits in its binary representation\ \", y); printf(\"but %x has an odd number of bits in its binary representation\ \", make_odd(y)); } printf(\"The half-nybbles of %d (in hex 0x%x) are:\", x, x); print_half_nybbles(x); printf(\"%x with reversed half-nybbles is %x\ \", x, reverse_half_nybbles(x)); printf(\"Enter a year: \"); scanf(\"%u\", &year); printf(\"Enter a month as an integer (1-12): \"); scanf(\"%u\", &month); printf(\"Enter a day as an integer (1-31): \"); scanf(\"%u\", &day); printf(\"Enter a zip code as an integer (0-99999): \"); scanf(\"%u\", &zip); printf(\"Enter a temperature as an integer: \"); scanf(\"%u\", &high_temp); printf(\"Enter another temperature as an integer: \"); scanf(\"%u\", &low_temp); printf(\"Enter rainfall amount as an integer (mm): \"); scanf(\"%u\", &precip); printf(\"Enter a as an integer (km/hr): \"); scanf(\"%u\", &avg_wind_speed); log_entry=pack_log_entry(year, month, day, zip, high_temp, low_temp, precip, avg_wind_speed); printf(\"You entered: %u/%u/%u for zip %5d: high %d F, low %d F, precip %d mm, wind speed %d km/hr\ \", get_day(log_entry), get_month(log_entry), get_year(log_entry), get_zip(log_entry), get_high(log_entry), get_low(log_entry), get_precip(log_entry), get_wind(log_entry)); system(\"pause\"); return 0; } unsigne.
AnswerNote Entire skeleton of the code is provided. The below co.pdf
AnswerNote Entire skeleton of the code is provided. The below co.pdf
sharnapiyush773
ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers. Solution ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers..
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
sharnapiyush773
a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. b). Cells becoming old or cells undergoing damage require replacement of cells by cell division. The age of cell and any changes causing damage in cells determine the occurrance of cell division in both the prokaryotes and eukaryotes. c). The prokaryotes have small genomes and may contain less than 500 to more than 5000 genes. The bacteria have a small circular chromosome and show specific variation with other bacterial species. The bacteria contain extra chromosomal circular DNA called a plasmid. The plasmid is capable of replicating on its own and often carries the non essential genes that can be transferred from one cell to another. The bacterial genes are organized in to clusters known as operons that can be transferred at a time. The genes of the operons contain protein encoding gene as well as regulatory sequences. The prokaryotes are single celled organism whereas the eukaryotes are multi cellular. The eukaryotic genomes are large in size than the prokaryotes and may range 10 MB to over 100,000MB. Prokaryotes contain a single chromosome and eukaryotes contain several. The eukaryotic genes exhibit special characteristics like presence of non-coding or intervening sequences or introns and repetitive sequences that are responsible for large genome size in eukaryotes. d). In unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. Solution a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. Th.
a). In both unicellular and multicellular organisms the purpose of c.pdf
a). In both unicellular and multicellular organisms the purpose of c.pdf
sharnapiyush773
A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust Solution A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust.
A chemical property describes a substance and its ability to change .pdf
A chemical property describes a substance and its ability to change .pdf
sharnapiyush773
APM webinar hosted by the Scotland Network on 14 May 2024. Speakers: Chris Drysdale and Peter Huggett An interactive session discussing how Project Managers can identify mental health symptoms, provide tools to help themselves and others, plus also increase the capabilities of the Project Management function. This webinar was held on 14 May 2024. The covid-19 pandemic led to concerns about a worsening of mental health & wellbeing across the world and increased awareness in both society and the workplace. This webinar looks to advise the benefits of having a Mental Health First Aid function in the workplace whilst also providing tools and techniques that can be readily used and applied to yourself and colleagues. Additionally, there are wider benefits to Project Management which will be proposed and discussed.
Including Mental Health Support in Project Delivery, 14 May.pdf
Including Mental Health Support in Project Delivery, 14 May.pdf
Association for Project Management
Transport (British English) or Transportation (American English) ransportation has developed along three basic Mode (Media):- 1. Land Transportation (way)– (a) Road Transportation (b) Rail Transportation 2. Water Transportation 3. Air Transportation Tramway Inland water transport Ocean transport These may be classified as under: (a). Liners (b). Tramps Liners Vs Tramps Figure- Layout airport runway design TRAFFIC SIGNS
Basic Civil Engineering notes on Transportation Engineering & Modes of Transport
Basic Civil Engineering notes on Transportation Engineering & Modes of Transport
Denish Jangid
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public class Storm { //Attributes private String stormName; private int stormYear; private String stormStart; private String stormEnd; private int stormMag; /** * Constructor * * @param stormName * @param stormYear * @param stormStart * @param stormEnd * @param stormMag */ public Storm(String stormName, int stormYear, String stormStart, String stormEnd, int stormMag) { this.stormName = stormName; this.stormYear = stormYear; this.stormStart = stormStart; this.stormEnd = stormEnd; this.stormMag = stormMag; } /** * @return the stormName */ public String getStormName() { return stormName; } /** * @param stormName the stormName to set */ public void setStormName(String stormName) { this.stormName = stormName; } /** * @return the stormYear */ public int getStormYear() { return stormYear; } /** * @param stormYear the stormYear to set */ public void setStormYear(int stormYear) { this.stormYear = stormYear; } /** * @return the stormStart */ public String getStormStart() { return stormStart; } /** * @param stormStart the stormStart to set */ public void setStormStart(String stormStart) { this.stormStart = stormStart; } /** * @return the stormEnd */ public String getStormEnd() { return stormEnd; } /** * @param stormEnd the stormEnd to set */ public void setStormEnd(String stormEnd) { this.stormEnd = stormEnd; } /** * @return the stormMag */ public int getStormMag() { return stormMag; } /** * @param stormMag the stormMag to set */ public void setStormMag(int stormMag) { this.stormMag = stormMag; } @Override public String toString() { return \"\ \" + getStormYear() + \": \" + getStormName() + \" \" + ((getStormMag() == -1) ? \"(no info)\" : ((getStormMag() == 0) ? \"(tropical storm)\" : \"(hurricane level \" + getStormMag() + \")\")) + \": \" + ((getStormStart().equals(\"\")) ? \"(no start)\" : getStormStart().substring(0, 2) + \"/\" + getStormStart().substring(2)) + \" - \" + ((getStormEnd().equals(\"\")) ? \"(no end)\" : getStormEnd().substring(0, 2) + \"/\" + getStormEnd().substring(2)); } } import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class Database { private static final int MAX_SIZE = 50; //Attributes private Storm[] stormArr; private int count; /** * Constructor * Accepts a file and attempts to read it * Fills the storm array with the data */ public Database(File fileName) { //Initialize array this.stormArr = new Storm[MAX_SIZE]; this.count = 0; //Scanner to read from the file Scanner in = null; try { in = new Scanner(fileName); //Read data from the file while(in.hasNextLine()) { //Year of storm/ Name of storm/ mmdd storm started/ mmdd storm ended/ magnitude of storm String line = in.nextLine(); String[] data = line.replaceAll(\"/\", \"/ \").split(\"/\"); if(data.length < 5) System.out.println(\"Database entry not in the correct format: \" + line); //Add data to array this.stormArr[this.count] = new Storm(data[1].trim(), Integer.parseInt(data[0].trim()), data[2].trim(), data[3].trim(), (data[4].trim()..
public class Storm { Attributes private String stormName;.pdf
public class Storm { Attributes private String stormName;.pdf
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picture is missing Solution picture is missing.
picture is missingSolutionpicture is missing.pdf
picture is missingSolutionpicture is missing.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 ) { setFirstName(firstName); setLastName(lastName); setMiddleInitial(middleInitial); setFulltime(fulltime); setGender(gender); setEmployeeNum(employeeNum); } public String getFirstName() { return firstName; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getLastName() { return lastName; } public void setLastName(String lastName) { this.lastName = lastName; } public char getMiddleInitial() { return middleInitial; } public void setMiddleInitial(char middleInitial) { this.middleInitial = middleInitial; } public boolean isFulltime() { return fulltime; } public void setFulltime(boolean fulltime) { this.fulltime = fulltime; } public char getGender() { return gender; } public void setGender(char gender) { if(gender==\'M\' || gender==\'F\') { this.gender = gender; } else { this.gender=\'F\'; } } public int getEmployeeNum() { return employeeNum; } public void setEmployeeNum(int employeeNum) { while(true) { if(employeeNum>=10000 && employeeNum<=99999) { this.employeeNum = employeeNum; } else { new IllegalArgumentException(\"Invalid Employee Num.should be between 10000 and 99999(inclusive) \"); continue; } } } Override(Have to keep at the rate symbol before the override where ever it is in this project) public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Employee other = (Employee) obj; if (employeeNum != other.employeeNum) return false; if (firstName == null) { if (other.firstName != null) return false; } else if (!firstName.equals(other.firstName)) return false; if (fulltime != other.fulltime) return false; if (gender != other.gender) return false; if (lastName == null) { if (other.lastName != null) return false; } else if (!lastName.equals(other.lastName)) return false; if (middleInitial != other.middleInitial) return false; return true; } Override public String toString() { System.out.println(getEmployeeNum()); System.out.println(getFirstName()+\", \"+getLastName()+\" \"+getMiddleInitial()); System.out.println(\"Gender: \"+getGender()); if(isFulltime()) { System.out.println(\"Status: Full Time\"); } else { System.out.println(\"Status: Part Time\"); } return \" \"; } } _____________________________________________________________________ package employeeType.subTypes; 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, ful.
package employeeType.employee;public class Employee { private.pdf
package employeeType.employee;public class Employee { private.pdf
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OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.In the Open Systems Interconnection (OSI) communications model, the Network layer knows the address of the neighboring nodes in the network, packages output with the correct network address information, selects routes and quality of service, and recognizes and forwards to the Transport layer incoming messages for local host domains.It is the third layer, situated in detween transport layerand data-link layer.The network layer is the third level of the Open Systems Interconnection Model (OSI Model) and the layer that provides data routing paths for network communication. Data is transferred in the form of packets via logical network paths in an ordered format controlled by the network layer. The main aim of this layer is to deliver packets from source to destination across multiple links (networks). If two computers (system) are connected on the same link then there is no need for a network layer. It routes the signal through different channels to the other end and acts as a network controller.It also divides the outgoing messages into packets and to assemble incoming packets into messages for higher levels.The Network Layer is responsible for end-to-end (source to destination) packet delivery including routing through intermediate hosts, whereas the Data Link Layer is responsible for node-to-node (hop-to-hop) frame delivery on the same link. Internet Protocol, or IP, is the method that governs how computers share data across the Internet. When one computer sends data, such as an email or a web form, its message gets parsed into small packets that contain the sending computer\'s Internet address, the receiving computer\'s address, and part of the message. Internet Protocol serves several basic functions. The Internet Protocol (IP) is the method or protocol by which data is sent from one computer to another on the Internet. Each computer (known as a host) on the Internet has at least one IP address that uniquely identifies it from all other computers on the Internet. .IP packet headers contain addresses that identify the sending computer and the receiving computer. Routers use this information to guide each packet across communication networks and connect the sending and receiving computers.TCP/IP stands for Transmission Control Protocol / Internet Protocol. It is the communication protocol used for Internet and similar networks such as Internet and Extranet. It controls and manages the data transmission over the Internet. It also defines a mechanism through which every computer on the Internet is identified separately. Every computer on the Internet must have this protocol. Solution OSI (Open Systems Interconnection) is reference model for how applications can communicate over a network. A reference model is a conceptual framework for understanding relationships.I.
OSI (Open Systems Interconnection) is reference model for how applic.pdf
OSI (Open Systems Interconnection) is reference model for how applic.pdf
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Note: Modified code code: #include #include #include #include #include using namespace std; #pragma warning(disable: 4996) typedef enum { male = 0, female = 1 } gender; // enumeration type gender struct dog { char name[30]; gender genderValue; char breed[30]; int age; float weight; }; int count = 0; // the amount of dogs currently stored in the list (initialized at 0) struct dog list[30]; // initialize list of dogs // forward declaration of functions void flush(); void branching(char); void helper(char); int add(char*, char*, char*, int, float, struct dog*); // 30 points char* search(char*, int, struct dog*); // 10 points void display(); void save(char* fileName); void load(char* fileName); // 10 points int main() { load(\"Dog_List.txt\"); // load list of dogs from file (if it exists) char ch = \'i\'; printf(\"Assignment 5: Array of Structs and Enum Types\ \ \"); printf(\"Dog Adoption Center\ \ \"); do { printf(\"Please enter your selection:\ \"); printf(\"\\ta: add a new dog to the list\ \"); printf(\"\\ts: search for a dog on the list\ \"); printf(\"\\td: display list of dogs\ \"); printf(\"\\tq: quit and save your list\ \"); ch = tolower(getchar()); flush(); branching(ch); } while (ch != \'q\'); save(\"Dog_List.txt\"); // save list of dogs to file (overwrite if it exists) return 0; } // consume leftover \'\ \' characters void flush() { int c; do c = getchar(); while (c != \'\ \' && c != EOF); } // branch to different tasks void branching(char c) { switch (c) { case \'a\': case \'s\': helper(c); break; case \'d\': display(); break; case \'q\': break; default: printf(\"Invalid input!\ \"); } } // The helper function is used to determine how much information is needed and which function to send that information to. // It uses values that are returned from some functions to produce the correct ouput. // There is no implementation needed here, but you should study this function and know how it works. // It is always helpful to understand how the code works before implementing new features. // Do not change anything in this function or you risk failing the automated test cases. void helper(char c) { char input[100]; if (c == \'a\') { printf(\"\ Please enter the dog\'s information in the following format:\ \"); printf(\"\\tname:gender:breed:age:weight\ \"); fgets(input, sizeof(input), stdin); // discard \'\ \' chars attached to input input[strlen(input) - 1] = \'\\0\'; char* name = strtok(input, \":\"); // strtok used to parse string char* genderValueString = strtok(NULL, \":\"); char* breed = strtok(NULL, \":\"); int age = atoi(strtok(NULL, \":\")); // atoi used to convert string to int float weight = atof(strtok(NULL, \":\")); // atof used to convert string to float int result = add(name, genderValueString, breed, age, weight, list); if (result == 0) printf(\"\ That dog is already on the list\ \ \"); else printf(\"\ Dog added to list successfully\ \ \"); } else // c = \'s\' { printf(\"\ Please enter the dog\'s information in the following forma.
Note Modified codecode#includeiostream #include stdio.h.pdf
Note Modified codecode#includeiostream #include stdio.h.pdf
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Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about. Solution Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about..
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
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n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million Solution n = N *2^(rt) n = no of people at a given time N = initial population r = rate of population growth t = time so 59 = 35 * 2^ (r*50) so r = 0.015 so n = 35 * 2^ (20r) = 43.13 million.
n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf
n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf
sharnapiyush773
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if he felt that the application was wrongly rejected. The situation should have been handled before Vitek hacking the system. The reason for this is because he affected the larger populace and environment by his act. The sewage spill was significant. It polluted over 500 metres of open drain in a residential area and flowed into a tidal canal. Cleaning up the spill and its effects took days and required the deployment of considerable resources. \"Marine life died, the creek water turned black and the stench was unbearable for residents,\" said Janelle Bryant, investigations manager for the Australian Environmental Protection Agency. Solution Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if.
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
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It lacks the origin of replication which is most important for the initiation of replication Solution It lacks the origin of replication which is most important for the initiation of replication.
It lacks the origin of replication which is most important for the i.pdf
It lacks the origin of replication which is most important for the i.pdf
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In developmental biology, an embryo is divided into two hemispheres: the animal pole and the vegetal pole within a blastula. The animal pole consists of small cells that divide rapidly, in contrast with the vegetal pole below it. In some cases, the animal pole is thought to differentiate into the later embryo itself, forming the three primary germ layers and participating in gastrulation. The vegetal pole contains large yolky cells that divide very slowly, in contrast with the animal pole above it. In some cases, the vegetal pole is thought to differentiate into the extraembryonic membranes that protect and nourish the developing embryo, such as the placentain mammals and the chorion in birds. The development of the animal-vegetal axis occurs prior to fertilisation. Sperm entry can occur anywhere in the animal hemisphere. The point of sperm entry defines the dorso-ventral axis - cells opposite the region of sperm entry will eventually form the dorsal portion of the body a. Males release so many sperm that the egg is covered by them. b. The egg has a plasma membrane, a vitelline envelope, and a jelly coat. c. Acrosome enzymes digest away the zona pellucida around the egg as it extrudes a filament that attaches to a receptor on the vitelline jelly layer envelope. d. This interaction between filament and receptor is a lock-and-key reaction that is species- specific. e. The egg plasma membrane and the sperm nuclear membrane fuse, allowing the nucleus to enter. f. Fusion takes place and the zygote begins development. g. As soon as the plasma membranes of sperm and egg fuse, the plasma membrane and the vitelline envelope undergo changes that prevent entrance of any other sperm. h. The vitelline envelope now becomes the fertilization envelope. Early Developmental Stages 1. Development includes events and processes that occur as a single cell becomes a complex organism. 2. All chordate embryos go through same early developmental stages: zygote, morula, blastula, early and late gastrula. 3. The presence of yolk, dense nutrient material, affects how the embryonic cells complete the first three stages. 4. Following fertilization, a zygote undergoes cleavage, cell division without growth. a. DNA replication and mitosis occur repeatedly; the cells get smaller each division. b. As deuterostomes, lancelets have a radial and indeterminate pattern of cleavage. 1) In radial cleavage, any plane passing through will divide the embryo into symmetrical halves. 2) In indeterminate cleavage, cells have not differentiated; if separated, each one develops a complete organism. 5. Because the lancelet has little yolk, the cell divisions are equal in the resulting morula. 6. A cavity called the blastocoel develops forming the hollow ball called the blastula. 7. Gastrulation is invagination of some cells of the blastocyst into blastocoel to form three primary germ layers. a. The outer layer of cells becomes the ectoderm. b. The inner layer of cells becomes the endoderm. c.
In developmental biology, an embryo is divided into two hemispheres.pdf
In developmental biology, an embryo is divided into two hemispheres.pdf
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Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water. Solution Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water..
Bryophytes- Development of primitive vasculature for water transport.pdf
Bryophytes- Development of primitive vasculature for water transport.pdf
sharnapiyush773
given Solution given.
givenSolutiongiven.pdf
givenSolutiongiven.pdf
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Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame. Solution Ethernet II framing (also known as DIX Ethernet, named after DEC, Intel and Xerox, the major participants in its design), defines the two-octet EtherType field in an Ethernet frame, preceded by destination and source MAC addresses, that identifies an upper layer protocol encapsulating the frame data. The Length/EtherType field is the only one which differs between 802.3 and Ethernet II. In 802.3 it indicates the number of bytes of data in the frames payload, and can be anything from 0 to 1500 bytes. Frames must be at least 64 bytes long, not including the preamble, so, if the data field is shorter than 46 bytes, it must be compensated by the Pad field. The reason for specifying a minimum length lies with the collision-detect mechanism. In CSMA/CD a station must never be allowed to believe it has transmitted a frame successfully if that frame has, in fact, experienced a collision. In the worst case it takes twice the maximum propagation delay across the network before a station can be sure that a transmission has been successful. If a station sends a really short frame, it may actually finish sending and release the Ether without realising that a collision has occurred. The 802.3 design rules specify an upper limit on the maximum propagation delay in any Ethernet installation, and the minimum frame size is set to be more than twice this figure. In Ethernet II, on the other hand, this field is used to indicate the type of payload carried by the frame..
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
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content analysis refers to an analytical process for measuring the semantic content for communication Solution content analysis refers to an analytical process for measuring the semantic content for communication.
content analysis refers to an analytical process for measuring the s.pdf
content analysis refers to an analytical process for measuring the s.pdf
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Because acetone is soluble in water.Since both are miscible They cannot be separated and hence are not used for liquid-liquid extraction Solution Because acetone is soluble in water.Since both are miscible They cannot be separated and hence are not used for liquid-liquid extraction.
Because acetone is soluble in water.Since both are miscible They c.pdf
Because acetone is soluble in water.Since both are miscible They c.pdf
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Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect. Solution Answer: Dichlorodiphenyltrichloroethane (DDT) is an organochlorine insecticide. DDT causes the sodium channels in insect neurons to open which causing them to fire uncontrollably (repeatedly generate an impulse). This accounts for the repetitive body tremors, spasms and eventual death of the insect..
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
sharnapiyush773
Answer: Note: Entire skeleton of the code is provided. The below code is implemented as per the declarations provided. Program code: #include #include #include typedef uint64_t weatherlog_t; unsigned int add(unsigned int, unsigned int); unsigned int sub(unsigned int, unsigned int); unsigned int mul(unsigned int, unsigned int); void print_half_nybbles(unsigned int); unsigned int reverse_half_nybbles(unsigned int); int has_odd(unsigned int); unsigned int make_odd(unsigned int); int is_negative(int); weatherlog_t pack_log_entry(unsigned int, unsigned int, unsigned int, unsigned int, int, int, unsigned int, unsigned int); unsigned int get_year(weatherlog_t entry); unsigned int get_month(weatherlog_t entry); unsigned int get_day(weatherlog_t entry); unsigned int get_zip(weatherlog_t entry); unsigned int get_high(weatherlog_t entry); unsigned int get_low(weatherlog_t entry); unsigned int get_precip(weatherlog_t entry); unsigned int get_wind(weatherlog_t entry); int main(int argc, char **argv) { unsigned int i, j; int x, y; unsigned int year, month, day, zip, high_temp, low_temp, precip, avg_wind_speed; weatherlog_t log_entry; printf(\"Enter an integer: \"); scanf(\"%u\", &i); printf(\"Enter another integer: \"); scanf(\"%u\", &j); printf(\"One more integer, please: \"); scanf(\"%d\", &x); printf(\"Please enter a positive integer: \"); scanf(\"%d\", &y); printf(\"i + j = %u\ \", add(i,j)); printf(\"i - j = %u\ \", sub(i,j)); printf(\"i * j = %u\ \", mul(i,j)); if (is_negative(x)) printf(\"%d is negative\ \", x); else printf(\"%d is non-negative\ \", x); if (has_odd(y)) { printf(\"%x has an odd number of bits in its binary representation\ \", y); } else { printf(\"%x has an even number of bits in its binary representation\ \", y); printf(\"but %x has an odd number of bits in its binary representation\ \", make_odd(y)); } printf(\"The half-nybbles of %d (in hex 0x%x) are:\", x, x); print_half_nybbles(x); printf(\"%x with reversed half-nybbles is %x\ \", x, reverse_half_nybbles(x)); printf(\"Enter a year: \"); scanf(\"%u\", &year); printf(\"Enter a month as an integer (1-12): \"); scanf(\"%u\", &month); printf(\"Enter a day as an integer (1-31): \"); scanf(\"%u\", &day); printf(\"Enter a zip code as an integer (0-99999): \"); scanf(\"%u\", &zip); printf(\"Enter a temperature as an integer: \"); scanf(\"%u\", &high_temp); printf(\"Enter another temperature as an integer: \"); scanf(\"%u\", &low_temp); printf(\"Enter rainfall amount as an integer (mm): \"); scanf(\"%u\", &precip); printf(\"Enter a as an integer (km/hr): \"); scanf(\"%u\", &avg_wind_speed); log_entry=pack_log_entry(year, month, day, zip, high_temp, low_temp, precip, avg_wind_speed); printf(\"You entered: %u/%u/%u for zip %5d: high %d F, low %d F, precip %d mm, wind speed %d km/hr\ \", get_day(log_entry), get_month(log_entry), get_year(log_entry), get_zip(log_entry), get_high(log_entry), get_low(log_entry), get_precip(log_entry), get_wind(log_entry)); system(\"pause\"); return 0; } unsigne.
AnswerNote Entire skeleton of the code is provided. The below co.pdf
AnswerNote Entire skeleton of the code is provided. The below co.pdf
sharnapiyush773
ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers. Solution ANS: D (ventral root) Ventral root: The spinal cord grows into ventral and dorsal sides. The Posterior median sulcus groove is called dorsal side and anterior median sulcus groove is called ventral side The ventral roots contain efferent fibers and dorsal root contains afferent fibers. The dorsal roots contain afferent fibers and the ventral roots contain efferent fibers..
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
sharnapiyush773
a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. b). Cells becoming old or cells undergoing damage require replacement of cells by cell division. The age of cell and any changes causing damage in cells determine the occurrance of cell division in both the prokaryotes and eukaryotes. c). The prokaryotes have small genomes and may contain less than 500 to more than 5000 genes. The bacteria have a small circular chromosome and show specific variation with other bacterial species. The bacteria contain extra chromosomal circular DNA called a plasmid. The plasmid is capable of replicating on its own and often carries the non essential genes that can be transferred from one cell to another. The bacterial genes are organized in to clusters known as operons that can be transferred at a time. The genes of the operons contain protein encoding gene as well as regulatory sequences. The prokaryotes are single celled organism whereas the eukaryotes are multi cellular. The eukaryotic genomes are large in size than the prokaryotes and may range 10 MB to over 100,000MB. Prokaryotes contain a single chromosome and eukaryotes contain several. The eukaryotic genes exhibit special characteristics like presence of non-coding or intervening sequences or introns and repetitive sequences that are responsible for large genome size in eukaryotes. d). In unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. The gemetes undergo meiosis or reduction division to produce half number of chromosomes from each parent cell and when both male and female gametes fuse, they produce complete set of chromosomes. Solution a). In both unicellular and multicellular organisms the purpose of cell division is growth, maintenance and reproduction. in unicellular organisms like bacteria, the cell division takes place by mitosis to produce two idential daughter cells. In multicellular organisms the all the cells except the gametes undergo mitosis to produce daughter cells that look like parent cells. The cells undergoing mitosis also maintain same number of chromosomes similar to parent cells. Th.
a). In both unicellular and multicellular organisms the purpose of c.pdf
a). In both unicellular and multicellular organisms the purpose of c.pdf
sharnapiyush773
A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust Solution A chemical property describes a substance and its ability to change into a new substance with different properties. Chemical properties of matter are : FLAMMABILITY: Flammability is the ability to burn. example:wood can be burned to create smoke and ash REACTIVITY: Reactivity is when 2 substances get together,something can happen like bubbling,fizzing or color change. example:Iron reacts with Oxygen to form rust.
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