Correct answer b. LiF Since F has highest electr.pdf
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Correct answer: b. LiF Since F has highest electronegative value of 4 and it forms a shorter and stable molecule with any electropositive atom. So it has highest melting point Solution Correct answer: b. LiF Since F has highest electronegative value of 4 and it forms a shorter and stable molecule with any electropositive atom. So it has highest melting point.
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“According to new research in the Journal of Research in Personality.pdf
“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..
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
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.
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 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)..
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 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 Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
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 main motivation behind software reuse is to avoid wastage of tim.pdf
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 expression evaluation is compiler dependent, and may vary. A g.pdf
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.
Recommended
“According to new research in the Journal of Research in Personality.pdf
“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..
Trueas each officer has fixed paroleelesratio = no of parollesn.pdf
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.
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 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)..
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 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 Sarbanes-Oxley , 2002 contains following provisions which determ.pdf
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 main motivation behind software reuse is to avoid wastage of tim.pdf
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 expression evaluation is compiler dependent, and may vary. A g.pdf
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.
public class Storm { Attributes private String stormName;.pdf
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()..
package employeeType.employee;public class Employee { private.pdf
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.
OSI (Open Systems Interconnection) is reference model for how applic.pdf
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.
Note Modified codecode#includeiostream #include stdio.h.pdf
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.
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
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..
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 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.
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
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.
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 initiation of replication
Solution
It lacks the origin of replication which is most important for the initiation of replication.
In developmental biology, an embryo is divided into two hemispheres.pdf
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.
Bryophytes- Development of primitive vasculature for water transport.pdf
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..
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
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..
content analysis refers to an analytical process for measuring the s.pdf
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.
Because acetone is soluble in water.Since both are miscible They c.pdf
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.
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
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..
AnswerNote Entire skeleton of the code is provided. The below co.pdf
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.
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
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..
a). In both unicellular and multicellular organisms the purpose of c.pdf
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 chemical property describes a substance and its ability to change .pdf
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 Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
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public class Storm { Attributes private String stormName;.pdf
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()..
package employeeType.employee;public class Employee { private.pdf
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.
OSI (Open Systems Interconnection) is reference model for how applic.pdf
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.
Note Modified codecode#includeiostream #include stdio.h.pdf
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.
Oligotrophic area Usually, an olidgotropic organism is a one that .pdf
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..
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 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.
Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf
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.
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 initiation of replication
Solution
It lacks the origin of replication which is most important for the initiation of replication.
In developmental biology, an embryo is divided into two hemispheres.pdf
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.
Bryophytes- Development of primitive vasculature for water transport.pdf
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..
Ethernet II framing (also known as DIX Ethernet, named after DEC, In.pdf
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..
content analysis refers to an analytical process for measuring the s.pdf
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.
Because acetone is soluble in water.Since both are miscible They c.pdf
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.
AnswerDichlorodiphenyltrichloroethane (DDT) is an organochlorine .pdf
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..
AnswerNote Entire skeleton of the code is provided. The below co.pdf
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.
ANS D (ventral root)Ventral rootThe spinal cord grows into ven.pdf
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..
a). In both unicellular and multicellular organisms the purpose of c.pdf
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 chemical property describes a substance and its ability to change .pdf
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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