What does this figure depict hydrogen bonding ionization, followed.pdf
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What does this figure depict? hydrogen bonding ionization, followed by ionic bonding nonpolar covalent bonding formation of a polar covalent bond Solution ANS: B. Ionization, followed by ionic bonding. Explanation: Ionic bonds: In this electrons are transferred from the metal atom to non-metal atom. Oppositely charged ions are formed which are held together by electro static attraction. The electrons on outside of the sodium are transferred to the chlorine molecule, so both atoms have a full outer shell and became ions..
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In 2012, the percent of American adults who owned cell phones and us.pdf
In 2012, the percent of American adults who owned cell phones and used their cell phone to send
or receive text messages was at an all-time high of 80%. Assume that 80% refers to the
population parameter. More recently in 2015, a polling firm contacts a SRS of 1200 people
chosen from the population of cell phone owners to confirm the percent who use their phone to
text. find mean, shape, standard deviation.
Solution.
In a multiple regression model, the error term e is assumed tohave.pdf
In a multiple regression model, the error term e is assumed to
have a mean of 1
have a variance of zero
have a standard deviation of 1
be normally distributeda.
have a mean of 1b.
have a variance of zeroc.
have a standard deviation of 1d.
be normally distributed
Solution
(D) normally distributed
with mean =0, variance = unknown.
I need help with this one method in java. Here are the guidelines. O.pdf
I need help with this one method in java. Here are the guidelines. Only public Employee[]
findAllBySubstring(String find).
EmployeeManager
EmployeeManager
- employees : Employee[]
- employeeMax : final int = 10
-currentEmployees : int
<> EmployeeManager
+ addEmployee( type : int, fn : String, ln : String, m : char, g : char, en : int, ft : boolean, amount
: double)
+ removeEmployee( index : int)
+ listAll()
+ listHourly()
+ listSalary()
+ listCommision()
+ resetWeek()
+ calculatePayout() : double
+ getIndex( empNum : int ) : int
+ annualRaises()
+ holidayBonuses() : double
+ increaseHours( index : int, amount : double)
+ increaseSales( index : int, amount : double)
+ findAllBySubstring(find : String) : Employee[]
- RabinKarp(name : String, find : String) : int
- stringHash(s : String) : int
- charNumericValue(c : char) : int
- RabinKarpHashes(s : String, hashes : int[], pos : int, length : int) : int
- linearSearchRecursive(nameHashes : int[], findHash : int, pos : int) : int
public EmployeeManager()
Constructor, creates the Employee array, sets currentEmployees to 0.
public void addEmployee(int, String, String, char, char, int, Boolean, double)
Takes an int representing the type of Employee to be added (1 – Hourly, 2 – Salary, 3 –
Commission) as well as the required data to create that Employee. If one of these values is not
passed output the line, “Invalid Employee Type, None Added”, and exit the method. If an
Employee with the given Employee Number already exists do not add the Employee and output
the line, “Duplicate Not Added”, and exit the method. If the array is at maximum capacity do not
add the new Employee, and output the line, \"Cannot add more Employees\".
public void removeEmployee(int)
Removes an Employee located at the given index from the Employee array.
public void listAll()
Lists all the current Employees. Outputs there are none if there are none.
public void listHourly()
Lists all the current HourlyEmployees. Outputs there are none if there are none.
public void listSalary()
Lists all the current SalaryEmployees. Outputs there are none if there are none.
public void listCommission()
Lists all the current CommissionEmployees. Outputs there are none if there are none.
public void resetWeek()
Resets the week for all Employees.
public double calculatePayout()
Returns the total weekly payout for all Employees.
public int getIndex(int)
Given an Employee Number, returns the index of that Employee in the array, if the Employee
doesn’t exist retuns -1.
public void annualRaises()
Applies annual raise to all current Employees.
public double holidayBonuses()
Outputs and returns the total holiday bonus of all Employees.
public void increaseHours(int, double)
Increase the hours worked of the Employee at the given index by the given double amount.
public void increaseSales(int, double)
Increase the sales of the Employee at the given index by the given double amount.
public Employee[] findAllBySubstring(String find)
This method will return.
help me Java projectI put problem and my own code in the linkmy .pdf
help me Java project
I put problem and my own code in the link
my code is long link : https://codeshare.io/YohNo
Solution
import javax.swing.*;
import java.awt.Font;
import javax.swing.JOptionPane;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Assignment21 {
/**
* @param args the command line arguments
*/
public static JFrame frame;
public static DecimalFormat fmt = new DecimalFormat (\"0.00;(0.00)\");
public static String message;
public static boolean firstTime = true;
public static boolean firstTime2 = true;
public static CheckingAccount CheckAcc;
public static Transaction newTrans;
public static void main(String[] args) {
// TODO code application logic here
frame = new JFrame (\"Transaction Options\");
frame.setDefaultCloseOperation (JFrame.EXIT_ON_CLOSE);
EOptionsPanel panel = new EOptionsPanel();
frame.getContentPane().add (panel);
frame.pack();
frame.setVisible(true);
}
public static int getTransCode()
{
String codeStr;
int code;
codeStr=JOptionPane.showInputDialog(\"\ Here are the Transaction
Codes:\ 1)Checking\ 2)Depositing\ 0)Terminate\");
code = Integer.parseInt(codeStr);
return code;
}
public static double getTransAmount(){
String tStr;
double tAmount;
tStr = JOptionPane.showInputDialog(\"Enter the trans amount: \");
tAmount = Double.parseDouble(tStr);
return tAmount;
}
public static double processCheck(double tAmount)
{
CheckAcc.setServiceCharge(0.15);
newTrans = new Transaction(CheckAcc.getTransCount(),3,0.15);
CheckAcc.addTrans(newTrans);
message += \"\ Transaction : Check in Amount of $\" + fmt.format(tAmount) +
\"\ Current Balance : $ \" + fmt.format(CheckAcc.getBalance()) +
\"\ Service Charge : Check --- charge: $0.15\";
if(CheckAcc.getBalance()<500 && firstTime2)
{
CheckAcc.setServiceCharge(5.00);
newTrans = new Transaction(CheckAcc.getTransCount(),3,5.00);
CheckAcc.addTrans(newTrans);
firstTime2=false;
message += \"\ Service Charge : Balance under $500 -- Charge: $5.00\";
}
if(CheckAcc.getBalance()<0)
{
CheckAcc.setServiceCharge(10.00);
newTrans = new Transaction(CheckAcc.getTransCount(),3,10.00);
CheckAcc.addTrans(newTrans);
message += \"\ Service Charge : Overdfart Balance --- Charge: $10.00\";
}
message += \"\ Total ServiceCharge : $ \" + fmt.format(CheckAcc.getServiceCharge());
JOptionPane.showMessageDialog(null, message);
return 0;
}
public static double processDeposit(double tAmount)
{
CheckAcc.setServiceCharge(0.10);
newTrans = new Transaction(CheckAcc.getTransCount(),3,0.10);
CheckAcc.addTrans(newTrans);
message += \"\ Transaction : Deposit in Amount of $\" + fmt.format(tAmount) +
\"\ Current Balance : $ \" + fmt.format(CheckAcc.getBalance()) +
\"\ Service Charge : Check --- charge: $0.10\"+
\"\ Total ServiceCharge : $ \" + fmt.format(CheckAcc.getServiceCharge());
JOptionPane.showMessageDialog(null, message);
return 0;
}
public static void inputTransactions()
{
// define local variables
String initialStr;
double initial, tAmount, check, deposit, amount=0;
int code, number=0, id=0;
// get in.
Heading ibe the following picture. Main computer UART DTE Serial chan.pdf
Heading ibe the following picture. Main computer UART DTE Serial channel Interface Modem
DCE UART
Solution
Straight cable connectivity
The diagam shows the straight cable (TX-Tx and Rx-Rx) connection between two dissimilar
devices, here DTE and DVD.
DTE : Data Terminal Unit
DVR : Data Communication unit.
Explain how TWO (2) different structural features can be used to dis.pdf
Explain how TWO (2) different structural features can be used to discriminate between the three
types of muscles. (2 marks)
Solution
Three types of muscles found in the human body are:
1)Skeletal muscle
2)Smooth
3)Cardiac
FUNCTION OF MUSCLES:
1)Regulate organs
2)Help us to move and maintain posture
FUNCTIONS OF SKELETAL MUSCLE TISSUE
FUNCTIONS OF CARDIAC MUSCLE
Pumping of blood through the heart: Alternate contraction and relaxation of cardiac muscle
pumps
FUNCTIONS OF SMOOTH MUSCLE TISSUE
Smooth muscle also contributes to moving fluids through the body and to the elimination of
indigestible matter from the gastrointestinal system.
Explain two (2) alternative ways in which plants can obtain nutrient.pdf
Explain two (2) alternative ways in which plants can obtain nutrients when they live in nutrient
poor soils.
Solution
When plants live in nutrient poor soils they develop specific adaptations to obtain nutrients. For
example, carnivorous plants such as Nepenthes, Dionea, Urticularia, that are growing in nitrogen
poor soils developed modified leaves into pitchers, fly traps and urticles that can trap insects and
enzymes produced in these structures digest them. Some plants like legumes develop symbiotic
association with nitrogen fixing bacteria like Rhizobium and form root nodules. The plants
provide shelter to bacteria and bacteria fix the atmosphric nitrogen and supply to the plants.
Mangrove plants like Avicennia growing in oxygen poor marshy soils develop aerial roots that
obtain oxygen. Some plants establish mutual relationship with Mycorrhiza (fungus) to obtain
nutrients like nitrogen and phosphorous. Therefore, plants develop adaptations for obtaining
nutrients and survival in nutrient poor soil environments..
Couldnt find the right subject that it belongs to...There ar.pdf
Couldn\'t find the right \"subject\" that it belongs to...
There are several big picture meta-theoretical themes used in this class—as presented in the
syllabus and textbook, and as explained by the instructor. Find the theme that is NOT part of the
assumptions from the meta-theory used in this class.
Question 4 options:
Development is promoted by person-centered contexts and interactions.
People carry the past forward into a dynamic present.
Genetics are at the root of behavioral development.
Development is influenced by an active changing context and an active developing person.
All people face fundamental developmental tasks, although there are many healthy pathways
through them.
Solution
reference
Yardley, L., Morrison, L., Bradbury, K., & Muller, I. (2015). The Person-Based Approach to
Intervention Development: Application to Digital Health-Related Behavior Change
Interventions. Journal of Medical Internet Research, 17(1), e30. http://doi.org/10.2196/jmir.4055
topic : The Person-Based Approach to Intervention Development...............Application to Digital
Health-Related Behavior Change Intervention.
The basic point of the individual based approach is to ground the improvement of conduct
change intercessions in a significant comprehension of the viewpoint and psychosocial setting of
the general population who will utilize them, increased through iterative top to bottom subjective
research. There is broad accord in the eHealth inquire about group that evoking and tending to
the necessities and point of view of the expected intercession client is a crucial piece of good
mediation improvement to guarantee (at any rate) that intercessions are usable and locks in. This
is a basic issue for eHealth in the event that it is to satisfy its potential and defeat the issues of
low take-up and adherence . It is troublesome notwithstanding for master mediation designers to
completely foresee the needs and needs of clients , thus intercession engineers as of now
routinely evoke the perspectives of target clients in an assortment of ways , yet there is
shockingly little level headed discussion and itemized direction concerning how best to do this .
The individual based approach gives a procedure that empowers designers to increase imperative
bits of knowledge into how diverse individuals experience and actualize mediations, and a
system to help engineers distinguish the key attributes that will make an intercession more
important, alluring, and valuable to the individuals who draw in with it.
The individual based approach was produced by our exploration group through down to earth
involvement of making and assessing various fruitful wellbeing related intercessions, including
general wellbeing mediations (eg, to oversee weight and stress, advance physical action and hand
cleanliness) and disease administration intercessions (to help clients adapt to dazedness, back
torment, weakness, respiratory conditions, hypertension, diabetes, growth, stroke, and nu.
Recommended
In 2012, the percent of American adults who owned cell phones and us.pdf
In 2012, the percent of American adults who owned cell phones and used their cell phone to send
or receive text messages was at an all-time high of 80%. Assume that 80% refers to the
population parameter. More recently in 2015, a polling firm contacts a SRS of 1200 people
chosen from the population of cell phone owners to confirm the percent who use their phone to
text. find mean, shape, standard deviation.
Solution.
In a multiple regression model, the error term e is assumed tohave.pdf
In a multiple regression model, the error term e is assumed to
have a mean of 1
have a variance of zero
have a standard deviation of 1
be normally distributeda.
have a mean of 1b.
have a variance of zeroc.
have a standard deviation of 1d.
be normally distributed
Solution
(D) normally distributed
with mean =0, variance = unknown.
I need help with this one method in java. Here are the guidelines. O.pdf
I need help with this one method in java. Here are the guidelines. Only public Employee[]
findAllBySubstring(String find).
EmployeeManager
EmployeeManager
- employees : Employee[]
- employeeMax : final int = 10
-currentEmployees : int
<> EmployeeManager
+ addEmployee( type : int, fn : String, ln : String, m : char, g : char, en : int, ft : boolean, amount
: double)
+ removeEmployee( index : int)
+ listAll()
+ listHourly()
+ listSalary()
+ listCommision()
+ resetWeek()
+ calculatePayout() : double
+ getIndex( empNum : int ) : int
+ annualRaises()
+ holidayBonuses() : double
+ increaseHours( index : int, amount : double)
+ increaseSales( index : int, amount : double)
+ findAllBySubstring(find : String) : Employee[]
- RabinKarp(name : String, find : String) : int
- stringHash(s : String) : int
- charNumericValue(c : char) : int
- RabinKarpHashes(s : String, hashes : int[], pos : int, length : int) : int
- linearSearchRecursive(nameHashes : int[], findHash : int, pos : int) : int
public EmployeeManager()
Constructor, creates the Employee array, sets currentEmployees to 0.
public void addEmployee(int, String, String, char, char, int, Boolean, double)
Takes an int representing the type of Employee to be added (1 – Hourly, 2 – Salary, 3 –
Commission) as well as the required data to create that Employee. If one of these values is not
passed output the line, “Invalid Employee Type, None Added”, and exit the method. If an
Employee with the given Employee Number already exists do not add the Employee and output
the line, “Duplicate Not Added”, and exit the method. If the array is at maximum capacity do not
add the new Employee, and output the line, \"Cannot add more Employees\".
public void removeEmployee(int)
Removes an Employee located at the given index from the Employee array.
public void listAll()
Lists all the current Employees. Outputs there are none if there are none.
public void listHourly()
Lists all the current HourlyEmployees. Outputs there are none if there are none.
public void listSalary()
Lists all the current SalaryEmployees. Outputs there are none if there are none.
public void listCommission()
Lists all the current CommissionEmployees. Outputs there are none if there are none.
public void resetWeek()
Resets the week for all Employees.
public double calculatePayout()
Returns the total weekly payout for all Employees.
public int getIndex(int)
Given an Employee Number, returns the index of that Employee in the array, if the Employee
doesn’t exist retuns -1.
public void annualRaises()
Applies annual raise to all current Employees.
public double holidayBonuses()
Outputs and returns the total holiday bonus of all Employees.
public void increaseHours(int, double)
Increase the hours worked of the Employee at the given index by the given double amount.
public void increaseSales(int, double)
Increase the sales of the Employee at the given index by the given double amount.
public Employee[] findAllBySubstring(String find)
This method will return.
help me Java projectI put problem and my own code in the linkmy .pdf
help me Java project
I put problem and my own code in the link
my code is long link : https://codeshare.io/YohNo
Solution
import javax.swing.*;
import java.awt.Font;
import javax.swing.JOptionPane;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Assignment21 {
/**
* @param args the command line arguments
*/
public static JFrame frame;
public static DecimalFormat fmt = new DecimalFormat (\"0.00;(0.00)\");
public static String message;
public static boolean firstTime = true;
public static boolean firstTime2 = true;
public static CheckingAccount CheckAcc;
public static Transaction newTrans;
public static void main(String[] args) {
// TODO code application logic here
frame = new JFrame (\"Transaction Options\");
frame.setDefaultCloseOperation (JFrame.EXIT_ON_CLOSE);
EOptionsPanel panel = new EOptionsPanel();
frame.getContentPane().add (panel);
frame.pack();
frame.setVisible(true);
}
public static int getTransCode()
{
String codeStr;
int code;
codeStr=JOptionPane.showInputDialog(\"\ Here are the Transaction
Codes:\ 1)Checking\ 2)Depositing\ 0)Terminate\");
code = Integer.parseInt(codeStr);
return code;
}
public static double getTransAmount(){
String tStr;
double tAmount;
tStr = JOptionPane.showInputDialog(\"Enter the trans amount: \");
tAmount = Double.parseDouble(tStr);
return tAmount;
}
public static double processCheck(double tAmount)
{
CheckAcc.setServiceCharge(0.15);
newTrans = new Transaction(CheckAcc.getTransCount(),3,0.15);
CheckAcc.addTrans(newTrans);
message += \"\ Transaction : Check in Amount of $\" + fmt.format(tAmount) +
\"\ Current Balance : $ \" + fmt.format(CheckAcc.getBalance()) +
\"\ Service Charge : Check --- charge: $0.15\";
if(CheckAcc.getBalance()<500 && firstTime2)
{
CheckAcc.setServiceCharge(5.00);
newTrans = new Transaction(CheckAcc.getTransCount(),3,5.00);
CheckAcc.addTrans(newTrans);
firstTime2=false;
message += \"\ Service Charge : Balance under $500 -- Charge: $5.00\";
}
if(CheckAcc.getBalance()<0)
{
CheckAcc.setServiceCharge(10.00);
newTrans = new Transaction(CheckAcc.getTransCount(),3,10.00);
CheckAcc.addTrans(newTrans);
message += \"\ Service Charge : Overdfart Balance --- Charge: $10.00\";
}
message += \"\ Total ServiceCharge : $ \" + fmt.format(CheckAcc.getServiceCharge());
JOptionPane.showMessageDialog(null, message);
return 0;
}
public static double processDeposit(double tAmount)
{
CheckAcc.setServiceCharge(0.10);
newTrans = new Transaction(CheckAcc.getTransCount(),3,0.10);
CheckAcc.addTrans(newTrans);
message += \"\ Transaction : Deposit in Amount of $\" + fmt.format(tAmount) +
\"\ Current Balance : $ \" + fmt.format(CheckAcc.getBalance()) +
\"\ Service Charge : Check --- charge: $0.10\"+
\"\ Total ServiceCharge : $ \" + fmt.format(CheckAcc.getServiceCharge());
JOptionPane.showMessageDialog(null, message);
return 0;
}
public static void inputTransactions()
{
// define local variables
String initialStr;
double initial, tAmount, check, deposit, amount=0;
int code, number=0, id=0;
// get in.
Heading ibe the following picture. Main computer UART DTE Serial chan.pdf
Heading ibe the following picture. Main computer UART DTE Serial channel Interface Modem
DCE UART
Solution
Straight cable connectivity
The diagam shows the straight cable (TX-Tx and Rx-Rx) connection between two dissimilar
devices, here DTE and DVD.
DTE : Data Terminal Unit
DVR : Data Communication unit.
Explain how TWO (2) different structural features can be used to dis.pdf
Explain how TWO (2) different structural features can be used to discriminate between the three
types of muscles. (2 marks)
Solution
Three types of muscles found in the human body are:
1)Skeletal muscle
2)Smooth
3)Cardiac
FUNCTION OF MUSCLES:
1)Regulate organs
2)Help us to move and maintain posture
FUNCTIONS OF SKELETAL MUSCLE TISSUE
FUNCTIONS OF CARDIAC MUSCLE
Pumping of blood through the heart: Alternate contraction and relaxation of cardiac muscle
pumps
FUNCTIONS OF SMOOTH MUSCLE TISSUE
Smooth muscle also contributes to moving fluids through the body and to the elimination of
indigestible matter from the gastrointestinal system.
Explain two (2) alternative ways in which plants can obtain nutrient.pdf
Explain two (2) alternative ways in which plants can obtain nutrients when they live in nutrient
poor soils.
Solution
When plants live in nutrient poor soils they develop specific adaptations to obtain nutrients. For
example, carnivorous plants such as Nepenthes, Dionea, Urticularia, that are growing in nitrogen
poor soils developed modified leaves into pitchers, fly traps and urticles that can trap insects and
enzymes produced in these structures digest them. Some plants like legumes develop symbiotic
association with nitrogen fixing bacteria like Rhizobium and form root nodules. The plants
provide shelter to bacteria and bacteria fix the atmosphric nitrogen and supply to the plants.
Mangrove plants like Avicennia growing in oxygen poor marshy soils develop aerial roots that
obtain oxygen. Some plants establish mutual relationship with Mycorrhiza (fungus) to obtain
nutrients like nitrogen and phosphorous. Therefore, plants develop adaptations for obtaining
nutrients and survival in nutrient poor soil environments..
Couldnt find the right subject that it belongs to...There ar.pdf
Couldn\'t find the right \"subject\" that it belongs to...
There are several big picture meta-theoretical themes used in this class—as presented in the
syllabus and textbook, and as explained by the instructor. Find the theme that is NOT part of the
assumptions from the meta-theory used in this class.
Question 4 options:
Development is promoted by person-centered contexts and interactions.
People carry the past forward into a dynamic present.
Genetics are at the root of behavioral development.
Development is influenced by an active changing context and an active developing person.
All people face fundamental developmental tasks, although there are many healthy pathways
through them.
Solution
reference
Yardley, L., Morrison, L., Bradbury, K., & Muller, I. (2015). The Person-Based Approach to
Intervention Development: Application to Digital Health-Related Behavior Change
Interventions. Journal of Medical Internet Research, 17(1), e30. http://doi.org/10.2196/jmir.4055
topic : The Person-Based Approach to Intervention Development...............Application to Digital
Health-Related Behavior Change Intervention.
The basic point of the individual based approach is to ground the improvement of conduct
change intercessions in a significant comprehension of the viewpoint and psychosocial setting of
the general population who will utilize them, increased through iterative top to bottom subjective
research. There is broad accord in the eHealth inquire about group that evoking and tending to
the necessities and point of view of the expected intercession client is a crucial piece of good
mediation improvement to guarantee (at any rate) that intercessions are usable and locks in. This
is a basic issue for eHealth in the event that it is to satisfy its potential and defeat the issues of
low take-up and adherence . It is troublesome notwithstanding for master mediation designers to
completely foresee the needs and needs of clients , thus intercession engineers as of now
routinely evoke the perspectives of target clients in an assortment of ways , yet there is
shockingly little level headed discussion and itemized direction concerning how best to do this .
The individual based approach gives a procedure that empowers designers to increase imperative
bits of knowledge into how diverse individuals experience and actualize mediations, and a
system to help engineers distinguish the key attributes that will make an intercession more
important, alluring, and valuable to the individuals who draw in with it.
The individual based approach was produced by our exploration group through down to earth
involvement of making and assessing various fruitful wellbeing related intercessions, including
general wellbeing mediations (eg, to oversee weight and stress, advance physical action and hand
cleanliness) and disease administration intercessions (to help clients adapt to dazedness, back
torment, weakness, respiratory conditions, hypertension, diabetes, growth, stroke, and nu.
Consider the following C code snippet C codevoid setArray(int.pdf
Consider the following C code snippet:
// C code
void setArray(int num) {
int i;
int array[10];
for (i=0; i < 10; i=i+1) {
array[i]=compare(num, i);
}
}
int compare(int a, int b) {
if (sub(a, b) >= 0)
return 1;
else
return 0;
}
int sub(int a, int b) {
return ab;
}
Implement the C code snippet above in MIPS assembly language. Use $s0 to hold the variable i.
Be sure to handle the stack pointer appropriately. The array is stored on the stack of the setArray
function (i.e. when setArray function is called it allocates space for the array on the stack.).
Solution
I have used the Test function i.e ain function to get this module of the fuction definition working.
Thw below is the MIPS code for the complete program:-
.file 1 \"\"
.section .mdebug.abi32
.previous
.gnu_attribute 4, 1
.abicalls
.text
.align 2
.globl _Z3subii
$LFB0 = .
.set nomips16
.ent _Z3subii
.type _Z3subii, @function
_Z3subii:
.frame $fp,8,$31 # vars= 0, regs= 1/0, args= 0, gp= 0
.mask 0x40000000,-4
.fmask 0x00000000,0
.set noreorder
.set nomacro
addiu $sp,$sp,-8
$LCFI0:
sw $fp,4($sp)
$LCFI1:
move $fp,$sp
movz $31,$31,$0
$LCFI2:
sw $4,8($fp)
sw $5,12($fp)
lw $3,8($fp)
lw $2,12($fp)
nop
subu $2,$3,$2
move $sp,$fp
lw $fp,4($sp)
addiu $sp,$sp,8
j $31
nop
.set macro
.set reorder
.end _Z3subii
$LFE0:
.size _Z3subii, .-_Z3subii
.align 2
.globl _Z7compareii
$LFB1 = .
.set nomips16
.ent _Z7compareii
.type _Z7compareii, @function
_Z7compareii:
.frame $fp,32,$31 # vars= 0, regs= 2/0, args= 16, gp= 8
.mask 0xc0000000,-4
.fmask 0x00000000,0
.set noreorder
.cpload $25
.set nomacro
addiu $sp,$sp,-32
$LCFI3:
sw $31,28($sp)
$LCFI4:
sw $fp,24($sp)
movz $31,$31,$0
$LCFI5:
move $fp,$sp
$LCFI6:
.cprestore 16
sw $4,32($fp)
sw $5,36($fp)
lw $4,32($fp)
lw $5,36($fp)
lw $2,%got(_Z3subii)($28)
nop
move $25,$2
jalr $25
nop
lw $28,16($fp)
nor $2,$0,$2
srl $2,$2,31
andi $2,$2,0x00ff
beq $2,$0,$L4
nop
li $2,1 # 0x1
b $L5
nop
$L4:
move $2,$0
$L5:
move $sp,$fp
lw $31,28($sp)
lw $fp,24($sp)
addiu $sp,$sp,32
j $31
nop
.set macro
.set reorder
.end _Z7compareii
$LFE1:
.size _Z7compareii, .-_Z7compareii
.align 2
.globl _Z8setArrayi
$LFB2 = .
.set nomips16
.ent _Z8setArrayi
.type _Z8setArrayi, @function
_Z8setArrayi:
.frame $fp,88,$31 # vars= 48, regs= 3/0, args= 16, gp= 8
.mask 0xc0010000,-4
.fmask 0x00000000,0
.set noreorder
.cpload $25
.set nomacro
addiu $sp,$sp,-88
$LCFI7:
sw $31,84($sp)
$LCFI8:
sw $fp,80($sp)
movz $31,$31,$0
$LCFI9:
sw $16,76($sp)
$LCFI10:
move $fp,$sp
$LCFI11:
.cprestore 16
sw $4,88($fp)
sw $0,24($fp)
b $L8
nop
$L9:
lw $16,24($fp)
lw $4,88($fp)
lw $5,24($fp)
lw $2,%got(_Z7compareii)($28)
nop
move $25,$2
jalr $25
nop
lw $28,16($fp)
move $3,$2
sll $2,$16,2
addiu $4,$fp,24
addu $2,$4,$2
sw $3,4($2)
lw $2,24($fp)
nop
addiu $2,$2,1
sw $2,24($fp)
$L8:
lw $2,24($fp)
nop
slt $2,$2,10
andi $2,$2,0x00ff
bne $2,$0,$L9
nop
move $sp,$fp
lw $31,84($sp)
lw $fp,80($sp)
lw $16,76($sp)
addiu $sp,$sp,88
j $31
nop
.set macro
.set reorder
.end _Z8setArrayi
$LFE2:
.size _Z8setArrayi, .-_Z8setArrayi
.align 2
.globl ma.
Describe a mechanism that the body uses to prevent a mutation from be.pdf
Describe a mechanism that the body uses to prevent a mutation from becoming an abnormal
cell.
Solution
The body has several mechanisms in place to ensure that mutations do not turn lethal. The first of
those steps is the proof reading activity of the DNA polymerase which reads the DNA strand and
corrects the sequence as it synthesizes the new DNA fragment. Thus, the daughter cells contain
the original non-mutated sequence. Secondly, organisms are mostly diploid i.e. they contain two
sets of chromosomes which includes two forms of the same gene. The mutation would be lethal
or cause loss of function only when the other copy of the gene is also mutated (recessive
mutation). Also, mutant cells are usually cleared by the immune system which can detect and
produce specific antibodies against these cells..
classify domian of life differencesSolutionClassify the domain.pdf
classify domian of life differences
Solution
Classify the domain of life differences: the domain of life is mainly classified based on the
following differences
1. Ribosomal RNAs (rRNA) inside the cell
2. The structure of cell membrane with lipid structure
3. Cell sensitivity to antibiotics
Three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae,
kingdom Animalia)\"
Explanation:
The mitochondria, histone (nucleosomes) are not evolved in archae domain & prokaryote
domain but evolved in the “kingdom of eukaryotes”, as an evolutionary adaptation difference.
This is a major differential aspect among the three domains of life in their classification
according to endo-symbiotic theory.
The aspects of cell wall are constructed differently in all three domains of life such as bacteria,
archaic species, and eukaryotes. The archae has S -layer in their cell wall but do not possess
peptidoglygan layer as bacteria do. Lipopolysaccharide, peptidoglycan layers are absent in
eukaryotic cell walls and their mechanism of synthesis is absent in eukaryotes. Plant cells are
eukaryotic cells and contain cell wall but animal cells do not contain cell wall. Bacteria are
prokaryotes & do not possess histone proteins but eukaryotes contain histone proteins associated
with DNA. Any morphological characteristics of group of species are completely different to that
of three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae,
kingdom Animalia)\". These morphological features are completely different to the ancestral
species and leading to new phylum evolvement. Internal structural features, embryological
development, nucleus structure, genome nature (RNA or DNA), type of cellular respiration &
energy synthesis.
The major domian of life differences with morphological relations in the evolution of vertebrate
animal life (at the phylum level & domain level) from invertebrates (from four -limb species):
The major features of embryonic vertebrate zygote formation & followed by eight-cell stage of
embryo finally blastula stage and gastrulation. This stage is absent in the archae & prokaryotes
species. There are total 2 types of groups of species that are divided based on presence of
morphological innovation that is \"presence of vertebral column\" in animal domain &
prokaryote. Those living species which do not have vertebra\" are referred as \"invertebrates\"
and those which possess vertebrae\" are referred as “chordates\" or \"vertebrates\"..
Compare and contrast the development of a WBS in traditional project.pdf
Compare and contrast the development of a WBS in traditional project management versus
development in an agile environment.
Solution
An Agile based WBS is organized around units of end user functionality - commonly referred to
as user stories. Each user story represents the work to deliver that increment of functionality. In
Agile based WBS, Features are decomposed into Epics (large user stories) and Epics into User
Stories. User stories are decomposed to represent functionality that can be completed within one
iteration - typically one to four weeks depending on the team/project. Each leaf level WBS
element represent the working software to be produced through analysis, design, coding and
testing of the requirement in the user story.
Because each user story is atomic, stories can be re-prioritized, added and or removed without
changing the scope, provided that the total size/effort of stories within the scope of the project
does not change. Additionally, iterative development facilitates multiple releases (deployments)
within a project. So, user stories are typically scheduled within an iteration and a release. Note,
that until an iteration has started, the stories within that iteration can be changed, as part of
iteration planning.
In case of traditional project management WBS is organized around the primary products of the
project (or planned outcomes) instead of the work needed to produce the products (planned
actions). Since the planned outcomes are the desired ends of the project, they form a relatively
stable set of categories in which the costs of the planned actions needed to achieve them can be
collected.
A well-designed WBS makes it easy to assign each project activity to one and only one terminal
element of the WBS. In addition to its function in cost accounting, the WBS also helps map
requirements from one level of system specification to another, for example a requirements cross
reference matrix mapping functional requirements to high level or low level design documents..
B. You are an evolutionary biologist in a heated argument with a cre.pdf
B. You are an evolutionary biologist in a heated argument with a creation scientist who makes
the statement—\"Mathematically, it is inconceivable that anything as complex as a protein, let
alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for
chance. 2. Now play the other side and respond to the supporting statement as you think a
creation scientist would respond.?
Solution
You are an evolutionary biologist in a heated argument with a creation scientist who makes the
statement—
\"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell
or a human, could spring up by chance\".
1. Make a supporting statement for chance.
Chance: It plays a part in evolution (for example, random mutations in a gene pool can give rise
to few new traits), but evolution does not depend ONLY on chance in creating new organisms,
proteins or any other entities. This is quite opposite to natural selection. Natural selection is the
principal which is known for the mechanism of evolution. By chance some mutation caused in
the gene pool and it gave to new acceptable traits. If they found adjustable to the environment,
they continues as a new species. The \"undesirable\" (nonadaptive) ones will be thus elimated
from that gene pool. As long as the forces of selection stay constant, natural selection pushes the
evolution in only one direction and produce new desired structures. Thus chance plays a part in
evolution. Nature plays a major role in this. Fossil studies support this theory/ process. Where we
can see many intermediate or failed species in between major species.
Archaeopteryx, a link between reptiles and birds, extinctinted soon is a good example of
intermediate species.
Haldane and Oparin proposed a theory of the origin of life on Earth through the transformation,
during the gradual chemical evolution of particles that contain carbon in the primordial soup
shows an example of Evolution.
2. Now play the other side and respond to the supporting statement as you think a creation
scientist would respond?
As a creation scientist i do not agree with science and evolution part, all this is created by God
and thus it follows biblical roots. All that it is in the nature is creation of God, all of a sudden it
appeared on the earth. According to us GOD created all species like birds,animals etc., later he
wanted to create a genius brained animal, so he created humans. And God creation according to
us all this is due to CHOICE not because of CHANCE.Thus it involved logical and systematic
process..
Assume Hashtable is a simple array of size 8, with indices 0..7. Num.pdf
Assume Hashtable is a simple array of size 8, with indices 0..7. Numeric keys are mapped by a
Hashfunction that gives the mod(8,n) value for any key “n” yielding the Hashtable index for that
key (0..7). A Hashtable entry is null unless a key exists with that index as its hashed index; if so,
the Hashtable entry points to the first node of a linked list of keys with that hash index. The last
node on this linked list has a null reference for the next referenced node. Assume the occurrence
of a linked list node is represented by the object “Node” and its “Data” and “NextRef” attributes.
Create pseudocode for the \"Remove\" operation.
Solution
Removal of a Node from a Singly Linked List
To complete deletion of firstNode in the list we have to change Head pointing to NextRef of
firstNode.
Pseudocode:
firstNode = Head
Head = firstNode->NextRef
free firstNode
Complexity:
Time Complexity: O(1)
Space Complexity: O(1)
Sourcecode:
int delNodeData(int num)
{
struct Node *prev_ptr, *node;
node=Head;
while(cur_ptr != NULL)
{
if(cur_ptr->Data == num)
{
if(cur_ptr==Head)
{
Head=cur_ptr->NextRef;
free(node);
return 0;
}
else
{
prev_ptr->NextRef=node->NextRef;
free(node);
return 0;
}
}
else
{
prev_ptr=node;
node=node->NextRef;
}
}
printf(\"\ Element %d is not found in the List\", num);
return 1;
}
Deleting Last Node in the Singly Linked List
Traverse to Last Node in the List using two pointers namely prevNode and node. Once node
reaches the last Node in the list point NextRef in prevNode to NULL and free the node.
Pseudocode:
node = head
forever:
if node->NextRef == NULL
break
prevNode = node
node = node->NextRef
prevNode->NextRef = NULL
free Node
Complexity:
Time Complexity: O(n)
Space Complexity: O(1)
Deleting Node from position \'p\' in the List
To delete a Node at the position \'p\' we have to first traverse the list until we reach the position
\'p\'.
For this case have to maintain two pointers namely prevNode and Node.
Since Singly Linked Lists are uni-directional we have to maintain the information about
previous Node in prevNode. Once we reach the position \'p\' we have to modify prevNode
NextRef pointing to Node NextRef and free Node.
Pseudocode:
curNode = head
curPos = 1
forever:
if curPos == P || Node == NULL
break
prevNode = node
node = node->NextRef
curPos++
if noode != NULL:
prevNode->NextRef = node->NextRef
free curNode
Complexity:
Time Complexity: O(n) worst case
Space Complexity: O(3)
Sourcecode:
int delNodeLoc(int loc)
{
struct Node *prev_ptr, *node;
int i;
node=Head;
if(loc > (length()) || loc <= 0)
{
printf(\"\ Deletion of Node at given location is not possible\ \");
}
else
{
// If the location is starting of the list
if (loc == 1)
{
Head=node->NextRef;
free(node);
return 0;
}
else
{
for(i=1;iNextRef;
}
prev_ptr->NextRef=node->NextRef;
free(node);
}
}
return 1;
}.
Amon the following, which has the lowest levels of dissolved iron.pdf
Amon the following, which has the lowest levels of dissolved iron?
a. Atlantic Ocean
b. Indian Ocean
c. Pacific Ocean
d. The temperate regions of all oceans
Solution
The correct option is Pacific ocean because of the low dust sedimentation and iron is readily
soluble in low oxygen regions and the continental margins. It was believed that iron remained in
the localized areas but didn\'t contribute to the overall iron content of the ocean. As it is
explained by Nielsen, as soon as the iron-rich fluids hit the seawater with high oxygen content,
the iron gets dumped out and is not contributed to the overall amount of the dissolved iron..
A box contains 10 red balls and 40 black balls. Two balls are drawn .pdf
A box contains 10 red balls and 40 black balls. Two balls are drawn successively without
replacement from the box. Is the probability of selecting a red ball in the first draw the same as
the probability of selecting a red ball in the second draw? Justify your answers by computing the
probabilities.
Solution.
As late as 1992, the United States was running budget deficits of ne.pdf
As late as 1992, the United States was running budget deficits of nearly $300 billion. During the
remainder of the 1990\'s, deficits declined and became surpluses. As the new century began,
these surpluses again turned into deficits. •Explain the decline in deficits and subsequent
surpluses in the late 1990\'s. •Explain the return to deficit spending since the turn of the century.
•Consider the causes of the deficits and surpluses and provide your own insight as to whether
these surpluses or deficits have a \"positive\" or \"negative\" effect on our economy
Solution
As the 1990s came to an end, virtually all financial-market observers and politicians rejoiced that
budget deficits were gone and that the U.S. had entered a period of fiscal stability. Politicians on
both sides of the aisle viewed the new world of budget surpluses as an opportunity to save Social
Security, to cut Federal debt, and to lower taxes.
Budget deficits or budget surpluses are neither good nor bad, they are fiscal policy options that
can stabilize or de-stabilize an economy. The U.S. economy, under Reagan, Bush, and early
Clinton, experienced higher not lower rates of economic growth at a time when budget deficits
were at record levels and tax rates were on the decline. Over the past 20 years a period of record
budget deficits the stock market increased from 950 to over 11,000, providing ample evidence
that budget deficits are not necessarily bad for an economy.On the other hand, budget surpluses
are not necessarily good. As the budget surpluses began to expand dramatically in late 1999 and
into 2000, the stock market appeared to be reversing the record up-trend of the 1980s and \'90s.
One possible reason for the reversal: the growing budget surplus was sapping the savings of the
private sector. Consumers were able to maintain their spending only through expanding debt to
record level.
In the late 90s, the confluence of strong economic growth, decelerating federal government
spending, and substantial increases in tax rates (revenues) had the unexpected effect of producing
enormous budget surpluses. As unexpected budget surpluses burst on the scene in the late 1990s,
most mainstream economists weren’t sensitive to the contractionary impact that surpluses can
have on the economy. The reason was simple that none of their economic models ever included
the forecast of a substantial surplus.
Ronald Reagan’s was the vice president of the USA for eight years and knew firsthand how
difficult it was to bring the federal budget deficit down from 6 percent of GDP in 1983 to 2.8
percent in Reagan’s last budget. Bush knew that the many budget deals of the 1980s had been
hard fought and always involved higher taxes as part of the deficit reduction, Reagan signed into
law 11 major tax increases that raised taxes by $133 billion in 1988 or 2.7 percent of GDP.Bush
administration economists believed that an easier monetary policy was the key to stimulating
growth. But Federal Reserve Board.
A bacteriophage population is introduced to a bacterial colony that .pdf
A bacteriophage population is introduced to a bacterial colony that is in stationary phase with
rapidly depleting nutrients. Appraise the following scenarios and select which one would favor
long-term survival of the bacteriophage.
A. The bacteriophages infect the remaining viable cells and integrate their genetic material into
the host cell genome thereby preserving their genome.
B. The bacteriophages do not infect the bacterial cells since they are dying; they remain inert
outside of the cells.
C. The bacteriophages infect the bacterial cells and go through all stages of development apart
from release, thus keeping the progeny phage from entering an inhospitable environment.
Solution
Answer :
C. The bacteriophages infect the bacterial cells and go through all stages of development apart
from release, thus keeping the progeny phage from entering an inhospitable environment..
Xavier and Yolanda both have classes that end at noon and they agree .pdf
Xavier and Yolanda both have classes that end at noon and they agree to meet every day after
class. They arrive at the coffee shop independently. Xavier\'s arrival time is X and Yolanda\'s
arrival time is Y, where X and Y are measured in minutes after noon. The individual density
functions are given. (Xavier arrives sometime after noon and is more likely to arrive promptly
than late. Yolanda always arrives by 12:10 PM and is more likely to arrive late than promptly.)
After Yolanda arrives, she\'ll wait for up to 20 minutes for Xavier, but he won\'t wait for her.
Find the probability that they meet. (Round your answer to three decimal places.) f1(x) = {e^-x if
x gt o and o if x
Solution.
1. Low platelet count is a recessively inherited trait. Reevaluation.pdf
1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates
that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her
husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia
(15pts).
a. What are the parental gametes? (2pts)
b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible
genotypes. (2 pts)
c. What is the gamete frequency if the two genotypes in (b) were crossed with each other?
(4pts)? (4pts)
d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts)
Solution
Both the conditions are recessive.
Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for
beta thalessemia. Genotype is LLbb
Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta-
thalessemia. Genotype is llBb
The possible gametes produced by these parents:
Nikoleta: Lb
Husband: lB, lb
Possible genotype of the F1 offspring: LlBb or Llbb
Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype
should be Llbb.
Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb).
Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2
The following is the phenotype frequency:
LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3
LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3
llBb (Low platelet count and no beta thallasemia) = 1
llbb (Low platelet count and beta thallasemia) = 1
Thus phenotypic ratio stands as:
Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low
platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1.
1. An important contribution of Fiedlers research on the contingen.pdf
1. An important contribution of Fiedler\'s research on the contingency model of leadership is that
it: a. uses weights which are determined in an arbitrary manner in order to determine situation
favorability. b. goes beyond the notion of leadership styles to try to show how styles fit a
situation. c. recommends medium least preferred coworker leaders as they are more effective in
many situations. d. is clear about how the model will work over time. 2. Which of the following
characteristics was explored by the studies evaluating the characteristics of the leader-member
exchange relationship? a. Income b. Educational qualification c. Value agreement d. Recruitment
process 3. Which of the following is a characteristic of entrepreneurial leaders? a. They are
concerned with innovation and creativity. b. They are interested in maintaining the status quo. c.
They avoid taking risks and are unwilling to stretch themselves. d. They tend to be dependent on
their followers. 4. An important aspect of leadership is: a. dictating changes required in an
organization. b. maintaining a passive relationship with followers. c. maintaining the status quo
and not creating change. d. influencing others to come together around a common vision. 5. It
becomes easier for people to see their own opportunities for leadership when they: a. start seeing
leaders as individuals who are above others. b. understand that the concept of leadership cannot
continue to develop. c. stop equating leadership with greatness and public visibility. d.
comprehend that leadership involves maintaining the status quo. 6. Erica, a team manager, was
known for heading projects with diligence and conscientiousness. However, she was insensitive
and excessively ambitious. She did not take feedback in a constructive manner. As a result, this
prevented her from moving forward in terms of her career since she had problems
communicating with her team members and motivating them. In the context of the given
scenario, which of the following phenomenon does Erica experience? a. Communication
apprehension b. Fear-based motivation c. Paradigm change d. Derailment Name: Class: Date:
Quiz 1: Chp 1-3 (Fall 2015) 2 7. Which of the following statements is true of the contingency
approach to leadership? a. A leadership style that works in one situation will work in all
situations. b. These approaches seek to investigate traits that can improve performance in all
situations. c. The needs, maturity, and cohesiveness of followers have been identified as a key
contingency. d. These approaches hold that behavior that is effective in some circumstances is
effective under all conditions. 8. Stage two in the development of the individualized leadership
theory explored the _____ in more detail. a. vertical dyad linkage model b. partnership building
c. leader–member exchange d. entrepreneurship model 9. Which of the following statements is
true of leaders who work collaboratively? a. They learn to keep the lines of commu.
^^^ Discuss about Header Node And also write a program for unorder.pdf
^^^ Discuss about Header Node? And also write a program for unordered single linked list and
linked implementation of QUEUE?
Solution
Header node:
Sometimes it is desire to keep an extra node at the front of a List. Such a node doesn’t represent
an item in the List and is called a Header Node or a List Header. The Information field of such
Header node might be unused but the Next field maintains the first node address. More often the
information of such a node could be used to keep Global Information about the Entire List.
One of the applications of Header node is the information portion of the Header node contains
the number of nodes (not including Header) in the List. In this Structure the Header must be
adjusted the total number of nodes when we apply addition or deletion operation on the List. We
can directly obtained the total number of nodes without traversing the entire List.
By using this application, we can develop Data Structure Stack easily. Since Header node
information field maintains the number of elements that the Stack contains.
Another application is we can simply develop the Data Structure Queue. Until now, two external
pointers, Front & Rear, were necessary for a List to represent a Queue. However, now only a
single external pointer is sufficient to maintain a Queue that is Head node. Because the Header
node next field behave like a Front and information field of Header node maintains last node
address so that it is behave like a Rear .
Another possibility for the use of the information portion of a List Header is as a pointer
to a Curr node in the List during a traversal process. This would be eliminate the node for an
external pointer during traversal.
#include
#include
#include
#include
struct node
{
char info;
struct node *next;
};
typedef struct node sl;
void display(sl *);
sl * create(sl *root)
{
sl *curr,*new;
char val;
printf(\"\ Enter $ to STOP --- Otherwise Continue : \");
fflush(stdin);
scanf(\"%c\",&val);
while(val != \'$\')
{
new=(sl *) malloc(sizeof(sl));
new->info = val;
new->next = NULL;
if(root == NULL)
root = new;
else
{
curr = root;
while( curr->next != NULL )
curr = curr->next;
curr->next = new;
}
printf(\" Enter $ to STOP --- Otherwise Continue : \");
fflush(stdin);
scanf(\"%c\",&val);
}
printf(\"\ The Created Linked List Is \ \");
display(root);
return root;
}
sl * delete(sl *root,char val)
{
sl *temp,*curr,*rear;
curr = root;
rear = NULL;
while( curr != NULL && val != curr->info )
{
rear = curr;
curr = curr->next;
}
if(curr == NULL)
{
if( rear == NULL)
{
printf(\"\ Linked List is Empty\");
printf(\"\ Deletion is not Possible \");
}
else
{
printf(\"\ deleted Node does not exist in List\");
printf(\"\ Deletion is not Possible \");
}
}
else
{
if( rear == NULL)
{
temp = root;
root = root->next;
}
else
{
temp = curr;
rear->next = curr->next;
}
printf(\"\ Node is deleted \");
printf(\"\ Deleted Node information is %c\",temp->info);
free(temp);
}
display(root);
return root;
}
sl * insert(sl *root,char val)
{.
Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf
Write the code above and the ones below in netbeans IDE 8.1
3. (Eliminate duplicates) Write a method that returns a new array by eliminating the duplicate
values in the array using the following method header: public static int[]
eliminateDuplicates(int[] list) Write a test program that reads in ten integers, invokes the method,
and displays the result. Here is the sample run of the program:
Enter ten numbers: 1 2 3 2 1 6 3 4 5 2 [Enter]
The distinct numbers are: 1 2 3 6 4 5
4. (Sort students) Write a program that prompts the user to enter the number of students, the
students’ names, and their scores, and prints student names in decreasing order of their scores.
Solution
Please follow the code and comments for description :
a)
CODE :
import java.util.Scanner;
public class MeanSD {
public static double deviation (double[] x) { // method to return the deviation
double sum = 0, mean = 0, deviation = 0; // required initialisations
double[] temp = new double[10]; // temporary array
for (int i = 0; i < 10; i++) //calculate the standard deviation
{
temp[i] = Math.pow((x[i] - mean), 2); // assigning the value to the array
sum += temp[i]; // adding up the values
}
mean = sum / 10; // getting the mean
deviation = Math.sqrt(mean); // calculating the deviation
return deviation; // returning the deviation
}
public static double mean (double[] x) { // method to return themean of the values entered
double sum = 0, mean; // required initialisations
for (int i = 0; i < 10; i++) //Take input in the array
{
sum += x[i]; //sum of all elements
}
mean = sum / 10; // calculating the mean
return mean; // returning the mean value
}
public static void main(String[] args) { // driver method
System.out.println(\"Enter the 10 numbers.\"); // prompt to enter the data
Scanner in = new Scanner(System.in); // scanner class to get the data from the user
double[] arr = new double[10]; // array that saves the data
double sum = 0, mean = 0, deviation = 0; // required initialisations
for (int i = 0; i < 10; i++) //Take input in the array
{
System.out.print(\"Enter a number : \"); // prompt to enter the number
arr[i] = in.nextDouble(); // getting the data from the console
}
mean = mean(arr); // calling the method to return the mean value
System.out.println(\"Mean : \" + mean); //Display mean of all elements
deviation = deviation(arr); // calling the method to return teh deviation value
System.out.println(\"Deviation : \" + deviation); // display the result
}
}
OUPTPUT :
Enter the 10 numbers.
Enter a number : 1
Enter a number : 2
Enter a number : 3
Enter a number : 4
Enter a number : 5
Enter a number : 6
Enter a number : 7
Enter a number : 8
Enter a number : 9
Enter a number : 10
Mean : 5.5
Deviation : 6.2048368229954285
b)
CODE :
import java.util.Scanner;
public class MyDuplicateElements {
public static int[] eliminateDuplicates(int[] input) { // method to remove the duplicates
int j = 0; // required initialisations
int i = 1;
//return if the array length is less than 2
if (input..
Which statement about gills is falseThey contain the environmenta.pdf
Which statement about gills is false?
They contain the environmental medium.
The respiratory membrane has a large surface area.
They are surrounded by the environmental medium.
Solution
Which statement about gills is false?
They contain the environmental medium.---------answer
The respiratory membrane has a large surface area.
They are surrounded by the environmental medium.
======================================================
EXPLANATION
GILLS
Gills significantly increment the surface area for exchange of gases. They happen in an
assortment of creature gatherings including arthropods , annelids, fish, and creatures of land and
water. Gills normally are convoluted outgrowths containing veins secured by a thin epithelial
layer. Commonly gills are sorted out into a progression of plates and might be interior (as in
crabs and fish) or outer to the body (as in a few creatures of land and water).
Gills are extremely proficient at expelling oxygen from water: there is just 1/20 the measure of
oxygen present in water as in a similar volume of air. Water streams over gills in one course
while blood streams the other way through gill vessels. This countercurrent stream augments
oxygen exchange..
Why do Boolean functions play a central role in hardware architectur.pdf
Why do Boolean functions play a central role in hardware architectures?
Describe Composite Gates.
Describe Multiplexor’s and Demultiplexor’s? What is the importance of the use of the selection
bit and the data bits?
Solution
In electronics, a multiplexer (or mux) is a device that selects one of several analog or digital
input signals and forwards the selected input into a single line. A multiplexer of 2n inputs has n
select lines, which are used to select which input line to send to the output.Multiplexers are
mainly used to increase the amount of data that can be sent over the network within a certain
amount of time and bandwidth. A multiplexer is also called a data selector. Multiplexers can also
be used to implement Boolean functions of multiple variables.
An electronic multiplexer makes it possible for several signals to share one device or resource,
for example one A/D converter or one communication line, instead of having one device per
input signal.
Conversely, a demultiplexer (or demux) is a device taking a single input signal and selecting one
of many data-output-lines, which is connected to the single input. A multiplexer is often used
with a complementary demultiplexer on the receiving end.
An electronic multiplexer can be considered as a multiple-input, single-output switch, and a
demultiplexer as a single-input, multiple-output switch.The schematic symbol for a multiplexer
is an isosceles trapezoid with the longer parallel side containing the input pins and the short
parallel side containing the output pin.The schematic on the right shows a 2-to-1 multiplexer on
the left and an equivalent switch on the right. The wire connects the desired input to the output..
Which of the following is not a primary tissue category a. Epitheli.pdf
Which of the following is not a primary tissue category? a. Epithelial b. Simple Squamous c.
Muscle d. Nerve What type of tissue in the liver functions to secrete bile? a. Epithelial b.
Connective c. Muscle d. Nerve What type of tissue in the pancreas holds epithelial cells in
place? a. Epithelial b. Connective c. Muscle d. Nerve What would be a good function of
stratified squamous epithelium? a. Absorption b. Secretion c. Contraction d. Protection Where
would you find transitional epithelium? a. The trachea b. The lungs c. The uterus d. The skin
If I saw multiple layers of tall cells, how could I describe them? a. Simple Squamous b.
Stratified cuboidal c. Stratified columnar d. Simple columnar
Solution
1. b. Simple squamous
The primary tissues are - Epithelial, connective, muscular and nervous.
2. a. Epithelial
Bile is secreted by Cholangiocytes which are epithelial cells.
3. b. Connective
Connective tissue in pancreas holds the epithelial cells.
4. d. Protection
The major function of stratified squamous epithelium is protection.
This problem illustrates one of the possible pitfalls of blindly appl.pdf
This problem illustrates one of the possible pitfalls of blindly applying numerical methods
without paying attention to the theoretical aspects of the differential equation itself. Consider the
equation ty\' + 3y - 9t^2 = 0. Use the MATLAB program myeuler.m from Chapter 8 to compute
the Euler Method approximation to the solution with initial condition y(-0.5) = 3.15, using step
size h = 0.2 and n = 10 steps The program will generate a lot of ordered pairs (x_i, .y_i). Use plot
to graph the piecewise linear function connecting the points (x_i, y_i). Now modify the program
to implement the Improved Euler Method. Can you make sense of your answers? Next, use
ode45 to find an approximate solution on the interval (-0.5, 0.5), and plot it with plot. Print out
the values of the solution at the points -0.06: 0.02: 0.06. What is the interval on which the
approximate solution is defined? Solve the equation explicitly and graph the solutions for the
initial conditions y(0) = 0, y(-0.5) = 3.15, y(0, 5) = 3.15, y(-0.5) = -3.45, and y(0.5) = -3.45. Now
explain your results in (a)-(c). Could we have known, without solving the equation, whether to
expect meaningful results in parts (a) and (b)? Why? Can you explain how ode45 avoids making
the same mistake?
Solution
clc
clear all
close all
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)
if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;
end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\');
subplot(2,2,2);stem(h1,h);
title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\');
subplot(2,2,[3 4]);stem(n,y);
title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\');
Comment only
X = input(\'Enter x: \'); %input vector X and H
H = input(\'Enter h: \') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end
for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesse.
Suppose a multiple regression model is given by y = - 4.09x1 + 8.59x,.pdf
Suppose a multiple regression model is given by y = - 4.09x1 + 8.59x, -23.91. What would an
interpretation of the coefficient of x1 be? Fill in the blank below An interpretation of the
coefficient of x1 would be, if x1 increases by 1 unit, then the response variable will decrease by
units, on average, while holding x2, constant
Solution
-4.09 Units as the Equations has the constant multipled -4.09 to x1 units.
More Related Content
More from arihantmum
Consider the following C code snippet C codevoid setArray(int.pdf
Consider the following C code snippet:
// C code
void setArray(int num) {
int i;
int array[10];
for (i=0; i < 10; i=i+1) {
array[i]=compare(num, i);
}
}
int compare(int a, int b) {
if (sub(a, b) >= 0)
return 1;
else
return 0;
}
int sub(int a, int b) {
return ab;
}
Implement the C code snippet above in MIPS assembly language. Use $s0 to hold the variable i.
Be sure to handle the stack pointer appropriately. The array is stored on the stack of the setArray
function (i.e. when setArray function is called it allocates space for the array on the stack.).
Solution
I have used the Test function i.e ain function to get this module of the fuction definition working.
Thw below is the MIPS code for the complete program:-
.file 1 \"\"
.section .mdebug.abi32
.previous
.gnu_attribute 4, 1
.abicalls
.text
.align 2
.globl _Z3subii
$LFB0 = .
.set nomips16
.ent _Z3subii
.type _Z3subii, @function
_Z3subii:
.frame $fp,8,$31 # vars= 0, regs= 1/0, args= 0, gp= 0
.mask 0x40000000,-4
.fmask 0x00000000,0
.set noreorder
.set nomacro
addiu $sp,$sp,-8
$LCFI0:
sw $fp,4($sp)
$LCFI1:
move $fp,$sp
movz $31,$31,$0
$LCFI2:
sw $4,8($fp)
sw $5,12($fp)
lw $3,8($fp)
lw $2,12($fp)
nop
subu $2,$3,$2
move $sp,$fp
lw $fp,4($sp)
addiu $sp,$sp,8
j $31
nop
.set macro
.set reorder
.end _Z3subii
$LFE0:
.size _Z3subii, .-_Z3subii
.align 2
.globl _Z7compareii
$LFB1 = .
.set nomips16
.ent _Z7compareii
.type _Z7compareii, @function
_Z7compareii:
.frame $fp,32,$31 # vars= 0, regs= 2/0, args= 16, gp= 8
.mask 0xc0000000,-4
.fmask 0x00000000,0
.set noreorder
.cpload $25
.set nomacro
addiu $sp,$sp,-32
$LCFI3:
sw $31,28($sp)
$LCFI4:
sw $fp,24($sp)
movz $31,$31,$0
$LCFI5:
move $fp,$sp
$LCFI6:
.cprestore 16
sw $4,32($fp)
sw $5,36($fp)
lw $4,32($fp)
lw $5,36($fp)
lw $2,%got(_Z3subii)($28)
nop
move $25,$2
jalr $25
nop
lw $28,16($fp)
nor $2,$0,$2
srl $2,$2,31
andi $2,$2,0x00ff
beq $2,$0,$L4
nop
li $2,1 # 0x1
b $L5
nop
$L4:
move $2,$0
$L5:
move $sp,$fp
lw $31,28($sp)
lw $fp,24($sp)
addiu $sp,$sp,32
j $31
nop
.set macro
.set reorder
.end _Z7compareii
$LFE1:
.size _Z7compareii, .-_Z7compareii
.align 2
.globl _Z8setArrayi
$LFB2 = .
.set nomips16
.ent _Z8setArrayi
.type _Z8setArrayi, @function
_Z8setArrayi:
.frame $fp,88,$31 # vars= 48, regs= 3/0, args= 16, gp= 8
.mask 0xc0010000,-4
.fmask 0x00000000,0
.set noreorder
.cpload $25
.set nomacro
addiu $sp,$sp,-88
$LCFI7:
sw $31,84($sp)
$LCFI8:
sw $fp,80($sp)
movz $31,$31,$0
$LCFI9:
sw $16,76($sp)
$LCFI10:
move $fp,$sp
$LCFI11:
.cprestore 16
sw $4,88($fp)
sw $0,24($fp)
b $L8
nop
$L9:
lw $16,24($fp)
lw $4,88($fp)
lw $5,24($fp)
lw $2,%got(_Z7compareii)($28)
nop
move $25,$2
jalr $25
nop
lw $28,16($fp)
move $3,$2
sll $2,$16,2
addiu $4,$fp,24
addu $2,$4,$2
sw $3,4($2)
lw $2,24($fp)
nop
addiu $2,$2,1
sw $2,24($fp)
$L8:
lw $2,24($fp)
nop
slt $2,$2,10
andi $2,$2,0x00ff
bne $2,$0,$L9
nop
move $sp,$fp
lw $31,84($sp)
lw $fp,80($sp)
lw $16,76($sp)
addiu $sp,$sp,88
j $31
nop
.set macro
.set reorder
.end _Z8setArrayi
$LFE2:
.size _Z8setArrayi, .-_Z8setArrayi
.align 2
.globl ma.
Describe a mechanism that the body uses to prevent a mutation from be.pdf
Describe a mechanism that the body uses to prevent a mutation from becoming an abnormal
cell.
Solution
The body has several mechanisms in place to ensure that mutations do not turn lethal. The first of
those steps is the proof reading activity of the DNA polymerase which reads the DNA strand and
corrects the sequence as it synthesizes the new DNA fragment. Thus, the daughter cells contain
the original non-mutated sequence. Secondly, organisms are mostly diploid i.e. they contain two
sets of chromosomes which includes two forms of the same gene. The mutation would be lethal
or cause loss of function only when the other copy of the gene is also mutated (recessive
mutation). Also, mutant cells are usually cleared by the immune system which can detect and
produce specific antibodies against these cells..
classify domian of life differencesSolutionClassify the domain.pdf
classify domian of life differences
Solution
Classify the domain of life differences: the domain of life is mainly classified based on the
following differences
1. Ribosomal RNAs (rRNA) inside the cell
2. The structure of cell membrane with lipid structure
3. Cell sensitivity to antibiotics
Three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae,
kingdom Animalia)\"
Explanation:
The mitochondria, histone (nucleosomes) are not evolved in archae domain & prokaryote
domain but evolved in the “kingdom of eukaryotes”, as an evolutionary adaptation difference.
This is a major differential aspect among the three domains of life in their classification
according to endo-symbiotic theory.
The aspects of cell wall are constructed differently in all three domains of life such as bacteria,
archaic species, and eukaryotes. The archae has S -layer in their cell wall but do not possess
peptidoglygan layer as bacteria do. Lipopolysaccharide, peptidoglycan layers are absent in
eukaryotic cell walls and their mechanism of synthesis is absent in eukaryotes. Plant cells are
eukaryotic cells and contain cell wall but animal cells do not contain cell wall. Bacteria are
prokaryotes & do not possess histone proteins but eukaryotes contain histone proteins associated
with DNA. Any morphological characteristics of group of species are completely different to that
of three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae,
kingdom Animalia)\". These morphological features are completely different to the ancestral
species and leading to new phylum evolvement. Internal structural features, embryological
development, nucleus structure, genome nature (RNA or DNA), type of cellular respiration &
energy synthesis.
The major domian of life differences with morphological relations in the evolution of vertebrate
animal life (at the phylum level & domain level) from invertebrates (from four -limb species):
The major features of embryonic vertebrate zygote formation & followed by eight-cell stage of
embryo finally blastula stage and gastrulation. This stage is absent in the archae & prokaryotes
species. There are total 2 types of groups of species that are divided based on presence of
morphological innovation that is \"presence of vertebral column\" in animal domain &
prokaryote. Those living species which do not have vertebra\" are referred as \"invertebrates\"
and those which possess vertebrae\" are referred as “chordates\" or \"vertebrates\"..
Compare and contrast the development of a WBS in traditional project.pdf
Compare and contrast the development of a WBS in traditional project management versus
development in an agile environment.
Solution
An Agile based WBS is organized around units of end user functionality - commonly referred to
as user stories. Each user story represents the work to deliver that increment of functionality. In
Agile based WBS, Features are decomposed into Epics (large user stories) and Epics into User
Stories. User stories are decomposed to represent functionality that can be completed within one
iteration - typically one to four weeks depending on the team/project. Each leaf level WBS
element represent the working software to be produced through analysis, design, coding and
testing of the requirement in the user story.
Because each user story is atomic, stories can be re-prioritized, added and or removed without
changing the scope, provided that the total size/effort of stories within the scope of the project
does not change. Additionally, iterative development facilitates multiple releases (deployments)
within a project. So, user stories are typically scheduled within an iteration and a release. Note,
that until an iteration has started, the stories within that iteration can be changed, as part of
iteration planning.
In case of traditional project management WBS is organized around the primary products of the
project (or planned outcomes) instead of the work needed to produce the products (planned
actions). Since the planned outcomes are the desired ends of the project, they form a relatively
stable set of categories in which the costs of the planned actions needed to achieve them can be
collected.
A well-designed WBS makes it easy to assign each project activity to one and only one terminal
element of the WBS. In addition to its function in cost accounting, the WBS also helps map
requirements from one level of system specification to another, for example a requirements cross
reference matrix mapping functional requirements to high level or low level design documents..
B. You are an evolutionary biologist in a heated argument with a cre.pdf
B. You are an evolutionary biologist in a heated argument with a creation scientist who makes
the statement—\"Mathematically, it is inconceivable that anything as complex as a protein, let
alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for
chance. 2. Now play the other side and respond to the supporting statement as you think a
creation scientist would respond.?
Solution
You are an evolutionary biologist in a heated argument with a creation scientist who makes the
statement—
\"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell
or a human, could spring up by chance\".
1. Make a supporting statement for chance.
Chance: It plays a part in evolution (for example, random mutations in a gene pool can give rise
to few new traits), but evolution does not depend ONLY on chance in creating new organisms,
proteins or any other entities. This is quite opposite to natural selection. Natural selection is the
principal which is known for the mechanism of evolution. By chance some mutation caused in
the gene pool and it gave to new acceptable traits. If they found adjustable to the environment,
they continues as a new species. The \"undesirable\" (nonadaptive) ones will be thus elimated
from that gene pool. As long as the forces of selection stay constant, natural selection pushes the
evolution in only one direction and produce new desired structures. Thus chance plays a part in
evolution. Nature plays a major role in this. Fossil studies support this theory/ process. Where we
can see many intermediate or failed species in between major species.
Archaeopteryx, a link between reptiles and birds, extinctinted soon is a good example of
intermediate species.
Haldane and Oparin proposed a theory of the origin of life on Earth through the transformation,
during the gradual chemical evolution of particles that contain carbon in the primordial soup
shows an example of Evolution.
2. Now play the other side and respond to the supporting statement as you think a creation
scientist would respond?
As a creation scientist i do not agree with science and evolution part, all this is created by God
and thus it follows biblical roots. All that it is in the nature is creation of God, all of a sudden it
appeared on the earth. According to us GOD created all species like birds,animals etc., later he
wanted to create a genius brained animal, so he created humans. And God creation according to
us all this is due to CHOICE not because of CHANCE.Thus it involved logical and systematic
process..
Assume Hashtable is a simple array of size 8, with indices 0..7. Num.pdf
Assume Hashtable is a simple array of size 8, with indices 0..7. Numeric keys are mapped by a
Hashfunction that gives the mod(8,n) value for any key “n” yielding the Hashtable index for that
key (0..7). A Hashtable entry is null unless a key exists with that index as its hashed index; if so,
the Hashtable entry points to the first node of a linked list of keys with that hash index. The last
node on this linked list has a null reference for the next referenced node. Assume the occurrence
of a linked list node is represented by the object “Node” and its “Data” and “NextRef” attributes.
Create pseudocode for the \"Remove\" operation.
Solution
Removal of a Node from a Singly Linked List
To complete deletion of firstNode in the list we have to change Head pointing to NextRef of
firstNode.
Pseudocode:
firstNode = Head
Head = firstNode->NextRef
free firstNode
Complexity:
Time Complexity: O(1)
Space Complexity: O(1)
Sourcecode:
int delNodeData(int num)
{
struct Node *prev_ptr, *node;
node=Head;
while(cur_ptr != NULL)
{
if(cur_ptr->Data == num)
{
if(cur_ptr==Head)
{
Head=cur_ptr->NextRef;
free(node);
return 0;
}
else
{
prev_ptr->NextRef=node->NextRef;
free(node);
return 0;
}
}
else
{
prev_ptr=node;
node=node->NextRef;
}
}
printf(\"\ Element %d is not found in the List\", num);
return 1;
}
Deleting Last Node in the Singly Linked List
Traverse to Last Node in the List using two pointers namely prevNode and node. Once node
reaches the last Node in the list point NextRef in prevNode to NULL and free the node.
Pseudocode:
node = head
forever:
if node->NextRef == NULL
break
prevNode = node
node = node->NextRef
prevNode->NextRef = NULL
free Node
Complexity:
Time Complexity: O(n)
Space Complexity: O(1)
Deleting Node from position \'p\' in the List
To delete a Node at the position \'p\' we have to first traverse the list until we reach the position
\'p\'.
For this case have to maintain two pointers namely prevNode and Node.
Since Singly Linked Lists are uni-directional we have to maintain the information about
previous Node in prevNode. Once we reach the position \'p\' we have to modify prevNode
NextRef pointing to Node NextRef and free Node.
Pseudocode:
curNode = head
curPos = 1
forever:
if curPos == P || Node == NULL
break
prevNode = node
node = node->NextRef
curPos++
if noode != NULL:
prevNode->NextRef = node->NextRef
free curNode
Complexity:
Time Complexity: O(n) worst case
Space Complexity: O(3)
Sourcecode:
int delNodeLoc(int loc)
{
struct Node *prev_ptr, *node;
int i;
node=Head;
if(loc > (length()) || loc <= 0)
{
printf(\"\ Deletion of Node at given location is not possible\ \");
}
else
{
// If the location is starting of the list
if (loc == 1)
{
Head=node->NextRef;
free(node);
return 0;
}
else
{
for(i=1;iNextRef;
}
prev_ptr->NextRef=node->NextRef;
free(node);
}
}
return 1;
}.
Amon the following, which has the lowest levels of dissolved iron.pdf
Amon the following, which has the lowest levels of dissolved iron?
a. Atlantic Ocean
b. Indian Ocean
c. Pacific Ocean
d. The temperate regions of all oceans
Solution
The correct option is Pacific ocean because of the low dust sedimentation and iron is readily
soluble in low oxygen regions and the continental margins. It was believed that iron remained in
the localized areas but didn\'t contribute to the overall iron content of the ocean. As it is
explained by Nielsen, as soon as the iron-rich fluids hit the seawater with high oxygen content,
the iron gets dumped out and is not contributed to the overall amount of the dissolved iron..
A box contains 10 red balls and 40 black balls. Two balls are drawn .pdf
A box contains 10 red balls and 40 black balls. Two balls are drawn successively without
replacement from the box. Is the probability of selecting a red ball in the first draw the same as
the probability of selecting a red ball in the second draw? Justify your answers by computing the
probabilities.
Solution.
As late as 1992, the United States was running budget deficits of ne.pdf
As late as 1992, the United States was running budget deficits of nearly $300 billion. During the
remainder of the 1990\'s, deficits declined and became surpluses. As the new century began,
these surpluses again turned into deficits. •Explain the decline in deficits and subsequent
surpluses in the late 1990\'s. •Explain the return to deficit spending since the turn of the century.
•Consider the causes of the deficits and surpluses and provide your own insight as to whether
these surpluses or deficits have a \"positive\" or \"negative\" effect on our economy
Solution
As the 1990s came to an end, virtually all financial-market observers and politicians rejoiced that
budget deficits were gone and that the U.S. had entered a period of fiscal stability. Politicians on
both sides of the aisle viewed the new world of budget surpluses as an opportunity to save Social
Security, to cut Federal debt, and to lower taxes.
Budget deficits or budget surpluses are neither good nor bad, they are fiscal policy options that
can stabilize or de-stabilize an economy. The U.S. economy, under Reagan, Bush, and early
Clinton, experienced higher not lower rates of economic growth at a time when budget deficits
were at record levels and tax rates were on the decline. Over the past 20 years a period of record
budget deficits the stock market increased from 950 to over 11,000, providing ample evidence
that budget deficits are not necessarily bad for an economy.On the other hand, budget surpluses
are not necessarily good. As the budget surpluses began to expand dramatically in late 1999 and
into 2000, the stock market appeared to be reversing the record up-trend of the 1980s and \'90s.
One possible reason for the reversal: the growing budget surplus was sapping the savings of the
private sector. Consumers were able to maintain their spending only through expanding debt to
record level.
In the late 90s, the confluence of strong economic growth, decelerating federal government
spending, and substantial increases in tax rates (revenues) had the unexpected effect of producing
enormous budget surpluses. As unexpected budget surpluses burst on the scene in the late 1990s,
most mainstream economists weren’t sensitive to the contractionary impact that surpluses can
have on the economy. The reason was simple that none of their economic models ever included
the forecast of a substantial surplus.
Ronald Reagan’s was the vice president of the USA for eight years and knew firsthand how
difficult it was to bring the federal budget deficit down from 6 percent of GDP in 1983 to 2.8
percent in Reagan’s last budget. Bush knew that the many budget deals of the 1980s had been
hard fought and always involved higher taxes as part of the deficit reduction, Reagan signed into
law 11 major tax increases that raised taxes by $133 billion in 1988 or 2.7 percent of GDP.Bush
administration economists believed that an easier monetary policy was the key to stimulating
growth. But Federal Reserve Board.
A bacteriophage population is introduced to a bacterial colony that .pdf
A bacteriophage population is introduced to a bacterial colony that is in stationary phase with
rapidly depleting nutrients. Appraise the following scenarios and select which one would favor
long-term survival of the bacteriophage.
A. The bacteriophages infect the remaining viable cells and integrate their genetic material into
the host cell genome thereby preserving their genome.
B. The bacteriophages do not infect the bacterial cells since they are dying; they remain inert
outside of the cells.
C. The bacteriophages infect the bacterial cells and go through all stages of development apart
from release, thus keeping the progeny phage from entering an inhospitable environment.
Solution
Answer :
C. The bacteriophages infect the bacterial cells and go through all stages of development apart
from release, thus keeping the progeny phage from entering an inhospitable environment..
Xavier and Yolanda both have classes that end at noon and they agree .pdf
Xavier and Yolanda both have classes that end at noon and they agree to meet every day after
class. They arrive at the coffee shop independently. Xavier\'s arrival time is X and Yolanda\'s
arrival time is Y, where X and Y are measured in minutes after noon. The individual density
functions are given. (Xavier arrives sometime after noon and is more likely to arrive promptly
than late. Yolanda always arrives by 12:10 PM and is more likely to arrive late than promptly.)
After Yolanda arrives, she\'ll wait for up to 20 minutes for Xavier, but he won\'t wait for her.
Find the probability that they meet. (Round your answer to three decimal places.) f1(x) = {e^-x if
x gt o and o if x
Solution.
1. Low platelet count is a recessively inherited trait. Reevaluation.pdf
1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates
that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her
husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia
(15pts).
a. What are the parental gametes? (2pts)
b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible
genotypes. (2 pts)
c. What is the gamete frequency if the two genotypes in (b) were crossed with each other?
(4pts)? (4pts)
d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts)
Solution
Both the conditions are recessive.
Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for
beta thalessemia. Genotype is LLbb
Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta-
thalessemia. Genotype is llBb
The possible gametes produced by these parents:
Nikoleta: Lb
Husband: lB, lb
Possible genotype of the F1 offspring: LlBb or Llbb
Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype
should be Llbb.
Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb).
Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2
The following is the phenotype frequency:
LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3
LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3
llBb (Low platelet count and no beta thallasemia) = 1
llbb (Low platelet count and beta thallasemia) = 1
Thus phenotypic ratio stands as:
Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low
platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1.
1. An important contribution of Fiedlers research on the contingen.pdf
1. An important contribution of Fiedler\'s research on the contingency model of leadership is that
it: a. uses weights which are determined in an arbitrary manner in order to determine situation
favorability. b. goes beyond the notion of leadership styles to try to show how styles fit a
situation. c. recommends medium least preferred coworker leaders as they are more effective in
many situations. d. is clear about how the model will work over time. 2. Which of the following
characteristics was explored by the studies evaluating the characteristics of the leader-member
exchange relationship? a. Income b. Educational qualification c. Value agreement d. Recruitment
process 3. Which of the following is a characteristic of entrepreneurial leaders? a. They are
concerned with innovation and creativity. b. They are interested in maintaining the status quo. c.
They avoid taking risks and are unwilling to stretch themselves. d. They tend to be dependent on
their followers. 4. An important aspect of leadership is: a. dictating changes required in an
organization. b. maintaining a passive relationship with followers. c. maintaining the status quo
and not creating change. d. influencing others to come together around a common vision. 5. It
becomes easier for people to see their own opportunities for leadership when they: a. start seeing
leaders as individuals who are above others. b. understand that the concept of leadership cannot
continue to develop. c. stop equating leadership with greatness and public visibility. d.
comprehend that leadership involves maintaining the status quo. 6. Erica, a team manager, was
known for heading projects with diligence and conscientiousness. However, she was insensitive
and excessively ambitious. She did not take feedback in a constructive manner. As a result, this
prevented her from moving forward in terms of her career since she had problems
communicating with her team members and motivating them. In the context of the given
scenario, which of the following phenomenon does Erica experience? a. Communication
apprehension b. Fear-based motivation c. Paradigm change d. Derailment Name: Class: Date:
Quiz 1: Chp 1-3 (Fall 2015) 2 7. Which of the following statements is true of the contingency
approach to leadership? a. A leadership style that works in one situation will work in all
situations. b. These approaches seek to investigate traits that can improve performance in all
situations. c. The needs, maturity, and cohesiveness of followers have been identified as a key
contingency. d. These approaches hold that behavior that is effective in some circumstances is
effective under all conditions. 8. Stage two in the development of the individualized leadership
theory explored the _____ in more detail. a. vertical dyad linkage model b. partnership building
c. leader–member exchange d. entrepreneurship model 9. Which of the following statements is
true of leaders who work collaboratively? a. They learn to keep the lines of commu.
^^^ Discuss about Header Node And also write a program for unorder.pdf
^^^ Discuss about Header Node? And also write a program for unordered single linked list and
linked implementation of QUEUE?
Solution
Header node:
Sometimes it is desire to keep an extra node at the front of a List. Such a node doesn’t represent
an item in the List and is called a Header Node or a List Header. The Information field of such
Header node might be unused but the Next field maintains the first node address. More often the
information of such a node could be used to keep Global Information about the Entire List.
One of the applications of Header node is the information portion of the Header node contains
the number of nodes (not including Header) in the List. In this Structure the Header must be
adjusted the total number of nodes when we apply addition or deletion operation on the List. We
can directly obtained the total number of nodes without traversing the entire List.
By using this application, we can develop Data Structure Stack easily. Since Header node
information field maintains the number of elements that the Stack contains.
Another application is we can simply develop the Data Structure Queue. Until now, two external
pointers, Front & Rear, were necessary for a List to represent a Queue. However, now only a
single external pointer is sufficient to maintain a Queue that is Head node. Because the Header
node next field behave like a Front and information field of Header node maintains last node
address so that it is behave like a Rear .
Another possibility for the use of the information portion of a List Header is as a pointer
to a Curr node in the List during a traversal process. This would be eliminate the node for an
external pointer during traversal.
#include
#include
#include
#include
struct node
{
char info;
struct node *next;
};
typedef struct node sl;
void display(sl *);
sl * create(sl *root)
{
sl *curr,*new;
char val;
printf(\"\ Enter $ to STOP --- Otherwise Continue : \");
fflush(stdin);
scanf(\"%c\",&val);
while(val != \'$\')
{
new=(sl *) malloc(sizeof(sl));
new->info = val;
new->next = NULL;
if(root == NULL)
root = new;
else
{
curr = root;
while( curr->next != NULL )
curr = curr->next;
curr->next = new;
}
printf(\" Enter $ to STOP --- Otherwise Continue : \");
fflush(stdin);
scanf(\"%c\",&val);
}
printf(\"\ The Created Linked List Is \ \");
display(root);
return root;
}
sl * delete(sl *root,char val)
{
sl *temp,*curr,*rear;
curr = root;
rear = NULL;
while( curr != NULL && val != curr->info )
{
rear = curr;
curr = curr->next;
}
if(curr == NULL)
{
if( rear == NULL)
{
printf(\"\ Linked List is Empty\");
printf(\"\ Deletion is not Possible \");
}
else
{
printf(\"\ deleted Node does not exist in List\");
printf(\"\ Deletion is not Possible \");
}
}
else
{
if( rear == NULL)
{
temp = root;
root = root->next;
}
else
{
temp = curr;
rear->next = curr->next;
}
printf(\"\ Node is deleted \");
printf(\"\ Deleted Node information is %c\",temp->info);
free(temp);
}
display(root);
return root;
}
sl * insert(sl *root,char val)
{.
Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf
Write the code above and the ones below in netbeans IDE 8.1
3. (Eliminate duplicates) Write a method that returns a new array by eliminating the duplicate
values in the array using the following method header: public static int[]
eliminateDuplicates(int[] list) Write a test program that reads in ten integers, invokes the method,
and displays the result. Here is the sample run of the program:
Enter ten numbers: 1 2 3 2 1 6 3 4 5 2 [Enter]
The distinct numbers are: 1 2 3 6 4 5
4. (Sort students) Write a program that prompts the user to enter the number of students, the
students’ names, and their scores, and prints student names in decreasing order of their scores.
Solution
Please follow the code and comments for description :
a)
CODE :
import java.util.Scanner;
public class MeanSD {
public static double deviation (double[] x) { // method to return the deviation
double sum = 0, mean = 0, deviation = 0; // required initialisations
double[] temp = new double[10]; // temporary array
for (int i = 0; i < 10; i++) //calculate the standard deviation
{
temp[i] = Math.pow((x[i] - mean), 2); // assigning the value to the array
sum += temp[i]; // adding up the values
}
mean = sum / 10; // getting the mean
deviation = Math.sqrt(mean); // calculating the deviation
return deviation; // returning the deviation
}
public static double mean (double[] x) { // method to return themean of the values entered
double sum = 0, mean; // required initialisations
for (int i = 0; i < 10; i++) //Take input in the array
{
sum += x[i]; //sum of all elements
}
mean = sum / 10; // calculating the mean
return mean; // returning the mean value
}
public static void main(String[] args) { // driver method
System.out.println(\"Enter the 10 numbers.\"); // prompt to enter the data
Scanner in = new Scanner(System.in); // scanner class to get the data from the user
double[] arr = new double[10]; // array that saves the data
double sum = 0, mean = 0, deviation = 0; // required initialisations
for (int i = 0; i < 10; i++) //Take input in the array
{
System.out.print(\"Enter a number : \"); // prompt to enter the number
arr[i] = in.nextDouble(); // getting the data from the console
}
mean = mean(arr); // calling the method to return the mean value
System.out.println(\"Mean : \" + mean); //Display mean of all elements
deviation = deviation(arr); // calling the method to return teh deviation value
System.out.println(\"Deviation : \" + deviation); // display the result
}
}
OUPTPUT :
Enter the 10 numbers.
Enter a number : 1
Enter a number : 2
Enter a number : 3
Enter a number : 4
Enter a number : 5
Enter a number : 6
Enter a number : 7
Enter a number : 8
Enter a number : 9
Enter a number : 10
Mean : 5.5
Deviation : 6.2048368229954285
b)
CODE :
import java.util.Scanner;
public class MyDuplicateElements {
public static int[] eliminateDuplicates(int[] input) { // method to remove the duplicates
int j = 0; // required initialisations
int i = 1;
//return if the array length is less than 2
if (input..
Which statement about gills is falseThey contain the environmenta.pdf
Which statement about gills is false?
They contain the environmental medium.
The respiratory membrane has a large surface area.
They are surrounded by the environmental medium.
Solution
Which statement about gills is false?
They contain the environmental medium.---------answer
The respiratory membrane has a large surface area.
They are surrounded by the environmental medium.
======================================================
EXPLANATION
GILLS
Gills significantly increment the surface area for exchange of gases. They happen in an
assortment of creature gatherings including arthropods , annelids, fish, and creatures of land and
water. Gills normally are convoluted outgrowths containing veins secured by a thin epithelial
layer. Commonly gills are sorted out into a progression of plates and might be interior (as in
crabs and fish) or outer to the body (as in a few creatures of land and water).
Gills are extremely proficient at expelling oxygen from water: there is just 1/20 the measure of
oxygen present in water as in a similar volume of air. Water streams over gills in one course
while blood streams the other way through gill vessels. This countercurrent stream augments
oxygen exchange..
Why do Boolean functions play a central role in hardware architectur.pdf
Why do Boolean functions play a central role in hardware architectures?
Describe Composite Gates.
Describe Multiplexor’s and Demultiplexor’s? What is the importance of the use of the selection
bit and the data bits?
Solution
In electronics, a multiplexer (or mux) is a device that selects one of several analog or digital
input signals and forwards the selected input into a single line. A multiplexer of 2n inputs has n
select lines, which are used to select which input line to send to the output.Multiplexers are
mainly used to increase the amount of data that can be sent over the network within a certain
amount of time and bandwidth. A multiplexer is also called a data selector. Multiplexers can also
be used to implement Boolean functions of multiple variables.
An electronic multiplexer makes it possible for several signals to share one device or resource,
for example one A/D converter or one communication line, instead of having one device per
input signal.
Conversely, a demultiplexer (or demux) is a device taking a single input signal and selecting one
of many data-output-lines, which is connected to the single input. A multiplexer is often used
with a complementary demultiplexer on the receiving end.
An electronic multiplexer can be considered as a multiple-input, single-output switch, and a
demultiplexer as a single-input, multiple-output switch.The schematic symbol for a multiplexer
is an isosceles trapezoid with the longer parallel side containing the input pins and the short
parallel side containing the output pin.The schematic on the right shows a 2-to-1 multiplexer on
the left and an equivalent switch on the right. The wire connects the desired input to the output..
Which of the following is not a primary tissue category a. Epitheli.pdf
Which of the following is not a primary tissue category? a. Epithelial b. Simple Squamous c.
Muscle d. Nerve What type of tissue in the liver functions to secrete bile? a. Epithelial b.
Connective c. Muscle d. Nerve What type of tissue in the pancreas holds epithelial cells in
place? a. Epithelial b. Connective c. Muscle d. Nerve What would be a good function of
stratified squamous epithelium? a. Absorption b. Secretion c. Contraction d. Protection Where
would you find transitional epithelium? a. The trachea b. The lungs c. The uterus d. The skin
If I saw multiple layers of tall cells, how could I describe them? a. Simple Squamous b.
Stratified cuboidal c. Stratified columnar d. Simple columnar
Solution
1. b. Simple squamous
The primary tissues are - Epithelial, connective, muscular and nervous.
2. a. Epithelial
Bile is secreted by Cholangiocytes which are epithelial cells.
3. b. Connective
Connective tissue in pancreas holds the epithelial cells.
4. d. Protection
The major function of stratified squamous epithelium is protection.
This problem illustrates one of the possible pitfalls of blindly appl.pdf
This problem illustrates one of the possible pitfalls of blindly applying numerical methods
without paying attention to the theoretical aspects of the differential equation itself. Consider the
equation ty\' + 3y - 9t^2 = 0. Use the MATLAB program myeuler.m from Chapter 8 to compute
the Euler Method approximation to the solution with initial condition y(-0.5) = 3.15, using step
size h = 0.2 and n = 10 steps The program will generate a lot of ordered pairs (x_i, .y_i). Use plot
to graph the piecewise linear function connecting the points (x_i, y_i). Now modify the program
to implement the Improved Euler Method. Can you make sense of your answers? Next, use
ode45 to find an approximate solution on the interval (-0.5, 0.5), and plot it with plot. Print out
the values of the solution at the points -0.06: 0.02: 0.06. What is the interval on which the
approximate solution is defined? Solve the equation explicitly and graph the solutions for the
initial conditions y(0) = 0, y(-0.5) = 3.15, y(0, 5) = 3.15, y(-0.5) = -3.45, and y(0.5) = -3.45. Now
explain your results in (a)-(c). Could we have known, without solving the equation, whether to
expect meaningful results in parts (a) and (b)? Why? Can you explain how ode45 avoids making
the same mistake?
Solution
clc
clear all
close all
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)
if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;
end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\');
subplot(2,2,2);stem(h1,h);
title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\');
subplot(2,2,[3 4]);stem(n,y);
title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\');
Comment only
X = input(\'Enter x: \'); %input vector X and H
H = input(\'Enter h: \') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end
for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesse.
Suppose a multiple regression model is given by y = - 4.09x1 + 8.59x,.pdf
Suppose a multiple regression model is given by y = - 4.09x1 + 8.59x, -23.91. What would an
interpretation of the coefficient of x1 be? Fill in the blank below An interpretation of the
coefficient of x1 would be, if x1 increases by 1 unit, then the response variable will decrease by
units, on average, while holding x2, constant
Solution
-4.09 Units as the Equations has the constant multipled -4.09 to x1 units.
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