SolutionNet cash provided by operating activities for the year end.pdf
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Solution Net cash provided by operating activities for the year ended December 31, 2014 Eliminate Depreciation expense +80000 Adjust revenue and expenses from accrual accounting to cashNet income151490 Eliminate Depreciation expense +80000 Adjust revenue and expenses from accrual accounting to cashIncrease in account receivable(20370)Decrease in inventory+12840Increase in prepaid expenses(4370)Decrease in account payable(6480)Increase in accrued expenses payable+11240Total cash generated by operating activities224360.
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pH = pKa + log (Sa) where S =moles of salt a = m.pdf
pH = pKa + log (S/a) where S =moles of salt a = moles of acid pH = 4.745 + log
(0.025 / 0.1) = 4.143
Solution
pH = pKa + log (S/a) where S =moles of salt a = moles of acid pH = 4.745 + log
(0.025 / 0.1) = 4.143.
H is reduced, while Fe is oxidized. Note the oxi.pdf
H is reduced, while Fe is oxidized. Note: the oxidation state of H changes from +1
In HI to 0 in H2. the oxidation state of Fe changes from 0 in Fe(CO)5 to +2 in Fe(CO)4I2. CO is
a neutral ligand.
Solution
H is reduced, while Fe is oxidized. Note: the oxidation state of H changes from +1
In HI to 0 in H2. the oxidation state of Fe changes from 0 in Fe(CO)5 to +2 in Fe(CO)4I2. CO is
a neutral ligand..
Elastic fluids Nonmetals Metals Earths .pdf
Elastic fluids Nonmetals Metals Earths
Solution
Elastic fluids Nonmetals Metals Earths.
Vehicle.javapublic class Vehicle { Declaring instance var.pdf
Vehicle.java
public class Vehicle
{
//Declaring instance variables
private int max_no_passengers;
private int top_speed;
private double miles_travelled;
//Parameterized constructor
public Vehicle(int max_no_passengers, int top_speed, double miles_travelled) {
super();
this.max_no_passengers = max_no_passengers;
this.top_speed = top_speed;
this.miles_travelled = miles_travelled;
}
//Getters and setters
public int getMax_no_passengers() {
return max_no_passengers;
}
public void setMax_no_passengers(int max_no_passengers) {
this.max_no_passengers = max_no_passengers;
}
public int getTop_speed() {
return top_speed;
}
public void setTop_speed(int top_speed) {
this.top_speed = top_speed;
}
public double getMiles_travelled() {
return miles_travelled;
}
public void setMiles_travelled(double miles_travelled) {
this.miles_travelled = miles_travelled;
}
/* This getInfor() method returns the information about this vehicle class object
* Params: void
* Return:String
*/
public String getInfo()
{
String str=\"\ Maximum No of Passengers :\"+getMax_no_passengers()+
\"\ Top Speed :\"+getTop_speed()+
\"\ Total Miles Travelled :\"+getMiles_travelled();
return str;
}
}
_____________________________________________
FoodTruck.java
import java.util.ArrayList;
public class FoodTruck extends Vehicle{
//Declaring instance variables
private String name;
private ArrayList menu;
//Parameterized constructor
public FoodTruck(int max_no_passengers, int top_speed,
double miles_travelled, String name, ArrayList menu) {
super(max_no_passengers, top_speed, miles_travelled);
this.name = name;
this.menu = menu;
}
//Getters and setters
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public ArrayList getMenu() {
return menu;
}
public void setMenu(ArrayList menu) {
this.menu = menu;
}
/* This getInfor() method returns the information about this Food truck class Object
* And also the information about the Vehicle class Object.
* Params: void
* Return:String
*/
@Override
public String getInfo() {
String str=\"Food Truck Name : \"+name+
\"\ Menu of items : \"+menu+super.getInfo();
return str;
}
}
________________________________________________
Test.java
import java.util.ArrayList;
public class Test {
public static void main(String[] args) {
//Creating an ArrayList Object which holds List of Menu Items
ArrayList menu1=new ArrayList();
//Adding menu items to the ArrayList
menu1.add(\"Burger\");
menu1.add(\"Pizza\");
menu1.add(\"Chips\");
//Creating FoodTruck Object by passing parameters.
FoodTruck truck1=new FoodTruck(15,100,10000,\"Mobile Food Court\",menu1);
//Displaying the Truck1 Information.
System.out.println(\"__________Truck 1 Details__________\");
System.out.println(truck1.getInfo());
//Creating an ArrayList Object which holds List of Menu Items
ArrayList menu2=new ArrayList();
//Adding menu items to the ArrayList
menu2.add(\"Tea\");
menu2.add(\"Coffe\");
menu2.add(\"Cool Drinks\");
//Creating FoodTruck Object by passing parameters.
Ionization energy is the energy required to remove theouter most ele.pdf
Ionization energy is the energy required to remove theouter most electron from the gaseous
atom.
The electronic configuration (E.C.) of O (8) is [He]2s2 2p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of P (15) is [Ne]3s2 3p3
The electronic configuration (E.C.) of Al (13) is [Ne]3s2 3p1
The electronic configuration (E.C.) of K(19) is [Ar]4s1
The electronic configuration (E.C.) of Na (11) is [Ne]3s1
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- from P is more than that of Alwhich in
ture is more than Na
4)Among P & S , P has half filled configuration of porbitals so more stable than S .Therefore
more energy for P thanS
So the overall order of the ionization energy is O >P > S > Al > Na > K
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- from P is more than that of Alwhich in
ture is more than Na
4)Among P & S , P has half filled configuration of porbitals so more stable than S .Therefore
more energy for P thanS
So the overall order of the ionization energy is O >P > S > Al > Na > K
Solution
Ionization energy is the energy required to remove theouter most electron from the gaseous
atom.
The electronic configuration (E.C.) of O (8) is [He]2s2 2p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of P (15) is [Ne]3s2 3p3
The electronic configuration (E.C.) of Al (13) is [Ne]3s2 3p1
The electronic configuration (E.C.) of K(19) is [Ar]4s1
The electronic configuration (E.C.) of Na (11) is [Ne]3s1
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- .
D. Both C and N are hybridized sp3 .pdf
D. Both C and N are hybridized sp3
Solution
D. Both C and N are hybridized sp3.
The presence of conjugated Carbon-carbon double bonds causes colorat.pdf
The presence of conjugated Carbon-carbon double bonds causes coloration. When those bonds
are broken, the color fades.
Solution
The presence of conjugated Carbon-carbon double bonds causes coloration. When those bonds
are broken, the color fades..
The amount to be deposited at the start of his studies is found as.pdf
The amount to be deposited at the start of his studies is found as
PV = C [ 1 - ( 1+r)-2 / r ] x ( 1+r)-t
PV = $ 12,300 [ 1 - ( 1+0.06)-2 / 0.06 ] x( 1+0.06) -1
PV = 21,274
Solution
The amount to be deposited at the start of his studies is found as
PV = C [ 1 - ( 1+r)-2 / r ] x ( 1+r)-t
PV = $ 12,300 [ 1 - ( 1+0.06)-2 / 0.06 ] x( 1+0.06) -1
PV = 21,274.
Recommended
pH = pKa + log (Sa) where S =moles of salt a = m.pdf
pH = pKa + log (S/a) where S =moles of salt a = moles of acid pH = 4.745 + log
(0.025 / 0.1) = 4.143
Solution
pH = pKa + log (S/a) where S =moles of salt a = moles of acid pH = 4.745 + log
(0.025 / 0.1) = 4.143.
H is reduced, while Fe is oxidized. Note the oxi.pdf
H is reduced, while Fe is oxidized. Note: the oxidation state of H changes from +1
In HI to 0 in H2. the oxidation state of Fe changes from 0 in Fe(CO)5 to +2 in Fe(CO)4I2. CO is
a neutral ligand.
Solution
H is reduced, while Fe is oxidized. Note: the oxidation state of H changes from +1
In HI to 0 in H2. the oxidation state of Fe changes from 0 in Fe(CO)5 to +2 in Fe(CO)4I2. CO is
a neutral ligand..
Elastic fluids Nonmetals Metals Earths .pdf
Elastic fluids Nonmetals Metals Earths
Solution
Elastic fluids Nonmetals Metals Earths.
Vehicle.javapublic class Vehicle { Declaring instance var.pdf
Vehicle.java
public class Vehicle
{
//Declaring instance variables
private int max_no_passengers;
private int top_speed;
private double miles_travelled;
//Parameterized constructor
public Vehicle(int max_no_passengers, int top_speed, double miles_travelled) {
super();
this.max_no_passengers = max_no_passengers;
this.top_speed = top_speed;
this.miles_travelled = miles_travelled;
}
//Getters and setters
public int getMax_no_passengers() {
return max_no_passengers;
}
public void setMax_no_passengers(int max_no_passengers) {
this.max_no_passengers = max_no_passengers;
}
public int getTop_speed() {
return top_speed;
}
public void setTop_speed(int top_speed) {
this.top_speed = top_speed;
}
public double getMiles_travelled() {
return miles_travelled;
}
public void setMiles_travelled(double miles_travelled) {
this.miles_travelled = miles_travelled;
}
/* This getInfor() method returns the information about this vehicle class object
* Params: void
* Return:String
*/
public String getInfo()
{
String str=\"\ Maximum No of Passengers :\"+getMax_no_passengers()+
\"\ Top Speed :\"+getTop_speed()+
\"\ Total Miles Travelled :\"+getMiles_travelled();
return str;
}
}
_____________________________________________
FoodTruck.java
import java.util.ArrayList;
public class FoodTruck extends Vehicle{
//Declaring instance variables
private String name;
private ArrayList menu;
//Parameterized constructor
public FoodTruck(int max_no_passengers, int top_speed,
double miles_travelled, String name, ArrayList menu) {
super(max_no_passengers, top_speed, miles_travelled);
this.name = name;
this.menu = menu;
}
//Getters and setters
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public ArrayList getMenu() {
return menu;
}
public void setMenu(ArrayList menu) {
this.menu = menu;
}
/* This getInfor() method returns the information about this Food truck class Object
* And also the information about the Vehicle class Object.
* Params: void
* Return:String
*/
@Override
public String getInfo() {
String str=\"Food Truck Name : \"+name+
\"\ Menu of items : \"+menu+super.getInfo();
return str;
}
}
________________________________________________
Test.java
import java.util.ArrayList;
public class Test {
public static void main(String[] args) {
//Creating an ArrayList Object which holds List of Menu Items
ArrayList menu1=new ArrayList();
//Adding menu items to the ArrayList
menu1.add(\"Burger\");
menu1.add(\"Pizza\");
menu1.add(\"Chips\");
//Creating FoodTruck Object by passing parameters.
FoodTruck truck1=new FoodTruck(15,100,10000,\"Mobile Food Court\",menu1);
//Displaying the Truck1 Information.
System.out.println(\"__________Truck 1 Details__________\");
System.out.println(truck1.getInfo());
//Creating an ArrayList Object which holds List of Menu Items
ArrayList menu2=new ArrayList();
//Adding menu items to the ArrayList
menu2.add(\"Tea\");
menu2.add(\"Coffe\");
menu2.add(\"Cool Drinks\");
//Creating FoodTruck Object by passing parameters.
Ionization energy is the energy required to remove theouter most ele.pdf
Ionization energy is the energy required to remove theouter most electron from the gaseous
atom.
The electronic configuration (E.C.) of O (8) is [He]2s2 2p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of P (15) is [Ne]3s2 3p3
The electronic configuration (E.C.) of Al (13) is [Ne]3s2 3p1
The electronic configuration (E.C.) of K(19) is [Ar]4s1
The electronic configuration (E.C.) of Na (11) is [Ne]3s1
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- from P is more than that of Alwhich in
ture is more than Na
4)Among P & S , P has half filled configuration of porbitals so more stable than S .Therefore
more energy for P thanS
So the overall order of the ionization energy is O >P > S > Al > Na > K
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- from P is more than that of Alwhich in
ture is more than Na
4)Among P & S , P has half filled configuration of porbitals so more stable than S .Therefore
more energy for P thanS
So the overall order of the ionization energy is O >P > S > Al > Na > K
Solution
Ionization energy is the energy required to remove theouter most electron from the gaseous
atom.
The electronic configuration (E.C.) of O (8) is [He]2s2 2p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of S (16) is [Ne]3s2 3p4
The electronic configuration (E.C.) of P (15) is [Ne]3s2 3p3
The electronic configuration (E.C.) of Al (13) is [Ne]3s2 3p1
The electronic configuration (E.C.) of K(19) is [Ar]4s1
The electronic configuration (E.C.) of Na (11) is [Ne]3s1
1)O & S belongs to same group so on moving from top tobottom atomc size increases ,so for the
removal of e- from S iseasier than O ==> ionization energy is more for O
2)Na &K belongs to same group so on moving from topto bottom atomc size increases ,so for the
removal of e- from K iseasier than Na ==> ionization energy is more for Na
3) Na, Al , & P beleong to the same period.On moving fromleft to right the atomic size will
decreases due the morenuclear charge.So to remove the e- .
D. Both C and N are hybridized sp3 .pdf
D. Both C and N are hybridized sp3
Solution
D. Both C and N are hybridized sp3.
The presence of conjugated Carbon-carbon double bonds causes colorat.pdf
The presence of conjugated Carbon-carbon double bonds causes coloration. When those bonds
are broken, the color fades.
Solution
The presence of conjugated Carbon-carbon double bonds causes coloration. When those bonds
are broken, the color fades..
The amount to be deposited at the start of his studies is found as.pdf
The amount to be deposited at the start of his studies is found as
PV = C [ 1 - ( 1+r)-2 / r ] x ( 1+r)-t
PV = $ 12,300 [ 1 - ( 1+0.06)-2 / 0.06 ] x( 1+0.06) -1
PV = 21,274
Solution
The amount to be deposited at the start of his studies is found as
PV = C [ 1 - ( 1+r)-2 / r ] x ( 1+r)-t
PV = $ 12,300 [ 1 - ( 1+0.06)-2 / 0.06 ] x( 1+0.06) -1
PV = 21,274.
SOA ( Service Oriented Architecture)SOA is a type of architectura.pdf
SOA ( Service Oriented Architecture):
SOA is a type of architectural pattern in computer software design where application
components provides services to other components via network.This removes the need for third
party product,vendor or technology.
The main functionality to this type of model is to make it easier for software components
connected over the internet to communicate with each other without human interaction.so every
computer can run any number of services and this can be archeived without changing the main
program itself.SOA provides a cost effective way to evolve and enhance Enterprise information
systems.
Web 2.0 on the other hand is world wide web websites that emphasize user-generated
content,interoperability for end users. Some have agreed that web 2.0 is the best form of
realizing SOA.
1)-2) Interoperability and portability:
SOA can be realized through the help of webservices, because the best characterstic of
webservices is the interoperability.Different distributed web services run on variety of software
platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples.
3) Platform independence:
For example take a case where XML,XML schema are there which are used to
publish,describe,search information. and there are multiple technologies and versions like SOAP,
WSDL, UDDI, and ebXML to do this. For the web services to be functional, we need an
environment of platform independent model.For example consider AJAX (asynchronous
javascript and XML) is not a technology, it is just a technique to createinteractive webpages for
this it supports XHTML,CSS,DOM,XML,XSLT,ATOM,SQL,REST,SOAP etc and uses
javascript to glue it all together.This is a perfect example of SOA present in current trend.
Solution
SOA ( Service Oriented Architecture):
SOA is a type of architectural pattern in computer software design where application
components provides services to other components via network.This removes the need for third
party product,vendor or technology.
The main functionality to this type of model is to make it easier for software components
connected over the internet to communicate with each other without human interaction.so every
computer can run any number of services and this can be archeived without changing the main
program itself.SOA provides a cost effective way to evolve and enhance Enterprise information
systems.
Web 2.0 on the other hand is world wide web websites that emphasize user-generated
content,interoperability for end users. Some have agreed that web 2.0 is the best form of
realizing SOA.
1)-2) Interoperability and portability:
SOA can be realized through the help of webservices, because the best characterstic of
webservices is the interoperability.Different distributed web services run on variety of software
platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples.
3) Platform independence:
For example take a case where XML,XML schema are there which are used to
publis.
P0 = 21.40D1 = 1.07D2 =1.1449D3 = 1.2250P3 = 26.22Growth r.pdf
P0 = 21.40
D1 = 1.07
D2 =1.1449
D3 = 1.2250
P3 = 26.22
Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round
to 7%.
Dividend yield = (d1/P0) = 1.07/21.40 = 0.05
Total return = 7% + 5% = 12%
Solution
P0 = 21.40
D1 = 1.07
D2 =1.1449
D3 = 1.2250
P3 = 26.22
Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round
to 7%.
Dividend yield = (d1/P0) = 1.07/21.40 = 0.05
Total return = 7% + 5% = 12%.
Nt = No er twhere density at time tNo= = starting densityNr=dN.pdf
Nt = No er t
where density at time t
No= = starting density
Nr=dN/dt so r=1/N*dN/dt
t =1
1):- Nt = No er t
if No =1
=1* e-.1*1
=9.047
2):- Nt = No er t
if No=3
=3* e0.1*1=1.105*3=3.3156
3):-Nt = No er t
if No=7=
7*e0.37*1=10.08
4):-Nt = No er t
if No=14
=14*e-0.3*1=10.36
Solution
Nt = No er t
where density at time t
No= = starting density
Nr=dN/dt so r=1/N*dN/dt
t =1
1):- Nt = No er t
if No =1
=1* e-.1*1
=9.047
2):- Nt = No er t
if No=3
=3* e0.1*1=1.105*3=3.3156
3):-Nt = No er t
if No=7=
7*e0.37*1=10.08
4):-Nt = No er t
if No=14
=14*e-0.3*1=10.36.
Normalization is necessay step in creation of database design becaus.pdf
Normalization is necessay step in creation of database design because:
Solution
Normalization is necessay step in creation of database design because:.
Hi, Please find my codeimport java.util.Random;public class Pro.pdf
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second array
public int[][] getFirstArray() { return firstArray;}
public int[][] getSecondArray() { return secondArray;}
public ProcessArray(int rows, int columns){ //Constructor of object ProcessArray
this.rows = rows;
this.columns = columns;
int[][] array = new int[rows][columns];
initializeArray(array);
randomlyFillArray();
computeArrayValues();
printArray(secondArray);
}
public void initializeArray(int[][] array){ //Initializes first and second arrays and sets each value
to 0
firstArray = new int[rows][columns];
secondArray = new int[rows][columns];
}
public void randomlyFillArray(){ //Fills first array with random numbers
for(int i = 0; i < firstArray.length;i++){
for(int j = 0; j < firstArray[0].length;j++){
Random r = new Random();
int num = r.nextInt(16);
firstArray[i][j] = num;
}
}
}
public void computeArrayValues(){
int col = firstArray[0].length;
int row = firstArray.length;
for(int i = 0; i < row; i++){
for(int j = 0; j< col; j++){
secondArray[i][j] = 0;
if((i - 1) >=0){
secondArray[i][j] += firstArray[i-1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i-1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i-1][j-1];
}
if((j+1) < col)
secondArray[i][j] += firstArray[i][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i][j-1];
if((i+1) < row){
secondArray[i][j] += firstArray[i+1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i+1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i+1][j-1];
}
}
}
}
public void printArray(int[][] Array){
System.out.println(\"\ Initial Array Filled With Random Numbers: \ \");
for(int a = 0; a < firstArray.length; a++){
for(int b = 0; b < firstArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", firstArray[a][b]);
else
System.out.printf(\"%d \", firstArray[a][b]);
}System.out.println();
}System.out.println();
System.out.println(\"Computed Array: \ \");
for(int a = 0; a < secondArray.length; a++){
for(int b = 0; b < secondArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", secondArray[a][b]);
else
System.out.printf(\"%d \", secondArray[a][b]);
}
System.out.println();
}System.out.println();
}
public static void main(String[] args) {
ProcessArray pr = new ProcessArray(3, 4);
}
}
/*
Sample Output:
Initial Array Filled With Random Numbers:
7 3 6 12
10 2 11 10
12 14 1 7
Computed Array:
15 36 38 27
38 64 55 37
26 36 44 22
*/
Solution
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second arr.
For sound intensity,ratio in dB is given byx = 10log(IIo)We.pdf
For sound intensity,
ratio in dB is given by:
x = 10*log(I/Io)
We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D
So, x = 10*log(D^2/(D+d)^2)
where D = original distance from speaker
d = distance he has moved from original distance from speaker
So, -14 = 10*log((D/(D+d))^2)
So, (D+d)/D = 5.01
So, 1 + d/D = 5.01
So, d/D = 4.01
So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D
distance away from speaker
So, answer is : 5.01*D
Solution
For sound intensity,
ratio in dB is given by:
x = 10*log(I/Io)
We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D
So, x = 10*log(D^2/(D+d)^2)
where D = original distance from speaker
d = distance he has moved from original distance from speaker
So, -14 = 10*log((D/(D+d))^2)
So, (D+d)/D = 5.01
So, 1 + d/D = 5.01
So, d/D = 4.01
So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D
distance away from speaker
So, answer is : 5.01*D.
Exp- Listeria monocytogenes is a facultative anaerobic bacteria tha.pdf
Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic
and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo
fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of
infection is ingestion of contaminated food.
Solution
Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic
and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo
fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of
infection is ingestion of contaminated food..
DNA controversy is a dispute about whether Rosalind Franklin and Wil.pdf
DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper
credit for their contribution to the determination of the structure of DNA. In January 1951,
Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she
worked as a research associate at King\'s College London in the Medical Research Council\'s
Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson
and Crick, without her knowledge. Her experimental results provided estimates of the water
content of DNA crystals and these reults were consistent with the two sugar-phosphate
backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that
hte X-ray diffraction images collected where the best evidence for the helical nature od DNA.
Crick and Waatson had been to the public seminar presented by Franklin, where they had spoken
to Wilkins about offering him a co-authorship on the article that first described the double helix
structure of DNA. On the completion of their model, Crick and Watson had invited Wilkins to be
a co-author of the paper describing the structure, he turned down this offer as he had taken no
part in building the model. Watson, Francis Crick and Maurice Wilkins were awarded the 1962
Nobel Prize for Physiology or Medicine for their discoveries concerning the molecular structure
of nucleic acids and its significance for information transfer in living material, with the help of
the research of Rosalind Franklin.
There is a little doubt that the significance of Franklin\'s data should have been acknowledged-
the evidence is overwhelming that it played an important role in Watson and Crick\'s modelling
process. Franklin could not share in the 1962 Nobel prize awarded to Watson and Crick because
she had died of ovarian cancer in 1958. Hence, it cannot be concluded that Watson and Crick
were not ethical about their research and also that they donot deserve a nobel prize for their
discovery. And the Historical records contain all the data of the work presented by Franklin and
also by Watson and Crick.
Solution
DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper
credit for their contribution to the determination of the structure of DNA. In January 1951,
Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she
worked as a research associate at King\'s College London in the Medical Research Council\'s
Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson
and Crick, without her knowledge. Her experimental results provided estimates of the water
content of DNA crystals and these reults were consistent with the two sugar-phosphate
backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that
hte X-ray diffraction images collected where the best evidence for the helical nature od DNA.
Crick and Waatson had .
Delta is a measures of the change in price of an option for a one po.pdf
Delta is a measures of the change in price of an option for a one point move in the underlying
asset. Delta measure the price sensitivity of underlying assets.
Option (A) is correct answer.
Solution
Delta is a measures of the change in price of an option for a one point move in the underlying
asset. Delta measure the price sensitivity of underlying assets.
Option (A) is correct answer..
consider the cross between White petals and dark blue petalsFlower.pdf
consider the cross between White petals and dark blue petals
Flower color in sweet peas.
Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied
the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was
9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in
first-generation had the purple flowers. And from crossing of these hybrids was obtained the
offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose,
that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So
obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes,
which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the
pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively
genotype \"CcPp\". The mechanism of genetic interaction of these genes has become known
recently. It was found that the color of sweet pea flowers depends on the synthesis of pigments -
anthocyanins. The synthesis of anthocyanins occurs in two stages. Gene \"C\" is responsible for
the first stage of the synthesis, while the gene \"P\" for second, and even if only one dominant
allele from these genes is lack, then synthesis of anthocyanins is not occurs. That is to say, that
the presence of a recessive allele in the homozygous state in the genotype of either of the two
genes is blocks the development of trait of flowers coloration. Such interaction of nonallelic
genes is called a double recessive epistasis. We can describe this interaction like this - recessive
allele of each gene in the homozygous state suppresses the expression of a dominant allele of
another ( \"cc\" suppresses \"P\", \"pp\" suppresses \"C\" ). Genetic traits file for this case should
be like this :
Solution
consider the cross between White petals and dark blue petals
Flower color in sweet peas.
Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied
the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was
9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in
first-generation had the purple flowers. And from crossing of these hybrids was obtained the
offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose,
that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So
obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes,
which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the
pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively
genotype \"CcPp\". The mechanism of genetic interaction of these genes has bec.
c. Aggregation is an element of sensitivity where data is summarized.pdf
c. Aggregation is an element of sensitivity where data is summarized to mask the details of a
subject matter. The other three choices are the goals of IT security.
Solution
c. Aggregation is an element of sensitivity where data is summarized to mask the details of a
subject matter. The other three choices are the goals of IT security..
Benfits, costs, and discount factors for calculating NPVCategory.pdf
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Solution
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
A.
B. Statistical based IDS compares normal to abnormal activity. Patte.pdf
B. Statistical based IDS compares normal to abnormal activity. Pattern, traffic, and protocol do
not therefore answers A, C, and D are incorrect.
Solution
B. Statistical based IDS compares normal to abnormal activity. Pattern, traffic, and protocol do
not therefore answers A, C, and D are incorrect..
3. more H2O(g) and CO(g) to form. as the equilib.pdf
3. more H2O(g) and CO(g) to form. as the equilibrium will shift towards right side
Solution
3. more H2O(g) and CO(g) to form. as the equilibrium will shift towards right side.
Answer .Ltext0 .section .rodata .LC0 0000.pdf
Answer:
.Ltext0:
.section .rodata
.LC0:
0000 0A0A0949 .string \"\ \ \\tIN DECIMAL FORM: %d\"
.text
.globl convert1
convert1:
.LFB0:
.cfi_startproc
0000 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
0001 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
0004 4883EC30 subq $48, %rsp
0008 48897DD8 movq %rdi, -40(%rbp)
000c 488975D0 movq %rsi, -48(%rbp)
0010 C745F800 movl $0, -8(%rbp)
0017 C745FC01 movl $1, -4(%rbp)
001e C745F020 movl $32, -16(%rbp)
0025 C745EC0F movl $15, -20(%rbp)
002c EB48 jmp .L2
.L6:
002e 836DF001 subl $1, -16(%rbp)
0032 8B45EC movl -20(%rbp), %eax
0035 4898 cltq
0037 488D1485 leaq 0(,%rax,4), %rdx
003f 488B45D8 movq -40(%rbp), %rax
0043 4801D0 addq %rdx, %rax
0046 8B00 movl (%rax), %eax
0048 83F801 cmpl $1, %eax
004b 7525 jne .L3
004d C745FC01 movl $1, -4(%rbp)
0054 C745F401 movl $1, -12(%rbp)
005b EB07 jmp .L4
.L5:
0060 8345F401 addl $1, -12(%rbp)
005d D165FC sall -4(%rbp)
.L4:
0064 8B45F4 movl -12(%rbp), %eax
0067 3B45F0 cmpl -16(%rbp), %eax
006a 7EF1 jle .L5
006c 8B45FC movl -4(%rbp), %eax
006f 0145F8 addl %eax, -8(%rbp)
.L3:
0072 836DEC01 subl $1, -20(%rbp)
.L2:
0076 837DEC00 cmpl $0, -20(%rbp)
007a 79B2 jns .L6
007c C745FC01 movl $1, -4(%rbp)
0083 C745EC0F movl $15, -20(%rbp)
008a EB44 jmp .L7
.L11:
008c 8B45EC movl -20(%rbp), %eax
008f 4898 cltq
0091 488D1485 leaq 0(,%rax,4), %rdx
0099 488B45D0 movq -48(%rbp), %rax
009d 4801D0 addq %rdx, %rax
00a0 8B00 movl (%rax), %eax
00a2 83F801 cmpl $1, %eax
00a5 7525 jne .L8
00a7 C745FC01 movl $1, -4(%rbp)
00ae C745F401 movl $1, -12(%rbp)
00b5 EB07 jmp .L9
.L10:
00ba 8345F401 addl $1, -12(%rbp)
00b7 D165FC sall -4(%rbp)
.L9:
00be 8B45F4 movl -12(%rbp), %eax
00c1 3B45EC cmpl -20(%rbp), %eax
00c4 7EF1 jle .L10
00c6 8B45FC movl -4(%rbp), %eax
00c9 0145F8 addl %eax, -8(%rbp)
.L8:
00cc 836DEC01 subl $1, -20(%rbp)
.L7:
00d0 837DEC00 cmpl $0, -20(%rbp)
00d4 79B6 jns .L11
00d6 8B45F8 movl -8(%rbp), %eax
00d9 89C6 movl %eax, %esi
00db BF000000 movl $.LC0, %edi
00e0 B8000000 movl $0, %eax
00e5 E8000000 call printf
00ea 90 nop
00eb C9 leave
.cfi_def_cfa 7, 8
00ec C3 ret
.cfi_endproc
.LFE0:
.globl convert2
convert2:
.LFB1:
.cfi_startproc
00ed 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
00ee 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
00f1 897DEC movl %edi, -20(%rbp)
00f4 488975E0 movq %rsi, -32(%rbp)
00f8 C745F800 movl $0, -8(%rbp)
000000
00ff 8B45EC movl -20(%rbp), %eax
0102 8945FC movl %eax, -4(%rbp)
0105 EB3F jmp .L13
.L15:
0107 8B45FC movl -4(%rbp), %eax
010a 99 cltd
010b C1EA1F shrl $31, %edx
010e 01D0 addl %edx, %eax
0110 83E001 andl $1, %eax
0113 29D0 subl %edx, %eax
0115 8945EC movl %eax, -20(%rbp)
0118 8B45F8 movl -8(%rbp), %eax
011b 8D5001 leal 1(%rax), %edx
011e 8955F8 movl %edx, -8(%rbp)
0121 4898 cltq
0123 488D1485 leaq 0(,%rax,4), %rdx
012b 488B45E0 movq -32(%rbp), %rax
012f 4801C2 addq %rax, %rdx
0132 8B45EC movl -20(%rbp), %eax
0135 8902 movl %eax, (%rdx)
0137 8B45FC movl -4(%rbp), %eax
013a 89C2 movl %eax, %edx
013c C1EA1F shrl $31, %edx
013f 01D0 addl .
Ans. Cotical reaction prevents more than one sperm from binding to a.pdf
Ans. Cotical reaction prevents more than one sperm from binding to an oocyte during
fertilization avoiding the situation of polyspermy.
Solution
Ans. Cotical reaction prevents more than one sperm from binding to an oocyte during
fertilization avoiding the situation of polyspermy..
ans 1net assets of the fund= total assets less outstanding expense.pdf
ans 1
net assets of the fund= total assets less outstanding expenses= 200-20 = $180m
NAV= net assets / number of units outstanding= $180m/20m shares= $9 per share
ans 2
Solution
ans 1
net assets of the fund= total assets less outstanding expenses= 200-20 = $180m
NAV= net assets / number of units outstanding= $180m/20m shares= $9 per share
ans 2.
Addition theorem of probability for two events states thatSoluti.pdf
Addition theorem of probability for two events states that
Solution
Addition theorem of probability for two events states that.
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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SOA ( Service Oriented Architecture)SOA is a type of architectura.pdf
SOA ( Service Oriented Architecture):
SOA is a type of architectural pattern in computer software design where application
components provides services to other components via network.This removes the need for third
party product,vendor or technology.
The main functionality to this type of model is to make it easier for software components
connected over the internet to communicate with each other without human interaction.so every
computer can run any number of services and this can be archeived without changing the main
program itself.SOA provides a cost effective way to evolve and enhance Enterprise information
systems.
Web 2.0 on the other hand is world wide web websites that emphasize user-generated
content,interoperability for end users. Some have agreed that web 2.0 is the best form of
realizing SOA.
1)-2) Interoperability and portability:
SOA can be realized through the help of webservices, because the best characterstic of
webservices is the interoperability.Different distributed web services run on variety of software
platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples.
3) Platform independence:
For example take a case where XML,XML schema are there which are used to
publish,describe,search information. and there are multiple technologies and versions like SOAP,
WSDL, UDDI, and ebXML to do this. For the web services to be functional, we need an
environment of platform independent model.For example consider AJAX (asynchronous
javascript and XML) is not a technology, it is just a technique to createinteractive webpages for
this it supports XHTML,CSS,DOM,XML,XSLT,ATOM,SQL,REST,SOAP etc and uses
javascript to glue it all together.This is a perfect example of SOA present in current trend.
Solution
SOA ( Service Oriented Architecture):
SOA is a type of architectural pattern in computer software design where application
components provides services to other components via network.This removes the need for third
party product,vendor or technology.
The main functionality to this type of model is to make it easier for software components
connected over the internet to communicate with each other without human interaction.so every
computer can run any number of services and this can be archeived without changing the main
program itself.SOA provides a cost effective way to evolve and enhance Enterprise information
systems.
Web 2.0 on the other hand is world wide web websites that emphasize user-generated
content,interoperability for end users. Some have agreed that web 2.0 is the best form of
realizing SOA.
1)-2) Interoperability and portability:
SOA can be realized through the help of webservices, because the best characterstic of
webservices is the interoperability.Different distributed web services run on variety of software
platforms and hardware architectures.JAVA j2EE and java WSDP are the best examples.
3) Platform independence:
For example take a case where XML,XML schema are there which are used to
publis.
P0 = 21.40D1 = 1.07D2 =1.1449D3 = 1.2250P3 = 26.22Growth r.pdf
P0 = 21.40
D1 = 1.07
D2 =1.1449
D3 = 1.2250
P3 = 26.22
Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round
to 7%.
Dividend yield = (d1/P0) = 1.07/21.40 = 0.05
Total return = 7% + 5% = 12%
Solution
P0 = 21.40
D1 = 1.07
D2 =1.1449
D3 = 1.2250
P3 = 26.22
Growth rate in dividends: PV= 1.07; FV = 1.2240; N = 2; CPT I/Y; I/Y = 6.9544% Let\'s round
to 7%.
Dividend yield = (d1/P0) = 1.07/21.40 = 0.05
Total return = 7% + 5% = 12%.
Nt = No er twhere density at time tNo= = starting densityNr=dN.pdf
Nt = No er t
where density at time t
No= = starting density
Nr=dN/dt so r=1/N*dN/dt
t =1
1):- Nt = No er t
if No =1
=1* e-.1*1
=9.047
2):- Nt = No er t
if No=3
=3* e0.1*1=1.105*3=3.3156
3):-Nt = No er t
if No=7=
7*e0.37*1=10.08
4):-Nt = No er t
if No=14
=14*e-0.3*1=10.36
Solution
Nt = No er t
where density at time t
No= = starting density
Nr=dN/dt so r=1/N*dN/dt
t =1
1):- Nt = No er t
if No =1
=1* e-.1*1
=9.047
2):- Nt = No er t
if No=3
=3* e0.1*1=1.105*3=3.3156
3):-Nt = No er t
if No=7=
7*e0.37*1=10.08
4):-Nt = No er t
if No=14
=14*e-0.3*1=10.36.
Normalization is necessay step in creation of database design becaus.pdf
Normalization is necessay step in creation of database design because:
Solution
Normalization is necessay step in creation of database design because:.
Hi, Please find my codeimport java.util.Random;public class Pro.pdf
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second array
public int[][] getFirstArray() { return firstArray;}
public int[][] getSecondArray() { return secondArray;}
public ProcessArray(int rows, int columns){ //Constructor of object ProcessArray
this.rows = rows;
this.columns = columns;
int[][] array = new int[rows][columns];
initializeArray(array);
randomlyFillArray();
computeArrayValues();
printArray(secondArray);
}
public void initializeArray(int[][] array){ //Initializes first and second arrays and sets each value
to 0
firstArray = new int[rows][columns];
secondArray = new int[rows][columns];
}
public void randomlyFillArray(){ //Fills first array with random numbers
for(int i = 0; i < firstArray.length;i++){
for(int j = 0; j < firstArray[0].length;j++){
Random r = new Random();
int num = r.nextInt(16);
firstArray[i][j] = num;
}
}
}
public void computeArrayValues(){
int col = firstArray[0].length;
int row = firstArray.length;
for(int i = 0; i < row; i++){
for(int j = 0; j< col; j++){
secondArray[i][j] = 0;
if((i - 1) >=0){
secondArray[i][j] += firstArray[i-1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i-1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i-1][j-1];
}
if((j+1) < col)
secondArray[i][j] += firstArray[i][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i][j-1];
if((i+1) < row){
secondArray[i][j] += firstArray[i+1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i+1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i+1][j-1];
}
}
}
}
public void printArray(int[][] Array){
System.out.println(\"\ Initial Array Filled With Random Numbers: \ \");
for(int a = 0; a < firstArray.length; a++){
for(int b = 0; b < firstArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", firstArray[a][b]);
else
System.out.printf(\"%d \", firstArray[a][b]);
}System.out.println();
}System.out.println();
System.out.println(\"Computed Array: \ \");
for(int a = 0; a < secondArray.length; a++){
for(int b = 0; b < secondArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", secondArray[a][b]);
else
System.out.printf(\"%d \", secondArray[a][b]);
}
System.out.println();
}System.out.println();
}
public static void main(String[] args) {
ProcessArray pr = new ProcessArray(3, 4);
}
}
/*
Sample Output:
Initial Array Filled With Random Numbers:
7 3 6 12
10 2 11 10
12 14 1 7
Computed Array:
15 36 38 27
38 64 55 37
26 36 44 22
*/
Solution
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second arr.
For sound intensity,ratio in dB is given byx = 10log(IIo)We.pdf
For sound intensity,
ratio in dB is given by:
x = 10*log(I/Io)
We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D
So, x = 10*log(D^2/(D+d)^2)
where D = original distance from speaker
d = distance he has moved from original distance from speaker
So, -14 = 10*log((D/(D+d))^2)
So, (D+d)/D = 5.01
So, 1 + d/D = 5.01
So, d/D = 4.01
So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D
distance away from speaker
So, answer is : 5.01*D
Solution
For sound intensity,
ratio in dB is given by:
x = 10*log(I/Io)
We know, I = P/(4*pi*D^2) <----- I is inversely proportional to D
So, x = 10*log(D^2/(D+d)^2)
where D = original distance from speaker
d = distance he has moved from original distance from speaker
So, -14 = 10*log((D/(D+d))^2)
So, (D+d)/D = 5.01
So, 1 + d/D = 5.01
So, d/D = 4.01
So, d = 4.01*D <-------- So, he is about 4.01 times D away from original distance or 5.01*D
distance away from speaker
So, answer is : 5.01*D.
Exp- Listeria monocytogenes is a facultative anaerobic bacteria tha.pdf
Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic
and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo
fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of
infection is ingestion of contaminated food.
Solution
Exp:- Listeria monocytogenes is a facultative anaerobic bacteria that can survive both aerobic
and anaerobic conditions and is responsible for listeriosis disease. The bacterium can undergo
fermentation in absence of oxygen and normal glycolysis and Krebs cycle.The main route of
infection is ingestion of contaminated food..
DNA controversy is a dispute about whether Rosalind Franklin and Wil.pdf
DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper
credit for their contribution to the determination of the structure of DNA. In January 1951,
Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she
worked as a research associate at King\'s College London in the Medical Research Council\'s
Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson
and Crick, without her knowledge. Her experimental results provided estimates of the water
content of DNA crystals and these reults were consistent with the two sugar-phosphate
backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that
hte X-ray diffraction images collected where the best evidence for the helical nature od DNA.
Crick and Waatson had been to the public seminar presented by Franklin, where they had spoken
to Wilkins about offering him a co-authorship on the article that first described the double helix
structure of DNA. On the completion of their model, Crick and Watson had invited Wilkins to be
a co-author of the paper describing the structure, he turned down this offer as he had taken no
part in building the model. Watson, Francis Crick and Maurice Wilkins were awarded the 1962
Nobel Prize for Physiology or Medicine for their discoveries concerning the molecular structure
of nucleic acids and its significance for information transfer in living material, with the help of
the research of Rosalind Franklin.
There is a little doubt that the significance of Franklin\'s data should have been acknowledged-
the evidence is overwhelming that it played an important role in Watson and Crick\'s modelling
process. Franklin could not share in the 1962 Nobel prize awarded to Watson and Crick because
she had died of ovarian cancer in 1958. Hence, it cannot be concluded that Watson and Crick
were not ethical about their research and also that they donot deserve a nobel prize for their
discovery. And the Historical records contain all the data of the work presented by Franklin and
also by Watson and Crick.
Solution
DNA controversy is a dispute about whether Rosalind Franklin and Wilkins were given proper
credit for their contribution to the determination of the structure of DNA. In January 1951,
Franklin discovered that there were two forms of DNA by using x-ray diffraction techniques, she
worked as a research associate at King\'s College London in the Medical Research Council\'s
Biophysics Unit. The controversy arose from claims that some of the data were shown to Watson
and Crick, without her knowledge. Her experimental results provided estimates of the water
content of DNA crystals and these reults were consistent with the two sugar-phosphate
backbones being on the outside of the molecule. Franklin personallytold Crick and Watson that
hte X-ray diffraction images collected where the best evidence for the helical nature od DNA.
Crick and Waatson had .
Delta is a measures of the change in price of an option for a one po.pdf
Delta is a measures of the change in price of an option for a one point move in the underlying
asset. Delta measure the price sensitivity of underlying assets.
Option (A) is correct answer.
Solution
Delta is a measures of the change in price of an option for a one point move in the underlying
asset. Delta measure the price sensitivity of underlying assets.
Option (A) is correct answer..
consider the cross between White petals and dark blue petalsFlower.pdf
consider the cross between White petals and dark blue petals
Flower color in sweet peas.
Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied
the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was
9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in
first-generation had the purple flowers. And from crossing of these hybrids was obtained the
offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose,
that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So
obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes,
which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the
pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively
genotype \"CcPp\". The mechanism of genetic interaction of these genes has become known
recently. It was found that the color of sweet pea flowers depends on the synthesis of pigments -
anthocyanins. The synthesis of anthocyanins occurs in two stages. Gene \"C\" is responsible for
the first stage of the synthesis, while the gene \"P\" for second, and even if only one dominant
allele from these genes is lack, then synthesis of anthocyanins is not occurs. That is to say, that
the presence of a recessive allele in the homozygous state in the genotype of either of the two
genes is blocks the development of trait of flowers coloration. Such interaction of nonallelic
genes is called a double recessive epistasis. We can describe this interaction like this - recessive
allele of each gene in the homozygous state suppresses the expression of a dominant allele of
another ( \"cc\" suppresses \"P\", \"pp\" suppresses \"C\" ). Genetic traits file for this case should
be like this :
Solution
consider the cross between White petals and dark blue petals
Flower color in sweet peas.
Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied
the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was
9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in
first-generation had the purple flowers. And from crossing of these hybrids was obtained the
offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose,
that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So
obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes,
which interact each other. For convenience, we mark these genes as \"C\" and \"P\". Thus, the
pure lines will have genotypes \"CCpp\" and \"ccPP\", and first generation hybrids respectively
genotype \"CcPp\". The mechanism of genetic interaction of these genes has bec.
c. Aggregation is an element of sensitivity where data is summarized.pdf
c. Aggregation is an element of sensitivity where data is summarized to mask the details of a
subject matter. The other three choices are the goals of IT security.
Solution
c. Aggregation is an element of sensitivity where data is summarized to mask the details of a
subject matter. The other three choices are the goals of IT security..
Benfits, costs, and discount factors for calculating NPVCategory.pdf
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Solution
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
Annual Expenses
($321,000)
($321,000)
($321,000)
($321,000)
($321,000)
Annual operational costs
$5,000
$5,000
$5,500
$5,500
$7,000
$8,000
4
Net benefits/costs
($1,581,000)
$725,000
$753,000
$783,240
$815,699
$851,171
5
Discount factor
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
6
Net Present value
($292,940.04)
$723,987.26
$770,797.68
$818,146.69
$869,562.18
$922,069.54
7
Cummulative NPV
(292,940.04)
$431,047
$1,201,845
$2,019,992
$2,889,554
$3,811,623
8
Payback period
1 year + 292940.40 /(292940.40 + 431047) = 0.40 or 1 year + 146 days(.40*365)
Benfits, costs, and discount factors for calculating NPV
Category
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
1
Benefits value
$1,046,000
$1,074,000
$1,104,240
$1,136,899
$1,172,171
yearly Benefits
$55,000
$60,000
$70,000
$75,000
$80,000
$80,000
2
Development Costs
($1,581,000)
($225,000)
($225,000)
($225,000)
($225,000)
($225,000)
3
A.
B. Statistical based IDS compares normal to abnormal activity. Patte.pdf
B. Statistical based IDS compares normal to abnormal activity. Pattern, traffic, and protocol do
not therefore answers A, C, and D are incorrect.
Solution
B. Statistical based IDS compares normal to abnormal activity. Pattern, traffic, and protocol do
not therefore answers A, C, and D are incorrect..
3. more H2O(g) and CO(g) to form. as the equilib.pdf
3. more H2O(g) and CO(g) to form. as the equilibrium will shift towards right side
Solution
3. more H2O(g) and CO(g) to form. as the equilibrium will shift towards right side.
Answer .Ltext0 .section .rodata .LC0 0000.pdf
Answer:
.Ltext0:
.section .rodata
.LC0:
0000 0A0A0949 .string \"\ \ \\tIN DECIMAL FORM: %d\"
.text
.globl convert1
convert1:
.LFB0:
.cfi_startproc
0000 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
0001 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
0004 4883EC30 subq $48, %rsp
0008 48897DD8 movq %rdi, -40(%rbp)
000c 488975D0 movq %rsi, -48(%rbp)
0010 C745F800 movl $0, -8(%rbp)
0017 C745FC01 movl $1, -4(%rbp)
001e C745F020 movl $32, -16(%rbp)
0025 C745EC0F movl $15, -20(%rbp)
002c EB48 jmp .L2
.L6:
002e 836DF001 subl $1, -16(%rbp)
0032 8B45EC movl -20(%rbp), %eax
0035 4898 cltq
0037 488D1485 leaq 0(,%rax,4), %rdx
003f 488B45D8 movq -40(%rbp), %rax
0043 4801D0 addq %rdx, %rax
0046 8B00 movl (%rax), %eax
0048 83F801 cmpl $1, %eax
004b 7525 jne .L3
004d C745FC01 movl $1, -4(%rbp)
0054 C745F401 movl $1, -12(%rbp)
005b EB07 jmp .L4
.L5:
0060 8345F401 addl $1, -12(%rbp)
005d D165FC sall -4(%rbp)
.L4:
0064 8B45F4 movl -12(%rbp), %eax
0067 3B45F0 cmpl -16(%rbp), %eax
006a 7EF1 jle .L5
006c 8B45FC movl -4(%rbp), %eax
006f 0145F8 addl %eax, -8(%rbp)
.L3:
0072 836DEC01 subl $1, -20(%rbp)
.L2:
0076 837DEC00 cmpl $0, -20(%rbp)
007a 79B2 jns .L6
007c C745FC01 movl $1, -4(%rbp)
0083 C745EC0F movl $15, -20(%rbp)
008a EB44 jmp .L7
.L11:
008c 8B45EC movl -20(%rbp), %eax
008f 4898 cltq
0091 488D1485 leaq 0(,%rax,4), %rdx
0099 488B45D0 movq -48(%rbp), %rax
009d 4801D0 addq %rdx, %rax
00a0 8B00 movl (%rax), %eax
00a2 83F801 cmpl $1, %eax
00a5 7525 jne .L8
00a7 C745FC01 movl $1, -4(%rbp)
00ae C745F401 movl $1, -12(%rbp)
00b5 EB07 jmp .L9
.L10:
00ba 8345F401 addl $1, -12(%rbp)
00b7 D165FC sall -4(%rbp)
.L9:
00be 8B45F4 movl -12(%rbp), %eax
00c1 3B45EC cmpl -20(%rbp), %eax
00c4 7EF1 jle .L10
00c6 8B45FC movl -4(%rbp), %eax
00c9 0145F8 addl %eax, -8(%rbp)
.L8:
00cc 836DEC01 subl $1, -20(%rbp)
.L7:
00d0 837DEC00 cmpl $0, -20(%rbp)
00d4 79B6 jns .L11
00d6 8B45F8 movl -8(%rbp), %eax
00d9 89C6 movl %eax, %esi
00db BF000000 movl $.LC0, %edi
00e0 B8000000 movl $0, %eax
00e5 E8000000 call printf
00ea 90 nop
00eb C9 leave
.cfi_def_cfa 7, 8
00ec C3 ret
.cfi_endproc
.LFE0:
.globl convert2
convert2:
.LFB1:
.cfi_startproc
00ed 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
00ee 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
00f1 897DEC movl %edi, -20(%rbp)
00f4 488975E0 movq %rsi, -32(%rbp)
00f8 C745F800 movl $0, -8(%rbp)
000000
00ff 8B45EC movl -20(%rbp), %eax
0102 8945FC movl %eax, -4(%rbp)
0105 EB3F jmp .L13
.L15:
0107 8B45FC movl -4(%rbp), %eax
010a 99 cltd
010b C1EA1F shrl $31, %edx
010e 01D0 addl %edx, %eax
0110 83E001 andl $1, %eax
0113 29D0 subl %edx, %eax
0115 8945EC movl %eax, -20(%rbp)
0118 8B45F8 movl -8(%rbp), %eax
011b 8D5001 leal 1(%rax), %edx
011e 8955F8 movl %edx, -8(%rbp)
0121 4898 cltq
0123 488D1485 leaq 0(,%rax,4), %rdx
012b 488B45E0 movq -32(%rbp), %rax
012f 4801C2 addq %rax, %rdx
0132 8B45EC movl -20(%rbp), %eax
0135 8902 movl %eax, (%rdx)
0137 8B45FC movl -4(%rbp), %eax
013a 89C2 movl %eax, %edx
013c C1EA1F shrl $31, %edx
013f 01D0 addl .
Ans. Cotical reaction prevents more than one sperm from binding to a.pdf
Ans. Cotical reaction prevents more than one sperm from binding to an oocyte during
fertilization avoiding the situation of polyspermy.
Solution
Ans. Cotical reaction prevents more than one sperm from binding to an oocyte during
fertilization avoiding the situation of polyspermy..
ans 1net assets of the fund= total assets less outstanding expense.pdf
ans 1
net assets of the fund= total assets less outstanding expenses= 200-20 = $180m
NAV= net assets / number of units outstanding= $180m/20m shares= $9 per share
ans 2
Solution
ans 1
net assets of the fund= total assets less outstanding expenses= 200-20 = $180m
NAV= net assets / number of units outstanding= $180m/20m shares= $9 per share
ans 2.
Addition theorem of probability for two events states thatSoluti.pdf
Addition theorem of probability for two events states that
Solution
Addition theorem of probability for two events states that.
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This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
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