pH = pKa + log (Sa) where S =moles of salt a = m.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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..
Vehicle.javapublic class Vehicle { Declaring instance var.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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- .
The presence of conjugated Carbon-carbon double bonds causes colorat.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
SolutionNet cash provided by operating activities for the year end.pdfanujsharmaanuj14
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.
pH = pKa + log (Sa) where S =moles of salt a = m.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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..
Vehicle.javapublic class Vehicle { Declaring instance var.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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- .
The presence of conjugated Carbon-carbon double bonds causes colorat.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
SolutionNet cash provided by operating activities for the year end.pdfanujsharmaanuj14
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.
SOA ( Service Oriented Architecture)SOA is a type of architectura.pdfanujsharmaanuj14
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.
Nt = No er twhere density at time tNo= = starting densityNr=dN.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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..
B. Statistical based IDS compares normal to abnormal activity. Patte.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
Ans. Cotical reaction prevents more than one sperm from binding to a.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
SOA ( Service Oriented Architecture)SOA is a type of architectura.pdfanujsharmaanuj14
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.
Nt = No er twhere density at time tNo= = starting densityNr=dN.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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..
B. Statistical based IDS compares normal to abnormal activity. Patte.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
Ans. Cotical reaction prevents more than one sperm from binding to a.pdfanujsharmaanuj14
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.pdfanujsharmaanuj14
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.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
A Strategic Approach: GenAI in EducationPeter Windle
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.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Embracing GenAI - A Strategic ImperativePeter Windle
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.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.