pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pO.pdf
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pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pOH = 14 - 1.21 = 12.79 Solution pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pOH = 14 - 1.21 = 12.79.
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![pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pOH = 14 - 1.21 = 12.79
Solution
pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pOH = 14 - 1.21 = 12.79](https://image.slidesharecdn.com/poh-logoh-log0-230412010218-75361cc2/85/pOH-log-OH-log-0-061-1-21-pH-14-pO-pdf-1-320.jpg)
Recommended
The key here is that the relative rates of appearancedisappearance .pdf
The key here is that the relative rates of appearance/disappearance of the materials is directly
related to their coefficients in the balanced equation. In other words, since each IO3(-) ion yields
3 I2 molecules, the rate of appearance of I2 must be 3 times the rate of disappearance of IO3(-).
Thus, the rate of appearance of I2 = 2.2 * 10^-3 M/s * 3 Rate of appearance of I2 = 6.6 * 10^-3
mol/(L*s)
Solution
The key here is that the relative rates of appearance/disappearance of the materials is directly
related to their coefficients in the balanced equation. In other words, since each IO3(-) ion yields
3 I2 molecules, the rate of appearance of I2 must be 3 times the rate of disappearance of IO3(-).
Thus, the rate of appearance of I2 = 2.2 * 10^-3 M/s * 3 Rate of appearance of I2 = 6.6 * 10^-3
mol/(L*s).
the molar ratio of C with O = 11, However, the a.pdf
the molar ratio of C with O = 1:1, However, the atomic mass of C = 12 , while the
atomic mass of O =16 so the percentage of O in CO = mass of O / mass of CO = 16/ (12+16) =
16/28 = 57.1%
Solution
the molar ratio of C with O = 1:1, However, the atomic mass of C = 12 , while the
atomic mass of O =16 so the percentage of O in CO = mass of O / mass of CO = 16/ (12+16) =
16/28 = 57.1%.
Ne and F2 are both uncharged molecules so there c.pdf
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here
Solution
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here.
x = 10 + (1p^2) Elasticity , E = (dxdp)(px) .pdf
x = 10 + (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p
= 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526
Solution
x = 10 + (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p
= 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526.
The second phenol can resonance stabilize once th.pdf
The second phenol can resonance stabilize once the H has been removed from the
O. The negative charge on the O would come down and create a double bond, the other double
bond would move over counterclockwise, the next double bond will move towards the nitrogen,
and finally the double bond will go from the N-O bond onto the O, giving it a negative charge.
Solution
The second phenol can resonance stabilize once the H has been removed from the
O. The negative charge on the O would come down and create a double bond, the other double
bond would move over counterclockwise, the next double bond will move towards the nitrogen,
and finally the double bond will go from the N-O bond onto the O, giving it a negative charge..
Value of the stock = Dividend Required returnValue of the stock .pdf
Value of the stock = Dividend / Required return
Value of the stock = $2.75 / 0.09
Value of the stock = $30.56
Solution
Value of the stock = Dividend / Required return
Value of the stock = $2.75 / 0.09
Value of the stock = $30.56.
The phenotypes of parental arePPSS - purple smoothppss - yellow.pdf
The phenotypes of parental are:
PPSS - purple smooth
ppss - yellow wrinkled
The two possible gametes of the parental generation are PS and ps. Since they are homozygous
at both the genes, only one genotype of gametes is possible for each.
Punnet square: All the F1 offspring are heterozygous at both the loci, and are phenotypically
purple smooth.
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Solution
The phenotypes of parental are:
PPSS - purple smooth
ppss - yellow wrinkled
The two possible gametes of the parental generation are PS and ps. Since they are homozygous
at both the genes, only one genotype of gametes is possible for each.
Punnet square: All the F1 offspring are heterozygous at both the loci, and are phenotypically
purple smooth.
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs.
Recommended
The key here is that the relative rates of appearancedisappearance .pdf
The key here is that the relative rates of appearance/disappearance of the materials is directly
related to their coefficients in the balanced equation. In other words, since each IO3(-) ion yields
3 I2 molecules, the rate of appearance of I2 must be 3 times the rate of disappearance of IO3(-).
Thus, the rate of appearance of I2 = 2.2 * 10^-3 M/s * 3 Rate of appearance of I2 = 6.6 * 10^-3
mol/(L*s)
Solution
The key here is that the relative rates of appearance/disappearance of the materials is directly
related to their coefficients in the balanced equation. In other words, since each IO3(-) ion yields
3 I2 molecules, the rate of appearance of I2 must be 3 times the rate of disappearance of IO3(-).
Thus, the rate of appearance of I2 = 2.2 * 10^-3 M/s * 3 Rate of appearance of I2 = 6.6 * 10^-3
mol/(L*s).
the molar ratio of C with O = 11, However, the a.pdf
the molar ratio of C with O = 1:1, However, the atomic mass of C = 12 , while the
atomic mass of O =16 so the percentage of O in CO = mass of O / mass of CO = 16/ (12+16) =
16/28 = 57.1%
Solution
the molar ratio of C with O = 1:1, However, the atomic mass of C = 12 , while the
atomic mass of O =16 so the percentage of O in CO = mass of O / mass of CO = 16/ (12+16) =
16/28 = 57.1%.
Ne and F2 are both uncharged molecules so there c.pdf
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here
Solution
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here.
x = 10 + (1p^2) Elasticity , E = (dxdp)(px) .pdf
x = 10 + (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p
= 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526
Solution
x = 10 + (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p
= 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526.
The second phenol can resonance stabilize once th.pdf
The second phenol can resonance stabilize once the H has been removed from the
O. The negative charge on the O would come down and create a double bond, the other double
bond would move over counterclockwise, the next double bond will move towards the nitrogen,
and finally the double bond will go from the N-O bond onto the O, giving it a negative charge.
Solution
The second phenol can resonance stabilize once the H has been removed from the
O. The negative charge on the O would come down and create a double bond, the other double
bond would move over counterclockwise, the next double bond will move towards the nitrogen,
and finally the double bond will go from the N-O bond onto the O, giving it a negative charge..
Value of the stock = Dividend Required returnValue of the stock .pdf
Value of the stock = Dividend / Required return
Value of the stock = $2.75 / 0.09
Value of the stock = $30.56
Solution
Value of the stock = Dividend / Required return
Value of the stock = $2.75 / 0.09
Value of the stock = $30.56.
The phenotypes of parental arePPSS - purple smoothppss - yellow.pdf
The phenotypes of parental are:
PPSS - purple smooth
ppss - yellow wrinkled
The two possible gametes of the parental generation are PS and ps. Since they are homozygous
at both the genes, only one genotype of gametes is possible for each.
Punnet square: All the F1 offspring are heterozygous at both the loci, and are phenotypically
purple smooth.
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Solution
The phenotypes of parental are:
PPSS - purple smooth
ppss - yellow wrinkled
The two possible gametes of the parental generation are PS and ps. Since they are homozygous
at both the genes, only one genotype of gametes is possible for each.
Punnet square: All the F1 offspring are heterozygous at both the loci, and are phenotypically
purple smooth.
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs
Gametes
ps
ps
PS
PpSs
PpSs
PS
PpSs
PpSs.
H3PO3 Phosphorous acid H3AsO4 Arsenic acid HC.pdf
H3PO3 :Phosphorous acid H3AsO4 :Arsenic acid HClO2 :Chlorous acid HBrO
:Hypobromous Acid HC2H3O2 :Acetic acid HIO4 :Periodic acid HNO3 : Nitric acid HF
:Hydrofluoric acid sulfur trioxide : SO3 dinitrogen pentaoxide :N2O5 sulfur hexafluoride : SF6
carbon disuflide :CS2
Solution
H3PO3 :Phosphorous acid H3AsO4 :Arsenic acid HClO2 :Chlorous acid HBrO
:Hypobromous Acid HC2H3O2 :Acetic acid HIO4 :Periodic acid HNO3 : Nitric acid HF
:Hydrofluoric acid sulfur trioxide : SO3 dinitrogen pentaoxide :N2O5 sulfur hexafluoride : SF6
carbon disuflide :CS2.
The active transport of sodium chloride by the ascending loop of Hen.pdf
The active transport of sodium chloride by the ascending loop of Henle is independent of
antidiuretic hormone (ADH) and causes a progressive concentration of the interstitial fluid of the
medulla but a dilutio nof the luminal fluid. ADH increases the permeability of the corical
collecting ducts to water, and so water is reabsorbed by this segment until the luminal fluid is
isoosmotic to cortical interstitial fluid. The luminal fluid then enters and flows through the
medullary collecting ducts and the concentrated medullary interstitium causes water to move out
of these ducts, made highly permeable to water by ADH. The result is the concentration of the
collecting duct fluid and the urine.
Renal sodium regulation:
Sodium excretion is the difference between the amount of sodium filtered and the amount
reabsorbed. The major control of tubular sodium reabsorption is the adrenal cortical hormone
aldosterone, which stimulates sodium reabsorption in the cortical collecting ducts.
Renal water regulation
Water excretion is the difference between the amount of water filtered and the amount
reabsorbed. ADH mediates water absorption which is controlled by baroreceptors and
osmoreceptors of the posterior pituitary. Via the baroreceptor reflexes, a low extracellular
volume stimulates ADH secretion and a high extracellular volume inhibits it. Via the
osmoreceptors, a high body-fluid osmolarity stimulates ADH secretion and a low osmolarity
inhibits it.
Homeostatic control of hydrogen ions by renal mechanisms
The kidneys maintain a stable plasma hydrogen ioin concentration by regulating plasma
bicarbonate concentration. They can either excrete bicarbonate or contribute new bicarbonate to
the blood. Bicarbonate is reabsorbed when hydrogen ions, generated in the tubular cells by a
process catalysed by carbonic anhydrase, are secreted into the lumen and combine with filtered
bicarbonate. The secreted hydrogen ions are not excreted in this situation.
In contrast, when the secreted hydrogen ions comine in the lumen with filtered phosphate or
other non-bicarbonate buffer, they are excreted and the kidneys have contributed new
bicarbonate to the blood. The kidneys also contribute new bicarbonate to the blood when they
produce and excrete ammonia.
Solution
The active transport of sodium chloride by the ascending loop of Henle is independent of
antidiuretic hormone (ADH) and causes a progressive concentration of the interstitial fluid of the
medulla but a dilutio nof the luminal fluid. ADH increases the permeability of the corical
collecting ducts to water, and so water is reabsorbed by this segment until the luminal fluid is
isoosmotic to cortical interstitial fluid. The luminal fluid then enters and flows through the
medullary collecting ducts and the concentrated medullary interstitium causes water to move out
of these ducts, made highly permeable to water by ADH. The result is the concentration of the
collecting duct fluid and the urine.
Renal sodium regulation:
S.
ENTROPY INCEREASES for liquid to gas solid to g.pdf
ENTROPY INCEREASES for liquid to gas solid to gas
Solution
ENTROPY INCEREASES for liquid to gas solid to gas.
Solution - Debt to assets ratio = 0.60Weight of debt in capital s.pdf
Solution
:- Debt to assets ratio = 0.60
Weight of debt in capital structure of company = 0.60
Weight of common equity in capital structure of company = 0.40 (1 - 0.60)
Debt-equity ratio = Debt / Equity
= 0.60 / 0.40
= 1.5 : 1
Equity multiplier = Total assets / Total equity
= 1 / 0.40
= 2.50 Times
Conclusion :-Debt-equity ratio1.5 : 1Equity multiplier2.50 Times..
d) More hydrogen bonds will form with water due t.pdf
d) More hydrogen bonds will form with water due to the carbonyl group and the
hydroxyl group
Solution
d) More hydrogen bonds will form with water due to the carbonyl group and the
hydroxyl group.
Plant is a dicot showing dichotomous branching.Growth form The p.pdf
Plant is a dicot showing dichotomous branching.
Growth form : The plant is a tree; the plant is a shrub (i.e., a woody plant with several stems
growing from the base)
Leaf type : The leaf blade is compound (i.e., made up of two or more discrete leaflets
Leaves per node : There is one leaf per node along the stem
Leaf blade edges : The edge of the leaf blade has teeth
Plant genus: Aralia
Species : spinosa
Family: Araliaceae (Ginseng Family)
The tree is commonly known as Devil\'s walking stick.
Solution
Plant is a dicot showing dichotomous branching.
Growth form : The plant is a tree; the plant is a shrub (i.e., a woody plant with several stems
growing from the base)
Leaf type : The leaf blade is compound (i.e., made up of two or more discrete leaflets
Leaves per node : There is one leaf per node along the stem
Leaf blade edges : The edge of the leaf blade has teeth
Plant genus: Aralia
Species : spinosa
Family: Araliaceae (Ginseng Family)
The tree is commonly known as Devil\'s walking stick..
Please observe the the code and validations of user inputs.import .pdf
Please observe the the code and validations of user inputs.
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Random;
public class SlotMachineSimulation
{
public static void main(String[]args) throws FileNotFoundException
{
//Open File
File file = new File(\"input.txt\");
int choice;
//Open File to write on
PrintWriter outputFile = new PrintWriter(\"output.txt\");
// Create a Scanner object for keyboard input.
Scanner keyboard=new Scanner(System.in);
// Create a Scanner object for input File data.
Scanner inputFile = new Scanner(file);
Scanner readFile=new Scanner(file);
//Declaring Variables
int fr1,fr2,fr3;
String fruit;
int num;
int select;
int amountBet;
int won,lost = 0;
String decision;
int total=0;
System.out.println(\"Welcome to Guanyu Tian\'s Slot Machine!\");
int n = inputFile.nextInt();
do{
System.out.println(\"You inserted \" + n + \" into the slot machine!\" );
System.out.println(\"You currently have \" + n + \", you can bet the multiple of $10.\");
System.out.println(\"If you enter 1, you will bet $10\ If you enter 2, you will bet $20, etc.\");
do{
//ask user to insert the bet amount
System.out.println(\"Please enter how much do you want to bet this time: \");
num=keyboard.nextInt();
if(num<1 || num>(n/10))
{
amountBet=num*10;
System.out.printf(\"Sorry, you don\'t have $\"+amountBet+\" now\\t\");
System.out.println(\"You can only bet less than the amount of money you have.\");
System.out.println(\"You currently have $\"+n+\", You can the multiple of $10.\");
System.out.println(\"If you enter 1, you will bet $10\ If you enter 2, you will bet $20, etc.\");
//System.exit(0);
}
}while(1>num || (n/10)6 ){
System.out.println(\"Invalid selection. You must enter a number between 1 and 6 inclusive\");
//System.exit(0);
}
} while(select<1 || select >6);
//Generating a random fruit
Random randNum=new Random();
fr1 = randNum.nextInt(6)+1;
fr2 = randNum.nextInt(6)+1;
fr3 = randNum.nextInt(6)+1;
if(select == fr1&&select==fr2&&select==fr3){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won = amountBet *3;
total=n+won;
System.out.println(\"You have won 3 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 3 times bet amount.\"+won);
outputFile.println(n);
}
else if(select==fr1&&select==fr2||select==fr2&&select==fr3||select==fr3&&select==fr1){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won= amountBet *2;
total=n+won;
System.out.println(\"You have won 2 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 2 times bet amount.\"+won);
outputFile.println(n);
}
else if(select == fr1||select == fr2||select == fr3){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won= amountBet ;
total=n+won;
System.out.println(\"You have won 1 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 1 times bet amount.
Non-accountant means financial statements in order to understand the.pdf
Non-accountant means financial statements in order to understand the companie\'s financial
results or to evaluate possible investments or expansion of the existing organisation.This is why
non-accountants should understand the accounting. Usually executives and board leaders will not
examine or have a clear understanding if all the details included in the notes. Even accountants
and CFOs can have some difficulty explaining the reporting requirements.Thus accountants and
non financial executives should work together to develop and explain key ratios to focus on, such
efficiency and liquidity ratios.
Solution
Non-accountant means financial statements in order to understand the companie\'s financial
results or to evaluate possible investments or expansion of the existing organisation.This is why
non-accountants should understand the accounting. Usually executives and board leaders will not
examine or have a clear understanding if all the details included in the notes. Even accountants
and CFOs can have some difficulty explaining the reporting requirements.Thus accountants and
non financial executives should work together to develop and explain key ratios to focus on, such
efficiency and liquidity ratios..
Cerium exhibits three oxidation states, +2, +3 an.pdf
Cerium exhibits three oxidation states, +2, +3 and +4. The +2 state is rare and is
observed in CeH2, CeI2 and CeS.[5] The most common compound of cerium is cerium(IV)
oxide (CeO2).\\ yes it can be used Ce^3+ (aq) + VO2^+ ------- Ce (s) + VO3 E^o =-4.68
Solution
Cerium exhibits three oxidation states, +2, +3 and +4. The +2 state is rare and is
observed in CeH2, CeI2 and CeS.[5] The most common compound of cerium is cerium(IV)
oxide (CeO2).\\ yes it can be used Ce^3+ (aq) + VO2^+ ------- Ce (s) + VO3 E^o =-4.68.
import java.awt.; victimisation AWTs Graphics and Color impo.pdf
import java.awt.*; // victimisation AWT\'s Graphics and Color
import java.awt.event.*; // victimisation AWT event categories and perceiver interfaces
import javax.swing.*; // victimisation Swing\'s parts and containers
/** Custom Drawing Code templet */
// A Swing application extends javax.swing.JFrame
public category CGTemplate extends JFrame outline constants
public static final int CANVAS_WIDTH = 640;
public static final int CANVAS_HEIGHT = 480;
// Declare AN instance of the drawing canvas,
// that is AN inner category known as DrawCanvas extending javax.swing.JPanel.
personal DrawCanvas canvas;
// builder to line up the graphical user interface parts and event handlers
public CGTemplate() because the JFrame\'s content-pane
instrumentation cp = getContentPane();
cp.add(canvas);
// or \"setContentPane(canvas);\"
setDefaultCloseOperation(EXIT_ON_CLOSE); // Handle the shut button
pack(); // Either pack() the components; or setSize()
setTitle(\"......\"); // \"super\" JFrame sets the title
setVisible(true); // \"super\" JFrame show
}
/**
* outline inner category DrawCanvas, that could be a JPanel used for custom drawing.
*/
personal category DrawCanvas extends JPanel for instance,
// Drawing primitive shapes
g.setColor(Color.YELLOW); // set the drawing color
g.drawLine(30, 40, 100, 200);
g.drawOval(150, 180, 10, 10);
g.drawRect(200, 210, 20, 30);
g.setColor(Color.RED); // modification the drawing color
g.fillOval(300, 310, 30, 50);
g.fillRect(400, 350, 60, 50);
// Printing texts
g.setColor(Color.WHITE);
g.setFont(new Font(\"Monospaced\", Font.PLAIN, 12));
g.drawString(\"Testing custom drawing ...\", 10, 20);
}
}
// The entry main methodology
public static void main(String[] args) {
// Run the graphical user interface codes on the Event-Dispatching thread for thread safety
SwingUtilities.invokeLater(new Runnable() {
@Override
public void run() {
new CGTemplate(); // Let the builder do the duty
}
});
}
}
Solution
import java.awt.*; // victimisation AWT\'s Graphics and Color
import java.awt.event.*; // victimisation AWT event categories and perceiver interfaces
import javax.swing.*; // victimisation Swing\'s parts and containers
/** Custom Drawing Code templet */
// A Swing application extends javax.swing.JFrame
public category CGTemplate extends JFrame outline constants
public static final int CANVAS_WIDTH = 640;
public static final int CANVAS_HEIGHT = 480;
// Declare AN instance of the drawing canvas,
// that is AN inner category known as DrawCanvas extending javax.swing.JPanel.
personal DrawCanvas canvas;
// builder to line up the graphical user interface parts and event handlers
public CGTemplate() because the JFrame\'s content-pane
instrumentation cp = getContentPane();
cp.add(canvas);
// or \"setContentPane(canvas);\"
setDefaultCloseOperation(EXIT_ON_CLOSE); // Handle the shut button
pack(); // Either pack() the components; or setSize()
setTitle(\"......\"); // \"super\" JFrame sets the title
setVisible(true); // \"super\" JFrame show
}
/**.
Federated identity management is mainly used by security leaders. It.pdf
Federated identity management is mainly used by security leaders. It is mainly DEALS with the
identifying the individuals and it permits secure access to resource. It is a secure arrangement
that can made among the enterprises.
Security benefits of Federated Identity Management:
1. Offer authentication for XML based open standard data format:
2. Security standpoint: The more organizations adopt it because it prevents hackers looking for
vulnerabilities. It is a simplified password management. Only authenticated persons have access
right.
3. Lose coupling of Directories: It does not require user information to be preserved and
coordinated between directories.
4. Platform neutrality:
In FIDM abstracts the security framework away from platform architecture and vendor
implementations, for more independent of application logic and it provide secure communication
in any platform.
5. Reduced cost for service providers:
In this re-use a single act of authentication many times across several services can reduce the
cost of maintaining account.
6. Best online experience for end users:
While end users enter into the particular site, perform sign on by allowing users to authenticate at
an identity provider and then access service providers without other verification.
Solution
Federated identity management is mainly used by security leaders. It is mainly DEALS with the
identifying the individuals and it permits secure access to resource. It is a secure arrangement
that can made among the enterprises.
Security benefits of Federated Identity Management:
1. Offer authentication for XML based open standard data format:
2. Security standpoint: The more organizations adopt it because it prevents hackers looking for
vulnerabilities. It is a simplified password management. Only authenticated persons have access
right.
3. Lose coupling of Directories: It does not require user information to be preserved and
coordinated between directories.
4. Platform neutrality:
In FIDM abstracts the security framework away from platform architecture and vendor
implementations, for more independent of application logic and it provide secure communication
in any platform.
5. Reduced cost for service providers:
In this re-use a single act of authentication many times across several services can reduce the
cost of maintaining account.
6. Best online experience for end users:
While end users enter into the particular site, perform sign on by allowing users to authenticate at
an identity provider and then access service providers without other verification..
e) 0.4Explanation-No. of weeks in which more than 3 orders w.pdf
e) 0.4
Explanation-
No. of weeks in which more than 3 orders were placed
= 4+3+3
=10
P(E) = 10/(total weeks)
= 10/25
= 0.4
Solution
e) 0.4
Explanation-
No. of weeks in which more than 3 orders were placed
= 4+3+3
=10
P(E) = 10/(total weeks)
= 10/25
= 0.4.
Asprin is insoluble in water. Water is added to .pdf
Asprin is insoluble in water. Water is added to precipitate out the asprin so that it
can be easily recovered from the reaction mixture by filtration.
Solution
Asprin is insoluble in water. Water is added to precipitate out the asprin so that it
can be easily recovered from the reaction mixture by filtration..
a. carbon is SP2 hybridised and oxygen is also SP.pdf
a. carbon is SP2 hybridised and oxygen is also SP2 hybridised b. C which is on the
left of C=O is sp3 hybridised c. O which is on the right of C = O bond is sp3 hybridised
Solution
a. carbon is SP2 hybridised and oxygen is also SP2 hybridised b. C which is on the
left of C=O is sp3 hybridised c. O which is on the right of C = O bond is sp3 hybridised.
c++ code#include iostream#include string#include stdlib.h.pdf
c++ code
#include
#include
#include
using namespace std;
int bomb_count, checkx, checky;
const int sizex = 10, sizey = 10;
bool lose;
class Grid {
public:
bool has_bomb, marked;
Grid(void) {
has_bomb = false;
marked = false;
}
};
void drawBoad(Grid board[sizex][sizey])
{
// _0_1_2_3_4_5_6_7_8_9_
// 0|_|_|_|_|_|_|_|_|_|_|
// 1|_|_|_|_|_|_|_|_|_|_|
// 2|_|_|_|_|_|_|_|_|_|_|
// 3|_|_|_|_|_|_|_|_|_|_|
// 4|_|_|_|_|_|_|_|_|_|_|
// 5|_|_|_|_|_|_|_|_|_|_|
// 6|_|_|_|_|_|_|_|_|_|_|
// 7|_|_|_|_|_|_|_|_|_|_|
// 8|_|_|_|_|_|_|_|_|_|_|
// 9|_|_|_|_|_|_|_|_|_|_|
//This draws the top line
cout << \" _\";
for (int i = 0; i < sizex; i++)
{
cout << i << \"_\";
}
cout << endl;
//This draws the body
for (int y = 0; y < sizey; y++)
{
cout << y << \"|\";
for (int x = 0; x < sizex; x++)
{
if (board[x][y].has_bomb && board[x][y].marked)
{
cout << \"!|\";
}
if (board[x][y].has_bomb)
{
cout << \"b|\";
}
else if (board[x][y].marked)
{
cout << \"x|\";
}
else
{
cout << \"_|\";
}
}
cout << endl;
}
}
int main()
{
{
lose = false;
bomb_count = 15;
checkx = 0;
checky = 0;
Grid gameboard[sizex][sizey];
for (int i = 0; i < bomb_count; i++)
{
int xpos = rand() % 10;
int ypos = rand() % 10;
if (gameboard[xpos][ypos].has_bomb == false) {
gameboard[xpos][ypos].has_bomb = true;
}
else
{
i--;
}
}
drawBoad(gameboard);
cout << endl;
while (lose != true)
{
cout << \"Input x grid to check.\" << endl;
cin >> checkx;
cout << endl << \"Input y grid to check.\" << endl;
cin >> checky;
if (gameboard[checkx - 1][checky - 1].has_bomb == true) {
cout << \"Boom!\";
lose = true;
}
else {
cout << \"Try again.\" << endl;
}
gameboard[checkx - 1][checky - 1].marked = true;
cout << endl;
drawBoad(gameboard);
cout << endl;
}
return 0;
}
ithe lengthr size of standard array in java immutable it will have a fixed length.
java code
package {
import flash.display.Sprite;
import flash.events.MouseEvent;
public class Main extends Sprite {
public var zones:Array;
public var board:Sprite;
public var boardWidth:int;
public var boardHeight:int;
public var numberOfMines:int;
public function Main() { //constructor----
zones = new Array();
board = new Sprite();
boardWidth = 10;
boardHeight = 10;
numberOfMines = 20;
gameover_btn.visible = false;
again_btn.visible = false;
minesleft_txt.text = String(numberOfMines);
addChildAt(board, 0);
createBoard();
placeMines();
placeNumbers();
again_btn.addEventListener(MouseEvent.CLICK, makeAnotherGame);
} //end of constructor
private function createBoard():void{
for(var i:Number = 0; i < boardHeight; i++){
var hzones:Array = new Array();
zones[i] = hzones;
for(var j:Number = 0; j < boardWidth; j++){
var newzone:Zone = new Zone(j, i);
zones[i][j] = newzone;
board.addChild(zones[i][j]);
zones[i][j].x = j*40 + 20;
zones[i][j].y = i*40 + 20;
}
}
}
private function placeMines():void{
var bombsCounter:uint = numberOfMines;
while(bombsCounter > 0){
var randomNumber:Number = Math.round(Math.random()* ((boardHeight *
boardWidth) - 1));
var rowCount:Number = Math.floor(randomNumber / bo.
B and C are correct. Hosts do not directly connect to core-or distri.pdf
B and C are correct. Hosts do not directly connect to core-or distribution-layer switches.
A, D, and E are incorrect. A is incorrect because access switches provide direct access to users.
D andE are incorrect because cut through and fragment free are switch modes, not design layers.
Solution
B and C are correct. Hosts do not directly connect to core-or distribution-layer switches.
A, D, and E are incorrect. A is incorrect because access switches provide direct access to users.
D andE are incorrect because cut through and fragment free are switch modes, not design layers..
AnswerVertex is a user defined class which has generic instance v.pdf
Answer:
Vertex is a user defined class which has generic instance variables . Deque, Set, Hashset etc are
standard java data structures provided ready made with Java.
Now, first one is a function named topsort, which receives a parameter Graph & returns a Deque
of vertex.
2nd is an instantiation of a deque variable named stack.
3rd is an instantiation of set variable named visited.
Solution
Answer:
Vertex is a user defined class which has generic instance variables . Deque, Set, Hashset etc are
standard java data structures provided ready made with Java.
Now, first one is a function named topsort, which receives a parameter Graph & returns a Deque
of vertex.
2nd is an instantiation of a deque variable named stack.
3rd is an instantiation of set variable named visited..
Answer is a.Prokaryote classified into Bacteria and Archaea, displ.pdf
Answer is a.
Prokaryote classified into Bacteria and Archaea, display diverse adaptations that allow them to
occupy many different habitats as they can adapted metabolically to grow under extreme
conditions, not possible for other organisms.
Solution
Answer is a.
Prokaryote classified into Bacteria and Archaea, display diverse adaptations that allow them to
occupy many different habitats as they can adapted metabolically to grow under extreme
conditions, not possible for other organisms..
3a.Error detection provides for accurate data transfer. Indeed, if.pdf
3a.
Error detection provides for accurate data transfer. Indeed, if the error can be identified in the
data sent over the network, corrective measures can be taken so as to correct the data, thus
maintaining the accuracy of the data. Basic approach used for error detection is the use of
redundancy, where additional bits are added to facilitate detection of errors. Out of various
techniques, Two-dimensional parity & Checksum are discussed as follows:
In one-dimensional parity check, a redundant bit known as parity bit is appended to every data
unit so that total number of 1’s in the data becomes even. The two-dimensional parity check is
used to enhance performance over 1-D parity check. Here, the block of bits are arranged in a
tabular format. For each row and for each column, 1D parity checks are being evaluated
respectively and sent along with the data, which is further checked at the receiving end. This
technique detects burst error of more than n bits with high probability.
7c.
Distance vector routing protocol is one of the main approaches to Intra-domain networking. It is
quite simple to implement. The name Distance vector implies “how far & in which direction” the
data packets need to be sent. \"Direction\" usually means the next hop address and the exit
interface. \"Distance\" is a measure of the cost to reach a certain node. The least cost route
between any two nodes is the route with minimum distance. Each node maintains a vector of
minimum distance to every node. As opposed to Link state protocol, where each node tells all
nodes about its neighbors, here each node tells its neighbors about its shortest route to all other
nodes in the network, on a periodic basis. It is based on Distributed Bellman-Ford Algorithm.
Routing Information Protocol (RIP) is one of the oldest distance-vector routing protocols which
employ the hop count as a routing metric. RIP is a routing protocol which exchanges network
information between routers dynamically. It uses local broadcast to share routing information.
RIP broadcasts routing updates in every 30 seconds, regardless something in network has
changed or not. Once 30 seconds expires, routers running RIP protocol will broadcast their
routing information to any devices connected to their interfaces. Before sending routing updates
router add a initiating metric to every routes which it has and increments the metric of incoming
routes in advertisements so the listing router can learn how far destination network is. While
sending broadcasts RIP does not care about who listens these broadcast updates or not. After
sending broadcast RIP does not care whether neighbors received these broadcast updates or not.
When router receives routing updates, it compares them with the routes which it already has in
its routing table. If update has information about a route which is not available in its routing
table, router will consider that route as a new route. Router will add all new routes in routing
table before up.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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H3PO3 Phosphorous acid H3AsO4 Arsenic acid HC.pdf
H3PO3 :Phosphorous acid H3AsO4 :Arsenic acid HClO2 :Chlorous acid HBrO
:Hypobromous Acid HC2H3O2 :Acetic acid HIO4 :Periodic acid HNO3 : Nitric acid HF
:Hydrofluoric acid sulfur trioxide : SO3 dinitrogen pentaoxide :N2O5 sulfur hexafluoride : SF6
carbon disuflide :CS2
Solution
H3PO3 :Phosphorous acid H3AsO4 :Arsenic acid HClO2 :Chlorous acid HBrO
:Hypobromous Acid HC2H3O2 :Acetic acid HIO4 :Periodic acid HNO3 : Nitric acid HF
:Hydrofluoric acid sulfur trioxide : SO3 dinitrogen pentaoxide :N2O5 sulfur hexafluoride : SF6
carbon disuflide :CS2.
The active transport of sodium chloride by the ascending loop of Hen.pdf
The active transport of sodium chloride by the ascending loop of Henle is independent of
antidiuretic hormone (ADH) and causes a progressive concentration of the interstitial fluid of the
medulla but a dilutio nof the luminal fluid. ADH increases the permeability of the corical
collecting ducts to water, and so water is reabsorbed by this segment until the luminal fluid is
isoosmotic to cortical interstitial fluid. The luminal fluid then enters and flows through the
medullary collecting ducts and the concentrated medullary interstitium causes water to move out
of these ducts, made highly permeable to water by ADH. The result is the concentration of the
collecting duct fluid and the urine.
Renal sodium regulation:
Sodium excretion is the difference between the amount of sodium filtered and the amount
reabsorbed. The major control of tubular sodium reabsorption is the adrenal cortical hormone
aldosterone, which stimulates sodium reabsorption in the cortical collecting ducts.
Renal water regulation
Water excretion is the difference between the amount of water filtered and the amount
reabsorbed. ADH mediates water absorption which is controlled by baroreceptors and
osmoreceptors of the posterior pituitary. Via the baroreceptor reflexes, a low extracellular
volume stimulates ADH secretion and a high extracellular volume inhibits it. Via the
osmoreceptors, a high body-fluid osmolarity stimulates ADH secretion and a low osmolarity
inhibits it.
Homeostatic control of hydrogen ions by renal mechanisms
The kidneys maintain a stable plasma hydrogen ioin concentration by regulating plasma
bicarbonate concentration. They can either excrete bicarbonate or contribute new bicarbonate to
the blood. Bicarbonate is reabsorbed when hydrogen ions, generated in the tubular cells by a
process catalysed by carbonic anhydrase, are secreted into the lumen and combine with filtered
bicarbonate. The secreted hydrogen ions are not excreted in this situation.
In contrast, when the secreted hydrogen ions comine in the lumen with filtered phosphate or
other non-bicarbonate buffer, they are excreted and the kidneys have contributed new
bicarbonate to the blood. The kidneys also contribute new bicarbonate to the blood when they
produce and excrete ammonia.
Solution
The active transport of sodium chloride by the ascending loop of Henle is independent of
antidiuretic hormone (ADH) and causes a progressive concentration of the interstitial fluid of the
medulla but a dilutio nof the luminal fluid. ADH increases the permeability of the corical
collecting ducts to water, and so water is reabsorbed by this segment until the luminal fluid is
isoosmotic to cortical interstitial fluid. The luminal fluid then enters and flows through the
medullary collecting ducts and the concentrated medullary interstitium causes water to move out
of these ducts, made highly permeable to water by ADH. The result is the concentration of the
collecting duct fluid and the urine.
Renal sodium regulation:
S.
ENTROPY INCEREASES for liquid to gas solid to g.pdf
ENTROPY INCEREASES for liquid to gas solid to gas
Solution
ENTROPY INCEREASES for liquid to gas solid to gas.
Solution - Debt to assets ratio = 0.60Weight of debt in capital s.pdf
Solution
:- Debt to assets ratio = 0.60
Weight of debt in capital structure of company = 0.60
Weight of common equity in capital structure of company = 0.40 (1 - 0.60)
Debt-equity ratio = Debt / Equity
= 0.60 / 0.40
= 1.5 : 1
Equity multiplier = Total assets / Total equity
= 1 / 0.40
= 2.50 Times
Conclusion :-Debt-equity ratio1.5 : 1Equity multiplier2.50 Times..
d) More hydrogen bonds will form with water due t.pdf
d) More hydrogen bonds will form with water due to the carbonyl group and the
hydroxyl group
Solution
d) More hydrogen bonds will form with water due to the carbonyl group and the
hydroxyl group.
Plant is a dicot showing dichotomous branching.Growth form The p.pdf
Plant is a dicot showing dichotomous branching.
Growth form : The plant is a tree; the plant is a shrub (i.e., a woody plant with several stems
growing from the base)
Leaf type : The leaf blade is compound (i.e., made up of two or more discrete leaflets
Leaves per node : There is one leaf per node along the stem
Leaf blade edges : The edge of the leaf blade has teeth
Plant genus: Aralia
Species : spinosa
Family: Araliaceae (Ginseng Family)
The tree is commonly known as Devil\'s walking stick.
Solution
Plant is a dicot showing dichotomous branching.
Growth form : The plant is a tree; the plant is a shrub (i.e., a woody plant with several stems
growing from the base)
Leaf type : The leaf blade is compound (i.e., made up of two or more discrete leaflets
Leaves per node : There is one leaf per node along the stem
Leaf blade edges : The edge of the leaf blade has teeth
Plant genus: Aralia
Species : spinosa
Family: Araliaceae (Ginseng Family)
The tree is commonly known as Devil\'s walking stick..
Please observe the the code and validations of user inputs.import .pdf
Please observe the the code and validations of user inputs.
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Random;
public class SlotMachineSimulation
{
public static void main(String[]args) throws FileNotFoundException
{
//Open File
File file = new File(\"input.txt\");
int choice;
//Open File to write on
PrintWriter outputFile = new PrintWriter(\"output.txt\");
// Create a Scanner object for keyboard input.
Scanner keyboard=new Scanner(System.in);
// Create a Scanner object for input File data.
Scanner inputFile = new Scanner(file);
Scanner readFile=new Scanner(file);
//Declaring Variables
int fr1,fr2,fr3;
String fruit;
int num;
int select;
int amountBet;
int won,lost = 0;
String decision;
int total=0;
System.out.println(\"Welcome to Guanyu Tian\'s Slot Machine!\");
int n = inputFile.nextInt();
do{
System.out.println(\"You inserted \" + n + \" into the slot machine!\" );
System.out.println(\"You currently have \" + n + \", you can bet the multiple of $10.\");
System.out.println(\"If you enter 1, you will bet $10\ If you enter 2, you will bet $20, etc.\");
do{
//ask user to insert the bet amount
System.out.println(\"Please enter how much do you want to bet this time: \");
num=keyboard.nextInt();
if(num<1 || num>(n/10))
{
amountBet=num*10;
System.out.printf(\"Sorry, you don\'t have $\"+amountBet+\" now\\t\");
System.out.println(\"You can only bet less than the amount of money you have.\");
System.out.println(\"You currently have $\"+n+\", You can the multiple of $10.\");
System.out.println(\"If you enter 1, you will bet $10\ If you enter 2, you will bet $20, etc.\");
//System.exit(0);
}
}while(1>num || (n/10)6 ){
System.out.println(\"Invalid selection. You must enter a number between 1 and 6 inclusive\");
//System.exit(0);
}
} while(select<1 || select >6);
//Generating a random fruit
Random randNum=new Random();
fr1 = randNum.nextInt(6)+1;
fr2 = randNum.nextInt(6)+1;
fr3 = randNum.nextInt(6)+1;
if(select == fr1&&select==fr2&&select==fr3){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won = amountBet *3;
total=n+won;
System.out.println(\"You have won 3 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 3 times bet amount.\"+won);
outputFile.println(n);
}
else if(select==fr1&&select==fr2||select==fr2&&select==fr3||select==fr3&&select==fr1){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won= amountBet *2;
total=n+won;
System.out.println(\"You have won 2 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 2 times bet amount.\"+won);
outputFile.println(n);
}
else if(select == fr1||select == fr2||select == fr3){
System.out.println(fr1+\" \"+fr2+\" \"+fr3);
won= amountBet ;
total=n+won;
System.out.println(\"You have won 1 times bet amount.\"+won);
System.out.println(\"Your current balance.\"+total);
outputFile.println(\"You have won 1 times bet amount.
Non-accountant means financial statements in order to understand the.pdf
Non-accountant means financial statements in order to understand the companie\'s financial
results or to evaluate possible investments or expansion of the existing organisation.This is why
non-accountants should understand the accounting. Usually executives and board leaders will not
examine or have a clear understanding if all the details included in the notes. Even accountants
and CFOs can have some difficulty explaining the reporting requirements.Thus accountants and
non financial executives should work together to develop and explain key ratios to focus on, such
efficiency and liquidity ratios.
Solution
Non-accountant means financial statements in order to understand the companie\'s financial
results or to evaluate possible investments or expansion of the existing organisation.This is why
non-accountants should understand the accounting. Usually executives and board leaders will not
examine or have a clear understanding if all the details included in the notes. Even accountants
and CFOs can have some difficulty explaining the reporting requirements.Thus accountants and
non financial executives should work together to develop and explain key ratios to focus on, such
efficiency and liquidity ratios..
Cerium exhibits three oxidation states, +2, +3 an.pdf
Cerium exhibits three oxidation states, +2, +3 and +4. The +2 state is rare and is
observed in CeH2, CeI2 and CeS.[5] The most common compound of cerium is cerium(IV)
oxide (CeO2).\\ yes it can be used Ce^3+ (aq) + VO2^+ ------- Ce (s) + VO3 E^o =-4.68
Solution
Cerium exhibits three oxidation states, +2, +3 and +4. The +2 state is rare and is
observed in CeH2, CeI2 and CeS.[5] The most common compound of cerium is cerium(IV)
oxide (CeO2).\\ yes it can be used Ce^3+ (aq) + VO2^+ ------- Ce (s) + VO3 E^o =-4.68.
import java.awt.; victimisation AWTs Graphics and Color impo.pdf
import java.awt.*; // victimisation AWT\'s Graphics and Color
import java.awt.event.*; // victimisation AWT event categories and perceiver interfaces
import javax.swing.*; // victimisation Swing\'s parts and containers
/** Custom Drawing Code templet */
// A Swing application extends javax.swing.JFrame
public category CGTemplate extends JFrame outline constants
public static final int CANVAS_WIDTH = 640;
public static final int CANVAS_HEIGHT = 480;
// Declare AN instance of the drawing canvas,
// that is AN inner category known as DrawCanvas extending javax.swing.JPanel.
personal DrawCanvas canvas;
// builder to line up the graphical user interface parts and event handlers
public CGTemplate() because the JFrame\'s content-pane
instrumentation cp = getContentPane();
cp.add(canvas);
// or \"setContentPane(canvas);\"
setDefaultCloseOperation(EXIT_ON_CLOSE); // Handle the shut button
pack(); // Either pack() the components; or setSize()
setTitle(\"......\"); // \"super\" JFrame sets the title
setVisible(true); // \"super\" JFrame show
}
/**
* outline inner category DrawCanvas, that could be a JPanel used for custom drawing.
*/
personal category DrawCanvas extends JPanel for instance,
// Drawing primitive shapes
g.setColor(Color.YELLOW); // set the drawing color
g.drawLine(30, 40, 100, 200);
g.drawOval(150, 180, 10, 10);
g.drawRect(200, 210, 20, 30);
g.setColor(Color.RED); // modification the drawing color
g.fillOval(300, 310, 30, 50);
g.fillRect(400, 350, 60, 50);
// Printing texts
g.setColor(Color.WHITE);
g.setFont(new Font(\"Monospaced\", Font.PLAIN, 12));
g.drawString(\"Testing custom drawing ...\", 10, 20);
}
}
// The entry main methodology
public static void main(String[] args) {
// Run the graphical user interface codes on the Event-Dispatching thread for thread safety
SwingUtilities.invokeLater(new Runnable() {
@Override
public void run() {
new CGTemplate(); // Let the builder do the duty
}
});
}
}
Solution
import java.awt.*; // victimisation AWT\'s Graphics and Color
import java.awt.event.*; // victimisation AWT event categories and perceiver interfaces
import javax.swing.*; // victimisation Swing\'s parts and containers
/** Custom Drawing Code templet */
// A Swing application extends javax.swing.JFrame
public category CGTemplate extends JFrame outline constants
public static final int CANVAS_WIDTH = 640;
public static final int CANVAS_HEIGHT = 480;
// Declare AN instance of the drawing canvas,
// that is AN inner category known as DrawCanvas extending javax.swing.JPanel.
personal DrawCanvas canvas;
// builder to line up the graphical user interface parts and event handlers
public CGTemplate() because the JFrame\'s content-pane
instrumentation cp = getContentPane();
cp.add(canvas);
// or \"setContentPane(canvas);\"
setDefaultCloseOperation(EXIT_ON_CLOSE); // Handle the shut button
pack(); // Either pack() the components; or setSize()
setTitle(\"......\"); // \"super\" JFrame sets the title
setVisible(true); // \"super\" JFrame show
}
/**.
Federated identity management is mainly used by security leaders. It.pdf
Federated identity management is mainly used by security leaders. It is mainly DEALS with the
identifying the individuals and it permits secure access to resource. It is a secure arrangement
that can made among the enterprises.
Security benefits of Federated Identity Management:
1. Offer authentication for XML based open standard data format:
2. Security standpoint: The more organizations adopt it because it prevents hackers looking for
vulnerabilities. It is a simplified password management. Only authenticated persons have access
right.
3. Lose coupling of Directories: It does not require user information to be preserved and
coordinated between directories.
4. Platform neutrality:
In FIDM abstracts the security framework away from platform architecture and vendor
implementations, for more independent of application logic and it provide secure communication
in any platform.
5. Reduced cost for service providers:
In this re-use a single act of authentication many times across several services can reduce the
cost of maintaining account.
6. Best online experience for end users:
While end users enter into the particular site, perform sign on by allowing users to authenticate at
an identity provider and then access service providers without other verification.
Solution
Federated identity management is mainly used by security leaders. It is mainly DEALS with the
identifying the individuals and it permits secure access to resource. It is a secure arrangement
that can made among the enterprises.
Security benefits of Federated Identity Management:
1. Offer authentication for XML based open standard data format:
2. Security standpoint: The more organizations adopt it because it prevents hackers looking for
vulnerabilities. It is a simplified password management. Only authenticated persons have access
right.
3. Lose coupling of Directories: It does not require user information to be preserved and
coordinated between directories.
4. Platform neutrality:
In FIDM abstracts the security framework away from platform architecture and vendor
implementations, for more independent of application logic and it provide secure communication
in any platform.
5. Reduced cost for service providers:
In this re-use a single act of authentication many times across several services can reduce the
cost of maintaining account.
6. Best online experience for end users:
While end users enter into the particular site, perform sign on by allowing users to authenticate at
an identity provider and then access service providers without other verification..
e) 0.4Explanation-No. of weeks in which more than 3 orders w.pdf
e) 0.4
Explanation-
No. of weeks in which more than 3 orders were placed
= 4+3+3
=10
P(E) = 10/(total weeks)
= 10/25
= 0.4
Solution
e) 0.4
Explanation-
No. of weeks in which more than 3 orders were placed
= 4+3+3
=10
P(E) = 10/(total weeks)
= 10/25
= 0.4.
Asprin is insoluble in water. Water is added to .pdf
Asprin is insoluble in water. Water is added to precipitate out the asprin so that it
can be easily recovered from the reaction mixture by filtration.
Solution
Asprin is insoluble in water. Water is added to precipitate out the asprin so that it
can be easily recovered from the reaction mixture by filtration..
a. carbon is SP2 hybridised and oxygen is also SP.pdf
a. carbon is SP2 hybridised and oxygen is also SP2 hybridised b. C which is on the
left of C=O is sp3 hybridised c. O which is on the right of C = O bond is sp3 hybridised
Solution
a. carbon is SP2 hybridised and oxygen is also SP2 hybridised b. C which is on the
left of C=O is sp3 hybridised c. O which is on the right of C = O bond is sp3 hybridised.
c++ code#include iostream#include string#include stdlib.h.pdf
c++ code
#include
#include
#include
using namespace std;
int bomb_count, checkx, checky;
const int sizex = 10, sizey = 10;
bool lose;
class Grid {
public:
bool has_bomb, marked;
Grid(void) {
has_bomb = false;
marked = false;
}
};
void drawBoad(Grid board[sizex][sizey])
{
// _0_1_2_3_4_5_6_7_8_9_
// 0|_|_|_|_|_|_|_|_|_|_|
// 1|_|_|_|_|_|_|_|_|_|_|
// 2|_|_|_|_|_|_|_|_|_|_|
// 3|_|_|_|_|_|_|_|_|_|_|
// 4|_|_|_|_|_|_|_|_|_|_|
// 5|_|_|_|_|_|_|_|_|_|_|
// 6|_|_|_|_|_|_|_|_|_|_|
// 7|_|_|_|_|_|_|_|_|_|_|
// 8|_|_|_|_|_|_|_|_|_|_|
// 9|_|_|_|_|_|_|_|_|_|_|
//This draws the top line
cout << \" _\";
for (int i = 0; i < sizex; i++)
{
cout << i << \"_\";
}
cout << endl;
//This draws the body
for (int y = 0; y < sizey; y++)
{
cout << y << \"|\";
for (int x = 0; x < sizex; x++)
{
if (board[x][y].has_bomb && board[x][y].marked)
{
cout << \"!|\";
}
if (board[x][y].has_bomb)
{
cout << \"b|\";
}
else if (board[x][y].marked)
{
cout << \"x|\";
}
else
{
cout << \"_|\";
}
}
cout << endl;
}
}
int main()
{
{
lose = false;
bomb_count = 15;
checkx = 0;
checky = 0;
Grid gameboard[sizex][sizey];
for (int i = 0; i < bomb_count; i++)
{
int xpos = rand() % 10;
int ypos = rand() % 10;
if (gameboard[xpos][ypos].has_bomb == false) {
gameboard[xpos][ypos].has_bomb = true;
}
else
{
i--;
}
}
drawBoad(gameboard);
cout << endl;
while (lose != true)
{
cout << \"Input x grid to check.\" << endl;
cin >> checkx;
cout << endl << \"Input y grid to check.\" << endl;
cin >> checky;
if (gameboard[checkx - 1][checky - 1].has_bomb == true) {
cout << \"Boom!\";
lose = true;
}
else {
cout << \"Try again.\" << endl;
}
gameboard[checkx - 1][checky - 1].marked = true;
cout << endl;
drawBoad(gameboard);
cout << endl;
}
return 0;
}
ithe lengthr size of standard array in java immutable it will have a fixed length.
java code
package {
import flash.display.Sprite;
import flash.events.MouseEvent;
public class Main extends Sprite {
public var zones:Array;
public var board:Sprite;
public var boardWidth:int;
public var boardHeight:int;
public var numberOfMines:int;
public function Main() { //constructor----
zones = new Array();
board = new Sprite();
boardWidth = 10;
boardHeight = 10;
numberOfMines = 20;
gameover_btn.visible = false;
again_btn.visible = false;
minesleft_txt.text = String(numberOfMines);
addChildAt(board, 0);
createBoard();
placeMines();
placeNumbers();
again_btn.addEventListener(MouseEvent.CLICK, makeAnotherGame);
} //end of constructor
private function createBoard():void{
for(var i:Number = 0; i < boardHeight; i++){
var hzones:Array = new Array();
zones[i] = hzones;
for(var j:Number = 0; j < boardWidth; j++){
var newzone:Zone = new Zone(j, i);
zones[i][j] = newzone;
board.addChild(zones[i][j]);
zones[i][j].x = j*40 + 20;
zones[i][j].y = i*40 + 20;
}
}
}
private function placeMines():void{
var bombsCounter:uint = numberOfMines;
while(bombsCounter > 0){
var randomNumber:Number = Math.round(Math.random()* ((boardHeight *
boardWidth) - 1));
var rowCount:Number = Math.floor(randomNumber / bo.
B and C are correct. Hosts do not directly connect to core-or distri.pdf
B and C are correct. Hosts do not directly connect to core-or distribution-layer switches.
A, D, and E are incorrect. A is incorrect because access switches provide direct access to users.
D andE are incorrect because cut through and fragment free are switch modes, not design layers.
Solution
B and C are correct. Hosts do not directly connect to core-or distribution-layer switches.
A, D, and E are incorrect. A is incorrect because access switches provide direct access to users.
D andE are incorrect because cut through and fragment free are switch modes, not design layers..
AnswerVertex is a user defined class which has generic instance v.pdf
Answer:
Vertex is a user defined class which has generic instance variables . Deque, Set, Hashset etc are
standard java data structures provided ready made with Java.
Now, first one is a function named topsort, which receives a parameter Graph & returns a Deque
of vertex.
2nd is an instantiation of a deque variable named stack.
3rd is an instantiation of set variable named visited.
Solution
Answer:
Vertex is a user defined class which has generic instance variables . Deque, Set, Hashset etc are
standard java data structures provided ready made with Java.
Now, first one is a function named topsort, which receives a parameter Graph & returns a Deque
of vertex.
2nd is an instantiation of a deque variable named stack.
3rd is an instantiation of set variable named visited..
Answer is a.Prokaryote classified into Bacteria and Archaea, displ.pdf
Answer is a.
Prokaryote classified into Bacteria and Archaea, display diverse adaptations that allow them to
occupy many different habitats as they can adapted metabolically to grow under extreme
conditions, not possible for other organisms.
Solution
Answer is a.
Prokaryote classified into Bacteria and Archaea, display diverse adaptations that allow them to
occupy many different habitats as they can adapted metabolically to grow under extreme
conditions, not possible for other organisms..
3a.Error detection provides for accurate data transfer. Indeed, if.pdf
3a.
Error detection provides for accurate data transfer. Indeed, if the error can be identified in the
data sent over the network, corrective measures can be taken so as to correct the data, thus
maintaining the accuracy of the data. Basic approach used for error detection is the use of
redundancy, where additional bits are added to facilitate detection of errors. Out of various
techniques, Two-dimensional parity & Checksum are discussed as follows:
In one-dimensional parity check, a redundant bit known as parity bit is appended to every data
unit so that total number of 1’s in the data becomes even. The two-dimensional parity check is
used to enhance performance over 1-D parity check. Here, the block of bits are arranged in a
tabular format. For each row and for each column, 1D parity checks are being evaluated
respectively and sent along with the data, which is further checked at the receiving end. This
technique detects burst error of more than n bits with high probability.
7c.
Distance vector routing protocol is one of the main approaches to Intra-domain networking. It is
quite simple to implement. The name Distance vector implies “how far & in which direction” the
data packets need to be sent. \"Direction\" usually means the next hop address and the exit
interface. \"Distance\" is a measure of the cost to reach a certain node. The least cost route
between any two nodes is the route with minimum distance. Each node maintains a vector of
minimum distance to every node. As opposed to Link state protocol, where each node tells all
nodes about its neighbors, here each node tells its neighbors about its shortest route to all other
nodes in the network, on a periodic basis. It is based on Distributed Bellman-Ford Algorithm.
Routing Information Protocol (RIP) is one of the oldest distance-vector routing protocols which
employ the hop count as a routing metric. RIP is a routing protocol which exchanges network
information between routers dynamically. It uses local broadcast to share routing information.
RIP broadcasts routing updates in every 30 seconds, regardless something in network has
changed or not. Once 30 seconds expires, routers running RIP protocol will broadcast their
routing information to any devices connected to their interfaces. Before sending routing updates
router add a initiating metric to every routes which it has and increments the metric of incoming
routes in advertisements so the listing router can learn how far destination network is. While
sending broadcasts RIP does not care about who listens these broadcast updates or not. After
sending broadcast RIP does not care whether neighbors received these broadcast updates or not.
When router receives routing updates, it compares them with the routes which it already has in
its routing table. If update has information about a route which is not available in its routing
table, router will consider that route as a new route. Router will add all new routes in routing
table before up.
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