The fraction for of .15% = 3/2000 I convereted .15% to a decimal which equals .0015.
Solution
The fraction for of .15% = 3/2000 I convereted .15% to a decimal which equals .0015..
1.i. Oxygen is the final electron acceptor for the electron transpor.pdfannaiwatertreatment
1.i. Oxygen is the final electron acceptor for the electron transport chain.
ii. the matrix
iii. the intermembrane space
iv. ATP synthase
v. It is present at numerous locations in the membrane of the mitochondria
vi. NADH+
vii. Thylakoid membrane
viii. Stroma of the chloroplast.
Solution
1.i. Oxygen is the final electron acceptor for the electron transport chain.
ii. the matrix
iii. the intermembrane space
iv. ATP synthase
v. It is present at numerous locations in the membrane of the mitochondria
vi. NADH+
vii. Thylakoid membrane
viii. Stroma of the chloroplast..
The transverse electric and transverse magnetic (TEM) wave traveling.pdfannaiwatertreatment
The transverse electric and transverse magnetic (TEM) wave traveling in the y-direction
charactertized by Ey=0,Hy=0.
this means that the electric and magnetic fields are completely transverse to the direction of wave
propagation. so electric field have Ez component. because Ez*Hx=Ay(direction of wave
propagation)
Solution
The transverse electric and transverse magnetic (TEM) wave traveling in the y-direction
charactertized by Ey=0,Hy=0.
this means that the electric and magnetic fields are completely transverse to the direction of wave
propagation. so electric field have Ez component. because Ez*Hx=Ay(direction of wave
propagation).
Solution You sound a tone and then squirt pickle juice in a child.pdfannaiwatertreatment
Solution
:
You sound a tone and then squirt pickle juice in a child\'s mouth causing salivation. After
several pairings of the tone and the pickle juice, the tone alone causes salivation. The tone is a
conditioned stimulus, and the pickle juice is a unconditioned stimulus.Firstly tone is working as
neutral stimulus and pickle as unconditioned stimulus.After the classical conditioning tone alone
acts as conditioned stimulus and cause for salivation alone..
Anisole is a weakly polar ether compound.Acetone is a polar ketone.pdfannaiwatertreatment
Anisole is a weakly polar ether compound.
Acetone is a polar ketone.
Ethanol is a polar alcohol.
Diethyl ether is a weakly polar ether.
Anisole has the highest affinity for diethyl ether which has the most similar intermolecular
bonding forces. Thus it will move farthest up the TLC plate in diethyl ether.
Solution
Anisole is a weakly polar ether compound.
Acetone is a polar ketone.
Ethanol is a polar alcohol.
Diethyl ether is a weakly polar ether.
Anisole has the highest affinity for diethyl ether which has the most similar intermolecular
bonding forces. Thus it will move farthest up the TLC plate in diethyl ether..
No , its located towards cytoplasmic side. Its used as precursor.pdfannaiwatertreatment
No , it\'s located towards cytoplasmic side. It\'s used as precursor for cardiolipin synthesis. It s
present in the inner mitochondrial membrane of heart, it\'s also present in prokaryotes
Solution
No , it\'s located towards cytoplasmic side. It\'s used as precursor for cardiolipin synthesis. It s
present in the inner mitochondrial membrane of heart, it\'s also present in prokaryotes.
HI,The image is not clear,but i tried my best to decode it.As far .pdfannaiwatertreatment
HI,
The image is not clear,but i tried my best to decode it.As far i know u are correct.
Solution
HI,
The image is not clear,but i tried my best to decode it.As far i know u are correct..
Let A,B and C be the events that coin of each type respectively were.pdfannaiwatertreatment
Let A,B and C be the events that coin of each type respectively were flipped
Let H be the event of getting a head
i) We need to detremine P(A|H)=P(A and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(A and H) = 2/5*0.2=.08
P(A|H)=.08/.54=.148148
ii)
P(B|H)=P(B and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(B and H) = 2/5*0.7=.28
P(B|H)=.28/.54=.518519
iii)
P(C|H)=P(C and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(C and H) = 1/5*0.9=.18
P(A|H)=.18/.54=.33333
2.) Probability that it will be heads if you flip it again = 0.54
Solution
Let A,B and C be the events that coin of each type respectively were flipped
Let H be the event of getting a head
i) We need to detremine P(A|H)=P(A and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(A and H) = 2/5*0.2=.08
P(A|H)=.08/.54=.148148
ii)
P(B|H)=P(B and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(B and H) = 2/5*0.7=.28
P(B|H)=.28/.54=.518519
iii)
P(C|H)=P(C and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(C and H) = 1/5*0.9=.18
P(A|H)=.18/.54=.33333
2.) Probability that it will be heads if you flip it again = 0.54.
1.i. Oxygen is the final electron acceptor for the electron transpor.pdfannaiwatertreatment
1.i. Oxygen is the final electron acceptor for the electron transport chain.
ii. the matrix
iii. the intermembrane space
iv. ATP synthase
v. It is present at numerous locations in the membrane of the mitochondria
vi. NADH+
vii. Thylakoid membrane
viii. Stroma of the chloroplast.
Solution
1.i. Oxygen is the final electron acceptor for the electron transport chain.
ii. the matrix
iii. the intermembrane space
iv. ATP synthase
v. It is present at numerous locations in the membrane of the mitochondria
vi. NADH+
vii. Thylakoid membrane
viii. Stroma of the chloroplast..
The transverse electric and transverse magnetic (TEM) wave traveling.pdfannaiwatertreatment
The transverse electric and transverse magnetic (TEM) wave traveling in the y-direction
charactertized by Ey=0,Hy=0.
this means that the electric and magnetic fields are completely transverse to the direction of wave
propagation. so electric field have Ez component. because Ez*Hx=Ay(direction of wave
propagation)
Solution
The transverse electric and transverse magnetic (TEM) wave traveling in the y-direction
charactertized by Ey=0,Hy=0.
this means that the electric and magnetic fields are completely transverse to the direction of wave
propagation. so electric field have Ez component. because Ez*Hx=Ay(direction of wave
propagation).
Solution You sound a tone and then squirt pickle juice in a child.pdfannaiwatertreatment
Solution
:
You sound a tone and then squirt pickle juice in a child\'s mouth causing salivation. After
several pairings of the tone and the pickle juice, the tone alone causes salivation. The tone is a
conditioned stimulus, and the pickle juice is a unconditioned stimulus.Firstly tone is working as
neutral stimulus and pickle as unconditioned stimulus.After the classical conditioning tone alone
acts as conditioned stimulus and cause for salivation alone..
Anisole is a weakly polar ether compound.Acetone is a polar ketone.pdfannaiwatertreatment
Anisole is a weakly polar ether compound.
Acetone is a polar ketone.
Ethanol is a polar alcohol.
Diethyl ether is a weakly polar ether.
Anisole has the highest affinity for diethyl ether which has the most similar intermolecular
bonding forces. Thus it will move farthest up the TLC plate in diethyl ether.
Solution
Anisole is a weakly polar ether compound.
Acetone is a polar ketone.
Ethanol is a polar alcohol.
Diethyl ether is a weakly polar ether.
Anisole has the highest affinity for diethyl ether which has the most similar intermolecular
bonding forces. Thus it will move farthest up the TLC plate in diethyl ether..
No , its located towards cytoplasmic side. Its used as precursor.pdfannaiwatertreatment
No , it\'s located towards cytoplasmic side. It\'s used as precursor for cardiolipin synthesis. It s
present in the inner mitochondrial membrane of heart, it\'s also present in prokaryotes
Solution
No , it\'s located towards cytoplasmic side. It\'s used as precursor for cardiolipin synthesis. It s
present in the inner mitochondrial membrane of heart, it\'s also present in prokaryotes.
HI,The image is not clear,but i tried my best to decode it.As far .pdfannaiwatertreatment
HI,
The image is not clear,but i tried my best to decode it.As far i know u are correct.
Solution
HI,
The image is not clear,but i tried my best to decode it.As far i know u are correct..
Let A,B and C be the events that coin of each type respectively were.pdfannaiwatertreatment
Let A,B and C be the events that coin of each type respectively were flipped
Let H be the event of getting a head
i) We need to detremine P(A|H)=P(A and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(A and H) = 2/5*0.2=.08
P(A|H)=.08/.54=.148148
ii)
P(B|H)=P(B and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(B and H) = 2/5*0.7=.28
P(B|H)=.28/.54=.518519
iii)
P(C|H)=P(C and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(C and H) = 1/5*0.9=.18
P(A|H)=.18/.54=.33333
2.) Probability that it will be heads if you flip it again = 0.54
Solution
Let A,B and C be the events that coin of each type respectively were flipped
Let H be the event of getting a head
i) We need to detremine P(A|H)=P(A and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(A and H) = 2/5*0.2=.08
P(A|H)=.08/.54=.148148
ii)
P(B|H)=P(B and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(B and H) = 2/5*0.7=.28
P(B|H)=.28/.54=.518519
iii)
P(C|H)=P(C and H)/P(H)
P(H) = (2*0.2+2*0.7+1*0.9)/5=2.7/5=.54
P(C and H) = 1/5*0.9=.18
P(A|H)=.18/.54=.33333
2.) Probability that it will be heads if you flip it again = 0.54.
International Portfoilo investment is affected by following few fact.pdfannaiwatertreatment
International Portfoilo investment is affected by following few factors
1) Foregin Exchage Rate: According to Purchasing power parity Interest rate is related to
exchange rate therfore Developed countries where interest rates are very low they look for
investment internationally where rates are high
2) Country Risk: It is the important factor which plays vital role of deciding investment
participation. Politically volatile countries attract low internatioanl investment in terms of FDI
3)Currency Risk: Highly depreciated risk in which fixed exchange rate dont support the
macroeconomic prudentials
Solution
International Portfoilo investment is affected by following few factors
1) Foregin Exchage Rate: According to Purchasing power parity Interest rate is related to
exchange rate therfore Developed countries where interest rates are very low they look for
investment internationally where rates are high
2) Country Risk: It is the important factor which plays vital role of deciding investment
participation. Politically volatile countries attract low internatioanl investment in terms of FDI
3)Currency Risk: Highly depreciated risk in which fixed exchange rate dont support the
macroeconomic prudentials.
Hi!For the first reaction, Iron (Fe) is going from a positive oxid.pdfannaiwatertreatment
Hi!
For the first reaction, Iron (Fe) is going from a positive oxidation state (bonded to 3 oxygen
molecules) to a 0 oxidation state on the product side. Therefore, Iron is being reduced. At the
same time, carbon goes from being bonded to a single oxygen molecule to being bonded to two
oxygen molecules on the product side (CO2). Therefore, Carbon is being oxidized.
For the second reaction, Hydrogen starts out on the rectant side as an H+ ion and gains a bond to
oxygen to form water (H2O) on the product side. Since it gains a bond to oxygen, Hydrogen is
being oxidized. At the same time, Nitrogen goes from being bonded to three oxygen molecules
(NO3) on the reactant side to being bonded to a single oxygen on the product side (NO).
Therefore, Nitrogen is being reduced.
I hope this helps! PLEASE dont forget to rate : )
Solution
Hi!
For the first reaction, Iron (Fe) is going from a positive oxidation state (bonded to 3 oxygen
molecules) to a 0 oxidation state on the product side. Therefore, Iron is being reduced. At the
same time, carbon goes from being bonded to a single oxygen molecule to being bonded to two
oxygen molecules on the product side (CO2). Therefore, Carbon is being oxidized.
For the second reaction, Hydrogen starts out on the rectant side as an H+ ion and gains a bond to
oxygen to form water (H2O) on the product side. Since it gains a bond to oxygen, Hydrogen is
being oxidized. At the same time, Nitrogen goes from being bonded to three oxygen molecules
(NO3) on the reactant side to being bonded to a single oxygen on the product side (NO).
Therefore, Nitrogen is being reduced.
I hope this helps! PLEASE dont forget to rate : ).
If we design any page or devoleped any code we use the word shield a.pdfannaiwatertreatment
If we design any page or devoleped any code we use the word shield as the privacy acts as a
shield to keep the data secured and not allowed to see by others. The main functios of thge shield
is:interference, scrutinity.
Data mining can threaten privacy by following ways they are
- They may collect our information in our systems throgh many techniques without consulting us
and taking proper permission from us.
- They may occur for transferring the data from one pc to anouther pc by using many mediums
wiothout contacting the personal users.
- If at all there will be having any groups combinely who are working on it there may be a
chance of entering in to that group and taking the data without any knowledge of others.
Solution
If we design any page or devoleped any code we use the word shield as the privacy acts as a
shield to keep the data secured and not allowed to see by others. The main functios of thge shield
is:interference, scrutinity.
Data mining can threaten privacy by following ways they are
- They may collect our information in our systems throgh many techniques without consulting us
and taking proper permission from us.
- They may occur for transferring the data from one pc to anouther pc by using many mediums
wiothout contacting the personal users.
- If at all there will be having any groups combinely who are working on it there may be a
chance of entering in to that group and taking the data without any knowledge of others..
import school.; import school.courses.;public class Main { p.pdfannaiwatertreatment
import school.*;
import school.courses.*;
public class Main {
public static void main(String[] args) {
Teacher phil = new Teacher(\"Phil\");
Teacher bill = new Teacher(\"Bill\");
Teacher lil = new Teacher(\"Lil\");
Teacher joe = new Teacher(\"Joe\");
Course[] courses = {
new NetworkCourse(15, phil),
new SwingCourse(30, bill),
new APIDesignCourse(50, lil),
new PerformanceCourse(5, joe)
};
School school = new School(courses);
Student ludwig = new Student(\"Ludwig\");
Student cam = new Student(\"Cam\");
Student daniel = new Student(\"Daniel\");
ludwig.setPreferredCourses(NetworkCourse.class, SwingCourse.class); //give students preferred
classes if they have them
cam.setPreferredCourses(APIDesignCourse.class, PerformanceCourse.class,
NetworkCourse.class);
school.register(ludwig, cam, daniel);
/*
* Below is where we test by printing things out
*/
test(school);
}
static void test(School school) {
/*
* Prints all the students in the school, all the courses in the school, and which course each
student has
*/
System.out.println(\"Students and their courses:\");
for(Student student : school.getStudents()) {
if(student != null) {
String message = student.getName() + \" is taking\"; //message will reset for each new student,
since we do = and not += here
for(Course course : student.getCourses())
message += \" - \" + course.getName();
System.out.println(message);
}
}
System.out.println(\"\ Courses and their students:\");
for(Course course : school.getCourses()) {
String message = course.getName() + \" is taken by\";
for(Student student : course.getStudents()) {
if(student != null)
message += \" - \" + student.getName();
}
System.out.println(message);
}
}
}
School.java
package school;
import java.util.*;
public class School {
private Course[] courses;
private Student[] students;
public School(Course[] courses) {
this.courses = courses;
//ive been told that constructors shouldnt contain logic. is there any other way to handle this?
int numOfStudents = 0;
for(Course course : courses)
numOfStudents += course.getStudents().length;
students = new Student[numOfStudents];
}
public void register(Student...students) { //this method is pretty messy, and loops quite a few
times. any suggestions?
if(isFull())
throw new IllegalStateException(\"Cannot register anymore students at this time\");
for(Student student : students) {
if(Arrays.asList(this.students).contains(student)) //wrapping the array every loop. is there any
better way to do this, without creating my own private contains method for students?
throw new IllegalArgumentException(\"You cannot add the same student to a school twice\");
//should I be throwing a runtime exception here? or should i just continue with the rest of the
students
for(Course course : courses) {
if(student.prefersCourse(course) && !course.isFull())
student.assignCourse(course);
}
verifyStudent(student); //make sure the student is ready for school
student.setSchool(this);
for(int i = 0; i < this.students.length; i++) {
if(this.students[i].
Det M of 2x2 matrix can be any value from -infty to +inftyE) B1+B2.pdfannaiwatertreatment
Det M of 2x2 matrix can be any value from -infty to +infty
E) B1+B2 will have a21 as 0 hence belong to the group
and Identity element is 0 matrix
Inverse of B = -B
Hence B is a subgroup
2)
Solution
Det M of 2x2 matrix can be any value from -infty to +infty
E) B1+B2 will have a21 as 0 hence belong to the group
and Identity element is 0 matrix
Inverse of B = -B
Hence B is a subgroup
2).
Aspirin is acetylsalicylic acid. As the compound ages, small amounts.pdfannaiwatertreatment
Aspirin is acetylsalicylic acid. As the compound ages, small amounts of water in the air allows
some the compound to hydrolyze. This give salicylic acid and acetic acid. Acetic acid is the main
ingredient in vinegar, therefor the smell.
Solution
Aspirin is acetylsalicylic acid. As the compound ages, small amounts of water in the air allows
some the compound to hydrolyze. This give salicylic acid and acetic acid. Acetic acid is the main
ingredient in vinegar, therefor the smell..
they are the atomic number of those element so from periodic table u have to guess
the element so L is sodium and M is florine
Solution
they are the atomic number of those element so from periodic table u have to guess
the element so L is sodium and M is florine.
Option II is correct ; in other 4 cases there is addition of hydrogen.
Solution
Option II is correct ; in other 4 cases there is addition of hydrogen..
look at a periodic table. The trend for ionization energy is as the groups increase
(left to right), ionization energy increases. As the periods decrease (bottom to top), ionization
energy increases. So helium is the most.
Solution
look at a periodic table. The trend for ionization energy is as the groups increase
(left to right), ionization energy increases. As the periods decrease (bottom to top), ionization
energy increases. So helium is the most..
Methyl Orange is a good indicator for pH\'s ranging from 3 to 5,with 3 being more
acidic than 5. When methyl orange is added to distilled water (pH 6.5-7), theindicator turns
yellow because the concentration of H+ ions indistilled water is low enough that the methyl
orange candisassociate it\'s H+ ions. When 6M HCl is added, H+ ions are added to the system,
disturbingthe equilibrium of methyl orange and making the solution moreacidic. Since H+ ions
are the product of the reaction, the methylorange will shift it\'s equilibrium to the left, therefore
makingmore of the reactant , HMO which turns the solution red. When 6M NAOH is added
after the HCl, the -OH ions react with the H+ions to create water (reaction 3 in your discussion
section), whichremoves H+ and -OH ions from the solution. The equilibrium ofmethyl orange is
disturbed again, and the equilibrium will shift tothe right to make more of the product. The
solution turns back toyellow as more reactants are turned into products, so HMO isconsumed and
H+ ions are created again. I think this covered all of your questions, I hope this helps! Steve
Solution
Methyl Orange is a good indicator for pH\'s ranging from 3 to 5,with 3 being more
acidic than 5. When methyl orange is added to distilled water (pH 6.5-7), theindicator turns
yellow because the concentration of H+ ions indistilled water is low enough that the methyl
orange candisassociate it\'s H+ ions. When 6M HCl is added, H+ ions are added to the system,
disturbingthe equilibrium of methyl orange and making the solution moreacidic. Since H+ ions
are the product of the reaction, the methylorange will shift it\'s equilibrium to the left, therefore
makingmore of the reactant , HMO which turns the solution red. When 6M NAOH is added
after the HCl, the -OH ions react with the H+ions to create water (reaction 3 in your discussion
section), whichremoves H+ and -OH ions from the solution. The equilibrium ofmethyl orange is
disturbed again, and the equilibrium will shift tothe right to make more of the product. The
solution turns back toyellow as more reactants are turned into products, so HMO isconsumed and
H+ ions are created again. I think this covered all of your questions, I hope this helps! Steve.
H2SO3 is strongest and CH3CH2CH2CH2OH is weakest. as H2SO4 is ionic
compound and CH3CH2CH2CH2OH is organic convalent compound hence H2SO4 will easily
donate H+ ion.
Solution
H2SO3 is strongest and CH3CH2CH2CH2OH is weakest. as H2SO4 is ionic
compound and CH3CH2CH2CH2OH is organic convalent compound hence H2SO4 will easily
donate H+ ion..
Hydrophilic substances are water loving. They have a tendency to mix with,
dissolve in, or be wetted by water.They can be useful to humans in many ways, especially for
keeping things clean in industrial settings. The hydrophilic portion of soap is the carboxylic acid
group -COONa
Solution
Hydrophilic substances are water loving. They have a tendency to mix with,
dissolve in, or be wetted by water.They can be useful to humans in many ways, especially for
keeping things clean in industrial settings. The hydrophilic portion of soap is the carboxylic acid
group -COONa.
Final traces of water are removed by treating the organic solution with a drying
agent such as NaSO4.
Solution
Final traces of water are removed by treating the organic solution with a drying
agent such as NaSO4..
Br-Br is the most nonpolar because it is a bond between two atoms of the same
element, so the the electrons are shared equally between the two atoms.
Solution
Br-Br is the most nonpolar because it is a bond between two atoms of the same
element, so the the electrons are shared equally between the two atoms..
International Portfoilo investment is affected by following few fact.pdfannaiwatertreatment
International Portfoilo investment is affected by following few factors
1) Foregin Exchage Rate: According to Purchasing power parity Interest rate is related to
exchange rate therfore Developed countries where interest rates are very low they look for
investment internationally where rates are high
2) Country Risk: It is the important factor which plays vital role of deciding investment
participation. Politically volatile countries attract low internatioanl investment in terms of FDI
3)Currency Risk: Highly depreciated risk in which fixed exchange rate dont support the
macroeconomic prudentials
Solution
International Portfoilo investment is affected by following few factors
1) Foregin Exchage Rate: According to Purchasing power parity Interest rate is related to
exchange rate therfore Developed countries where interest rates are very low they look for
investment internationally where rates are high
2) Country Risk: It is the important factor which plays vital role of deciding investment
participation. Politically volatile countries attract low internatioanl investment in terms of FDI
3)Currency Risk: Highly depreciated risk in which fixed exchange rate dont support the
macroeconomic prudentials.
Hi!For the first reaction, Iron (Fe) is going from a positive oxid.pdfannaiwatertreatment
Hi!
For the first reaction, Iron (Fe) is going from a positive oxidation state (bonded to 3 oxygen
molecules) to a 0 oxidation state on the product side. Therefore, Iron is being reduced. At the
same time, carbon goes from being bonded to a single oxygen molecule to being bonded to two
oxygen molecules on the product side (CO2). Therefore, Carbon is being oxidized.
For the second reaction, Hydrogen starts out on the rectant side as an H+ ion and gains a bond to
oxygen to form water (H2O) on the product side. Since it gains a bond to oxygen, Hydrogen is
being oxidized. At the same time, Nitrogen goes from being bonded to three oxygen molecules
(NO3) on the reactant side to being bonded to a single oxygen on the product side (NO).
Therefore, Nitrogen is being reduced.
I hope this helps! PLEASE dont forget to rate : )
Solution
Hi!
For the first reaction, Iron (Fe) is going from a positive oxidation state (bonded to 3 oxygen
molecules) to a 0 oxidation state on the product side. Therefore, Iron is being reduced. At the
same time, carbon goes from being bonded to a single oxygen molecule to being bonded to two
oxygen molecules on the product side (CO2). Therefore, Carbon is being oxidized.
For the second reaction, Hydrogen starts out on the rectant side as an H+ ion and gains a bond to
oxygen to form water (H2O) on the product side. Since it gains a bond to oxygen, Hydrogen is
being oxidized. At the same time, Nitrogen goes from being bonded to three oxygen molecules
(NO3) on the reactant side to being bonded to a single oxygen on the product side (NO).
Therefore, Nitrogen is being reduced.
I hope this helps! PLEASE dont forget to rate : ).
If we design any page or devoleped any code we use the word shield a.pdfannaiwatertreatment
If we design any page or devoleped any code we use the word shield as the privacy acts as a
shield to keep the data secured and not allowed to see by others. The main functios of thge shield
is:interference, scrutinity.
Data mining can threaten privacy by following ways they are
- They may collect our information in our systems throgh many techniques without consulting us
and taking proper permission from us.
- They may occur for transferring the data from one pc to anouther pc by using many mediums
wiothout contacting the personal users.
- If at all there will be having any groups combinely who are working on it there may be a
chance of entering in to that group and taking the data without any knowledge of others.
Solution
If we design any page or devoleped any code we use the word shield as the privacy acts as a
shield to keep the data secured and not allowed to see by others. The main functios of thge shield
is:interference, scrutinity.
Data mining can threaten privacy by following ways they are
- They may collect our information in our systems throgh many techniques without consulting us
and taking proper permission from us.
- They may occur for transferring the data from one pc to anouther pc by using many mediums
wiothout contacting the personal users.
- If at all there will be having any groups combinely who are working on it there may be a
chance of entering in to that group and taking the data without any knowledge of others..
import school.; import school.courses.;public class Main { p.pdfannaiwatertreatment
import school.*;
import school.courses.*;
public class Main {
public static void main(String[] args) {
Teacher phil = new Teacher(\"Phil\");
Teacher bill = new Teacher(\"Bill\");
Teacher lil = new Teacher(\"Lil\");
Teacher joe = new Teacher(\"Joe\");
Course[] courses = {
new NetworkCourse(15, phil),
new SwingCourse(30, bill),
new APIDesignCourse(50, lil),
new PerformanceCourse(5, joe)
};
School school = new School(courses);
Student ludwig = new Student(\"Ludwig\");
Student cam = new Student(\"Cam\");
Student daniel = new Student(\"Daniel\");
ludwig.setPreferredCourses(NetworkCourse.class, SwingCourse.class); //give students preferred
classes if they have them
cam.setPreferredCourses(APIDesignCourse.class, PerformanceCourse.class,
NetworkCourse.class);
school.register(ludwig, cam, daniel);
/*
* Below is where we test by printing things out
*/
test(school);
}
static void test(School school) {
/*
* Prints all the students in the school, all the courses in the school, and which course each
student has
*/
System.out.println(\"Students and their courses:\");
for(Student student : school.getStudents()) {
if(student != null) {
String message = student.getName() + \" is taking\"; //message will reset for each new student,
since we do = and not += here
for(Course course : student.getCourses())
message += \" - \" + course.getName();
System.out.println(message);
}
}
System.out.println(\"\ Courses and their students:\");
for(Course course : school.getCourses()) {
String message = course.getName() + \" is taken by\";
for(Student student : course.getStudents()) {
if(student != null)
message += \" - \" + student.getName();
}
System.out.println(message);
}
}
}
School.java
package school;
import java.util.*;
public class School {
private Course[] courses;
private Student[] students;
public School(Course[] courses) {
this.courses = courses;
//ive been told that constructors shouldnt contain logic. is there any other way to handle this?
int numOfStudents = 0;
for(Course course : courses)
numOfStudents += course.getStudents().length;
students = new Student[numOfStudents];
}
public void register(Student...students) { //this method is pretty messy, and loops quite a few
times. any suggestions?
if(isFull())
throw new IllegalStateException(\"Cannot register anymore students at this time\");
for(Student student : students) {
if(Arrays.asList(this.students).contains(student)) //wrapping the array every loop. is there any
better way to do this, without creating my own private contains method for students?
throw new IllegalArgumentException(\"You cannot add the same student to a school twice\");
//should I be throwing a runtime exception here? or should i just continue with the rest of the
students
for(Course course : courses) {
if(student.prefersCourse(course) && !course.isFull())
student.assignCourse(course);
}
verifyStudent(student); //make sure the student is ready for school
student.setSchool(this);
for(int i = 0; i < this.students.length; i++) {
if(this.students[i].
Det M of 2x2 matrix can be any value from -infty to +inftyE) B1+B2.pdfannaiwatertreatment
Det M of 2x2 matrix can be any value from -infty to +infty
E) B1+B2 will have a21 as 0 hence belong to the group
and Identity element is 0 matrix
Inverse of B = -B
Hence B is a subgroup
2)
Solution
Det M of 2x2 matrix can be any value from -infty to +infty
E) B1+B2 will have a21 as 0 hence belong to the group
and Identity element is 0 matrix
Inverse of B = -B
Hence B is a subgroup
2).
Aspirin is acetylsalicylic acid. As the compound ages, small amounts.pdfannaiwatertreatment
Aspirin is acetylsalicylic acid. As the compound ages, small amounts of water in the air allows
some the compound to hydrolyze. This give salicylic acid and acetic acid. Acetic acid is the main
ingredient in vinegar, therefor the smell.
Solution
Aspirin is acetylsalicylic acid. As the compound ages, small amounts of water in the air allows
some the compound to hydrolyze. This give salicylic acid and acetic acid. Acetic acid is the main
ingredient in vinegar, therefor the smell..
they are the atomic number of those element so from periodic table u have to guess
the element so L is sodium and M is florine
Solution
they are the atomic number of those element so from periodic table u have to guess
the element so L is sodium and M is florine.
Option II is correct ; in other 4 cases there is addition of hydrogen.
Solution
Option II is correct ; in other 4 cases there is addition of hydrogen..
look at a periodic table. The trend for ionization energy is as the groups increase
(left to right), ionization energy increases. As the periods decrease (bottom to top), ionization
energy increases. So helium is the most.
Solution
look at a periodic table. The trend for ionization energy is as the groups increase
(left to right), ionization energy increases. As the periods decrease (bottom to top), ionization
energy increases. So helium is the most..
Methyl Orange is a good indicator for pH\'s ranging from 3 to 5,with 3 being more
acidic than 5. When methyl orange is added to distilled water (pH 6.5-7), theindicator turns
yellow because the concentration of H+ ions indistilled water is low enough that the methyl
orange candisassociate it\'s H+ ions. When 6M HCl is added, H+ ions are added to the system,
disturbingthe equilibrium of methyl orange and making the solution moreacidic. Since H+ ions
are the product of the reaction, the methylorange will shift it\'s equilibrium to the left, therefore
makingmore of the reactant , HMO which turns the solution red. When 6M NAOH is added
after the HCl, the -OH ions react with the H+ions to create water (reaction 3 in your discussion
section), whichremoves H+ and -OH ions from the solution. The equilibrium ofmethyl orange is
disturbed again, and the equilibrium will shift tothe right to make more of the product. The
solution turns back toyellow as more reactants are turned into products, so HMO isconsumed and
H+ ions are created again. I think this covered all of your questions, I hope this helps! Steve
Solution
Methyl Orange is a good indicator for pH\'s ranging from 3 to 5,with 3 being more
acidic than 5. When methyl orange is added to distilled water (pH 6.5-7), theindicator turns
yellow because the concentration of H+ ions indistilled water is low enough that the methyl
orange candisassociate it\'s H+ ions. When 6M HCl is added, H+ ions are added to the system,
disturbingthe equilibrium of methyl orange and making the solution moreacidic. Since H+ ions
are the product of the reaction, the methylorange will shift it\'s equilibrium to the left, therefore
makingmore of the reactant , HMO which turns the solution red. When 6M NAOH is added
after the HCl, the -OH ions react with the H+ions to create water (reaction 3 in your discussion
section), whichremoves H+ and -OH ions from the solution. The equilibrium ofmethyl orange is
disturbed again, and the equilibrium will shift tothe right to make more of the product. The
solution turns back toyellow as more reactants are turned into products, so HMO isconsumed and
H+ ions are created again. I think this covered all of your questions, I hope this helps! Steve.
H2SO3 is strongest and CH3CH2CH2CH2OH is weakest. as H2SO4 is ionic
compound and CH3CH2CH2CH2OH is organic convalent compound hence H2SO4 will easily
donate H+ ion.
Solution
H2SO3 is strongest and CH3CH2CH2CH2OH is weakest. as H2SO4 is ionic
compound and CH3CH2CH2CH2OH is organic convalent compound hence H2SO4 will easily
donate H+ ion..
Hydrophilic substances are water loving. They have a tendency to mix with,
dissolve in, or be wetted by water.They can be useful to humans in many ways, especially for
keeping things clean in industrial settings. The hydrophilic portion of soap is the carboxylic acid
group -COONa
Solution
Hydrophilic substances are water loving. They have a tendency to mix with,
dissolve in, or be wetted by water.They can be useful to humans in many ways, especially for
keeping things clean in industrial settings. The hydrophilic portion of soap is the carboxylic acid
group -COONa.
Final traces of water are removed by treating the organic solution with a drying
agent such as NaSO4.
Solution
Final traces of water are removed by treating the organic solution with a drying
agent such as NaSO4..
Br-Br is the most nonpolar because it is a bond between two atoms of the same
element, so the the electrons are shared equally between the two atoms.
Solution
Br-Br is the most nonpolar because it is a bond between two atoms of the same
element, so the the electrons are shared equally between the two atoms..
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
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.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
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.
Cambridge International AS A Level Biology Coursebook - EBook (MaryFosbery J...
The fraction for of .15 = 32000 I convereted .15 to a decimal whi.pdf
1. The fraction for of .15% = 3/2000 I convereted .15% to a decimal which equals .0015.
Solution
The fraction for of .15% = 3/2000 I convereted .15% to a decimal which equals .0015.