Det M of 2x2 matrix can be any value from -infty to +inftyE) B1+B2.pdf
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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).
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1.i. Oxygen is the final electron acceptor for the electron transpor.pdf
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.pdf
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).
The fraction for of .15 = 32000 I convereted .15 to a decimal whi.pdf
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..
Solution You sound a tone and then squirt pickle juice in a child.pdf
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.pdf
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.pdf
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 .pdf
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..
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1.i. Oxygen is the final electron acceptor for the electron transpor.pdf
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.pdf
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).
The fraction for of .15 = 32000 I convereted .15 to a decimal whi.pdf
The fraction for of .15% = 3/2000 I convereted .15% to a decimal which equals .0015.
Solution
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Solution You sound a tone and then squirt pickle juice in a child.pdf
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.pdf
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.pdf
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 .pdf
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.pdf
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.pdf
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.pdf
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.pdf
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.pdf
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].
Aspirin is acetylsalicylic acid. As the compound ages, small amounts.pdf
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..
B. The solution has more hydroxide ions than hydronium ions. .pdf
B. The solution has more hydroxide ions than hydronium ions.
Solution
B. The solution has more hydroxide ions than hydronium ions..
they are the atomic number of those element so fr.pdf
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.
s(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7 .pdf
s*(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7
Solution
s*(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7.
Option II is correct ; in other 4 cases there is .pdf
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..
NaOH ==== Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H.pdf
NaOH ====> Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/0.25 = 4.0*10^-14 pH = -log[H+] = -log(4.0*10^-14) = 13.397 = 13.40
pH+pOH=14 pOh = 14-pH = 14-13.40=0.6
Solution
NaOH ====> Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/0.25 = 4.0*10^-14 pH = -log[H+] = -log(4.0*10^-14) = 13.397 = 13.40
pH+pOH=14 pOh = 14-pH = 14-13.40=0.6.
look at a periodic table. The trend for ionizatio.pdf
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 pHs rangi.pdf
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..pdf
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 hav.pdf
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.pdf
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 b.pdf
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..
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Unit 8 - Information and Communication Technology (Paper I).pdf
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.
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Let A,B and C be the events that coin of each type respectively were.pdf
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.pdf
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.pdf
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.pdf
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.pdf
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].
Aspirin is acetylsalicylic acid. As the compound ages, small amounts.pdf
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..
B. The solution has more hydroxide ions than hydronium ions. .pdf
B. The solution has more hydroxide ions than hydronium ions.
Solution
B. The solution has more hydroxide ions than hydronium ions..
they are the atomic number of those element so fr.pdf
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.
s(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7 .pdf
s*(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7
Solution
s*(s+1.7x10-3)=2.1x10-10 s=1.24x10^-7.
Option II is correct ; in other 4 cases there is .pdf
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..
NaOH ==== Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H.pdf
NaOH ====> Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/0.25 = 4.0*10^-14 pH = -log[H+] = -log(4.0*10^-14) = 13.397 = 13.40
pH+pOH=14 pOh = 14-pH = 14-13.40=0.6
Solution
NaOH ====> Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H+][OH-] = 1.0*10^-14
[H+] = (1.0*10^-14)/0.25 = 4.0*10^-14 pH = -log[H+] = -log(4.0*10^-14) = 13.397 = 13.40
pH+pOH=14 pOh = 14-pH = 14-13.40=0.6.
look at a periodic table. The trend for ionizatio.pdf
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 pHs rangi.pdf
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..pdf
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 hav.pdf
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.pdf
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 b.pdf
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..
More from annaiwatertreatment (20)
Let A,B and C be the events that coin of each type respectively were.pdf
Let A,B and C be the events that coin of each type respectively were.pdf
International Portfoilo investment is affected by following few fact.pdf
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Hi!For the first reaction, Iron (Fe) is going from a positive oxid.pdf
Hi!For the first reaction, Iron (Fe) is going from a positive oxid.pdf
If we design any page or devoleped any code we use the word shield a.pdf
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import school.; import school.courses.;public class Main { p.pdf
import school.; import school.courses.;public class Main { p.pdf
Aspirin is acetylsalicylic acid. As the compound ages, small amounts.pdf
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B. The solution has more hydroxide ions than hydronium ions. .pdf
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they are the atomic number of those element so fr.pdf
they are the atomic number of those element so fr.pdf
Option II is correct ; in other 4 cases there is .pdf
Option II is correct ; in other 4 cases there is .pdf
NaOH ==== Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H.pdf
NaOH ==== Na+ (aq) + OH-(aq) [OH-] = 0.25 M [H.pdf
look at a periodic table. The trend for ionizatio.pdf
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Methyl Orange is a good indicator for pHs rangi.pdf
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H2SO3 is strongest and CH3CH2CH2CH2OH is weakest..pdf
H2SO3 is strongest and CH3CH2CH2CH2OH is weakest..pdf
Hydrophilic substances are water loving. They hav.pdf
Hydrophilic substances are water loving. They hav.pdf
Final traces of water are removed by treating the.pdf
Final traces of water are removed by treating the.pdf
Br-Br is the most nonpolar because it is a bond b.pdf
Br-Br is the most nonpolar because it is a bond b.pdf
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TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Det M of 2x2 matrix can be any value from -infty to +inftyE) B1+B2.pdf
- 1. 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)