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S monoclinic is most stable since entropy is the .pdf
S monoclinic is most stable since entropy is the largest
Solution
S monoclinic is most stable since entropy is the largest.
Option D is correct. B is limiting reagent .pdf
Option D is correct. B is limiting reagent
Solution
Option D is correct. B is limiting reagent.
Let us assume the conc of Ba(OH)2 is known to us .pdf
Let us assume the conc of Ba(OH)2 is known to us and initially we have known
amount of Ba(OH)2 in the flask which is completely dissociated in form of Ba2+ and OH- ions.
Now H2SO4 is added. H2SO4 dissociates to form H+ and SO42- ions. H+ and OH- ions form
water while Ba2+ and SO42- form insoluble BaSO4 when enough H2SO4 is added to consume
the Ba(OH)2 completely the conductance of the analyte which is being monitored continuosly
shows a sharp decrease and is at its lowest point. at this point H2SO4 addition is stopped and
titration ends. Now you know the molarity of Ba(OH)2 ...(M1) volume of Ba(OH)2 taken
initially ...(V1) and volume of H2SO4 consumed ...(V2) use M1V1 = M2V2 only unknown M2
which is molarity of H2SO4 can be calculated !!
Solution
Let us assume the conc of Ba(OH)2 is known to us and initially we have known
amount of Ba(OH)2 in the flask which is completely dissociated in form of Ba2+ and OH- ions.
Now H2SO4 is added. H2SO4 dissociates to form H+ and SO42- ions. H+ and OH- ions form
water while Ba2+ and SO42- form insoluble BaSO4 when enough H2SO4 is added to consume
the Ba(OH)2 completely the conductance of the analyte which is being monitored continuosly
shows a sharp decrease and is at its lowest point. at this point H2SO4 addition is stopped and
titration ends. Now you know the molarity of Ba(OH)2 ...(M1) volume of Ba(OH)2 taken
initially ...(V1) and volume of H2SO4 consumed ...(V2) use M1V1 = M2V2 only unknown M2
which is molarity of H2SO4 can be calculated !!.
Methods a, b, c, and e are formal additions of H2.pdf
Methods a, b, c, and e are formal additions of H2O or OH to the double bond, so no
carbonyl compound. Method d splits molecule by the double bond and usually produces two
carbonyl compounds (one in a few cases). So it\'s d)
Solution
Methods a, b, c, and e are formal additions of H2O or OH to the double bond, so no
carbonyl compound. Method d splits molecule by the double bond and usually produces two
carbonyl compounds (one in a few cases). So it\'s d).
h2s4 structure is H-S-S-S-S-H there for two middl.pdf
h2s4 structure is H-S-S-S-S-H there for two middle two s atoms are in 0 oxidation
state and and sidd s atoms have -1 oxidation state
Solution
h2s4 structure is H-S-S-S-S-H there for two middle two s atoms are in 0 oxidation
state and and sidd s atoms have -1 oxidation state.
freezing point decreases with increase in imp.pdf
freezing point decreases with increase in impurity.
Solution
freezing point decreases with increase in impurity..
We are currently living in the so-called information age which can b.pdf
We are currently living in the so-called information age which can be described as an era were
economic activities are mainly information based (an age of informationalization). This is due to
the development and use of technology. The main characteristics of this era can be summarized
as a rise in the number of knowledge workers, a world that has become more open - in the sense
of communication (global village/Gutenberg galaxy) and internationalization (trans-border flow
of data).
This paradigm shift brings new ethical and juridical problems which are mainly related to issues
such as the right of access to information, the right of privacy which is threatened by the
emphasis on the free flow of information, and the protection of the economic interest of the
owners of intellectual property.
In this paper the ethical questions related to the right to privacy of the individual which is
threatened by the use of technology will be discussed. Specific attention will be given to the
challenges these ethical problems pose to the information professional. A number of practical
guidelines, based on ethical norms will be laid down.
Definition of Privacy
Privacy can be defined as an individual condition of life characterized by exclusion from
publicity (Neetling et al., 1996, p. 36). The concept follows from the right to be left alone (Stair,
1992, p. 635; Shank, 1986, p. 12)1 . Shank (1986, p. 13) states that such a perception of privacy
set the course for passing of privacy laws in the United States for the ninety years that followed.
As such privacy could be regarded as a natural right which provides the foundation for the legal
right. The right to privacy is therefore protected under private law.
The legal right to privacy is constitutionally protected in most democratic societies. This
constitutional right is expressed in a variety of legislative forms. Examples include the Privacy
Act (1974) in the USA, the proposed Open Democracy Act in South Africa (1996) and the Data
Protection Act in England. During 1994 Australia also accepted a Privacy Charter containing 18
privacy principles which describe the right of a citizen concerning personal privacy as effected
by handling of information by the state (Collier, 1994, p. 44-45). The Organization for Economic
and Coordination and Development (OECD) also accepted in 1980 the Guidelines for the
Protection of Privacy and Transborder Flow of Personal Data (Collier, 1994, p. 41).
Privacy is an important right because it is a necessary condition for other rights such as freedom
and personal autonomy. There is thus a relationship between privacy, freedom and human
dignity. Respecting a person\'s privacy is to acknowledge such a person\'s right to freedom and to
recognize that individual as an autonomous human being.
Private communications. This category of privacy concerns all forms of personal communication
which a person wishes to keep private. The information exchanged during a reference interview
between t.
Recommended
S monoclinic is most stable since entropy is the .pdf
S monoclinic is most stable since entropy is the largest
Solution
S monoclinic is most stable since entropy is the largest.
Option D is correct. B is limiting reagent .pdf
Option D is correct. B is limiting reagent
Solution
Option D is correct. B is limiting reagent.
Let us assume the conc of Ba(OH)2 is known to us .pdf
Let us assume the conc of Ba(OH)2 is known to us and initially we have known
amount of Ba(OH)2 in the flask which is completely dissociated in form of Ba2+ and OH- ions.
Now H2SO4 is added. H2SO4 dissociates to form H+ and SO42- ions. H+ and OH- ions form
water while Ba2+ and SO42- form insoluble BaSO4 when enough H2SO4 is added to consume
the Ba(OH)2 completely the conductance of the analyte which is being monitored continuosly
shows a sharp decrease and is at its lowest point. at this point H2SO4 addition is stopped and
titration ends. Now you know the molarity of Ba(OH)2 ...(M1) volume of Ba(OH)2 taken
initially ...(V1) and volume of H2SO4 consumed ...(V2) use M1V1 = M2V2 only unknown M2
which is molarity of H2SO4 can be calculated !!
Solution
Let us assume the conc of Ba(OH)2 is known to us and initially we have known
amount of Ba(OH)2 in the flask which is completely dissociated in form of Ba2+ and OH- ions.
Now H2SO4 is added. H2SO4 dissociates to form H+ and SO42- ions. H+ and OH- ions form
water while Ba2+ and SO42- form insoluble BaSO4 when enough H2SO4 is added to consume
the Ba(OH)2 completely the conductance of the analyte which is being monitored continuosly
shows a sharp decrease and is at its lowest point. at this point H2SO4 addition is stopped and
titration ends. Now you know the molarity of Ba(OH)2 ...(M1) volume of Ba(OH)2 taken
initially ...(V1) and volume of H2SO4 consumed ...(V2) use M1V1 = M2V2 only unknown M2
which is molarity of H2SO4 can be calculated !!.
Methods a, b, c, and e are formal additions of H2.pdf
Methods a, b, c, and e are formal additions of H2O or OH to the double bond, so no
carbonyl compound. Method d splits molecule by the double bond and usually produces two
carbonyl compounds (one in a few cases). So it\'s d)
Solution
Methods a, b, c, and e are formal additions of H2O or OH to the double bond, so no
carbonyl compound. Method d splits molecule by the double bond and usually produces two
carbonyl compounds (one in a few cases). So it\'s d).
h2s4 structure is H-S-S-S-S-H there for two middl.pdf
h2s4 structure is H-S-S-S-S-H there for two middle two s atoms are in 0 oxidation
state and and sidd s atoms have -1 oxidation state
Solution
h2s4 structure is H-S-S-S-S-H there for two middle two s atoms are in 0 oxidation
state and and sidd s atoms have -1 oxidation state.
freezing point decreases with increase in imp.pdf
freezing point decreases with increase in impurity.
Solution
freezing point decreases with increase in impurity..
We are currently living in the so-called information age which can b.pdf
We are currently living in the so-called information age which can be described as an era were
economic activities are mainly information based (an age of informationalization). This is due to
the development and use of technology. The main characteristics of this era can be summarized
as a rise in the number of knowledge workers, a world that has become more open - in the sense
of communication (global village/Gutenberg galaxy) and internationalization (trans-border flow
of data).
This paradigm shift brings new ethical and juridical problems which are mainly related to issues
such as the right of access to information, the right of privacy which is threatened by the
emphasis on the free flow of information, and the protection of the economic interest of the
owners of intellectual property.
In this paper the ethical questions related to the right to privacy of the individual which is
threatened by the use of technology will be discussed. Specific attention will be given to the
challenges these ethical problems pose to the information professional. A number of practical
guidelines, based on ethical norms will be laid down.
Definition of Privacy
Privacy can be defined as an individual condition of life characterized by exclusion from
publicity (Neetling et al., 1996, p. 36). The concept follows from the right to be left alone (Stair,
1992, p. 635; Shank, 1986, p. 12)1 . Shank (1986, p. 13) states that such a perception of privacy
set the course for passing of privacy laws in the United States for the ninety years that followed.
As such privacy could be regarded as a natural right which provides the foundation for the legal
right. The right to privacy is therefore protected under private law.
The legal right to privacy is constitutionally protected in most democratic societies. This
constitutional right is expressed in a variety of legislative forms. Examples include the Privacy
Act (1974) in the USA, the proposed Open Democracy Act in South Africa (1996) and the Data
Protection Act in England. During 1994 Australia also accepted a Privacy Charter containing 18
privacy principles which describe the right of a citizen concerning personal privacy as effected
by handling of information by the state (Collier, 1994, p. 44-45). The Organization for Economic
and Coordination and Development (OECD) also accepted in 1980 the Guidelines for the
Protection of Privacy and Transborder Flow of Personal Data (Collier, 1994, p. 41).
Privacy is an important right because it is a necessary condition for other rights such as freedom
and personal autonomy. There is thus a relationship between privacy, freedom and human
dignity. Respecting a person\'s privacy is to acknowledge such a person\'s right to freedom and to
recognize that individual as an autonomous human being.
Private communications. This category of privacy concerns all forms of personal communication
which a person wishes to keep private. The information exchanged during a reference interview
between t.
The balanced equation is2 H2O2 = 2 H2O + O2SolutionThe bal.pdf
The balanced equation is:
2 H2O2 => 2 H2O + O2
Solution
The balanced equation is:
2 H2O2 => 2 H2O + O2.
Tay-Sachs disease is caused by a mutation (abnormal change) in the g.pdf
Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A,
and it is a recessive trait. This means that people will have the disease if they have two copies of
the defective gene, but they will not have the disease if they have at least one unaffected copy.
People with one normal copy and one defective copy are called carriers, because they can pass
the disease on to their children.
Seizures are sudden altacks of disease, often referring to some type of violent spasms.
Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population,
about 1 in 250 people carries the gene.
People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting
in two defective genes that make the body unable to produce Hex-A correctly. People who have
only one defective gene are called carriers. Carriers do not have the disease, because they have
inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their
children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the
disease.
However, some populations of people include more carriers than others. For example, 1 in 27
people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People
of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also
have a higher than usual risk of carrying the Tay-Sachs gene.
Solution
Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A,
and it is a recessive trait. This means that people will have the disease if they have two copies of
the defective gene, but they will not have the disease if they have at least one unaffected copy.
People with one normal copy and one defective copy are called carriers, because they can pass
the disease on to their children.
Seizures are sudden altacks of disease, often referring to some type of violent spasms.
Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population,
about 1 in 250 people carries the gene.
People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting
in two defective genes that make the body unable to produce Hex-A correctly. People who have
only one defective gene are called carriers. Carriers do not have the disease, because they have
inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their
children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the
disease.
However, some populations of people include more carriers than others. For example, 1 in 27
people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People
of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also
have a higher than usual risk of carrying the Tay-Sachs gene..
E.) is less acidic in other A ,B triple bond an.pdf
E.) is less acidic in other A ,B triple bond and double bond are responsible for
acidity In C,D steric hinderance from -CH3 is responsible for more acidicity than rest thus E is
the answer
Solution
E.) is less acidic in other A ,B triple bond and double bond are responsible for
acidity In C,D steric hinderance from -CH3 is responsible for more acidicity than rest thus E is
the answer.
Solution when user sending the email, then user should select eithe.pdf
Solution
: when user sending the email, then user should select either POP3 or IMAP protocol to send the
mails. Some of the differences are giving below.
POP3 e-mail protocol:
IMAP e-mail protocol:
SMTP (simple mail transfer protocol):
SMTP is not normally used as intermediate mail servers for sending mail. Because SMTP cannot
transmit the executable files or any binary objects.
Similarly it cannot transmit the text data that contains the national language characters. It will
have certain size limit to have an e-mail. So if the mail size is over the limit then it will be
rejected..
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
please give me points as nobody ecen noticed the qusetion as it is u.pdf
please give me points as nobody ecen noticed the qusetion as it is ununderstandable
Solution
please give me points as nobody ecen noticed the qusetion as it is ununderstandable.
Part-IQuestion 1. What is Dr. Warren’s hypothesis regarding the ba.pdf
Part-I
Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach?
Give at least three alternative hypotheses that could explain these findings.
Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active,
chronic gastritis and the presence of the bacteria found in stomachs.
hypothesis 1. Dr. Warren noticed a thin blue line on the surface of the tissue.
hypothesis 2: Warren found numerous spiral shaped bacteria when stained with a Warthin-Starry
silver stain of the histological section.
hypothesis 3: Dr. Warren also observed that the number of bacteria observed seemed to correlate
with the degree of inflammation of the stomach lining—the more severe the inflammation, the
more abundant the bacteria.
Question 2.What should Dr. Warren do next?
Answer: Dr. Warren started Warthin-Starry silver stain for all the gastric biopsies he examined.
Part-II:
Question: At this stage of the story, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: Dr. Warren and Marshall had the quantitative data based on the observations made by
various researchers, such as Italian pathologist Giulio Bizzozero, Polish clinical researcher, Dr.
W. Jaworski. The results that obtained from the previous reports are purely observational, not
experimental.
Part-III:
Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: Now, at this stage, Warren and Marshall had electron microscopy data of the bacteria
that was colonized in the lab culture media, and therefore, the data was now experimentally
proven qualitative data.
Question: What other information would you like to have about this bacterium now that it can be
successfully cultured in the lab?
Answer: The bacteria that was isolated from the biopsy of patient\'s stomach for lab culture, was
a slow grower, and it took 5-6 days for colonization instead of 2 days which has taken by many
identified bacteria.
Part IV:
Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: The data was generated by pilot studies from 100 patients and the data was
experimental and also statistically analyzed, therefore it was qualitative as well as quantitative
data.
CANNOT SEE FIGURE 3. UNABLE TO ANSWER FURTHER.
Solution
Part-I
Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach?
Give at least three alternative hypotheses that could explain these findings.
Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active,
chronic gastritis and the presence of the bacteria found in stomachs.
hypothesis 1. Dr. W.
Peru is not a part of the Southern South America.SolutionPeru .pdf
Peru is not a part of the Southern South America.
Solution
Peru is not a part of the Southern South America..
Combustion of glucose is respiration. The equatio.pdf
Combustion of glucose is respiration. The equation for the combustion of glucose
is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or
oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need
ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from
related molecules. A lot of oxygen is required for this process! The sugar AND the oxygen are
delivered to your cells via your bloodstream. This process occurs partially in the cytoplasm, and
partially in the mitochondria. The mitochondria is another organelle in eukaryotic cells. like the
chloroplast, the mitochondria has two lipid bilayers around it, and its own genome (indicating
that it may be the result of endosymbiosis long ago). In some ways similar to the chloroplast, the
mitochondria also has two main sites for the reactions: The matrix, a liquidy part of the
mitochondrion, and the christae, the folded membranes in the mitochondrion. 1: Glycolysis
(\"splitting of sugar\"): This step happens in the cytoplasm. One Glucose (C6H12O6) is broken
down to 2 molecules of pyruvic acid. Results in the production of 2 ATPs for every glucose.
(page 113 of your book goes into painful detail about this process, but all YOU need to know is
that glucose is split into to 2 molecules of pyruvate!) 2: Transition Reaction: Pyruvic Acid is
shuttled into the mitochondria, where it is converyed to a molecule called Acetyl CoA for further
breakdown. 3: The Krebs Cycle, or Citric Acid Cycle: Occurs in the mitochondrial matrix, the
liquid-y part of the mitochondria. In the presence of Oxygen gas (O2), all the hydrogens (H2)
are stripped off the Acetyl CoA, two by two, to extract the electrons for making ATP, until there
are no hydrogens left - and all that is left of the sugar is CO2 - a waste product - and H2O
(exhale). The Krebs cycle results in the production of only ~4 ATPs, but produces a lot of
NADH, which will go on to the next step... Hans Krebs won the Nobel Prize in 1953 for his
discovery of the Citric Acid Cycle. 4: The Electron Transport Chain and Chemiosmosis (\"the
big ATP payoff\"). Occurs in the christae of the mirochondria, the folded membranes inside the
chloroplast. Electrons from Hydrogen are carried by NADH and passed down an electron
transport chain to result in the production of ATP. Results in the production of ~32 ATPs for
every glucose. Peter Mitchell won the Nobel Prize in 1978 for his work on energy production in
mitochondria, called the Chemiosmotic Theory.
Solution
Combustion of glucose is respiration. The equation for the combustion of glucose
is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or
oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need
ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from
related molecules. A lot of oxygen is required for this process! The sugar A.
in truss one...the forces in the member BC and DE have zero.in tru.pdf
in truss one...the forces in the member BC and DE have zero.
in truss two ... the forces in the members EF and GH have zero..
in truss three ... the forces in the members CD and EF have zero..
Solution
in truss one...the forces in the member BC and DE have zero.
in truss two ... the forces in the members EF and GH have zero..
in truss three ... the forces in the members CD and EF have zero...
iam giving you entire process of forensc duplication;the response.pdf
iam giving you entire process of forensc duplication;
the response strategy for forensc duplication is
Decision of when to perform a forensic duplication based is based, in part, on existing response
strategy for the instant situation
For example, many organizations have a policy of creating forensic HD duplicates of all PCs
used by executives that leave the organization.
Forensic Duplicates as Admissible Evidence;
Existing legal standards define minimum criteria for an item to be admitted into evidence
Collection process usually under scrutiny as well
Federal Rules of Evidence ;
Federal Rules of Evidence (FRE) 1002 state that the item or information presented in court must
be the original.
Exceptions: Definitions and Duplicates
If data are stored by computer or similar device, any printout or other output readable by sight,
shown to reflect the data accurately, is an original. Admissibility of Duplicates
A duplicate is admissible to the same extent as an original unless: • A genuine question is raised
as to the authenticity of the original, or • In the circumstances it would be unfair to admit the
duplicate in lieu of the original
first of all what is forensics duplication
A file that contains every bit of information from the source in a raw bitstream format.
Tools that create forensic duplicates:
1. dd
2. FTK Imager, Access Data
3. Dfcldd, US DOD Computer Forensics Lab version of the dd comand
Qualified Forensic Duplicate?
A file that contains every bit of information from the source, but may be stored in a altered form.
Tools that create qualified forensic duplicate output files:
1. SafeBack
2. EnCase
3. FTK Imager
Restored Image;
A restored image is what you get when you restore a forensic duplicate or a qualified forensic
duplicate to another storage medium.
Mismatched drive geometries can cause problems.
HD Development;
When hard drives grew beyond 512MB, the PC-BIOS needed to be updated (to recognize larger
drives). …software emulated a modern BIOS. Software pushed all of the real data on the drive
down one sector and stored its program and information in sector 2.
The real partition table would be at cylinder 0, head 0, sector2. Safeback, EnCase, FTK Imager,
and dd will create a restored image from the qualified forensic duplicate. EnCase and dd images
may not need to be restored. Treat images as virtual disks, eliminating the need for restoration.
Note, FTK Imager can create images in the EnCase Format
Mirror Image ;
Created from hardware that does at bit for bit copy from one hard drive to another.
Requires two identical hard drives
Doesn’t happen very often.
Tool Requirements: Forensic Duplication ;
Tool must: Create a forensic duplicate or mirror image of the original.
Handle read errors in a robust and graceful manner.
Not make any changes to source medium.
Capable of scientific and peer review.
Results must be third party repeatable and verifiable.
Legal Issues
Tools used for forensic duplication must pass the legal tests for r.
How does traffic analysis work Internet data packets have two parts.pdf
How does traffic analysis work? Internet data packets have two parts: a data payload and a
header used for routing. The data payload is whatever is being sent, whether that\'s an email
message, a web page, or an audio file. Even if you encrypt the data payload of your
communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly,
what you\'re saying. That\'s because it focuses on the header, which discloses source, destination,
size, timing, and so on.
A basic problem for the privacy minded is that the recipient of your communications can see that
you sent it by looking at headers. So can authorized intermediaries like Internet service
providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic
analysis might involve sitting somewhere between sender and recipient on the network, looking
at headers.
But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts
of the Internet and use sophisticated statistical techniques to track the communications patterns
of many different organizations and individuals. Encryption does not help against these attackers,
since it only hides the content of Internet traffic, not the headers.
Solution
How does traffic analysis work? Internet data packets have two parts: a data payload and a
header used for routing. The data payload is whatever is being sent, whether that\'s an email
message, a web page, or an audio file. Even if you encrypt the data payload of your
communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly,
what you\'re saying. That\'s because it focuses on the header, which discloses source, destination,
size, timing, and so on.
A basic problem for the privacy minded is that the recipient of your communications can see that
you sent it by looking at headers. So can authorized intermediaries like Internet service
providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic
analysis might involve sitting somewhere between sender and recipient on the network, looking
at headers.
But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts
of the Internet and use sophisticated statistical techniques to track the communications patterns
of many different organizations and individuals. Encryption does not help against these attackers,
since it only hides the content of Internet traffic, not the headers..
Br2 is the answer. note for nonpolar molecules,.pdf
Br2 is the answer. note: for nonpolar molecules, the dispersion force is the only
intermolecular force. Br2 is nonpolar while HF and HCl are polar.
Solution
Br2 is the answer. note: for nonpolar molecules, the dispersion force is the only
intermolecular force. Br2 is nonpolar while HF and HCl are polar..
Because leucine is hydrophobic and does not have .pdf
Because leucine is hydrophobic and does not have bulky groups, similar amino
acids can be substituted without changing the protein character very much. These amino acids
are isoleucine, valine, and alanine.
Solution
Because leucine is hydrophobic and does not have bulky groups, similar amino
acids can be substituted without changing the protein character very much. These amino acids
are isoleucine, valine, and alanine..
Designing affective Human Computer-Interfaces such as Embodied Conve.pdf
Designing affective Human Computer-Interfaces such as Embodied Conversational Agents
requires modeling the relations between spontaneous emotions and behaviors in several
modalities. There have been a lot of psychological researches on emotion and nonverbal
communication. Yet, these studies were based mostly on acted basic emotions. This paper
explores how manual annotation and image processing might cooperate towards the
representation of spontaneous emotional behavior in low resolution videos from TV. We
describe a corpus of TV interviews and the manual annotations that have been defined. We
explain the image processing algorithms that have been designed for the automatic estimation of
movement quantity. Finally, we explore several ways to compare the manual annotations and the
cues extracted by image processing.
Solution
Designing affective Human Computer-Interfaces such as Embodied Conversational Agents
requires modeling the relations between spontaneous emotions and behaviors in several
modalities. There have been a lot of psychological researches on emotion and nonverbal
communication. Yet, these studies were based mostly on acted basic emotions. This paper
explores how manual annotation and image processing might cooperate towards the
representation of spontaneous emotional behavior in low resolution videos from TV. We
describe a corpus of TV interviews and the manual annotations that have been defined. We
explain the image processing algorithms that have been designed for the automatic estimation of
movement quantity. Finally, we explore several ways to compare the manual annotations and the
cues extracted by image processing..
Dispersal of populationindividuals is described as movement of indi.pdf
Dispersal of population/individuals is described as movement of individuals from one
geographical residential location to another where they can occupy the new place, survive, settle
and reproduce appropriately. These set of individuals can thus colonize at the new geographical
location in a similar comfortable way as they used to colonize in the previous location. This
technically does not resemble to fragmentation or generation of a sub-set of parent population
because these methods give rise to a smaller subset of individuals which are impaired in some of
the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents
the high efficacy and survivability of the parent population even under new locations.
Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat
Solution
Dispersal of population/individuals is described as movement of individuals from one
geographical residential location to another where they can occupy the new place, survive, settle
and reproduce appropriately. These set of individuals can thus colonize at the new geographical
location in a similar comfortable way as they used to colonize in the previous location. This
technically does not resemble to fragmentation or generation of a sub-set of parent population
because these methods give rise to a smaller subset of individuals which are impaired in some of
the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents
the high efficacy and survivability of the parent population even under new locations.
Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat.
ConversionDemo.javaimport java.util.;package for scanner class.pdf
ConversionDemo.java
import java.util.*;//package for scanner class
class Tempconversion//temperature conversion class
{
public void farenheattocelcius(double temp)
{//method for farenheat to celcius conversion
System.out.println(\"Celcius Farenheat\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>30;i--)
{
temp = ((9.0*i)/5+32);//formula for converting into celcius
System.out.println(i+\" \" +temp);
}
}
public void celciustofarenheat(double temp)
{//method for converting celcius to farenheat
System.out.println(\"Farenheat Celcius\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>=30;i=i-10)
{
temp = 5.0*(i - 32)/9;//formula for converting into celcius
String temp1 = String.format(\"%.2f\", temp);
System.out.println(i+\" \" +temp1);
}
}
}
class ConversionDemo {//main class
public static void main(String[] args) {//main method
Tempconversion f=new Tempconversion();//object creation
f.farenheattocelcius(40);//calling method
f.celciustofarenheat(120);
}
}
output
Celcius Farenheat
-----------------------------
40 104.0
39 102.2
38 100.4
37 98.6
36 96.8
35 95.0
34 93.2
33 91.4
32 89.6
31 87.8
Farenheat Celcius
-----------------------------
120 48.89
110 43.33
100 37.78
90 32.22
80 26.67
70 21.11
60 15.56
50 10.00
40 4.44
30 -1.11
Solution
ConversionDemo.java
import java.util.*;//package for scanner class
class Tempconversion//temperature conversion class
{
public void farenheattocelcius(double temp)
{//method for farenheat to celcius conversion
System.out.println(\"Celcius Farenheat\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>30;i--)
{
temp = ((9.0*i)/5+32);//formula for converting into celcius
System.out.println(i+\" \" +temp);
}
}
public void celciustofarenheat(double temp)
{//method for converting celcius to farenheat
System.out.println(\"Farenheat Celcius\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>=30;i=i-10)
{
temp = 5.0*(i - 32)/9;//formula for converting into celcius
String temp1 = String.format(\"%.2f\", temp);
System.out.println(i+\" \" +temp1);
}
}
}
class ConversionDemo {//main class
public static void main(String[] args) {//main method
Tempconversion f=new Tempconversion();//object creation
f.farenheattocelcius(40);//calling method
f.celciustofarenheat(120);
}
}
output
Celcius Farenheat
-----------------------------
40 104.0
39 102.2
38 100.4
37 98.6
36 96.8
35 95.0
34 93.2
33 91.4
32 89.6
31 87.8
Farenheat Celcius
-----------------------------
120 48.89
110 43.33
100 37.78
90 32.22
80 26.67
70 21.11
60 15.56
50 10.00
40 4.44
30 -1.11.
C. Yes, YN + NY is greater than or equal to 30.SolutionC. Yes,.pdf
C. Yes, YN + NY is greater than or equal to 30.
Solution
C. Yes, YN + NY is greater than or equal to 30..
AnswerTotal asset turnover = salestotal assetTotal asset = cur.pdf
Answer:
Total asset turnover = sales/total asset
Total asset = current + non current asset
=( $100 + $142 ) million
= $242 million
Total asset turnover = $371/$242
= 1.5331 :1 or 153.31% (answer)
Solution
Answer:
Total asset turnover = sales/total asset
Total asset = current + non current asset
=( $100 + $142 ) million
= $242 million
Total asset turnover = $371/$242
= 1.5331 :1 or 153.31% (answer).
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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The balanced equation is2 H2O2 = 2 H2O + O2SolutionThe bal.pdf
The balanced equation is:
2 H2O2 => 2 H2O + O2
Solution
The balanced equation is:
2 H2O2 => 2 H2O + O2.
Tay-Sachs disease is caused by a mutation (abnormal change) in the g.pdf
Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A,
and it is a recessive trait. This means that people will have the disease if they have two copies of
the defective gene, but they will not have the disease if they have at least one unaffected copy.
People with one normal copy and one defective copy are called carriers, because they can pass
the disease on to their children.
Seizures are sudden altacks of disease, often referring to some type of violent spasms.
Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population,
about 1 in 250 people carries the gene.
People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting
in two defective genes that make the body unable to produce Hex-A correctly. People who have
only one defective gene are called carriers. Carriers do not have the disease, because they have
inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their
children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the
disease.
However, some populations of people include more carriers than others. For example, 1 in 27
people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People
of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also
have a higher than usual risk of carrying the Tay-Sachs gene.
Solution
Tay-Sachs disease is caused by a mutation (abnormal change) in the gene that codes for Hex-A,
and it is a recessive trait. This means that people will have the disease if they have two copies of
the defective gene, but they will not have the disease if they have at least one unaffected copy.
People with one normal copy and one defective copy are called carriers, because they can pass
the disease on to their children.
Seizures are sudden altacks of disease, often referring to some type of violent spasms.
Just about anyone can be a carrier of the gene for Tay-Sachs disease. In the general population,
about 1 in 250 people carries the gene.
People inherit Tay-Sachs disease when they inherit a defective gene from both parents, resulting
in two defective genes that make the body unable to produce Hex-A correctly. People who have
only one defective gene are called carriers. Carriers do not have the disease, because they have
inherited one healthy gene to code for Hex-A, but they may pass the defective gene on to their
children. If both parents are carriers, each child born to them has a 1 in 4 liklihood of having the
disease.
However, some populations of people include more carriers than others. For example, 1 in 27
people of eastern European Jewish (Ashkenazi) descent in the United States is a carrier. People
of French-Canadian ancestry from one part of Quebec and the Cajunpopulation in Louisiana also
have a higher than usual risk of carrying the Tay-Sachs gene..
E.) is less acidic in other A ,B triple bond an.pdf
E.) is less acidic in other A ,B triple bond and double bond are responsible for
acidity In C,D steric hinderance from -CH3 is responsible for more acidicity than rest thus E is
the answer
Solution
E.) is less acidic in other A ,B triple bond and double bond are responsible for
acidity In C,D steric hinderance from -CH3 is responsible for more acidicity than rest thus E is
the answer.
Solution when user sending the email, then user should select eithe.pdf
Solution
: when user sending the email, then user should select either POP3 or IMAP protocol to send the
mails. Some of the differences are giving below.
POP3 e-mail protocol:
IMAP e-mail protocol:
SMTP (simple mail transfer protocol):
SMTP is not normally used as intermediate mail servers for sending mail. Because SMTP cannot
transmit the executable files or any binary objects.
Similarly it cannot transmit the text data that contains the national language characters. It will
have certain size limit to have an e-mail. So if the mail size is over the limit then it will be
rejected..
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
please give me points as nobody ecen noticed the qusetion as it is u.pdf
please give me points as nobody ecen noticed the qusetion as it is ununderstandable
Solution
please give me points as nobody ecen noticed the qusetion as it is ununderstandable.
Part-IQuestion 1. What is Dr. Warren’s hypothesis regarding the ba.pdf
Part-I
Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach?
Give at least three alternative hypotheses that could explain these findings.
Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active,
chronic gastritis and the presence of the bacteria found in stomachs.
hypothesis 1. Dr. Warren noticed a thin blue line on the surface of the tissue.
hypothesis 2: Warren found numerous spiral shaped bacteria when stained with a Warthin-Starry
silver stain of the histological section.
hypothesis 3: Dr. Warren also observed that the number of bacteria observed seemed to correlate
with the degree of inflammation of the stomach lining—the more severe the inflammation, the
more abundant the bacteria.
Question 2.What should Dr. Warren do next?
Answer: Dr. Warren started Warthin-Starry silver stain for all the gastric biopsies he examined.
Part-II:
Question: At this stage of the story, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: Dr. Warren and Marshall had the quantitative data based on the observations made by
various researchers, such as Italian pathologist Giulio Bizzozero, Polish clinical researcher, Dr.
W. Jaworski. The results that obtained from the previous reports are purely observational, not
experimental.
Part-III:
Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: Now, at this stage, Warren and Marshall had electron microscopy data of the bacteria
that was colonized in the lab culture media, and therefore, the data was now experimentally
proven qualitative data.
Question: What other information would you like to have about this bacterium now that it can be
successfully cultured in the lab?
Answer: The bacteria that was isolated from the biopsy of patient\'s stomach for lab culture, was
a slow grower, and it took 5-6 days for colonization instead of 2 days which has taken by many
identified bacteria.
Part IV:
Question: At this stage of the study, what type(s) of data do Warren and Marshall have on the
spiral bacteria found in stomachs? Is it quantitative or qualitative? Is it experimental of
observational?
Answer: The data was generated by pilot studies from 100 patients and the data was
experimental and also statistically analyzed, therefore it was qualitative as well as quantitative
data.
CANNOT SEE FIGURE 3. UNABLE TO ANSWER FURTHER.
Solution
Part-I
Question 1. What is Dr. Warren’s hypothesis regarding the bacteria he found in the stomach?
Give at least three alternative hypotheses that could explain these findings.
Answer: Dr. Warren\'s hypothesis was that there was a definite correlation between active,
chronic gastritis and the presence of the bacteria found in stomachs.
hypothesis 1. Dr. W.
Peru is not a part of the Southern South America.SolutionPeru .pdf
Peru is not a part of the Southern South America.
Solution
Peru is not a part of the Southern South America..
Combustion of glucose is respiration. The equatio.pdf
Combustion of glucose is respiration. The equation for the combustion of glucose
is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or
oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need
ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from
related molecules. A lot of oxygen is required for this process! The sugar AND the oxygen are
delivered to your cells via your bloodstream. This process occurs partially in the cytoplasm, and
partially in the mitochondria. The mitochondria is another organelle in eukaryotic cells. like the
chloroplast, the mitochondria has two lipid bilayers around it, and its own genome (indicating
that it may be the result of endosymbiosis long ago). In some ways similar to the chloroplast, the
mitochondria also has two main sites for the reactions: The matrix, a liquidy part of the
mitochondrion, and the christae, the folded membranes in the mitochondrion. 1: Glycolysis
(\"splitting of sugar\"): This step happens in the cytoplasm. One Glucose (C6H12O6) is broken
down to 2 molecules of pyruvic acid. Results in the production of 2 ATPs for every glucose.
(page 113 of your book goes into painful detail about this process, but all YOU need to know is
that glucose is split into to 2 molecules of pyruvate!) 2: Transition Reaction: Pyruvic Acid is
shuttled into the mitochondria, where it is converyed to a molecule called Acetyl CoA for further
breakdown. 3: The Krebs Cycle, or Citric Acid Cycle: Occurs in the mitochondrial matrix, the
liquid-y part of the mitochondria. In the presence of Oxygen gas (O2), all the hydrogens (H2)
are stripped off the Acetyl CoA, two by two, to extract the electrons for making ATP, until there
are no hydrogens left - and all that is left of the sugar is CO2 - a waste product - and H2O
(exhale). The Krebs cycle results in the production of only ~4 ATPs, but produces a lot of
NADH, which will go on to the next step... Hans Krebs won the Nobel Prize in 1953 for his
discovery of the Citric Acid Cycle. 4: The Electron Transport Chain and Chemiosmosis (\"the
big ATP payoff\"). Occurs in the christae of the mirochondria, the folded membranes inside the
chloroplast. Electrons from Hydrogen are carried by NADH and passed down an electron
transport chain to result in the production of ATP. Results in the production of ~32 ATPs for
every glucose. Peter Mitchell won the Nobel Prize in 1978 for his work on energy production in
mitochondria, called the Chemiosmotic Theory.
Solution
Combustion of glucose is respiration. The equation for the combustion of glucose
is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g) Essentially, sugar (C6H12O6) is burned, or
oxidized, down to CO2 and H2O, releasing energy (ATP) in the process. Why do cells need
ATP? ALL cellular work -all the activities of life - requires energy, either from ATP or from
related molecules. A lot of oxygen is required for this process! The sugar A.
in truss one...the forces in the member BC and DE have zero.in tru.pdf
in truss one...the forces in the member BC and DE have zero.
in truss two ... the forces in the members EF and GH have zero..
in truss three ... the forces in the members CD and EF have zero..
Solution
in truss one...the forces in the member BC and DE have zero.
in truss two ... the forces in the members EF and GH have zero..
in truss three ... the forces in the members CD and EF have zero...
iam giving you entire process of forensc duplication;the response.pdf
iam giving you entire process of forensc duplication;
the response strategy for forensc duplication is
Decision of when to perform a forensic duplication based is based, in part, on existing response
strategy for the instant situation
For example, many organizations have a policy of creating forensic HD duplicates of all PCs
used by executives that leave the organization.
Forensic Duplicates as Admissible Evidence;
Existing legal standards define minimum criteria for an item to be admitted into evidence
Collection process usually under scrutiny as well
Federal Rules of Evidence ;
Federal Rules of Evidence (FRE) 1002 state that the item or information presented in court must
be the original.
Exceptions: Definitions and Duplicates
If data are stored by computer or similar device, any printout or other output readable by sight,
shown to reflect the data accurately, is an original. Admissibility of Duplicates
A duplicate is admissible to the same extent as an original unless: • A genuine question is raised
as to the authenticity of the original, or • In the circumstances it would be unfair to admit the
duplicate in lieu of the original
first of all what is forensics duplication
A file that contains every bit of information from the source in a raw bitstream format.
Tools that create forensic duplicates:
1. dd
2. FTK Imager, Access Data
3. Dfcldd, US DOD Computer Forensics Lab version of the dd comand
Qualified Forensic Duplicate?
A file that contains every bit of information from the source, but may be stored in a altered form.
Tools that create qualified forensic duplicate output files:
1. SafeBack
2. EnCase
3. FTK Imager
Restored Image;
A restored image is what you get when you restore a forensic duplicate or a qualified forensic
duplicate to another storage medium.
Mismatched drive geometries can cause problems.
HD Development;
When hard drives grew beyond 512MB, the PC-BIOS needed to be updated (to recognize larger
drives). …software emulated a modern BIOS. Software pushed all of the real data on the drive
down one sector and stored its program and information in sector 2.
The real partition table would be at cylinder 0, head 0, sector2. Safeback, EnCase, FTK Imager,
and dd will create a restored image from the qualified forensic duplicate. EnCase and dd images
may not need to be restored. Treat images as virtual disks, eliminating the need for restoration.
Note, FTK Imager can create images in the EnCase Format
Mirror Image ;
Created from hardware that does at bit for bit copy from one hard drive to another.
Requires two identical hard drives
Doesn’t happen very often.
Tool Requirements: Forensic Duplication ;
Tool must: Create a forensic duplicate or mirror image of the original.
Handle read errors in a robust and graceful manner.
Not make any changes to source medium.
Capable of scientific and peer review.
Results must be third party repeatable and verifiable.
Legal Issues
Tools used for forensic duplication must pass the legal tests for r.
How does traffic analysis work Internet data packets have two parts.pdf
How does traffic analysis work? Internet data packets have two parts: a data payload and a
header used for routing. The data payload is whatever is being sent, whether that\'s an email
message, a web page, or an audio file. Even if you encrypt the data payload of your
communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly,
what you\'re saying. That\'s because it focuses on the header, which discloses source, destination,
size, timing, and so on.
A basic problem for the privacy minded is that the recipient of your communications can see that
you sent it by looking at headers. So can authorized intermediaries like Internet service
providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic
analysis might involve sitting somewhere between sender and recipient on the network, looking
at headers.
But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts
of the Internet and use sophisticated statistical techniques to track the communications patterns
of many different organizations and individuals. Encryption does not help against these attackers,
since it only hides the content of Internet traffic, not the headers.
Solution
How does traffic analysis work? Internet data packets have two parts: a data payload and a
header used for routing. The data payload is whatever is being sent, whether that\'s an email
message, a web page, or an audio file. Even if you encrypt the data payload of your
communications, traffic analysis still reveals a great deal about what you\'re doing and, possibly,
what you\'re saying. That\'s because it focuses on the header, which discloses source, destination,
size, timing, and so on.
A basic problem for the privacy minded is that the recipient of your communications can see that
you sent it by looking at headers. So can authorized intermediaries like Internet service
providers, and sometimes unauthorized intermediaries as well. A very simple form of traffic
analysis might involve sitting somewhere between sender and recipient on the network, looking
at headers.
But there are also more powerful kinds of traffic analysis. Some attackers spy on multiple parts
of the Internet and use sophisticated statistical techniques to track the communications patterns
of many different organizations and individuals. Encryption does not help against these attackers,
since it only hides the content of Internet traffic, not the headers..
Br2 is the answer. note for nonpolar molecules,.pdf
Br2 is the answer. note: for nonpolar molecules, the dispersion force is the only
intermolecular force. Br2 is nonpolar while HF and HCl are polar.
Solution
Br2 is the answer. note: for nonpolar molecules, the dispersion force is the only
intermolecular force. Br2 is nonpolar while HF and HCl are polar..
Because leucine is hydrophobic and does not have .pdf
Because leucine is hydrophobic and does not have bulky groups, similar amino
acids can be substituted without changing the protein character very much. These amino acids
are isoleucine, valine, and alanine.
Solution
Because leucine is hydrophobic and does not have bulky groups, similar amino
acids can be substituted without changing the protein character very much. These amino acids
are isoleucine, valine, and alanine..
Designing affective Human Computer-Interfaces such as Embodied Conve.pdf
Designing affective Human Computer-Interfaces such as Embodied Conversational Agents
requires modeling the relations between spontaneous emotions and behaviors in several
modalities. There have been a lot of psychological researches on emotion and nonverbal
communication. Yet, these studies were based mostly on acted basic emotions. This paper
explores how manual annotation and image processing might cooperate towards the
representation of spontaneous emotional behavior in low resolution videos from TV. We
describe a corpus of TV interviews and the manual annotations that have been defined. We
explain the image processing algorithms that have been designed for the automatic estimation of
movement quantity. Finally, we explore several ways to compare the manual annotations and the
cues extracted by image processing.
Solution
Designing affective Human Computer-Interfaces such as Embodied Conversational Agents
requires modeling the relations between spontaneous emotions and behaviors in several
modalities. There have been a lot of psychological researches on emotion and nonverbal
communication. Yet, these studies were based mostly on acted basic emotions. This paper
explores how manual annotation and image processing might cooperate towards the
representation of spontaneous emotional behavior in low resolution videos from TV. We
describe a corpus of TV interviews and the manual annotations that have been defined. We
explain the image processing algorithms that have been designed for the automatic estimation of
movement quantity. Finally, we explore several ways to compare the manual annotations and the
cues extracted by image processing..
Dispersal of populationindividuals is described as movement of indi.pdf
Dispersal of population/individuals is described as movement of individuals from one
geographical residential location to another where they can occupy the new place, survive, settle
and reproduce appropriately. These set of individuals can thus colonize at the new geographical
location in a similar comfortable way as they used to colonize in the previous location. This
technically does not resemble to fragmentation or generation of a sub-set of parent population
because these methods give rise to a smaller subset of individuals which are impaired in some of
the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents
the high efficacy and survivability of the parent population even under new locations.
Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat
Solution
Dispersal of population/individuals is described as movement of individuals from one
geographical residential location to another where they can occupy the new place, survive, settle
and reproduce appropriately. These set of individuals can thus colonize at the new geographical
location in a similar comfortable way as they used to colonize in the previous location. This
technically does not resemble to fragmentation or generation of a sub-set of parent population
because these methods give rise to a smaller subset of individuals which are impaired in some of
the qualitites which help them in surviving at new locations. Contrary to it, dispersal represents
the high efficacy and survivability of the parent population even under new locations.
Thus, the correct statement which describes dispersal is: Individuals colonize a novel habitat.
ConversionDemo.javaimport java.util.;package for scanner class.pdf
ConversionDemo.java
import java.util.*;//package for scanner class
class Tempconversion//temperature conversion class
{
public void farenheattocelcius(double temp)
{//method for farenheat to celcius conversion
System.out.println(\"Celcius Farenheat\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>30;i--)
{
temp = ((9.0*i)/5+32);//formula for converting into celcius
System.out.println(i+\" \" +temp);
}
}
public void celciustofarenheat(double temp)
{//method for converting celcius to farenheat
System.out.println(\"Farenheat Celcius\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>=30;i=i-10)
{
temp = 5.0*(i - 32)/9;//formula for converting into celcius
String temp1 = String.format(\"%.2f\", temp);
System.out.println(i+\" \" +temp1);
}
}
}
class ConversionDemo {//main class
public static void main(String[] args) {//main method
Tempconversion f=new Tempconversion();//object creation
f.farenheattocelcius(40);//calling method
f.celciustofarenheat(120);
}
}
output
Celcius Farenheat
-----------------------------
40 104.0
39 102.2
38 100.4
37 98.6
36 96.8
35 95.0
34 93.2
33 91.4
32 89.6
31 87.8
Farenheat Celcius
-----------------------------
120 48.89
110 43.33
100 37.78
90 32.22
80 26.67
70 21.11
60 15.56
50 10.00
40 4.44
30 -1.11
Solution
ConversionDemo.java
import java.util.*;//package for scanner class
class Tempconversion//temperature conversion class
{
public void farenheattocelcius(double temp)
{//method for farenheat to celcius conversion
System.out.println(\"Celcius Farenheat\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>30;i--)
{
temp = ((9.0*i)/5+32);//formula for converting into celcius
System.out.println(i+\" \" +temp);
}
}
public void celciustofarenheat(double temp)
{//method for converting celcius to farenheat
System.out.println(\"Farenheat Celcius\");
System.out.println(\"-----------------------------\");
for(int i=(int) temp;i>=30;i=i-10)
{
temp = 5.0*(i - 32)/9;//formula for converting into celcius
String temp1 = String.format(\"%.2f\", temp);
System.out.println(i+\" \" +temp1);
}
}
}
class ConversionDemo {//main class
public static void main(String[] args) {//main method
Tempconversion f=new Tempconversion();//object creation
f.farenheattocelcius(40);//calling method
f.celciustofarenheat(120);
}
}
output
Celcius Farenheat
-----------------------------
40 104.0
39 102.2
38 100.4
37 98.6
36 96.8
35 95.0
34 93.2
33 91.4
32 89.6
31 87.8
Farenheat Celcius
-----------------------------
120 48.89
110 43.33
100 37.78
90 32.22
80 26.67
70 21.11
60 15.56
50 10.00
40 4.44
30 -1.11.
C. Yes, YN + NY is greater than or equal to 30.SolutionC. Yes,.pdf
C. Yes, YN + NY is greater than or equal to 30.
Solution
C. Yes, YN + NY is greater than or equal to 30..
AnswerTotal asset turnover = salestotal assetTotal asset = cur.pdf
Answer:
Total asset turnover = sales/total asset
Total asset = current + non current asset
=( $100 + $142 ) million
= $242 million
Total asset turnover = $371/$242
= 1.5331 :1 or 153.31% (answer)
Solution
Answer:
Total asset turnover = sales/total asset
Total asset = current + non current asset
=( $100 + $142 ) million
= $242 million
Total asset turnover = $371/$242
= 1.5331 :1 or 153.31% (answer).
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