1. Sternum.Bone by which ribs are attached on frontal side .2. R.pdf
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1. Sternum. Bone by which ribs are attached on frontal side . 2. Ribs. 3. Liver. It is the largest gland in the body located on the right side below diaphragm . Solution 1. Sternum. Bone by which ribs are attached on frontal side . 2. Ribs. 3. Liver. It is the largest gland in the body located on the right side below diaphragm ..
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Recommended
steps are as follows Dissolve the (pressumably.pdf
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene
Solution
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene.
the two features of the asynchronus connection that allow the receiv.pdf
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces
Solution
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces.
Telomerase is the enzyme responsible for maintenance of the length o.pdf
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand.
Solution
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand..
Suppose Bob sends an encrypted document to Alice. To be verifiable, .pdf
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document.
Solution
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document..
probablity = 212 100 = 16.6 Solutionprobablity = 212 10.pdf
probablity = 2/12 * 100 = 16.6 %
Solution
probablity = 2/12 * 100 = 16.6 %.
Platinum complexes like Asplastin are powerful antitumor medications.pdf
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and four ligands (which includes two neighboring chloride ions and two
neighboring amino groups). The Ctr1 will displaces these four ligands from Asplastin and will
bind to the platinum ion itself and it will transfer it through the channel and released in the
cytosol of the cell. In addition to this few amount of Asplastin molecules will interact with the
methionine-rich extracellular end of the copper transporter to form a trimeric channel structure to
get transported as a little interior bubble filled with the outer medium and will released into
various organelles including nucleus.
How it acts on cancer cells: The 4 steps how Asplastin will act is (i) Cellular uptake of drug, (ii)
aquation/activation of Asplastin, (iii) Drug binding to DNA and (iv) Lesions formed on DNA
will make the cell to opt for apoptosis.
N7 sites of deoxyguanosine residues on the DNA are the most nucleophilic positions and are
preferentially platinated. Asplastin will form a monofunctional adducts on DNA (You can see
only one covalent bond to the genomic polymer). Like in second reaction second guanine base
will get substituted by chloride ligand, thus forming a cross-link on the DNA (either on the same
strand or it may be on different strands). The most common lesion are formed from 1,2-d(GpG)
intrastrand cross-links and finally due to drug action we can see bending and unwinding of the
double helix and cells with damaged DNA will be arrested at the G2/M transition of the cell
cycle and undergoes apoptosis.
Solution
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and f.
Part 1 1)#include stdio.hint testWhileLoop() ; int testFo.pdf
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int count=0;
int n;
int sum=0;
for (count=0;count<10;count++)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Solution
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int c.
none of them are correct!Solutionnone of them are correct!.pdf
none of them are correct!
Solution
none of them are correct!.
Recommended
steps are as follows Dissolve the (pressumably.pdf
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene
Solution
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene.
the two features of the asynchronus connection that allow the receiv.pdf
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces
Solution
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces.
Telomerase is the enzyme responsible for maintenance of the length o.pdf
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand.
Solution
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand..
Suppose Bob sends an encrypted document to Alice. To be verifiable, .pdf
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document.
Solution
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document..
probablity = 212 100 = 16.6 Solutionprobablity = 212 10.pdf
probablity = 2/12 * 100 = 16.6 %
Solution
probablity = 2/12 * 100 = 16.6 %.
Platinum complexes like Asplastin are powerful antitumor medications.pdf
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and four ligands (which includes two neighboring chloride ions and two
neighboring amino groups). The Ctr1 will displaces these four ligands from Asplastin and will
bind to the platinum ion itself and it will transfer it through the channel and released in the
cytosol of the cell. In addition to this few amount of Asplastin molecules will interact with the
methionine-rich extracellular end of the copper transporter to form a trimeric channel structure to
get transported as a little interior bubble filled with the outer medium and will released into
various organelles including nucleus.
How it acts on cancer cells: The 4 steps how Asplastin will act is (i) Cellular uptake of drug, (ii)
aquation/activation of Asplastin, (iii) Drug binding to DNA and (iv) Lesions formed on DNA
will make the cell to opt for apoptosis.
N7 sites of deoxyguanosine residues on the DNA are the most nucleophilic positions and are
preferentially platinated. Asplastin will form a monofunctional adducts on DNA (You can see
only one covalent bond to the genomic polymer). Like in second reaction second guanine base
will get substituted by chloride ligand, thus forming a cross-link on the DNA (either on the same
strand or it may be on different strands). The most common lesion are formed from 1,2-d(GpG)
intrastrand cross-links and finally due to drug action we can see bending and unwinding of the
double helix and cells with damaged DNA will be arrested at the G2/M transition of the cell
cycle and undergoes apoptosis.
Solution
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and f.
Part 1 1)#include stdio.hint testWhileLoop() ; int testFo.pdf
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int count=0;
int n;
int sum=0;
for (count=0;count<10;count++)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Solution
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int c.
none of them are correct!Solutionnone of them are correct!.pdf
none of them are correct!
Solution
none of them are correct!.
Negative control plate with no DNA transforms competent cells withou.pdf
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly.
Solution
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly..
Modifications highlighted in bold lettersDropOutStack.javaim.pdf
//Modifications highlighted in bold letters
//DropOutStack.java
import java.util.Arrays;
public class DropOutStack {
/**
* Defines the interface to a stack collection.
*
* @author Java Foundations
* @version 4.0
*/
public interface StackADT {
/**
* Adds the specified element to the top of this stack.
*
* @param element
* element to be pushed onto the stack
*/
public void push(T element);
/**
* Removes and returns the top element from this stack.
*
* @return the element removed from the stack
*/
public T pop();
/**
* Returns without removing the top element of this stack.
*
* @return the element on top of the stack
*/
public T peek();
/**
* Returns true if this stack contains no elements.
*
* @return true if the stack is empty
*/
public boolean isEmpty();
/**
* Returns the number of elements in this stack.
*
* @return the number of elements in the stack
*/
public int size();
/**
* Returns a string representation of this stack.
*
* @return a string representation of the stack
*/
public String toString();
}
/**
* Represents the situation in which a collection is empty.
*
* @author Java Foundations
* @version 4.0
*/
public static class EmptyCollectionException extends RuntimeException {
/**
* Sets up this exception with an appropriate message.
*
* @param collection
* the name of the collection
*/
public EmptyCollectionException(String collection) {
super(\"The \" + collection + \" is empty.\");
}
}
/**
* Program entry point for stack testing.
*
* @param args
* Argument list.
*/
public static void main(String[] args) {
ArrayDropOutStack stack = new ArrayDropOutStack(4);
System.out.println(\"DROP-OUT STACK TESTING\");
stack.push(1);
stack.push(2);
stack.push(3);
stack.push(4);
System.out.println(\"Stack : \"+stack.toString());
System.out.println(\"pushing 5 element \");
stack.push(5);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
stack.pop();
System.out.println(\"poping element 5\"+stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
System.out.println(\"pushing 7 element \");
stack.push(7);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"pushing 8 element \");
stack.push(8);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
}
/**
* An array implementation of a stack in which the bottom of the stack is
* fixed at index 0.
*
* @author Java Foundations
* @version 4.0
*/
public static class ArrayDropOutStack implements StackADT {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack;
/**
* Creates an empty stack using the default capacity.
*/
public ArrayDropOutStack() {
this(DEFAULT_CAPACITY);
}
/**
* Creates an empty stack using.
Message types used by the DHCP boot sequenceop Message type.h.pdf
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
Solution
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority.
Law of conservation of matter(mass)In any closed system subjected .pdf
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another.
Solution
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another..
It is a unit of measurement for the amount of substance.The mole is .pdf
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023.
Solution
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023..
In insurance, the term risk pooling refers to the spreading of f.pdf
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that demand variability is reduced if one aggregates
demand across locations because, as demand is aggregated across different locations, it becomes
more likely that high demand from one customer will be offset by low demand from another.
However,consider the 1)\"high risk pools\", which simply means separating out the sickest, most
expensive people in the country and dumping them into a separate program.With the \"bad
apples\" (ie, human beings with terrible medical problems) safely tucked out of the way, the
average cost of treating everyone else supposedly suddenly becomes less pricey.This is the exact
opposite of the entire point of health insurance in the first place...spreading the risk and hence
risk pooling fails here.
2)Liability risk includes economic losses from being held responsible for harming others or their
property. This cannot be covered by risk pooling.
Solution
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that.
Some CO is Added The Pressure of CH4 and H2O will.pdf
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left
Solution
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left.
Comments Figures are not properly numbered. Further signal of figur.pdf
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated.
Solution
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated..
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.j.pdf
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.java
import java.util.*;
import java.util.LinkedList;
class Test {
HashSet uniqueStates;
void letsRoll() {
/*
* The capacity of jug1(jug1 capacity is greater than jug2 capacity)
*/
int jug1 = 3;
/*
* The capacity of jug two
*/
int jug2 = 2;
/*
* Final Amount needed in jug1
*/
int amtNeeded = 2;
State initialState = new State(0, 0);
State finalState = new State(amtNeeded, 0);
State finalPath = null;
uniqueStates = new HashSet();
/*
* Change queue to stack to convert to DFS
*/
LinkedList queue = new LinkedList();
queue.add(initialState);
while (!queue.isEmpty()) {
State currState = queue.poll();
if (currState.equals(finalState)) {
finalPath = currState;
break;
}
// if x is zero fill it
if (currState.x == 0) {
State nextState = new State(jug1, currState.y, currState);
//System.out.println(currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is empty fill it
if (currState.y == 0) {
State nextState = new State(currState.x, jug2, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if x is not empty , empty it
if (currState.x > 0) {
State nextState = new State(0, currState.y, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is not empty, empty it
if (currState.y > 0) {
State nextState = new State(currState.x, 0, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from x to y, when x non empty and y is not full
if (currState.x > 0 && currState.y < jug2) {
int amtToTransfer = Math.min(currState.x, jug2 - currState.y);
State nextState = new State(currState.x - amtToTransfer, currState.y
+ amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from y to x, when y is not empty and x is not full
if (currState.y > 0 && currState.x < jug1) {
int amtToTransfer = Math.min(currState.y, jug1 - currState.x);
State nextState = new State(currState.x + amtToTransfer, currState.y
- amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
}
if (finalPath != null) {
System.out.println(\"Final Result\");
System.out.println(\"J1 J2\");
System.out.println(finalPath);
}
else{
System.out.println(\"Not Possible\");
}
}
/*
* Checks whether State toCheck has been generated before, if not it is
* added to queue and uniqueStates Set
*/
void checkUniqueStates(HashSet uniqueStates, State toCheck,
LinkedList queue) {
if (!uniqueStates.contains(toCheck)) {
uniqueStates.add(toCheck);
queue.add(toCheck);
}
}
public static void main(String[] args) {
new Test().letsRoll();
}
}
2 State.java
class State {
/*
* Amount in jug1 for current State
*/
int x;
/*
* Amount in jug2 for current State
*/
int y;
/*
* Parent of current State
*/
State pre;
public State(int x, int y) {
super();
this.x = x;
this.y = y;
}
public State(int x, int y, State pre) {
super();
this.x = x;
this.y = y;
this.pre = pre;
System.out.println(\"(\"+x+\",\"+y+\")\");
}
@Override
public int hashCode() {
final int .
Balance sheetBalance sheet is the statement of financial position .pdf
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be said as the resources of the company and are classified as: Current assets, long
term investment, property, plant and equipment, intangible assets and other assets
Assets includes:
•Current assets – Cash, accounts receivables, prepaid expenses, inventories
•Long term investments – investment in stock
•Plant property and equipment – Land, building, equipment
•Intangible assets – Goodwill, patents, trademarks
Liabilities are the claims and can be classified as current liabilities, long term liabilities
Liabilities includes:
•Current liabilities: Accounts payable, income tax payable. Salaries payable, accrued expenses,
unearned revenue, current portion of long term debt due
•Long term liabilities- bonds payable, notes payable
Owner;s equity can be classified as the Capital stock which includes common stock and
preferred stock
It also includes Additional paid in capital, Retained earnings
Retained earnings are the accumulated earnings of the company. It is calculated as Beginning
balance, Add: Net income less: Dividends
Importance:
Balance sheet is important in every business. Considering the hospitality firm, it is important to
analyze different aspects of teh firm to know its financial postion. As the hospitality firm
emphasize on services, it is important to know the financial position of the company through its
financial statements. the hospitality firms like hotel needs to report their assets (building) held by
the company along with other assets, liabilities and equity.
Balance sheet is the statement which helps in analysing the financial position of the company, it
can also help in predicting the cash flows and the financial position of the company. With the
help of the balance sheet we can help in analysing the ability of the firm to pay off its short term
and long term debts.
Balance sheet reports its assets and liabilities on historical basis and uses different estimates to
analyse the amounts like uncollectibles
Solution
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be .
Answer BaSO4K+ is always soluble because it is an alkaline metal..pdf
Answer: BaSO4
K+ is always soluble because it is an alkaline metal. NO3- is also always soluble.
SO4 is usually soluble except when it is paired with Sr, Ba, Ca, or Pb.
Solution
Answer: BaSO4
K+ is always soluble because it is an alkaline metal. NO3- is also always soluble.
SO4 is usually soluble except when it is paired with Sr, Ba, Ca, or Pb..
Answer is B because he copied the member A and say that he is innoce.pdf
Answer is B because he copied the member A and say that he is innocent
and also here we can\'t give you a exact soln. because of the in appropriate format of the question
details
Solution
Answer is B because he copied the member A and say that he is innocent
and also here we can\'t give you a exact soln. because of the in appropriate format of the question
details.
A nephron consists of a renal corpuscle and a renal tubule. So the a.pdf
A nephron consists of a renal corpuscle and a renal tubule. So the answer is option 1.
Solution
A nephron consists of a renal corpuscle and a renal tubule. So the answer is option 1..
30). e). All of theseDifferent biotechnological methods such as ge.pdf
30). e). All of these
Different biotechnological methods such as gene cloning, DNA amplification, recombinant DNA
technology, are used during the process of DNA isolation and sequencing. For example, the
copies of genome are amplified by the process of DNA amplification, which we observe in PCR
(polymerase chain reaction). The resulting genome can be sequenced.
31). b). Cloning vectors
The purpose of cloning vectors is gene amplification. Cloning vectors are small pieces of DNA
isolated from a virus or any other higher organism. A close is inserted with a foreign DNA that
can be used for the purpose of cloning. Thus, a cloning vector must allow the gene editing by
using the restriction enzymes.
Solution
30). e). All of these
Different biotechnological methods such as gene cloning, DNA amplification, recombinant DNA
technology, are used during the process of DNA isolation and sequencing. For example, the
copies of genome are amplified by the process of DNA amplification, which we observe in PCR
(polymerase chain reaction). The resulting genome can be sequenced.
31). b). Cloning vectors
The purpose of cloning vectors is gene amplification. Cloning vectors are small pieces of DNA
isolated from a virus or any other higher organism. A close is inserted with a foreign DNA that
can be used for the purpose of cloning. Thus, a cloning vector must allow the gene editing by
using the restriction enzymes..
1) surrounds the meninges and CNSAllows the nutrients to pass from.pdf
1) surrounds the meninges and CNS
Allows the nutrients to pass from the general cirulation to CNS
Comprises endothelial cells in capillaries
2)ventricles- production of cerebrospinal fluid
Cerebellum- maintaining balance
Brain stem - regulation of digestion
Cognition - cerebral cortex
Midbrain- sleep cycles
Diencephalon - initial processing of vision
Solution
1) surrounds the meninges and CNS
Allows the nutrients to pass from the general cirulation to CNS
Comprises endothelial cells in capillaries
2)ventricles- production of cerebrospinal fluid
Cerebellum- maintaining balance
Brain stem - regulation of digestion
Cognition - cerebral cortex
Midbrain- sleep cycles
Diencephalon - initial processing of vision.
-Included in the capital account are a. No short term assets are i.pdf
-Included in the capital account are
a. No short term assets are included in capital account
b. Long-term financial assets only- correct
c. No short term assets are included in capital account
d. No short term assets are included in capital account
e. It also includes other investments apart from FDI
The current account balance is not directly affected by
a. Government restrictions. It affects because it affects the rates of goods
b. Exchange rate movements.It affects the pricing of imports and exports
c. National income- It does not have direct effect
d. Inflation- It affects because again it affects currency exchange rates
e. Interest rates-It affects because again it affects currency exchange rates
Which of the following factors does probably not directly affect a country\'s capital account and
its components?
a. Exchange rate movements- It affects because it affects value of foreign currency reserves
b. Interest rates- It affect exchange rates
c. Withholding taxes on foreign income- It has a direct effect
d. Inflation- It does not have direct effect
Which of the following statements is not true?
a. Sometimes, trade policies are used to punish countries for various actions.
b. Tariffs imposed by the EU have caused some friction between EU countries that commonly
import products and other EU countries.
c. Exporters may complain that they are being mistreated because the currency of their country is
too weak. For exporters it is better to have a weak currency as their exports becaome cheaper
d. Outsourcing affects the balance of trade because it means that a service is purchased in
another country.
e. All of these choices are true
Transfer payments represent aid, grants, and gifts from one country to another. Hence option d is
correct
7-A special drawing right (SDR) is a unit of account established by a. The Asian governments. b.
The Federal Reserve. c. The Bank of England. d. MNCs. e. The IMF
option e is correct
8)Like the International Monetary Fund (IMF), the ________ is composed of a collection of
nations as members. However, unlike the IMF, it uses the private rather than the government
sector to achieve its objectives.
a. International Financial Corporation (IFC)
Solution
-Included in the capital account are
a. No short term assets are included in capital account
b. Long-term financial assets only- correct
c. No short term assets are included in capital account
d. No short term assets are included in capital account
e. It also includes other investments apart from FDI
The current account balance is not directly affected by
a. Government restrictions. It affects because it affects the rates of goods
b. Exchange rate movements.It affects the pricing of imports and exports
c. National income- It does not have direct effect
d. Inflation- It affects because again it affects currency exchange rates
e. Interest rates-It affects because again it affects currency exchange rates
Which of the following factors does probably not directly affect a .
C++ Program to Implement Doubly Linked List #includei.pdf
/*
* C++ Program to Implement Doubly Linked List
*/
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*head;
/*
Class Declaration
*/
class dll
{
public:
void create(int value);
void append(int value);
void insert(int value, int index);
void del(int value);
void display();
dll()
{
head = NULL;
}
/* destructor*/
~dll()
{
node *ptr;
ptr = head;
while(ptr!=0){
node *next = ptr->next;
delete ptr;
ptr = next;
}
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int option, element, index;
dll dl;
while (1)
{
cout<>option;
switch ( option )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create(element);
cout<>element;
dl.append(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>index;
dl.insert(element, index);
cout<>element;
dl.del(element);
cout<info = value;
temp->next = NULL;
if (head == NULL)
{
temp->prev = NULL;
head = temp;
}
else
{
s = head;
while (s->next != NULL)
s = s->next;
s->next = temp;
temp->prev = s;
}
}
/*
* Insertion at the beginning
*/
void dll::append(int value)
{
if (head == NULL)
{
cout<<\"First Create the list.\"<prev = NULL;
temp->info = value;
temp->next = head;
head->prev = temp;
head = temp;
cout<<\"Element Inserted\"<next;
if (q == NULL)
{
cout<<\"There are less than \";
cout<info = value;
if (q->next == NULL)
{
q->next = tmp;
tmp->next = NULL;
tmp->prev = q;
}
else
{
tmp->next = q->next;
tmp->next->prev = tmp;
q->next = tmp;
tmp->prev = q;
}
cout<<\"Element Inserted\"<info == value)
{
tmp = head;
head = head->next;
head->prev = NULL;
cout<<\"Element Deleted\"<next->next != NULL)
{
/*Element deleted in between*/
if (q->next->info == value)
{
tmp = q->next;
q->next = tmp->next;
tmp->next->prev = q;
cout<<\"Element Deleted\"<next;
}
/*last element deleted*/
if (q->next->info == value)
{
tmp = q->next;
free(tmp);
q->next = NULL;
cout<<\"Element Deleted\"<info<<\" <-> \";
q = q->next;
}
cout<<\"NULL\"<
Solution
/*
* C++ Program to Implement Doubly Linked List
*/
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*head;
/*
Class Declaration
*/
class dll
{
public:
void create(int value);
void append(int value);
void insert(int value, int index);
void del(int value);
void display();
dll()
{
head = NULL;
}
/* destructor*/
~dll()
{
node *ptr;
ptr = head;
while(ptr!=0){
node *next = ptr->next;
delete ptr;
ptr = next;
}
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int option, element, index;
dll dl;
while (1)
{
cout<>option;
switch ( option )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create(element);
cout<>element;
dl.append(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>index;
dl.insert(element, index);
cout<>element;
dl.del(element);
cout<info = value;
temp->next = NULL;
if (head == NULL)
{
temp->prev = NULL;
head = temp;
}
else
{
s = head;
while (s->next != NULL)
s = s->next;
s->next = temp;
t.
678020731-Sumas-y-Restas-Para-Colorear.pdf
KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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Negative control plate with no DNA transforms competent cells withou.pdf
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly.
Solution
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly..
Modifications highlighted in bold lettersDropOutStack.javaim.pdf
//Modifications highlighted in bold letters
//DropOutStack.java
import java.util.Arrays;
public class DropOutStack {
/**
* Defines the interface to a stack collection.
*
* @author Java Foundations
* @version 4.0
*/
public interface StackADT {
/**
* Adds the specified element to the top of this stack.
*
* @param element
* element to be pushed onto the stack
*/
public void push(T element);
/**
* Removes and returns the top element from this stack.
*
* @return the element removed from the stack
*/
public T pop();
/**
* Returns without removing the top element of this stack.
*
* @return the element on top of the stack
*/
public T peek();
/**
* Returns true if this stack contains no elements.
*
* @return true if the stack is empty
*/
public boolean isEmpty();
/**
* Returns the number of elements in this stack.
*
* @return the number of elements in the stack
*/
public int size();
/**
* Returns a string representation of this stack.
*
* @return a string representation of the stack
*/
public String toString();
}
/**
* Represents the situation in which a collection is empty.
*
* @author Java Foundations
* @version 4.0
*/
public static class EmptyCollectionException extends RuntimeException {
/**
* Sets up this exception with an appropriate message.
*
* @param collection
* the name of the collection
*/
public EmptyCollectionException(String collection) {
super(\"The \" + collection + \" is empty.\");
}
}
/**
* Program entry point for stack testing.
*
* @param args
* Argument list.
*/
public static void main(String[] args) {
ArrayDropOutStack stack = new ArrayDropOutStack(4);
System.out.println(\"DROP-OUT STACK TESTING\");
stack.push(1);
stack.push(2);
stack.push(3);
stack.push(4);
System.out.println(\"Stack : \"+stack.toString());
System.out.println(\"pushing 5 element \");
stack.push(5);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
stack.pop();
System.out.println(\"poping element 5\"+stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
System.out.println(\"pushing 7 element \");
stack.push(7);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"pushing 8 element \");
stack.push(8);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
}
/**
* An array implementation of a stack in which the bottom of the stack is
* fixed at index 0.
*
* @author Java Foundations
* @version 4.0
*/
public static class ArrayDropOutStack implements StackADT {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack;
/**
* Creates an empty stack using the default capacity.
*/
public ArrayDropOutStack() {
this(DEFAULT_CAPACITY);
}
/**
* Creates an empty stack using.
Message types used by the DHCP boot sequenceop Message type.h.pdf
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
Solution
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority.
Law of conservation of matter(mass)In any closed system subjected .pdf
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another.
Solution
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another..
It is a unit of measurement for the amount of substance.The mole is .pdf
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023.
Solution
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023..
In insurance, the term risk pooling refers to the spreading of f.pdf
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that demand variability is reduced if one aggregates
demand across locations because, as demand is aggregated across different locations, it becomes
more likely that high demand from one customer will be offset by low demand from another.
However,consider the 1)\"high risk pools\", which simply means separating out the sickest, most
expensive people in the country and dumping them into a separate program.With the \"bad
apples\" (ie, human beings with terrible medical problems) safely tucked out of the way, the
average cost of treating everyone else supposedly suddenly becomes less pricey.This is the exact
opposite of the entire point of health insurance in the first place...spreading the risk and hence
risk pooling fails here.
2)Liability risk includes economic losses from being held responsible for harming others or their
property. This cannot be covered by risk pooling.
Solution
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that.
Some CO is Added The Pressure of CH4 and H2O will.pdf
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left
Solution
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left.
Comments Figures are not properly numbered. Further signal of figur.pdf
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated.
Solution
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated..
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.j.pdf
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.java
import java.util.*;
import java.util.LinkedList;
class Test {
HashSet uniqueStates;
void letsRoll() {
/*
* The capacity of jug1(jug1 capacity is greater than jug2 capacity)
*/
int jug1 = 3;
/*
* The capacity of jug two
*/
int jug2 = 2;
/*
* Final Amount needed in jug1
*/
int amtNeeded = 2;
State initialState = new State(0, 0);
State finalState = new State(amtNeeded, 0);
State finalPath = null;
uniqueStates = new HashSet();
/*
* Change queue to stack to convert to DFS
*/
LinkedList queue = new LinkedList();
queue.add(initialState);
while (!queue.isEmpty()) {
State currState = queue.poll();
if (currState.equals(finalState)) {
finalPath = currState;
break;
}
// if x is zero fill it
if (currState.x == 0) {
State nextState = new State(jug1, currState.y, currState);
//System.out.println(currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is empty fill it
if (currState.y == 0) {
State nextState = new State(currState.x, jug2, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if x is not empty , empty it
if (currState.x > 0) {
State nextState = new State(0, currState.y, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is not empty, empty it
if (currState.y > 0) {
State nextState = new State(currState.x, 0, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from x to y, when x non empty and y is not full
if (currState.x > 0 && currState.y < jug2) {
int amtToTransfer = Math.min(currState.x, jug2 - currState.y);
State nextState = new State(currState.x - amtToTransfer, currState.y
+ amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from y to x, when y is not empty and x is not full
if (currState.y > 0 && currState.x < jug1) {
int amtToTransfer = Math.min(currState.y, jug1 - currState.x);
State nextState = new State(currState.x + amtToTransfer, currState.y
- amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
}
if (finalPath != null) {
System.out.println(\"Final Result\");
System.out.println(\"J1 J2\");
System.out.println(finalPath);
}
else{
System.out.println(\"Not Possible\");
}
}
/*
* Checks whether State toCheck has been generated before, if not it is
* added to queue and uniqueStates Set
*/
void checkUniqueStates(HashSet uniqueStates, State toCheck,
LinkedList queue) {
if (!uniqueStates.contains(toCheck)) {
uniqueStates.add(toCheck);
queue.add(toCheck);
}
}
public static void main(String[] args) {
new Test().letsRoll();
}
}
2 State.java
class State {
/*
* Amount in jug1 for current State
*/
int x;
/*
* Amount in jug2 for current State
*/
int y;
/*
* Parent of current State
*/
State pre;
public State(int x, int y) {
super();
this.x = x;
this.y = y;
}
public State(int x, int y, State pre) {
super();
this.x = x;
this.y = y;
this.pre = pre;
System.out.println(\"(\"+x+\",\"+y+\")\");
}
@Override
public int hashCode() {
final int .
Balance sheetBalance sheet is the statement of financial position .pdf
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be said as the resources of the company and are classified as: Current assets, long
term investment, property, plant and equipment, intangible assets and other assets
Assets includes:
•Current assets – Cash, accounts receivables, prepaid expenses, inventories
•Long term investments – investment in stock
•Plant property and equipment – Land, building, equipment
•Intangible assets – Goodwill, patents, trademarks
Liabilities are the claims and can be classified as current liabilities, long term liabilities
Liabilities includes:
•Current liabilities: Accounts payable, income tax payable. Salaries payable, accrued expenses,
unearned revenue, current portion of long term debt due
•Long term liabilities- bonds payable, notes payable
Owner;s equity can be classified as the Capital stock which includes common stock and
preferred stock
It also includes Additional paid in capital, Retained earnings
Retained earnings are the accumulated earnings of the company. It is calculated as Beginning
balance, Add: Net income less: Dividends
Importance:
Balance sheet is important in every business. Considering the hospitality firm, it is important to
analyze different aspects of teh firm to know its financial postion. As the hospitality firm
emphasize on services, it is important to know the financial position of the company through its
financial statements. the hospitality firms like hotel needs to report their assets (building) held by
the company along with other assets, liabilities and equity.
Balance sheet is the statement which helps in analysing the financial position of the company, it
can also help in predicting the cash flows and the financial position of the company. With the
help of the balance sheet we can help in analysing the ability of the firm to pay off its short term
and long term debts.
Balance sheet reports its assets and liabilities on historical basis and uses different estimates to
analyse the amounts like uncollectibles
Solution
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be .
Answer BaSO4K+ is always soluble because it is an alkaline metal..pdf
Answer: BaSO4
K+ is always soluble because it is an alkaline metal. NO3- is also always soluble.
SO4 is usually soluble except when it is paired with Sr, Ba, Ca, or Pb.
Solution
Answer: BaSO4
K+ is always soluble because it is an alkaline metal. NO3- is also always soluble.
SO4 is usually soluble except when it is paired with Sr, Ba, Ca, or Pb..
Answer is B because he copied the member A and say that he is innoce.pdf
Answer is B because he copied the member A and say that he is innocent
and also here we can\'t give you a exact soln. because of the in appropriate format of the question
details
Solution
Answer is B because he copied the member A and say that he is innocent
and also here we can\'t give you a exact soln. because of the in appropriate format of the question
details.
A nephron consists of a renal corpuscle and a renal tubule. So the a.pdf
A nephron consists of a renal corpuscle and a renal tubule. So the answer is option 1.
Solution
A nephron consists of a renal corpuscle and a renal tubule. So the answer is option 1..
30). e). All of theseDifferent biotechnological methods such as ge.pdf
30). e). All of these
Different biotechnological methods such as gene cloning, DNA amplification, recombinant DNA
technology, are used during the process of DNA isolation and sequencing. For example, the
copies of genome are amplified by the process of DNA amplification, which we observe in PCR
(polymerase chain reaction). The resulting genome can be sequenced.
31). b). Cloning vectors
The purpose of cloning vectors is gene amplification. Cloning vectors are small pieces of DNA
isolated from a virus or any other higher organism. A close is inserted with a foreign DNA that
can be used for the purpose of cloning. Thus, a cloning vector must allow the gene editing by
using the restriction enzymes.
Solution
30). e). All of these
Different biotechnological methods such as gene cloning, DNA amplification, recombinant DNA
technology, are used during the process of DNA isolation and sequencing. For example, the
copies of genome are amplified by the process of DNA amplification, which we observe in PCR
(polymerase chain reaction). The resulting genome can be sequenced.
31). b). Cloning vectors
The purpose of cloning vectors is gene amplification. Cloning vectors are small pieces of DNA
isolated from a virus or any other higher organism. A close is inserted with a foreign DNA that
can be used for the purpose of cloning. Thus, a cloning vector must allow the gene editing by
using the restriction enzymes..
1) surrounds the meninges and CNSAllows the nutrients to pass from.pdf
1) surrounds the meninges and CNS
Allows the nutrients to pass from the general cirulation to CNS
Comprises endothelial cells in capillaries
2)ventricles- production of cerebrospinal fluid
Cerebellum- maintaining balance
Brain stem - regulation of digestion
Cognition - cerebral cortex
Midbrain- sleep cycles
Diencephalon - initial processing of vision
Solution
1) surrounds the meninges and CNS
Allows the nutrients to pass from the general cirulation to CNS
Comprises endothelial cells in capillaries
2)ventricles- production of cerebrospinal fluid
Cerebellum- maintaining balance
Brain stem - regulation of digestion
Cognition - cerebral cortex
Midbrain- sleep cycles
Diencephalon - initial processing of vision.
-Included in the capital account are a. No short term assets are i.pdf
-Included in the capital account are
a. No short term assets are included in capital account
b. Long-term financial assets only- correct
c. No short term assets are included in capital account
d. No short term assets are included in capital account
e. It also includes other investments apart from FDI
The current account balance is not directly affected by
a. Government restrictions. It affects because it affects the rates of goods
b. Exchange rate movements.It affects the pricing of imports and exports
c. National income- It does not have direct effect
d. Inflation- It affects because again it affects currency exchange rates
e. Interest rates-It affects because again it affects currency exchange rates
Which of the following factors does probably not directly affect a country\'s capital account and
its components?
a. Exchange rate movements- It affects because it affects value of foreign currency reserves
b. Interest rates- It affect exchange rates
c. Withholding taxes on foreign income- It has a direct effect
d. Inflation- It does not have direct effect
Which of the following statements is not true?
a. Sometimes, trade policies are used to punish countries for various actions.
b. Tariffs imposed by the EU have caused some friction between EU countries that commonly
import products and other EU countries.
c. Exporters may complain that they are being mistreated because the currency of their country is
too weak. For exporters it is better to have a weak currency as their exports becaome cheaper
d. Outsourcing affects the balance of trade because it means that a service is purchased in
another country.
e. All of these choices are true
Transfer payments represent aid, grants, and gifts from one country to another. Hence option d is
correct
7-A special drawing right (SDR) is a unit of account established by a. The Asian governments. b.
The Federal Reserve. c. The Bank of England. d. MNCs. e. The IMF
option e is correct
8)Like the International Monetary Fund (IMF), the ________ is composed of a collection of
nations as members. However, unlike the IMF, it uses the private rather than the government
sector to achieve its objectives.
a. International Financial Corporation (IFC)
Solution
-Included in the capital account are
a. No short term assets are included in capital account
b. Long-term financial assets only- correct
c. No short term assets are included in capital account
d. No short term assets are included in capital account
e. It also includes other investments apart from FDI
The current account balance is not directly affected by
a. Government restrictions. It affects because it affects the rates of goods
b. Exchange rate movements.It affects the pricing of imports and exports
c. National income- It does not have direct effect
d. Inflation- It affects because again it affects currency exchange rates
e. Interest rates-It affects because again it affects currency exchange rates
Which of the following factors does probably not directly affect a .
C++ Program to Implement Doubly Linked List #includei.pdf
/*
* C++ Program to Implement Doubly Linked List
*/
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*head;
/*
Class Declaration
*/
class dll
{
public:
void create(int value);
void append(int value);
void insert(int value, int index);
void del(int value);
void display();
dll()
{
head = NULL;
}
/* destructor*/
~dll()
{
node *ptr;
ptr = head;
while(ptr!=0){
node *next = ptr->next;
delete ptr;
ptr = next;
}
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int option, element, index;
dll dl;
while (1)
{
cout<>option;
switch ( option )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create(element);
cout<>element;
dl.append(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>index;
dl.insert(element, index);
cout<>element;
dl.del(element);
cout<info = value;
temp->next = NULL;
if (head == NULL)
{
temp->prev = NULL;
head = temp;
}
else
{
s = head;
while (s->next != NULL)
s = s->next;
s->next = temp;
temp->prev = s;
}
}
/*
* Insertion at the beginning
*/
void dll::append(int value)
{
if (head == NULL)
{
cout<<\"First Create the list.\"<prev = NULL;
temp->info = value;
temp->next = head;
head->prev = temp;
head = temp;
cout<<\"Element Inserted\"<next;
if (q == NULL)
{
cout<<\"There are less than \";
cout<info = value;
if (q->next == NULL)
{
q->next = tmp;
tmp->next = NULL;
tmp->prev = q;
}
else
{
tmp->next = q->next;
tmp->next->prev = tmp;
q->next = tmp;
tmp->prev = q;
}
cout<<\"Element Inserted\"<info == value)
{
tmp = head;
head = head->next;
head->prev = NULL;
cout<<\"Element Deleted\"<next->next != NULL)
{
/*Element deleted in between*/
if (q->next->info == value)
{
tmp = q->next;
q->next = tmp->next;
tmp->next->prev = q;
cout<<\"Element Deleted\"<next;
}
/*last element deleted*/
if (q->next->info == value)
{
tmp = q->next;
free(tmp);
q->next = NULL;
cout<<\"Element Deleted\"<info<<\" <-> \";
q = q->next;
}
cout<<\"NULL\"<
Solution
/*
* C++ Program to Implement Doubly Linked List
*/
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*head;
/*
Class Declaration
*/
class dll
{
public:
void create(int value);
void append(int value);
void insert(int value, int index);
void del(int value);
void display();
dll()
{
head = NULL;
}
/* destructor*/
~dll()
{
node *ptr;
ptr = head;
while(ptr!=0){
node *next = ptr->next;
delete ptr;
ptr = next;
}
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int option, element, index;
dll dl;
while (1)
{
cout<>option;
switch ( option )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create(element);
cout<>element;
dl.append(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>index;
dl.insert(element, index);
cout<>element;
dl.del(element);
cout<info = value;
temp->next = NULL;
if (head == NULL)
{
temp->prev = NULL;
head = temp;
}
else
{
s = head;
while (s->next != NULL)
s = s->next;
s->next = temp;
t.
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June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
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