Let S = I + TT H rightarrow H, where T is linear and bounded. Show.pdf
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Let S = I + T*T: H rightarrow H, where T is linear and bounded. Show that S^-1: S(H) rightarrow H exists. Solution T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of two bounded linear operator is a bounded linear operator. I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded linear operator as sum of two bounded linear operator is a bounded linear operator. Now, S : H --> H is a bounded linear operator. The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S. Let, h H. Then I(h) = h This can be written as S-1S(h) = h Or, S-1(S(h)) = h As h H and S : H --> H is a bounded linear operator, S(h) S(H). Let S(h) = h1 S(H) So, S-1(h1) = h where h H and h1 S(H) As h is arbitrary, we can say that S-1 : S(H) --> H exists. Proved.
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It is a nanotechnology question. In which you are able to pick one m.pdf
It is a nanotechnology question. In which you are able to pick one material for both questions or
one material for question one and another for question two.
Solution
Energy is of great importance in human life because of its benefits as the main resource for
human activity.
The application of nano technology or nano material in the field of energy, which involves
lithium-ion battery, fuel cell, light emitting diode (LED), ultra-capacitor, and solar cell
(including Grätzel cell).
Nanotechnology will bring significant benefits to the energy sector, especially to energy storage
and solar energy. Improved materials efficiency and reduced manufacturing costs are just two of
the real economic benefits that nanotechnology already brings these fields and that’s only the
beginning. Battery storage capacity could be extended, solar cells could be produced cheaper,
and the lifetime of solar cells or batteries for electric cars could be increased, all thanks to
continued development of nanotechnology.
The following nanomaterial technologies will be of particular importance: \"organic and printed
electronics\", \"nano-coatings\", \"nano-composites\", \"nano-fluids\", \"nano-catalysts\",
\"nanocarbons\" and \"nano-electrodes\".
Nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) . Nanocomposites are also used for the capture of energy in
silicon-based solar cells, typically in the form of nanowires Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) In the fi eld of energy storage the advantages of nanocomposites are
needed in thin, fl exible energy storage devices with nanocomposite units, e.g. supercapacitors,
Li-ion batteries and hybrid devices Nanotechnology in the sectors of solar energy and energy
storage Additionally, nanocomposites can be used in the process of semiconductor-assisted
photocatalysis for fuel production in order to obtain solar hydrogen or methanol For example, the
nanocomposites applied here are TiO2 -Gold composite nanoparticles or semiconductor-
semiconductor composites Another fi eld of application lies in the wind energy sector, where
lightweight nanocomposite materials are mainly used for rotor blades.
nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) Nanocomposites are also used for the capture of energy in silicon-
based solar cells, typically in the form of nanowires . Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) . In the fi eld of energy storage the advantages of nanocomposites .
I have this problem that I was given in class, but I cant for the .pdf
I have this problem that I was given in class, but I can\'t for the life of me figure out how to do
part (d). Could someone please do it for me if it\'s possible? If the problem is leaving out some
information, then let me know. I am having a problem finding a way to do this part with the
information given, so I will tell my teacher if this problem can\'t be answered with what is in the
problem statement. 1. During the Apollo program, the U.S. landed 12 men on the Moon and
returned them safely to Earth. It was an amazing feat and an exciting time to be an aerospace
engineer. The last Apollo mission was accomplished in December, 1972. The descent stage (DS)
separated from the command module (CM), which remained in lunar orbit with the CM pilot.
The DS carried two astronauts to a landing on the Moon. When the lunar expedition ended, the
ascent stage (AS) rendezvoused with the CM for the return trip to Earth. Here\'s your chance to
analyze the AS trajectory using actual values. 5.322 ft/sec (acceleration due to gravity on the
Moon) 5,187 lb. (Earth weightas 10,295 lb. (Earth weight including propellant); weight of AS
propellant- w weight). a) What is the mass of the AS including propellant? b) What does the AS
and propellant weigh on the Moon? c) What is the propellant mass? The thrust of the ascent
propulsion system (T-3,617 lb. The thrust duration is called action time (tama teton = 446 sec. Of
course, the mass of the AS varies as propellant is expended. d) Assuming the propellant mass
flow rate is constant, what is the mass flow rate during the motor burn (from t-0 to t-taction)?
Note that the AS mass, m, is a function of time such that m(t)=m() + m (t. t.) where m=-. Also let
t.-0, so that m(t)=m(0)+m t . Note that mc0 Now, the total force acting on the AS in the vertical
direction is T-mt)gThe vertical acceleration (up is positive) is T-m(t)g m(t) a(t) (From F ma). e)
What is the speed of the AS at the end of the motor burn, that is, when t? For a first-order
approximation, assume the AS trajectory is vertical and does not pitch over to rendezvous with
the CM. Further assumegarconstant. dt v()-sal)dt jdb.bm(a + bt) Some useful information: =-ln (a
+ bt
Solution
Autocorrect is almost as old as personal computers. Even some of the earliest word processors
had a spellcheck feature that suggested alternative words if a word you typed did not appear in
its internal dictionary. Certainly, today, autocorrect has come a long way since the Cupertino
effect. What’s the Cupertino effect? Well, in its early days, word processors would replace the
word “cooperation” with “Cupertino” (the city in Northern California where Apple and other
computercompanies are headquartered). This spelling suggestion made its way to many
documents published by the United Nations, NATO, and other official bodies, so the Cupertino
In the beginning autocorrect was primarily a smart spellchecker. If you typed “bfeore,” it
replaced it with “before.” Then in 2007, Microsoft int.
Implement this in Java Method to determine if a particular elemen.pdf
Implement this in Java
* Method to determine if a particular element is in this set.
* @param target * the element that needs to be found in this set
* @return * true if the target element is in this set, false otherwise
public boolean contains(int target)
{
}
Solution
public boolean contains(int target)
{
boolean flag=false;
for(Integer i:(this))
{
if(target==i.intValue())
{
flag=true;
break;
}
}
return flag;
}.
If an image is represented using 400 x 300 pixels and each pixel is .pdf
If an image is represented using 400 x 300 pixels and each pixel is represented using 8 bits, what
will be the exact amount of data needed in kilobytes to represent in the image?
Solution
there are 1200 pixels in total.
Each pixel is represented requires 8 bit for its representation which 1 byte of data.
Now for 1200 pixels we require 1200 bytes of data to represent a 400x300 pixel image.
1 byte = 1/1024kb
1200 byte = 1200*(1/1024) = 1.17 kb data is required to present an image..
Implement the unsorted single linked list as we did in the class and .pdf
Implement the unsorted single linked list as we did in the class and implement the following
operations: 1.DeletelastDuplicat(): For any element in the linked list, if there are multiple copies
(> = 2 copies), delete the last copy. DeleteSecondlastDuplicat(): For any element in the linked
list, if there are multiple copies (> = 2 copies), delete the second last copy Test your program
with the following operations: a)insert 5 b)Insert 7 c)Insert 11 d)Insert 5 e)Insert f) Insert 5 g)
Print out the list h) Delete the last duplicate of 5 i)Print out the list i) Delete the last duplicate
of 11 k) Print out the list I) Insert 11 m) Insert 7 n) Print out the list O) Delete the second last
duplicate of 5 p) Print out the list q) Delete the second last duplicate of 7 r) Print out the list
Solution
#include
#include
#include
using namespace std;
struct node
{
int data;
struct node *next;
}*start;
class LinkedList
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos(int);
void delete_lastdup();
void delete_seclast();
int search(int);
void display();
int get_count(int);
LinkedList()
{
start = NULL;
}
};
int main()
{
int choice, pos, value, res=0;
LinkedList sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<>pos;
sl.delete_pos(pos);
break;
case 5:
cout<<\"Search element in Link List: \"<>value;
res=sl.search(value);
cout<<\"Element \"<data = value;
temp->next = NULL;
return temp;
}
}
//function to insert element at beginning
void LinkedList::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<next != NULL)
{
if(s->data == value)
count++;
s = s->next;
}
return count;
}
//function to insert node at last
void LinkedList::insert_last()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<>value;
c=get_count(value);
if(c<2)
{
cout<<\"duplicate does not exist\"<>value;
c=get_count(value);
x=c-1;
if(c<2)
{
cout<<\"duplicate does not exist\"<data == value)
{
f++;
res=pos;
if(f==x)
break;
}
s = s->next;
}
delete_pos(res);
}
}
//delete element at p.
Hyposecretion of insulin is referred to as Type I diabetes insipidus.pdf
Hyposecretion of insulin is referred to as Type I diabetes insipidus. Type II diabetes mellitus
Type I diabetes mellitus sugarbeetes pancreatic dwarfism
Solution
33. Hyposecretion of insulin is referred to as -
C) Type I diabetes mellitus.
Type I diabetes mellitus : It is happened due to deficient amount of insulin production , by
pancreatic beta cells of islets of Langerhans..
How are the forces of diffusion and electrical force responsible for.pdf
How are the forces of diffusion and electrical force responsible for both the resting potential and
the action potential?
Solution
Ans). In a resting state, diffusion and electrical force balance each other to attain an equilibrium
for potassium and chloride ions. Large number of protein molecules cannot move through the
membrane due to their size because of their negative charge, they contribute to the relative
negativity of the intracellular fluid..
How does each of the differences between prokaryotes and eukaryotes .pdf
How does each of the differences between prokaryotes and eukaryotes influence bacterial
infection and treatment
Solution
The differences between prokaryotes and eukaryotes such as cell wall, nucleus, flagella,
membrane-bound organelles, genetic makeup, metabolic differences and antigenicity are
responsible for the infection and treatment of bacterial diseases. For example, the bacteria
contains cell wall, which is absence in animal cells. So, cell wall synthesis inhibition is a strategy
in the treatment of bacterial infections..
Recommended
It is a nanotechnology question. In which you are able to pick one m.pdf
It is a nanotechnology question. In which you are able to pick one material for both questions or
one material for question one and another for question two.
Solution
Energy is of great importance in human life because of its benefits as the main resource for
human activity.
The application of nano technology or nano material in the field of energy, which involves
lithium-ion battery, fuel cell, light emitting diode (LED), ultra-capacitor, and solar cell
(including Grätzel cell).
Nanotechnology will bring significant benefits to the energy sector, especially to energy storage
and solar energy. Improved materials efficiency and reduced manufacturing costs are just two of
the real economic benefits that nanotechnology already brings these fields and that’s only the
beginning. Battery storage capacity could be extended, solar cells could be produced cheaper,
and the lifetime of solar cells or batteries for electric cars could be increased, all thanks to
continued development of nanotechnology.
The following nanomaterial technologies will be of particular importance: \"organic and printed
electronics\", \"nano-coatings\", \"nano-composites\", \"nano-fluids\", \"nano-catalysts\",
\"nanocarbons\" and \"nano-electrodes\".
Nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) . Nanocomposites are also used for the capture of energy in
silicon-based solar cells, typically in the form of nanowires Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) In the fi eld of energy storage the advantages of nanocomposites are
needed in thin, fl exible energy storage devices with nanocomposite units, e.g. supercapacitors,
Li-ion batteries and hybrid devices Nanotechnology in the sectors of solar energy and energy
storage Additionally, nanocomposites can be used in the process of semiconductor-assisted
photocatalysis for fuel production in order to obtain solar hydrogen or methanol For example, the
nanocomposites applied here are TiO2 -Gold composite nanoparticles or semiconductor-
semiconductor composites Another fi eld of application lies in the wind energy sector, where
lightweight nanocomposite materials are mainly used for rotor blades.
nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) Nanocomposites are also used for the capture of energy in silicon-
based solar cells, typically in the form of nanowires . Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) . In the fi eld of energy storage the advantages of nanocomposites .
I have this problem that I was given in class, but I cant for the .pdf
I have this problem that I was given in class, but I can\'t for the life of me figure out how to do
part (d). Could someone please do it for me if it\'s possible? If the problem is leaving out some
information, then let me know. I am having a problem finding a way to do this part with the
information given, so I will tell my teacher if this problem can\'t be answered with what is in the
problem statement. 1. During the Apollo program, the U.S. landed 12 men on the Moon and
returned them safely to Earth. It was an amazing feat and an exciting time to be an aerospace
engineer. The last Apollo mission was accomplished in December, 1972. The descent stage (DS)
separated from the command module (CM), which remained in lunar orbit with the CM pilot.
The DS carried two astronauts to a landing on the Moon. When the lunar expedition ended, the
ascent stage (AS) rendezvoused with the CM for the return trip to Earth. Here\'s your chance to
analyze the AS trajectory using actual values. 5.322 ft/sec (acceleration due to gravity on the
Moon) 5,187 lb. (Earth weightas 10,295 lb. (Earth weight including propellant); weight of AS
propellant- w weight). a) What is the mass of the AS including propellant? b) What does the AS
and propellant weigh on the Moon? c) What is the propellant mass? The thrust of the ascent
propulsion system (T-3,617 lb. The thrust duration is called action time (tama teton = 446 sec. Of
course, the mass of the AS varies as propellant is expended. d) Assuming the propellant mass
flow rate is constant, what is the mass flow rate during the motor burn (from t-0 to t-taction)?
Note that the AS mass, m, is a function of time such that m(t)=m() + m (t. t.) where m=-. Also let
t.-0, so that m(t)=m(0)+m t . Note that mc0 Now, the total force acting on the AS in the vertical
direction is T-mt)gThe vertical acceleration (up is positive) is T-m(t)g m(t) a(t) (From F ma). e)
What is the speed of the AS at the end of the motor burn, that is, when t? For a first-order
approximation, assume the AS trajectory is vertical and does not pitch over to rendezvous with
the CM. Further assumegarconstant. dt v()-sal)dt jdb.bm(a + bt) Some useful information: =-ln (a
+ bt
Solution
Autocorrect is almost as old as personal computers. Even some of the earliest word processors
had a spellcheck feature that suggested alternative words if a word you typed did not appear in
its internal dictionary. Certainly, today, autocorrect has come a long way since the Cupertino
effect. What’s the Cupertino effect? Well, in its early days, word processors would replace the
word “cooperation” with “Cupertino” (the city in Northern California where Apple and other
computercompanies are headquartered). This spelling suggestion made its way to many
documents published by the United Nations, NATO, and other official bodies, so the Cupertino
In the beginning autocorrect was primarily a smart spellchecker. If you typed “bfeore,” it
replaced it with “before.” Then in 2007, Microsoft int.
Implement this in Java Method to determine if a particular elemen.pdf
Implement this in Java
* Method to determine if a particular element is in this set.
* @param target * the element that needs to be found in this set
* @return * true if the target element is in this set, false otherwise
public boolean contains(int target)
{
}
Solution
public boolean contains(int target)
{
boolean flag=false;
for(Integer i:(this))
{
if(target==i.intValue())
{
flag=true;
break;
}
}
return flag;
}.
If an image is represented using 400 x 300 pixels and each pixel is .pdf
If an image is represented using 400 x 300 pixels and each pixel is represented using 8 bits, what
will be the exact amount of data needed in kilobytes to represent in the image?
Solution
there are 1200 pixels in total.
Each pixel is represented requires 8 bit for its representation which 1 byte of data.
Now for 1200 pixels we require 1200 bytes of data to represent a 400x300 pixel image.
1 byte = 1/1024kb
1200 byte = 1200*(1/1024) = 1.17 kb data is required to present an image..
Implement the unsorted single linked list as we did in the class and .pdf
Implement the unsorted single linked list as we did in the class and implement the following
operations: 1.DeletelastDuplicat(): For any element in the linked list, if there are multiple copies
(> = 2 copies), delete the last copy. DeleteSecondlastDuplicat(): For any element in the linked
list, if there are multiple copies (> = 2 copies), delete the second last copy Test your program
with the following operations: a)insert 5 b)Insert 7 c)Insert 11 d)Insert 5 e)Insert f) Insert 5 g)
Print out the list h) Delete the last duplicate of 5 i)Print out the list i) Delete the last duplicate
of 11 k) Print out the list I) Insert 11 m) Insert 7 n) Print out the list O) Delete the second last
duplicate of 5 p) Print out the list q) Delete the second last duplicate of 7 r) Print out the list
Solution
#include
#include
#include
using namespace std;
struct node
{
int data;
struct node *next;
}*start;
class LinkedList
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos(int);
void delete_lastdup();
void delete_seclast();
int search(int);
void display();
int get_count(int);
LinkedList()
{
start = NULL;
}
};
int main()
{
int choice, pos, value, res=0;
LinkedList sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<>pos;
sl.delete_pos(pos);
break;
case 5:
cout<<\"Search element in Link List: \"<>value;
res=sl.search(value);
cout<<\"Element \"<data = value;
temp->next = NULL;
return temp;
}
}
//function to insert element at beginning
void LinkedList::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<next != NULL)
{
if(s->data == value)
count++;
s = s->next;
}
return count;
}
//function to insert node at last
void LinkedList::insert_last()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<>value;
c=get_count(value);
if(c<2)
{
cout<<\"duplicate does not exist\"<>value;
c=get_count(value);
x=c-1;
if(c<2)
{
cout<<\"duplicate does not exist\"<data == value)
{
f++;
res=pos;
if(f==x)
break;
}
s = s->next;
}
delete_pos(res);
}
}
//delete element at p.
Hyposecretion of insulin is referred to as Type I diabetes insipidus.pdf
Hyposecretion of insulin is referred to as Type I diabetes insipidus. Type II diabetes mellitus
Type I diabetes mellitus sugarbeetes pancreatic dwarfism
Solution
33. Hyposecretion of insulin is referred to as -
C) Type I diabetes mellitus.
Type I diabetes mellitus : It is happened due to deficient amount of insulin production , by
pancreatic beta cells of islets of Langerhans..
How are the forces of diffusion and electrical force responsible for.pdf
How are the forces of diffusion and electrical force responsible for both the resting potential and
the action potential?
Solution
Ans). In a resting state, diffusion and electrical force balance each other to attain an equilibrium
for potassium and chloride ions. Large number of protein molecules cannot move through the
membrane due to their size because of their negative charge, they contribute to the relative
negativity of the intracellular fluid..
How does each of the differences between prokaryotes and eukaryotes .pdf
How does each of the differences between prokaryotes and eukaryotes influence bacterial
infection and treatment
Solution
The differences between prokaryotes and eukaryotes such as cell wall, nucleus, flagella,
membrane-bound organelles, genetic makeup, metabolic differences and antigenicity are
responsible for the infection and treatment of bacterial diseases. For example, the bacteria
contains cell wall, which is absence in animal cells. So, cell wall synthesis inhibition is a strategy
in the treatment of bacterial infections..
For MIMO system, (a) Please talk about the advantage and disadvantag.pdf
For MIMO system, (a) Please talk about the advantage and disadvantage of MIMO technique.
(b) Please explain why we need Cooperative Diversity to replace MIMO in cellular network.
Solution
A)MIMO advantages
• Capacity scales linearly with number of antennas
Channel knowledge/estimation at Rx needed
• MIMO offers potential for
larger data rate
larger spectral efficiency
larger number of users
improved range/coverage
better interference suppression
better quality of service (QoS), lower bit-error rate (BER)
lower Tx power
MIMO disavantages
• Hardware complexity:
Each antenna needs a radio-frequency (RF) unit
Powerful digital signal processing (DSP) unit required
• Software complexity:
Most signal processing algorithms are computationally intensive
Power consumption:
Battery lifetime of mobile devices
Thermal problems
Antennas:
Antenna spacing (electromagnetic mutual coupling-e.g. mobile handsets)
RF interference and antenna correlation
B)
Most of the present researches on cooperative
network in which the user nodes are equipped with a single
antenna or multiple, there have been some new results which
exploit the benefits of multiple antenna deployment.
Cooperative MIMO technology allows a wireless network
system to attain better performance gains than provided by
either usual MIMO or cooperative systems. It promised
significant improvement in spectral efficiency and network
coverage phenomena for different next generation wireless
communication systems. In wireless communication, the path
towards the various techniques that gives high service quality
and data rate has been through the use of the cooperative
network provided by the rich scattering wireless channels.
Due to their great aspects, MIMO and cooperative systems
have found their way into several standards for future
wireless communication systems, especially in cellular
networks and wireless local area networks (LAN) in this
review article we are presenting the comparative analysis..
Explain why and give examples. In OO programming Polymorphism is on.pdf
Explain why and give examples. \"In OO programming Polymorphism is only possible when
using Interface or Inheritance\".
Solution
Polymorphism is the ability of an object to take on many forms. The most common use of
polymorphism in OOP occurs when a parent class reference is used to refer to a child class
object.
Any Java object that can pass more than one IS-A test is considered to be polymorphic. In Java,
all Java objects are polymorphic since any object will pass the IS-A test for their own type and
for the class Object.
It is important to know that the only possible way to access an object is through a reference
variable. A reference variable can be of only one type. Once declared, the type of a reference
variable cannot be changed.
The reference variable can be reassigned to other objects provided that it is not declared final.
The type of the reference variable would determine the methods that it can invoke on the object.
A reference variable can refer to any object of its declared type or any subtype of its declared
type. A reference variable can be declared as a class or interface type.
Polymorphism using inheritance:
Polymorphism using interface:
class Main
{
}.
Eukaryotic cells modify RNA after transcription What critical RNA pr.pdf
Eukaryotic cells modify RNA after transcription What critical RNA processing events usually
happen to pre-mRNA\'s before they sent to the cytoplasm for translation? What is a 5\' cap?
What is a poly A tail? What do these end modifications do for the mRNA transcript? What is
RNA splicing? What are introns? What are exons? How are intros spliced out? What is a
spliceosome? What is a \"snurp\" (snRNA)?
Solution
After its synthesis, the eukaryotic mRNA will undergo extensive modification like capping,
polyadenylation and splicing to enter into the process of translation.
(B)
Capping: Here the 5\' end of the mRNA is modified by the addition of 7-methylguanosine (m7G)
and the main function of this cap is to protect the 5\' end of the primary RNA transcript from
attack by ribonucleases and this 5’ cap will be recognized by eukaryotic initiation factors, so that
it can assemble the mature mRNA with the ribosome to start the process of translation.
At the 3\' end of the RNA, we will have polyadenylation signal and during transcription itself
this sequence will be chopped by an enzyme and another enzyme will add about 100100100 -
200200200 adenine (A) nucleotides to the 3’ end and this will form the poly-A tail.
The main function of this poly A tail is proving stability to the transcript and also helping it to
get exported from the nucleus to the cytosol.RNA splicing
(C )Splicing is the third big RNA processing event and the pre-mRNA will have two sequences,
exons and introns.
Introns are the non-coding sequences and exons are the coding sequences, here in this step the
through splicing, the introns will be removed and exons will be attached together.
In RNA splicing, specific parts of the pre-mRNA (introns) will be recognized and removed by a
protein-and-RNA complex called the spliceosome. Mature mRNA will have only exons, but no
introns.
(D)The splicing signal exon/GU-intron-AG/exon will be present in nuclear mRNA precursors
and 5\' and 3\' splice sites always have consensus sequences extending beyond GU and AG
motifs. During splicing, the exon-intron boundaries will be recognized by snRNA and the
consensus sequences within introns will get hybridized and now the proteins other snRNAs will
assemble the spliceosome on the transcript, the unpaired A present at 3\' side of the introns will
attacks the 5\' exon -intron boundary with the help of 2\' OH and this will give rise to lariat
structure. The free 3\' OH of the upstream exon will displace the downstream junctional
nucleotide, like this introns will be removed and exons will be attached together.
(E)A spliceosome is a large complex formed due to the assembly of snRNAs and protein
complexes, and plays an important role in splicing of pre mRNA.
snRNPs (snurps) is the small nuclear ribonucleic proteins and this is a RNA-protein complexes
and they will form a larger complex with the unmodified pre-mRNA and various other proteins
to form the structure called spliceosome.(A)Critical RNA processing events.
Driverimport java.util.Scanner;A class that keeps a f.pdf
Driver:
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i
Solution
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i 0)
this.cylinders = cylinders;
else
this.cylinders = 0;
}
/**
* @return the ownersName
*/
public String getOwnersName() {
return ownersName;
}
/**
* @param ownersName
* the ownersName to set
*/
public void setOwnersName(String ownersName) {
this.ownersName = ownersName;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Vehicle) {
Vehicle vehicle = (Vehicle) obj;
if ((this.getManuName().equals(vehicle.getManuName()))
&& (this.getOwnersName().equals(vehicle.getOwnersName()))
&& (this.getCylinders() == vehicle.getCylinders()))
return true;
else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return getManuName() + \" \" + getCylinders() + \" \" + getOwnersName();
}
}
public class Truck extends Vehicle {
double loadCap, towCap;
public Truck() {
// TODO Auto-generated constructor stub
super();
setLoadCap(0);
setTowCap(0);
}
/**
* @param manuName
* @param cylinders
* @param ownersName
* @param loadCap
* @param towCap
*/
public Truck(String manuName, int cylinders, String ownersName,
double loadCap, double towCap) {
super(manuName, cylinders, ownersName);
setLoadCap(loadCap);
setTowCap(towCap);
}
/**
* @return the loadCap
*/
public double getLoadCap() {
return loadCap;
}
/**
* @param loadCap
* the loadCap to set
*/
public void setLoadCap(double loadCap) {
if (loadCap > 0)
this.loadCap = loadCap;
else
this.loadCap = 0;
}
/**
* @return the towCap
*/
public double getTowCap() {
return towCap;
}
/**
* @param towCap
* the towCap to set
*/
public void setTowCap(double towCap) {
if (towCap > 0)
this.towCap = towCap;
else
this.towCap = 0;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Truck) {
if (super.equals(obj)) {
Truck vehicle = (Truck) obj;
if (this.getLoadCap() == vehicle.getLoadCap()
&& this.getTowCap() == vehicle.getTowCap())
return true;
else
return false;
} else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return super.toString() + \" \" + g.
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Solution
The idea here is to use the concept of two binary heaps along with the concept of list(doubly
linked list).
The doubly linked list contains all input items and indexes of corresponding min and max heap
nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root
node of min heap stores the address of minimum item in doubly linked list. Similarly, root of
max heap stores address of maximum item in doubly linked list. Following are the details of
operations.
1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a
O(1) operation.
1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a
O(1) operation.
3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this
address to find the node in doubly linked list. From the doubly linked list, we get node of Max
Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time.
delete() operations for max and min heaps take O(Logn) time.
4) removeMax(): is similar to deleteMin()
5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in
Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn)
#include
#include
#include
// A node of doubly linked list
struct LNode
{
int data;
int minHeapIndex;
int maxHeapIndex;
struct LNode *next, *prev;
};
// Structure for a doubly linked list
struct List
{
struct LNode *head;
};
// Structure for min heap
struct MinHeap
{
int size;
int capacity;
struct LNode* *array;
};
// Structure for max heap
struct MaxHeap
{
int size;
int capacity;
struct LNode* *array;
};
// The required data structure
struct CDS
{
struct MinHeap* minHeap;
struct MaxHeap* maxHeap;
struct List* list;
};
// A utility function to create a new List node
struct LNode* newLNode(int data)
{
struct LNode* node =
(struct LNode*) malloc(sizeof(struct LNode));
node->minHeapIndex = node->maxHeapIndex = -1;
node->data = data;
node->prev = node->next = NULL;
retu.
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.pdf
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.e.,D0). The dividend is
expected to grow at a constant rate of 6% a year. The required rate of return on the stock, r, is
11%. What is the value per share of the company\'s stock? Round your answer to two decimal
places.
Solution
Recent Dividend, D0 = $4.30
Growth Rate, g = 6%
Required Return, rs = 11%
D1 = D0 * (1 + g)
D1 = $4.30 * 1.06
D1 = $4.558
Value per share = D1 / (rs - g)
Value per share = $4.558 / (0.11 - 0.06)
Value per share = $4.558 / 0.05
Value per share = $91.16.
Anatomy Question Please Answer them all.A. Using a punnett square.pdf
Anatomy Question: Please Answer them all.
A. Using a punnett square, show the possible genotypes for John and Andrea\'s children? (John
has O blood type and Andrea has A blood type)
B. Based on the blood typing results, could Daryl be John and Andrea\'s Son? How Do you know
this? ( Daryl has O blood type)
Solution
A. Andrea :Mother has A blood type , her genotype will be either AO or AA
John :Father has O blood type ,his genotype will be OO.
O allele is masked by the presence of A allele or B allele.
OO blood group is expressed only in presence of OO alleles.
Father
Father
B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have
O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the
son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood
groupOO O blood group.
A person is lost in the desert. (S)he has no water and after a coupl.pdf
A person is lost in the desert. (S)he has no water and after a couple of days loses so much water
through perspiration that his/her body fluids become hypertonic. What happens to his cells?
What happens to his proteins? What happens to his carbohydrates?
Solution
Early sign of dehydration there is no symptom but later on there is dryness of mouth and tongue,
dark yellow urine,dryness of skin cramp in arm and leg. When fluid become hypertonic i.e
sodium is lost from cell this result in cell shrinkage.
After two days of starvation body will first utilize carbohydrate and then glycogen from liver and
muscle for energy. Protein from cell / muscle will be next to be used as source of energy for
body.
2. Describe an advantage and a limitation to tracing ancestry with b.pdf
Tracing ancestry with mitochondrial DNA and Y-DNA has the advantage of providing information about historical human evolution over thousands of years, but is limited in its ability to unravel full familial relationships. Autosomal DNA provides a more complete picture of ancestral inheritance patterns, but has ethical limitations when used for prenatal ancestral tracing.
Write your thought on the following1. petroski notes that most en.pdf
Write your thought on the following:
1. petroski notes that most engineering failures are much more involved than simple technical
mis-calculations and involve the failure of the design process or management culture.However,
not all engineering failures involve ethical issue.The infamous collapse of the first Tacoma
Narrow Brodge, and the losses of the Mars polar lander and Mars Climate Orbiter were technical
and design process failures.
Solution
In the face of new technology, how do we balance public welfare and progress? If Moisseiff had
designed a bridge similar to the ones which had already proven their stability, Tacoma Narrows
Bridge would never have collapsed costing thousands of dollars and endangering many lives. It
would also have been significantly more expensive. On the other hand, if engineers had never
tried innovative techniques, suspension bridges may never have been built at all. At the time of
their introduction, no one believed that a suspension bridge could safely accommodate trains.
Roebling, however, took a gamble, pushed the limits of the current technology, and built a
suspension bridge that he believed could safely support rail traffic. Luckily he was correct, and
suspension bridges soon became widely accepted (Petroski. Moisseiff also took a gamble, trying
to create a longer, sleeker, less expensive bridge, by pushing the limits of technology. He,
however, was not as lucky, and what could have been a breakthrough in technology turned into a
catastrophic failure. Every time engineers push the limits of technology they risk a similar loss
like losses of mas polar lander and mass climate orbiter in which there were technical error like
Mass polar lander
1. Inadequate specifications which includes what the software was supposed to do, but no
mention of what it must not to do.
2. Flawed review process: Violation of basic engineering safety pracitices in the digital part of
the system.
Mass Climate Orbiter
1. Units communication and attention to detail failure. The primary cause of this discrepancy
was that one piece of ground software supplied by Lockheed Martin produced results in a United
States customary unit, contrary to its Software Interface Specification (SIS), while a second
system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS.
Specifically, software that calculated the total impulse produced by thruster firings calculated
results in pound-seconds. The trajectory calculation software then used these results - expected to
be in newton-seconds - to update the predicted position of the spacecraft.
sometimes even It can cause a loss of life. How much is too much? When is a possible advance
worth a risk to public safety? What can the engineering profession do to make the
implementation of new technology safer? Do our current peer review and building code
committee processes adequately protect public safety?.
You have two cell lines, a macrophage cell line and a CD4 T-cell lin.pdf
You have two cell lines, a macrophage cell line and a CD4 T-cell line. The macrophage cell line
is resistant to infection by HIV, whereas the T-cell line is fully susceptible and fully permissive.
Why are macrophage cell line resistant to infection by HIV? Why are T-cell fully susceptible and
fully permissive? Please explain.
Also, A heterokaryon cell made from the fusion of these two cells is able to be infected by HIV.
Explain why? Please explain in detail.
Solution
Macrophages are antigen- presenting cells so they process the viral peptides and present it to
uninfected CD4 T-Cell which is the important transmission process. Also HIV-infected
macrophages release soluble cytotoxic factors which induce apoptosis in CD4 T-Cells.
Macrophages produce certain host restriction factors which inhibits the release of viral progeny
from infected cells. So macrophages are resistant to lysis because of its genetic make up and its
phenotypes (restriction factors).
So when two cell lines are fused to form heterokaryon cell, the genes of macrophages are
suppressed or it is recessive or the T cell genes are dominant to make the heterokaryon
susceptible to infection..
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and
females. In X-linked dominant traits, males hemizygous for trait will express the trait, but
females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected
female will show the trait. For autosomal dominant, two affected parents may produce an
unaffected child.
Solution
Answer:
A. True
Most X-linked conditions are recessive. This means that in a person with two X chromosomes
(most females), both copies of a gene (i.e., one on each X chromosome) must have a change or
mutation whereas in a person with one X chromosome (most males), only one copy of a gene
must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is
said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X
chromosome is typically affected with the condition.
The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome
to the son and other to the daughter.
B. False
For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a
female with two X chromosomes or males with on X chromosome must have a change or
mutation for an individual to be affected with the condition. For this reason, X-linked disorders
are often seen with similar frequency in males and females. However, since females also have
one normal X chromosome as well as an X chromosome with a mutation, the condition is often
more “mild.”
C.False
For X-linked recessive disorders, an affected father who has a mutation in a gene on the X
chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his
children. If the mother is not affected or a carrier, none of his sons will be affected since they can
only inherit a normal X chromosome from their mother and they inherit a Y chromosome from
their father.
But if the female is affected, it will receive one of the affected X chromosomes, which is enough
to express the trait in the son.
D. True
A person affected by an autosomal dominant disorder has a 50 percent chance of passing the
mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50
percent..
When examining the central tendency of a probability distribution, u.pdf
When examining the central tendency of a probability distribution, under what circumstance
would you be indifferent to using the median vs the mean?
Solution
If distribution is symmetric, then mean and median would coincide, and we would be indifferent.
If distribution is skewed, or has outliers, we prefer median because it is not affected by extreme
values..
What will organizations have to do to adapt to the influx of needed .pdf
What will organizations have to do to adapt to the influx of needed Generation Y individuals?
Solution
simple thing is choose the right person for a work and let them free, they will create miracles.
here generation Y people are those who can perform variaous tasks at a time and with more
accuracy. these are multi talented, always carries courage, never hides their talent and ready to
accept challenges.
first to attract those kind of people organisations should change their existing rules, policies and
regulations. all they were writen decades ago and still we are follwoing them. they need to
redesign the work and the structure. if we are succede at recruiting the correct people for our
work, then the remaining changes can automatically comes into the picture.
the organisations should create the work environment to the Y generation people
There are no precise dates when the generation starts and ends; most researchers and
commentators use birth years ranging from the early 1980s to early 2000s. Generation Y people
may be known as echo boomers because they are the children of baby boomers, or someone who
was born during the period of increased birth rates that occurred between 1946 and 1964. The
Generation Y cohort is also referred to as Millennials, the Internet generation, iGen and/or the
Net generation.
What are the main functions of the components of prokaryotic vs. euk.pdf
What are the main functions of the components of prokaryotic vs. eukaryotic cells? Could either
cell type survive without any of them? (Answer must be at least a page long, very detailed and in
paragraph form please)
Solution
Prokaryotes are unicellular living beings that need organelles or other interior film bound
structures . In this manner, they don\'t have a core, at the same time, rather, by and large have a
solitary chromosome: a bit of roundabout, twofold stranded DNA situated in a territory of the
phone called the nucleoid. Most prokaryotes have a cell divider outside the plasma film.
Like a prokaryotic cell, an eukaryotic cell has a plasma film, cytoplasm, and ribosomes.
Nonetheless, not at all like prokaryotic cells, eukaryotic cells have:
a film bound core
various film bound organelles (counting the endoplasmic reticulum, Golgi contraption,
chloroplasts, and mitochondria)
a few bar formed chromosomes
Since an eukaryotic cell\'s core is encompassed by a film, it is regularly said to have a \"genuine
core. \" Organelles (signifying \"little organ\") have specific cell parts, similarly as the organs of
your body have particular parts. They permit diverse capacities to be compartmentalized in
various territories of the cell.
The piece of the cell divider varies altogether between the areas Bacteria and Archaea, the two
spaces of life into which prokaryotes are separated . The structure of their cell dividers
additionally varies from the eukaryotic cell dividers found in plants (cellulose) or organisms and
bugs (chitin). The cell divider works as a defensive layer and is in charge of the life form\'s
shape. A few microorganisms have a container outside the cell divider. Different structures are
available in some prokaryotic species, yet not in others. For instance, the case found in a few
animal types empowers the living being to connect to surfaces, shields it from lack of hydration
and assault by phagocytic cells, and expands its imperviousness to our invulnerable reactions. A
few animal categories likewise have flagella utilized for velocity and pili utilized for connection
to surfaces. Plasmids, which comprise of additional chromosomal DNA, are likewise present in
numerous types of microorganisms and archaea.
Commonly, the core is the most conspicuous organelle in a cell. Eukaryotic cells have a genuine
core, which implies the cell\'s DNA is encompassed by a layer. Thusly, the core houses the cell\'s
DNA and coordinates the blend of proteins and ribosomes, the cell organelles in charge of
protein union. The atomic envelope is a twofold layer structure that constitutes the furthest bit of
the core. Both the internal and external layers of the atomic envelope are phospholipid bilayers.
The atomic envelope is punctuated with pores that control the section of particles, particles, and
RNA between the nucleoplasm and cytoplasm. The nucleoplasm is the semi-strong liquid inside
the core where we discover the chromatin and the nucleolus. M.
There is a requirement to design a system to sense the presence of gl.pdf
There is a requirement to design a system to sense the presence of glass bottles leaving an
automatic washer on a production line conveyor belt. The production line is for food products
and the environment is clean and maintained at a temperature of 15 degree C. The system must
meet the following criteria: The switching distance will lie between 5 and 20 mm. The output
from the sensor is to be of binary form, i.e. logical 1 when the part is present and logical 0 when
no part is present. It is necessary to ensure that a change of sensor state occurs closer to the
sensor when a bottle approaches than when it moves away. Bottles will pass the sensor at the
rate of 10 every second. There is no requirement for protection from contact with objects but it
must be proof against water drips. Recommend a suitable sensor and justify your
recommendation. Sketch a diagram that shows the functional blocks within the sensor and
explain each function.
Solution
lock Diagram of the system
Basic Elements Used
MICRO CONTROLLER
RESISTORS
CAPACITOR
INDUCTOR
POWER DIODE
WATER PUMP
IR SENSORS
DC MOTOR
CONVEYOR BELT
Resistore
A part from resister and inductor, a capacitor is the other basic component used in electronic
circuits. It is a device which
1. Has the ability to store charge which neither a resistor nor an inductor can do;
2. opposes any change of voltage in the circuit in which it is connected;
3. Blocks the passage of direct current through it.
CAPACITOR
The electronic component (mostly a coil) which opposes the change of current in circuit is
called an inductor. The ability of coil or inductor due to which it opposes the change of current
flowing through it is known as its inductance. The inductance of an inductor may also be defined
as the ability of an inductor to produce induced voltage when the current varies through it. The
unit of inductance is Henry which is named after Joseph Henry.
INDUCTOR
Diodes allow electricity to flow in only one direction. The arrow of the circuit symbol shows the
direction in which the current can flow. Diodes are the electrical version of a valve and early
diodes were actually called valves. The construction and the fundamental working of the power
diode and signal diode are same but the power handling capacity of the power diode is higher
than the signal diodes
POWER DIODE
Sensors
IR Sensors work by using a specific light sensor to detect a select light wavelength in the Infra-
Red (IR) spectrum. By using an LED which produces light at the same wavelength as what the
sensor is looking for, you can look at the intensity of the received light. When an object is close
to the sensor, the light from the LED bounces off the object and into the light sensor. This results
in a large jump in the intensity, which can be detected using a threshold
DRIVER CIRCUIT FOR DC MOTOR
We have used the dc motors to drive conveyer belt and to. The driver circuit for dc motors is
shown below.
COMPARATOR CIRCUIT
Comparator circuit is used in thi.
1. What are the characteristics of financial assets that have highly.pdf
1. What are the characteristics of financial assets that have highly developed secondary markets?
2. What is the difference between the secondary market in mortgages and the secondary market
in mortgage-backed securities?
Solution
1. Financial assets in the developed world are more complex and there are many options
available for investors. The financial assets include equity and derivative instruments and a fully
developed bond market. In developing nations, since most of the assets held by individuals are in
form of property and other hard assets, the financial assets are not that well developed. A
developing nation may have a stock market in the weak form and will not have a fully developed
bond market.
2. A secondary market in mortgages is where the home loans and their servicing rights and
bought and sold between lenders and investors.
In mortgage backed security, several primary mortgages which are sold in the secondary marked
are pooled into large mortgage back investor units and are sold to investors once again..
The tools that cyber criminals often use, including computer viruses.pdf
The tools that cyber criminals often use, including computer viruses, worms, trojan horses and
spyware, are called .
Solution
most of the cyber crimes are committed by the individuals or small groups and their attack
techniques ie there are few type of attacks that cyber criminal use to commit crimes ie botnet -a
network of software robots or bots that automatically spread malware , fast flux- moving data
quickly among the computers in botnet to make it difficult to trace the source of malware or
phishing websites , zoombie computer - a computer that has been hacked into and is used to
launch malacious attacks or to become part of botnet , social engineering - using lies and
manipulation to trick people into revealing their personal information and phishing is form of
social engineering and this attack include ie sending a victims an email that claims theirs
problem with their account and has a link to fake the website and entering their account
information into the site sends it atraight to the cyber criminal and trying to convince victims to
open an email attachments that contain malware by claiming it is something that might enjoy or
need and pretending to be network or account administrator and asking for the victims password
to perfom maintainance and claiming that victim has won prize must give the credit card
informaition in order to receive it and asking for the victims password for an internet service and
then using the same password to access other accounts and servicessince many people reuse the
same password and promising the victim that they wil send the millions of dollarsif they help out
sender by giving the money or the bank account information
there are some gey hat hackers who wants to come clean and turn away from crime an option is
to work for the people they used to torment by becoming security consultant in past they were
black hat hackers who use their computer expertise to break into the computer systems and steal
information illegally .
therefore the tools that cyber criminals often use including computer viruses , worms torjan
horses and spyware are called spam denial of service hacking malware. typically a spam is sent
to multiple receipients who didnot ask to recieve it and it includes email spam and instant
messaging spam , websearch engine spam spam inn blogs and mobile phone messaging spam
and it includes misleading advertisements and phishing messages and designed to trick
reciepients in giving personal information and financial information .these are the tools that
cyber criminals use..
The Poisson distribution analysis can be used to calculate the titer.pdf
The Poisson distribution analysis can be used to calculate the titer of a given suspension of virus.
I ran an experiment dealing with egg infectious dose using a newly isolated paramyxovirus in
suspension. I would like to know the titer of the virus suspension in egg infectious dose units.
100 eggs were inoculated with ten microliter of a 1: million dilution of the virus suspension.
After five days of incubation I discovered that ten eggs were not infected.
Calculate the titer of my virus in the undiluted suspension. What are the units ?
Solution
titre(pfu/ml) = number of plaques/(dilution*volume of diluted virus added to the egg)
titre= 90/(0.000001*0.01)
titre=9*109 pfu/ml.
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
Save
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
here is my code: ******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private Graphics.
Assessment and Planning in Educational technology.pptx
In an education system, it is understood that assessment is only for the students, but on the other hand, the Assessment of teachers is also an important aspect of the education system that ensures teachers are providing high-quality instruction to students. The assessment process can be used to provide feedback and support for professional development, to inform decisions about teacher retention or promotion, or to evaluate teacher effectiveness for accountability purposes.
More Related Content
More from arihantstoneart
For MIMO system, (a) Please talk about the advantage and disadvantag.pdf
For MIMO system, (a) Please talk about the advantage and disadvantage of MIMO technique.
(b) Please explain why we need Cooperative Diversity to replace MIMO in cellular network.
Solution
A)MIMO advantages
• Capacity scales linearly with number of antennas
Channel knowledge/estimation at Rx needed
• MIMO offers potential for
larger data rate
larger spectral efficiency
larger number of users
improved range/coverage
better interference suppression
better quality of service (QoS), lower bit-error rate (BER)
lower Tx power
MIMO disavantages
• Hardware complexity:
Each antenna needs a radio-frequency (RF) unit
Powerful digital signal processing (DSP) unit required
• Software complexity:
Most signal processing algorithms are computationally intensive
Power consumption:
Battery lifetime of mobile devices
Thermal problems
Antennas:
Antenna spacing (electromagnetic mutual coupling-e.g. mobile handsets)
RF interference and antenna correlation
B)
Most of the present researches on cooperative
network in which the user nodes are equipped with a single
antenna or multiple, there have been some new results which
exploit the benefits of multiple antenna deployment.
Cooperative MIMO technology allows a wireless network
system to attain better performance gains than provided by
either usual MIMO or cooperative systems. It promised
significant improvement in spectral efficiency and network
coverage phenomena for different next generation wireless
communication systems. In wireless communication, the path
towards the various techniques that gives high service quality
and data rate has been through the use of the cooperative
network provided by the rich scattering wireless channels.
Due to their great aspects, MIMO and cooperative systems
have found their way into several standards for future
wireless communication systems, especially in cellular
networks and wireless local area networks (LAN) in this
review article we are presenting the comparative analysis..
Explain why and give examples. In OO programming Polymorphism is on.pdf
Explain why and give examples. \"In OO programming Polymorphism is only possible when
using Interface or Inheritance\".
Solution
Polymorphism is the ability of an object to take on many forms. The most common use of
polymorphism in OOP occurs when a parent class reference is used to refer to a child class
object.
Any Java object that can pass more than one IS-A test is considered to be polymorphic. In Java,
all Java objects are polymorphic since any object will pass the IS-A test for their own type and
for the class Object.
It is important to know that the only possible way to access an object is through a reference
variable. A reference variable can be of only one type. Once declared, the type of a reference
variable cannot be changed.
The reference variable can be reassigned to other objects provided that it is not declared final.
The type of the reference variable would determine the methods that it can invoke on the object.
A reference variable can refer to any object of its declared type or any subtype of its declared
type. A reference variable can be declared as a class or interface type.
Polymorphism using inheritance:
Polymorphism using interface:
class Main
{
}.
Eukaryotic cells modify RNA after transcription What critical RNA pr.pdf
Eukaryotic cells modify RNA after transcription What critical RNA processing events usually
happen to pre-mRNA\'s before they sent to the cytoplasm for translation? What is a 5\' cap?
What is a poly A tail? What do these end modifications do for the mRNA transcript? What is
RNA splicing? What are introns? What are exons? How are intros spliced out? What is a
spliceosome? What is a \"snurp\" (snRNA)?
Solution
After its synthesis, the eukaryotic mRNA will undergo extensive modification like capping,
polyadenylation and splicing to enter into the process of translation.
(B)
Capping: Here the 5\' end of the mRNA is modified by the addition of 7-methylguanosine (m7G)
and the main function of this cap is to protect the 5\' end of the primary RNA transcript from
attack by ribonucleases and this 5’ cap will be recognized by eukaryotic initiation factors, so that
it can assemble the mature mRNA with the ribosome to start the process of translation.
At the 3\' end of the RNA, we will have polyadenylation signal and during transcription itself
this sequence will be chopped by an enzyme and another enzyme will add about 100100100 -
200200200 adenine (A) nucleotides to the 3’ end and this will form the poly-A tail.
The main function of this poly A tail is proving stability to the transcript and also helping it to
get exported from the nucleus to the cytosol.RNA splicing
(C )Splicing is the third big RNA processing event and the pre-mRNA will have two sequences,
exons and introns.
Introns are the non-coding sequences and exons are the coding sequences, here in this step the
through splicing, the introns will be removed and exons will be attached together.
In RNA splicing, specific parts of the pre-mRNA (introns) will be recognized and removed by a
protein-and-RNA complex called the spliceosome. Mature mRNA will have only exons, but no
introns.
(D)The splicing signal exon/GU-intron-AG/exon will be present in nuclear mRNA precursors
and 5\' and 3\' splice sites always have consensus sequences extending beyond GU and AG
motifs. During splicing, the exon-intron boundaries will be recognized by snRNA and the
consensus sequences within introns will get hybridized and now the proteins other snRNAs will
assemble the spliceosome on the transcript, the unpaired A present at 3\' side of the introns will
attacks the 5\' exon -intron boundary with the help of 2\' OH and this will give rise to lariat
structure. The free 3\' OH of the upstream exon will displace the downstream junctional
nucleotide, like this introns will be removed and exons will be attached together.
(E)A spliceosome is a large complex formed due to the assembly of snRNAs and protein
complexes, and plays an important role in splicing of pre mRNA.
snRNPs (snurps) is the small nuclear ribonucleic proteins and this is a RNA-protein complexes
and they will form a larger complex with the unmodified pre-mRNA and various other proteins
to form the structure called spliceosome.(A)Critical RNA processing events.
Driverimport java.util.Scanner;A class that keeps a f.pdf
Driver:
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i
Solution
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i 0)
this.cylinders = cylinders;
else
this.cylinders = 0;
}
/**
* @return the ownersName
*/
public String getOwnersName() {
return ownersName;
}
/**
* @param ownersName
* the ownersName to set
*/
public void setOwnersName(String ownersName) {
this.ownersName = ownersName;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Vehicle) {
Vehicle vehicle = (Vehicle) obj;
if ((this.getManuName().equals(vehicle.getManuName()))
&& (this.getOwnersName().equals(vehicle.getOwnersName()))
&& (this.getCylinders() == vehicle.getCylinders()))
return true;
else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return getManuName() + \" \" + getCylinders() + \" \" + getOwnersName();
}
}
public class Truck extends Vehicle {
double loadCap, towCap;
public Truck() {
// TODO Auto-generated constructor stub
super();
setLoadCap(0);
setTowCap(0);
}
/**
* @param manuName
* @param cylinders
* @param ownersName
* @param loadCap
* @param towCap
*/
public Truck(String manuName, int cylinders, String ownersName,
double loadCap, double towCap) {
super(manuName, cylinders, ownersName);
setLoadCap(loadCap);
setTowCap(towCap);
}
/**
* @return the loadCap
*/
public double getLoadCap() {
return loadCap;
}
/**
* @param loadCap
* the loadCap to set
*/
public void setLoadCap(double loadCap) {
if (loadCap > 0)
this.loadCap = loadCap;
else
this.loadCap = 0;
}
/**
* @return the towCap
*/
public double getTowCap() {
return towCap;
}
/**
* @param towCap
* the towCap to set
*/
public void setTowCap(double towCap) {
if (towCap > 0)
this.towCap = towCap;
else
this.towCap = 0;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Truck) {
if (super.equals(obj)) {
Truck vehicle = (Truck) obj;
if (this.getLoadCap() == vehicle.getLoadCap()
&& this.getTowCap() == vehicle.getTowCap())
return true;
else
return false;
} else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return super.toString() + \" \" + g.
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Solution
The idea here is to use the concept of two binary heaps along with the concept of list(doubly
linked list).
The doubly linked list contains all input items and indexes of corresponding min and max heap
nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root
node of min heap stores the address of minimum item in doubly linked list. Similarly, root of
max heap stores address of maximum item in doubly linked list. Following are the details of
operations.
1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a
O(1) operation.
1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a
O(1) operation.
3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this
address to find the node in doubly linked list. From the doubly linked list, we get node of Max
Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time.
delete() operations for max and min heaps take O(Logn) time.
4) removeMax(): is similar to deleteMin()
5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in
Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn)
#include
#include
#include
// A node of doubly linked list
struct LNode
{
int data;
int minHeapIndex;
int maxHeapIndex;
struct LNode *next, *prev;
};
// Structure for a doubly linked list
struct List
{
struct LNode *head;
};
// Structure for min heap
struct MinHeap
{
int size;
int capacity;
struct LNode* *array;
};
// Structure for max heap
struct MaxHeap
{
int size;
int capacity;
struct LNode* *array;
};
// The required data structure
struct CDS
{
struct MinHeap* minHeap;
struct MaxHeap* maxHeap;
struct List* list;
};
// A utility function to create a new List node
struct LNode* newLNode(int data)
{
struct LNode* node =
(struct LNode*) malloc(sizeof(struct LNode));
node->minHeapIndex = node->maxHeapIndex = -1;
node->data = data;
node->prev = node->next = NULL;
retu.
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.pdf
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.e.,D0). The dividend is
expected to grow at a constant rate of 6% a year. The required rate of return on the stock, r, is
11%. What is the value per share of the company\'s stock? Round your answer to two decimal
places.
Solution
Recent Dividend, D0 = $4.30
Growth Rate, g = 6%
Required Return, rs = 11%
D1 = D0 * (1 + g)
D1 = $4.30 * 1.06
D1 = $4.558
Value per share = D1 / (rs - g)
Value per share = $4.558 / (0.11 - 0.06)
Value per share = $4.558 / 0.05
Value per share = $91.16.
Anatomy Question Please Answer them all.A. Using a punnett square.pdf
Anatomy Question: Please Answer them all.
A. Using a punnett square, show the possible genotypes for John and Andrea\'s children? (John
has O blood type and Andrea has A blood type)
B. Based on the blood typing results, could Daryl be John and Andrea\'s Son? How Do you know
this? ( Daryl has O blood type)
Solution
A. Andrea :Mother has A blood type , her genotype will be either AO or AA
John :Father has O blood type ,his genotype will be OO.
O allele is masked by the presence of A allele or B allele.
OO blood group is expressed only in presence of OO alleles.
Father
Father
B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have
O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the
son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood
groupOO O blood group.
A person is lost in the desert. (S)he has no water and after a coupl.pdf
A person is lost in the desert. (S)he has no water and after a couple of days loses so much water
through perspiration that his/her body fluids become hypertonic. What happens to his cells?
What happens to his proteins? What happens to his carbohydrates?
Solution
Early sign of dehydration there is no symptom but later on there is dryness of mouth and tongue,
dark yellow urine,dryness of skin cramp in arm and leg. When fluid become hypertonic i.e
sodium is lost from cell this result in cell shrinkage.
After two days of starvation body will first utilize carbohydrate and then glycogen from liver and
muscle for energy. Protein from cell / muscle will be next to be used as source of energy for
body.
2. Describe an advantage and a limitation to tracing ancestry with b.pdf
Tracing ancestry with mitochondrial DNA and Y-DNA has the advantage of providing information about historical human evolution over thousands of years, but is limited in its ability to unravel full familial relationships. Autosomal DNA provides a more complete picture of ancestral inheritance patterns, but has ethical limitations when used for prenatal ancestral tracing.
Write your thought on the following1. petroski notes that most en.pdf
Write your thought on the following:
1. petroski notes that most engineering failures are much more involved than simple technical
mis-calculations and involve the failure of the design process or management culture.However,
not all engineering failures involve ethical issue.The infamous collapse of the first Tacoma
Narrow Brodge, and the losses of the Mars polar lander and Mars Climate Orbiter were technical
and design process failures.
Solution
In the face of new technology, how do we balance public welfare and progress? If Moisseiff had
designed a bridge similar to the ones which had already proven their stability, Tacoma Narrows
Bridge would never have collapsed costing thousands of dollars and endangering many lives. It
would also have been significantly more expensive. On the other hand, if engineers had never
tried innovative techniques, suspension bridges may never have been built at all. At the time of
their introduction, no one believed that a suspension bridge could safely accommodate trains.
Roebling, however, took a gamble, pushed the limits of the current technology, and built a
suspension bridge that he believed could safely support rail traffic. Luckily he was correct, and
suspension bridges soon became widely accepted (Petroski. Moisseiff also took a gamble, trying
to create a longer, sleeker, less expensive bridge, by pushing the limits of technology. He,
however, was not as lucky, and what could have been a breakthrough in technology turned into a
catastrophic failure. Every time engineers push the limits of technology they risk a similar loss
like losses of mas polar lander and mass climate orbiter in which there were technical error like
Mass polar lander
1. Inadequate specifications which includes what the software was supposed to do, but no
mention of what it must not to do.
2. Flawed review process: Violation of basic engineering safety pracitices in the digital part of
the system.
Mass Climate Orbiter
1. Units communication and attention to detail failure. The primary cause of this discrepancy
was that one piece of ground software supplied by Lockheed Martin produced results in a United
States customary unit, contrary to its Software Interface Specification (SIS), while a second
system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS.
Specifically, software that calculated the total impulse produced by thruster firings calculated
results in pound-seconds. The trajectory calculation software then used these results - expected to
be in newton-seconds - to update the predicted position of the spacecraft.
sometimes even It can cause a loss of life. How much is too much? When is a possible advance
worth a risk to public safety? What can the engineering profession do to make the
implementation of new technology safer? Do our current peer review and building code
committee processes adequately protect public safety?.
You have two cell lines, a macrophage cell line and a CD4 T-cell lin.pdf
You have two cell lines, a macrophage cell line and a CD4 T-cell line. The macrophage cell line
is resistant to infection by HIV, whereas the T-cell line is fully susceptible and fully permissive.
Why are macrophage cell line resistant to infection by HIV? Why are T-cell fully susceptible and
fully permissive? Please explain.
Also, A heterokaryon cell made from the fusion of these two cells is able to be infected by HIV.
Explain why? Please explain in detail.
Solution
Macrophages are antigen- presenting cells so they process the viral peptides and present it to
uninfected CD4 T-Cell which is the important transmission process. Also HIV-infected
macrophages release soluble cytotoxic factors which induce apoptosis in CD4 T-Cells.
Macrophages produce certain host restriction factors which inhibits the release of viral progeny
from infected cells. So macrophages are resistant to lysis because of its genetic make up and its
phenotypes (restriction factors).
So when two cell lines are fused to form heterokaryon cell, the genes of macrophages are
suppressed or it is recessive or the T cell genes are dominant to make the heterokaryon
susceptible to infection..
With X-linked inheritance, the X chromosome is transmitted in a diffe.pdf
With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and
females. In X-linked dominant traits, males hemizygous for trait will express the trait, but
females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected
female will show the trait. For autosomal dominant, two affected parents may produce an
unaffected child.
Solution
Answer:
A. True
Most X-linked conditions are recessive. This means that in a person with two X chromosomes
(most females), both copies of a gene (i.e., one on each X chromosome) must have a change or
mutation whereas in a person with one X chromosome (most males), only one copy of a gene
must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is
said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X
chromosome is typically affected with the condition.
The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome
to the son and other to the daughter.
B. False
For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a
female with two X chromosomes or males with on X chromosome must have a change or
mutation for an individual to be affected with the condition. For this reason, X-linked disorders
are often seen with similar frequency in males and females. However, since females also have
one normal X chromosome as well as an X chromosome with a mutation, the condition is often
more “mild.”
C.False
For X-linked recessive disorders, an affected father who has a mutation in a gene on the X
chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his
children. If the mother is not affected or a carrier, none of his sons will be affected since they can
only inherit a normal X chromosome from their mother and they inherit a Y chromosome from
their father.
But if the female is affected, it will receive one of the affected X chromosomes, which is enough
to express the trait in the son.
D. True
A person affected by an autosomal dominant disorder has a 50 percent chance of passing the
mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50
percent..
When examining the central tendency of a probability distribution, u.pdf
When examining the central tendency of a probability distribution, under what circumstance
would you be indifferent to using the median vs the mean?
Solution
If distribution is symmetric, then mean and median would coincide, and we would be indifferent.
If distribution is skewed, or has outliers, we prefer median because it is not affected by extreme
values..
What will organizations have to do to adapt to the influx of needed .pdf
What will organizations have to do to adapt to the influx of needed Generation Y individuals?
Solution
simple thing is choose the right person for a work and let them free, they will create miracles.
here generation Y people are those who can perform variaous tasks at a time and with more
accuracy. these are multi talented, always carries courage, never hides their talent and ready to
accept challenges.
first to attract those kind of people organisations should change their existing rules, policies and
regulations. all they were writen decades ago and still we are follwoing them. they need to
redesign the work and the structure. if we are succede at recruiting the correct people for our
work, then the remaining changes can automatically comes into the picture.
the organisations should create the work environment to the Y generation people
There are no precise dates when the generation starts and ends; most researchers and
commentators use birth years ranging from the early 1980s to early 2000s. Generation Y people
may be known as echo boomers because they are the children of baby boomers, or someone who
was born during the period of increased birth rates that occurred between 1946 and 1964. The
Generation Y cohort is also referred to as Millennials, the Internet generation, iGen and/or the
Net generation.
What are the main functions of the components of prokaryotic vs. euk.pdf
What are the main functions of the components of prokaryotic vs. eukaryotic cells? Could either
cell type survive without any of them? (Answer must be at least a page long, very detailed and in
paragraph form please)
Solution
Prokaryotes are unicellular living beings that need organelles or other interior film bound
structures . In this manner, they don\'t have a core, at the same time, rather, by and large have a
solitary chromosome: a bit of roundabout, twofold stranded DNA situated in a territory of the
phone called the nucleoid. Most prokaryotes have a cell divider outside the plasma film.
Like a prokaryotic cell, an eukaryotic cell has a plasma film, cytoplasm, and ribosomes.
Nonetheless, not at all like prokaryotic cells, eukaryotic cells have:
a film bound core
various film bound organelles (counting the endoplasmic reticulum, Golgi contraption,
chloroplasts, and mitochondria)
a few bar formed chromosomes
Since an eukaryotic cell\'s core is encompassed by a film, it is regularly said to have a \"genuine
core. \" Organelles (signifying \"little organ\") have specific cell parts, similarly as the organs of
your body have particular parts. They permit diverse capacities to be compartmentalized in
various territories of the cell.
The piece of the cell divider varies altogether between the areas Bacteria and Archaea, the two
spaces of life into which prokaryotes are separated . The structure of their cell dividers
additionally varies from the eukaryotic cell dividers found in plants (cellulose) or organisms and
bugs (chitin). The cell divider works as a defensive layer and is in charge of the life form\'s
shape. A few microorganisms have a container outside the cell divider. Different structures are
available in some prokaryotic species, yet not in others. For instance, the case found in a few
animal types empowers the living being to connect to surfaces, shields it from lack of hydration
and assault by phagocytic cells, and expands its imperviousness to our invulnerable reactions. A
few animal categories likewise have flagella utilized for velocity and pili utilized for connection
to surfaces. Plasmids, which comprise of additional chromosomal DNA, are likewise present in
numerous types of microorganisms and archaea.
Commonly, the core is the most conspicuous organelle in a cell. Eukaryotic cells have a genuine
core, which implies the cell\'s DNA is encompassed by a layer. Thusly, the core houses the cell\'s
DNA and coordinates the blend of proteins and ribosomes, the cell organelles in charge of
protein union. The atomic envelope is a twofold layer structure that constitutes the furthest bit of
the core. Both the internal and external layers of the atomic envelope are phospholipid bilayers.
The atomic envelope is punctuated with pores that control the section of particles, particles, and
RNA between the nucleoplasm and cytoplasm. The nucleoplasm is the semi-strong liquid inside
the core where we discover the chromatin and the nucleolus. M.
There is a requirement to design a system to sense the presence of gl.pdf
There is a requirement to design a system to sense the presence of glass bottles leaving an
automatic washer on a production line conveyor belt. The production line is for food products
and the environment is clean and maintained at a temperature of 15 degree C. The system must
meet the following criteria: The switching distance will lie between 5 and 20 mm. The output
from the sensor is to be of binary form, i.e. logical 1 when the part is present and logical 0 when
no part is present. It is necessary to ensure that a change of sensor state occurs closer to the
sensor when a bottle approaches than when it moves away. Bottles will pass the sensor at the
rate of 10 every second. There is no requirement for protection from contact with objects but it
must be proof against water drips. Recommend a suitable sensor and justify your
recommendation. Sketch a diagram that shows the functional blocks within the sensor and
explain each function.
Solution
lock Diagram of the system
Basic Elements Used
MICRO CONTROLLER
RESISTORS
CAPACITOR
INDUCTOR
POWER DIODE
WATER PUMP
IR SENSORS
DC MOTOR
CONVEYOR BELT
Resistore
A part from resister and inductor, a capacitor is the other basic component used in electronic
circuits. It is a device which
1. Has the ability to store charge which neither a resistor nor an inductor can do;
2. opposes any change of voltage in the circuit in which it is connected;
3. Blocks the passage of direct current through it.
CAPACITOR
The electronic component (mostly a coil) which opposes the change of current in circuit is
called an inductor. The ability of coil or inductor due to which it opposes the change of current
flowing through it is known as its inductance. The inductance of an inductor may also be defined
as the ability of an inductor to produce induced voltage when the current varies through it. The
unit of inductance is Henry which is named after Joseph Henry.
INDUCTOR
Diodes allow electricity to flow in only one direction. The arrow of the circuit symbol shows the
direction in which the current can flow. Diodes are the electrical version of a valve and early
diodes were actually called valves. The construction and the fundamental working of the power
diode and signal diode are same but the power handling capacity of the power diode is higher
than the signal diodes
POWER DIODE
Sensors
IR Sensors work by using a specific light sensor to detect a select light wavelength in the Infra-
Red (IR) spectrum. By using an LED which produces light at the same wavelength as what the
sensor is looking for, you can look at the intensity of the received light. When an object is close
to the sensor, the light from the LED bounces off the object and into the light sensor. This results
in a large jump in the intensity, which can be detected using a threshold
DRIVER CIRCUIT FOR DC MOTOR
We have used the dc motors to drive conveyer belt and to. The driver circuit for dc motors is
shown below.
COMPARATOR CIRCUIT
Comparator circuit is used in thi.
1. What are the characteristics of financial assets that have highly.pdf
1. What are the characteristics of financial assets that have highly developed secondary markets?
2. What is the difference between the secondary market in mortgages and the secondary market
in mortgage-backed securities?
Solution
1. Financial assets in the developed world are more complex and there are many options
available for investors. The financial assets include equity and derivative instruments and a fully
developed bond market. In developing nations, since most of the assets held by individuals are in
form of property and other hard assets, the financial assets are not that well developed. A
developing nation may have a stock market in the weak form and will not have a fully developed
bond market.
2. A secondary market in mortgages is where the home loans and their servicing rights and
bought and sold between lenders and investors.
In mortgage backed security, several primary mortgages which are sold in the secondary marked
are pooled into large mortgage back investor units and are sold to investors once again..
The tools that cyber criminals often use, including computer viruses.pdf
The tools that cyber criminals often use, including computer viruses, worms, trojan horses and
spyware, are called .
Solution
most of the cyber crimes are committed by the individuals or small groups and their attack
techniques ie there are few type of attacks that cyber criminal use to commit crimes ie botnet -a
network of software robots or bots that automatically spread malware , fast flux- moving data
quickly among the computers in botnet to make it difficult to trace the source of malware or
phishing websites , zoombie computer - a computer that has been hacked into and is used to
launch malacious attacks or to become part of botnet , social engineering - using lies and
manipulation to trick people into revealing their personal information and phishing is form of
social engineering and this attack include ie sending a victims an email that claims theirs
problem with their account and has a link to fake the website and entering their account
information into the site sends it atraight to the cyber criminal and trying to convince victims to
open an email attachments that contain malware by claiming it is something that might enjoy or
need and pretending to be network or account administrator and asking for the victims password
to perfom maintainance and claiming that victim has won prize must give the credit card
informaition in order to receive it and asking for the victims password for an internet service and
then using the same password to access other accounts and servicessince many people reuse the
same password and promising the victim that they wil send the millions of dollarsif they help out
sender by giving the money or the bank account information
there are some gey hat hackers who wants to come clean and turn away from crime an option is
to work for the people they used to torment by becoming security consultant in past they were
black hat hackers who use their computer expertise to break into the computer systems and steal
information illegally .
therefore the tools that cyber criminals often use including computer viruses , worms torjan
horses and spyware are called spam denial of service hacking malware. typically a spam is sent
to multiple receipients who didnot ask to recieve it and it includes email spam and instant
messaging spam , websearch engine spam spam inn blogs and mobile phone messaging spam
and it includes misleading advertisements and phishing messages and designed to trick
reciepients in giving personal information and financial information .these are the tools that
cyber criminals use..
The Poisson distribution analysis can be used to calculate the titer.pdf
The Poisson distribution analysis can be used to calculate the titer of a given suspension of virus.
I ran an experiment dealing with egg infectious dose using a newly isolated paramyxovirus in
suspension. I would like to know the titer of the virus suspension in egg infectious dose units.
100 eggs were inoculated with ten microliter of a 1: million dilution of the virus suspension.
After five days of incubation I discovered that ten eggs were not infected.
Calculate the titer of my virus in the undiluted suspension. What are the units ?
Solution
titre(pfu/ml) = number of plaques/(dilution*volume of diluted virus added to the egg)
titre= 90/(0.000001*0.01)
titre=9*109 pfu/ml.
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
Save
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
here is my code: ******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private Graphics.
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