Representatives from four different colleges of UM- Dearborn will be .pdf
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Representatives from four different colleges of UM- Dearborn will be seated in a row. There are 4 representatives from College of \'Art Sciences, and Letter\', 3 from college of \'Business\', 5 from College of \'Education, Health and Human Services\', and 6 from \'College of Engineering and Computer Science\'. Representatives will be seated according to their colleges (identities of each representative is not important). In how many different ways will these representatives be seated? Solution 4 from college of \'arts sciences and letter\' arranged in 4! ways =24 ways 3 from college of \'business\' arranged in 3! ways = 6 ways 5 from college of \'education health and human sevices\' arranged in 5! ways = 120 ways 6 from college of \'engineering and computer science\' arranged in 6! ways = 720 ways And these four college can be arranged in 4! ways i.e., 24 ways So total = 24*24*6*120*720 = 298598400 ways.
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D. Evolutionary adaptation
E. Cellular organization
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Main.java
--------------
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#ifndef endLocation
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can use atoi
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a. A 3:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype?
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Solution
Ans:
This is the result of a problem common in trying to determine the nature of inheritance patterns
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Solution
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1.) Detterent to crime:- Death penalty is given to criminals who do first degree crimes ike murder
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2) Fear of death:- Though death is inevitable we all still fear it. Imposing death punishment for
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committing a crime.
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Do you think that knowing about aggression between groups of chimpanzees is useful in
understanding conflicts between human societies? Why or why not?
Solution
In many primates where the range is small , contact with one or more groups of the same species
is a very common feature. However the nature of these encounters varies from species to species.
Primate groups remain permanently in a specific locality which is described as there home range.
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resources and represent the region where the group is more likely to be found.It also represents
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territory.
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primates is that the difference in behavior is merely quantitative and not qualitative.
Interactions between groups of chimpanzees include aggression, display chasing and sometimes
violent fighters also. Male chimpanzees are highly intolerable to unfamiliar chimpanzees,
specially males and aggressively defend their territories and resources.
Lethal aggression is a common feature among the chimpanzees. It is common in groups of male
chimpanzees which do patrolling around their territory. Sometimes they are also accompanied by
one or two females. During this patrolling, they exhibit silent movement, stop frequently, sniff
around and climb tall trees and sit on them for hours together to survey the area. They appear to
be tense during such situations and any sudden sound causes them to exhibit nervous behavior
like touching each other for reassurance or embrace each other.
When they encounter members of another community, They exhibit some form of aggression till
the other group retreats. Chimpanzees are also known to exhibit cruelty, and aggression
combined with compassion, tenderness and altruism which is comparable to human behavior, but
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is F5. A mutation of this gene where Guanine is substituted by Adenine which results in the
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Create a new java class called ListNode. Implement ListNode as a gen.pdf
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}
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current.next = newNode;
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public E get(int index) {
ListNode current = goTo(index);
return current.data;
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public int indexOf(E value) {
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ListNode current = front;
while (current != null) {
if (current.data == value) {
return index;
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index++;
current = current.next;
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return new LinkedListIterator();
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ListNode current = goTo(index);
current.data = value;
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public int size() {
int count = 0;
ListNode current = front;
while (current != null) {
current = current.next;
count++;
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return count;
}
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if (front == null) {
return \"[]\";
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String result = \"[\" + front.data;
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while (current != null) {
result += \", \" + current.data;
current = current.next;
}
result += \"]\";
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}
private ListNode goTo(int index) {
ListNode current = front;
for (int i = 0; i < index; i++) {
current = current.next;
}
return current;
}
public class ListNode {
public E data;
public ListNode next;
public ListNode(E data) {
this.data = data;
this.next = null;
}
public ListNode(E data, ListNode next) {
this.data = data;
this.next = next;
}
}
private class LinkedListIterator implements Iterator {
private ListNode current; // current position in list
public LinkedListIterator() {
current = f.
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2. What is social engineering?
3. How will an SPI firewall handle a packet containing a TCP segment which is an
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5. How will the ACL in Figure 3-23 (page 119 in the textbook) handle a packet that attempts to
open a connection to an FTP server? Explain.
6. For each of the following passwords, first state the kind of attack that would be necessary to
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specific reasons.
a) password
b) Winter1
c) SpringBreaK
d) 2!T*d (00-10)
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Solution
Answers:
1)How do the viruses propagate between computers?
-> The browsers gain more features and functions more methods for virus propagation appear
directly to a computer across the network.
Web pages :
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propagation appear.
Messages:
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-> Whether the virus is spread through links, embedded code, or an attachment depends on the
type of message and the client software.
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-> Running an infected program that you copied from a CD or other media (or downloaded) is a
means of infection. Sometimes the virus is a Trojan, where it appears to be a useful program, but
is in fact a virus.
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-> Social engineering is an attack vector that relies heavily on human interaction and often
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-> Social engineering is the art of manipulating people so they give up confidential information.
-> A social engineer runs what used to be called a \"con game.
Popular types of social engineering attacks :
1)Phishing:
-> Phishing is when a malicious party sends a fraudulent email disguised as a legitimate email,
often purporting to be from a trusted source.
2) Pretexting:
->Pretexting is another form of social engineering where attackers focus on creating a good
pretext, or a fabricated scenario, that they can use to try and steal their victims’ personal
information.
3. How will an SPI firewall handle a packet containing a TCP segment which is an
acknowledgement?
-->Process it through the Access Control List (ACL)
-> Stateful Packet Inspection: Connections have distinct states or stages.
->Different states are subject to different attacks. Stateful firewalls use different filtering rules
for different states. Different because it uses filter rules in order to accept or deny traffic.
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York, Providence, Washington, D.C., and Philadelphia. 1. Describe the types of networks you
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Solution
1.
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themselves.In the same city ,if they have another office,they can use metropolitian area
network(MAN),which is used to communicate in a city or within a cities.For their offices in
other cities,they will use WAN(Wide area network)which is used for communication worldwide
.
2.
The standard protocols and technologies used are TCP-IP for the wired transmission and UDP
for wireless Transmission.
4. Show that Prefix(L)-x for some y, xy E L is regular if L is regul.pdf
4. Show that Prefix(L)-x: for some y, xy E L is regular if L is regular
Solution
As L is a regular language it will have a DFA. The new language prefix(L) should accept x if xy
(for some y) belongs to L. As we are given that y can be anything, this can be achieved by taking
all states which have a path from start and will reach a final state for some string, we make all
those states as final states. This DFA will accept a substring of any string accepted by L. Now as
prefix(L) has a DFA, it is a regular language..
26) Which is a function of an operating system O Browse the Internet.pdf
26) Which is a function of an operating system? O Browse the Internet O Edit images O Protect
against malware O Provide user interface Next Question
Solution
26.
Provide user interface
All other mentioned functionality can be achieved by other software
27.
Defender
28.
APIs
29.
device driver
Without device drivers you can use your hardware functionality optimally
30.
Application Software:
User can not directly interact with hardware directly,
they need some kind of application software to interact with them.
1.Declare a student structure that contains - Students first an.pdf
1.Declare a student structure that contains :
- Student\'s first and last name
-Student ID
2. Create the following functions:
-getStudentInfo(void)
Declares a single student, uses printf()/scanf() to get input
Returns the single student back
- printStudentInfo(student_t *st_ptr)
Takes the pointer to a student (to avoid making a copy)
-Prints out all of the student information
3. Declare an array of five students
4. Using a for loop and getStudentInfo() function, get input of all the students
5. Using a for loop and printStudentInfo() function, print all the output of all students.
Solution
getStudentInfo(void)
{
printStudentInfo(student_t *st_ptr)
{
str_ptr.getStudentInfo();
display(stud);
return 0;
}
int main()
{
printStudentInfo(str_ptr);
}
return 0;
}.
1. Describe how plant vasculature is optimized for water transport i.pdf
1. Describe how plant vasculature is optimized for water transport in the tallest trees
2. What property of water molecules allow water to stay together while being transported to
extreme heights? And how does temperature affect this?
3.Water and minerals enter the plant through _______, and are then transported through plant\'s
vasculature, via the _________. This flow is maintained through constant ________, where
water exists through __________ in the leaves. The sugars that are made during photosynthesis
in the ____________ are transported downwards through the vasculature via the _______.
4. Give an example of ways plant have evolved to cope with drought,
Solution
Water transport in tree occurs by cappilary action .Water is trasported by root to leaves through
xylem . xylem is made up by tube like cells tracheids and vessel . These join together from root
to leaves . Two forces are usful to flow water upward 1. root pressure and 2. transpiration . Root
pressure occure when water flow into root due to osmotic pressure . Transpiration occures due to
evoporation from leaf surface . The adhesive nature of water molecules to inner cappilary wall
help to keep water column against gravitational force .
2 Adhesive nature of water to inner cappilary wall . Temprature increases transpiration and low
osmatic pressure in leaves .
3. water and mineral entered the plant through ROOT and are then trasported through plant
vasculature , via the XYLEM These flow is maintained through constant TRANSPIRATION
where water exist through STOMETA in the leaves . Thesuger that are made during
photosynthesis in the leaf are transported downward through the vasculature via PHLOEM.
4 Plant have evolved to cope with drought as xerophytic plant have well develop root system ,
fleshy leaves and stometa sunk in fleshy leave to avoid tranpiration ..
A19730 Which, ofthe foliowrgconsidered a tusa.a.for restructuring ro.pdf
A19730 Which, ofthe foliowrgconsidered a tusa.a.for restructuring? road sector More road
engreers ) Increased donor funding Use of contractors V) None of the above (9) 12) What is the
function of a ) Planning for extensions 1o the rod network Providing contractiors for carrying out
road works i) Providing finance for road maintenance iv) Establishing road policy v) None of the
above (h) What other tem can be used for a supplier 12) i) Client i) Manager iv) Producer v)
None of the above (i) 121 Where is road ownership defined? i) Ministry procedures i) Ministerial
memo i) Legislation iv) Contract v) None of the above 0) How is official notice given that a road
has been designated? 121 i) Publication in the road administration asset plan ili) Letter circulated
to residents of the area served by the road iv) Road listed in a policy document v) None of the
above What action may be required by the road administration following road 12 designation?
(k) roviding length marker posts on the road ii) Provision of a maintenance depot on the road ili)
Guaranteed minimum level of maintenance funding for the road iv) Road will be paved v) None
of the above (1) Which is considered a functional classification? i) Secondary road ii) Collector
road ii) Class B road iv) Municipal road v) None of the above
Solution
Question number (f) Answer is (ii) Strenthened management.
Explanation: There are 3 bulding blocks of Road sector restructring :
1) continuous and stable finance.
2) Strenthened management.
3) Creating Ownership and Establishing Responsibility.
Question number (g) Answer is (i) Planning for extension to the road network.
Question number (h) Answer is (v) None of the above.
Question number (i) Answer is (iii) Legislation.
Question number (g) Answer is (v) None of the above.
Explanation when a road is designated a notice is published in a government gazette or other
official publication.
Question number (k) Answer is (i) Providing length marker posts on the road.
Question number (l) Answer is (ii) Collector road..
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...Nguyen Thanh Tu Collection
https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiMore Related Content
More from mohamednihalshahru
In accordance to the PLSS land description system, locate the section.pdf
In accordance to the PLSS land description system, locate the section of land in the township;
locate the parcel land in the section; and determine the acreage of the land parcel. Given the
following: W 1/2, NW 1/4, SE 1/4, SE 1/4, Section 28, R2W, T7S, New Mexico Mexico
Meridian Area of Land parcel acres Calculate the end area (meters^2) of the land cross section
below using the alternative coordinate method.
Solution.
How can a topologically associating domain (TAD) mutation effect mor.pdf
How can a topologically associating domain (TAD) mutation effect more that one gene? Explain
why a TAD mutation shouldn\'t have any effect on the normal shape of a protein? And what
seems to be one of the normal functions of a TAD.?
Solution
Topological associated domain are genome region whose DNA sequence prefrentially contact
each other.
TAD is a fundamental structural unit that regulates elements to their cognate
promoters.Function: TAD are critical for accurate gene expression Any aleration can lead to
disease.
Hello, I need help writing a code in C. It wants to do things with f.pdf
Hello, I need help writing a code in C. It wants to do things with files. It gives us two files. The
code must first ask which file they want, then open the file. Once opened, the file will have three
colums: the names alphabetically, the rank of the name (a name with a rank of 232 is the 232nd
most popular name in 2014), and the total number of recorded instances. The program then
prompts the user to input a name, when inputed, If the name is found in the database, display the
name rank, and number of the name in sequence with the prior two names and the following two
names. If the name is not shown, show only the two entries before and after the position where
the name would be. My major is mechanical so i have never been very good at coding. Any help
would be greatly appreciated. Thank you for your time.
Solution
#include
int main()
{
int c;
char str[30];
printf(\"Enter your choice \ 1.First file\ 2.Second file\ \");
scanf(\"%d\",&c);
switch(c)
{
case 1: FILE *fptr=fopen(\"First file\",rb);
break;
case 2:FILE *fptr=fopen(\"Second file\",rb);
break;
default: printf(\"Enter the correct option\ \");
return 1;
}
struct emp
{
char name[20];
int pos;
int num;
}e1;
printf(\"Enter the name for search the name in that file:\ \");
scanf(\"%s\",str);
while(1)
{
fread(fptr,1,&e1);
if(strcmp(e1->name,str)==1)
{
printf(\"%s\\t%d\\t%d\ \",e1->name,e1->pos,e1->num);
break;
}
}
fclose(fptr);
return 0;
}.
Fix the following C program in linux. When it is compiled it gives t.pdf
Fix the following C program in linux. When it is compiled it gives the following error: warning
cast to pointer from integer of different size.
Solution
#ifndef endLocation
#define endLocation
#include \"myLinkedlist.h\"
#include \"myString.h\"
class endLocation
can use atoi
int travelCost; // can use atoi
public:
endLocation();
endLocation(String, int, int);
endLocation(endLocation& rhs);
void setName(String);
void setTime(int);
void setCost(int);
String getName();
int getTime();
int getCost();
};
#endif // endLocation
//.cpp implementation
endLocation::endLocation()
endLocation::endLocation(endLocation& rhs)
void endLocation::setName(String inputName)
void endLocation::setTime(int inputTime)
void endLocation::setCost(int inputCost)
String endLocation::getName()
come locationName;
}
int endLocation::getTime()
come travelTime;
}
int endLocation::getCost()
come travelCost;
}
Here is my main
int main( int argc, char* const argv[] )
range of lines here
for(int i = 0; i < num_lines; ++i)dissect and store all strings
size = findIndex(line, \'|\');
strncpy(name, line, size);
index = size + 1;
size = findIndex(line + index, \'|\');
strncpy(time, line + index, size);
index += size + 1;
size = findIndex(line + index, \'|\');
strncpy(cost, line + index, size);
}
in.close();
}
else.
7. Wet ear wax (W) is dominant over dry ear wax (w)W=wet ear wax.pdf
7. Wet ear wax (W) is dominant over dry ear wax (w)
W=wet ear wax
w=dry ear wax
a. A 3:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype?
b. A 1:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype?
Solution
Ans:
This is the result of a problem common in trying to determine the nature of inheritance patterns
in humans.
The results are combined from offspring from several families.
Thus the sticky x sticky cross includes families in which parents may be (AA x AA; Aa X AA).
aa x aa..
Examine your opinion on the death penalty. Defend why you are for th.pdf
Examine your opinion on the death penalty. Defend why you are for the death penalty or against
it. Be sure to be specific in your defense on your stance.
Solution
Death penalty or execution is govt. sanctioned punishment by death.Iam in favour of death
penalty as:-
1.) Detterent to crime:- Death penalty is given to criminals who do first degree crimes ike murder
, terrorism and espionage.It produces a strong detterent effect on crime and criminals.
2) Fear of death:- Though death is inevitable we all still fear it. Imposing death punishment for
brutal crimes like rapes and murders will definately make the criminal think twice before
committing a crime.
3) No escape:- A criminal murderer or serial killer can escape imprisonment in the long run
leading to more crime and threat to society.. Death penalty puts an end to crime.
4) Saves lifes:- Death penalty produces a strong detterent effect on crime and criminals and
hence saves lifes..
Do you think that knowing about aggression between groups of chimpan.pdf
Do you think that knowing about aggression between groups of chimpanzees is useful in
understanding conflicts between human societies? Why or why not?
Solution
In many primates where the range is small , contact with one or more groups of the same species
is a very common feature. However the nature of these encounters varies from species to species.
Primate groups remain permanently in a specific locality which is described as there home range.
Members of the same group exist in this range however it is not necessary that the individuals of
the group should remain in the same range but may change over to other ranges and join a new
community.
This home range consists of a central core area with highest concentration of the predictable
resources and represent the region where the group is more likely to be found.It also represents
the territory of the group which is defended against intrusion. There may be overlapping of the
home range between the groups but not the core area.
Territoriality is a behavior exhibited by many primates to protect their home range. This is seen
extensively where the home ranges are small and permits patrolling and protection of the
territory.
Biological continuum refers to the fact that the organisms are related through common ancestry
and the behavioral traits seen in one organisms are seen in others also but in varying degrees.
The man is considered to be a highly evolved social organism with a huge brain and thinking
capacity. He is more related to chimpanzees and the main difference between us and the other
primates is that the difference in behavior is merely quantitative and not qualitative.
Interactions between groups of chimpanzees include aggression, display chasing and sometimes
violent fighters also. Male chimpanzees are highly intolerable to unfamiliar chimpanzees,
specially males and aggressively defend their territories and resources.
Lethal aggression is a common feature among the chimpanzees. It is common in groups of male
chimpanzees which do patrolling around their territory. Sometimes they are also accompanied by
one or two females. During this patrolling, they exhibit silent movement, stop frequently, sniff
around and climb tall trees and sit on them for hours together to survey the area. They appear to
be tense during such situations and any sudden sound causes them to exhibit nervous behavior
like touching each other for reassurance or embrace each other.
When they encounter members of another community, They exhibit some form of aggression till
the other group retreats. Chimpanzees are also known to exhibit cruelty, and aggression
combined with compassion, tenderness and altruism which is comparable to human behavior, but
the main difference is that chimpanzees are not aware of the suffering they inflict on others while
humans do. Thus, the behavior of chimpanzees can be more related to human behavior, but the
difference lies in quantitative than qualitative.
Diana’s Death UnraveledHemophilia causes a lack of blood clotting,.pdf
Diana’s Death Unraveled
Hemophilia causes a lack of blood clotting, but Diana died from overactive blood clotting. Factor
V Leiden is the most common hereditary hypercoagulability disorder amongst Eurasians. About
5% of Caucasians in North America are affected, and it is inherited as an autosomal dominant
mutation that causes high levels of clotting Factor V. It can cause deep vein thrombosis (blood
clot in the vein), pain, life-threatening strokes, and heart attacks.
1. Based on the types of mutations we’ve discussed so far in class, which does Factor V Leiden
fall into? (1pt)
2. Kevin Keller’s aunt’s husband and his girl cousin have died of heart attacks, but his mother,
sister, and brother are fine. Lisa Keller’s uncle is fine, but her aunt had multiple miscarriages
before she died. Her mother suffered from deep vein thrombosis, and her grandfather died of a
heart attack. Recall that Lisa has a brother with hemophilia C and an unaffected sister. Create a
pedigree diagramming these relationships. (3 pts)
b. Who do you think is more likely to carry the Factor V Leiden mutation? (1 pt)
a. Kevin Keller
b. Lisa Keller
c. Both
c. What do you think happened with the Morrison’s and Keller’s when Olivia and Diana were
young? (2 pt)
3. There is enough research evidence to support phenotypic changes in hemophiliacs with Factor
V Leiden mutations. Based on what we’ve discussed in chapter 4, what are some suggestions as
to why this might happen? Be sure to describe the normal symptoms of hemophilia and how
protein levels are affected to support your answer. (3pts)
Solution
1. Missense substitution mutation leads to Factor V Leiden . The gene which codes to the factor
is F5. A mutation of this gene where Guanine is substituted by Adenine which results in the
changing of amino acid Arginine to Glutamine.
I am not sure of the answers for the remaining questions. Please post those questions again for
other experts to answer. Thank you..
Create a new java class called ListNode. Implement ListNode as a gen.pdf
Create a new java class called ListNode. Implement ListNode as a generic version of IntNode.
public class ListNode { }
Create a generic class called Queue, which is a linked list implementation of a queue data type.
Queue must have the following methods:
• public void enqueue(T value){} This method should add a new node to the back of the queue.
• public T dequeue(){} This method should remove a node from the front of the queue and return
its value
• public T peek(){} This method should return the value of the front node without removing it
from the queue
Create a new generic class called Stack, which is a linked list implementation of a stack data
type. Stack must have the following methods:
• public void push(T value){} This method should add a new node to the top of the stack.
• public T pop(){} This method should remove a node from the top of the stack and return its
integer value
• public T peek(){} This method should return the integer value of the top node without
removing it from the stack
Solution
import java.util.Iterator;
public class LinkedList extends AbstractList {
private ListNode front; // refers to first node in list (null if empty)
public LinkedList() {
front = null; // null front means empty
}
public void add(int index, E value) {
if (index == 0) {
// insert at the front
front = new ListNode(value, front);
} else {
ListNode current = goTo(index - 1);
ListNode newNode = new ListNode(value, current.next);
current.next = newNode;
}
}
public E get(int index) {
ListNode current = goTo(index);
return current.data;
}
public int indexOf(E value) {
int index = 0;
ListNode current = front;
while (current != null) {
if (current.data == value) {
return index;
}
index++;
current = current.next;
}
return -1;
}
public Iterator iterator() {
return new LinkedListIterator();
}
public void remove(int index) {
if (index == 0) {
// removing from the front
front = front.next;
} else {
ListNode current = goTo(index - 1);
current.next = current.next.next; // deletes the node
}
}
public void set(int index, E value) {
ListNode current = goTo(index);
current.data = value;
}
public int size() {
int count = 0;
ListNode current = front;
while (current != null) {
current = current.next;
count++;
}
return count;
}
public String toString() {
if (front == null) {
return \"[]\";
} else {
String result = \"[\" + front.data;
ListNode current = front.next;
while (current != null) {
result += \", \" + current.data;
current = current.next;
}
result += \"]\";
return result;
}
}
private ListNode goTo(int index) {
ListNode current = front;
for (int i = 0; i < index; i++) {
current = current.next;
}
return current;
}
public class ListNode {
public E data;
public ListNode next;
public ListNode(E data) {
this.data = data;
this.next = null;
}
public ListNode(E data, ListNode next) {
this.data = data;
this.next = next;
}
}
private class LinkedListIterator implements Iterator {
private ListNode current; // current position in list
public LinkedListIterator() {
current = f.
Complete the following short answer questions1. How do the virus.pdf
Complete the following short answer questions:
1. How do the viruses propagate between computers?
2. What is social engineering?
3. How will an SPI firewall handle a packet containing a TCP segment which is an
acknowledgement?
4. Revise the access control list (ACL) in Figure 3-23 (page 119 in the textbook) to permit
access to an FTP server with IP address 10.32.67.112.
5. How will the ACL in Figure 3-23 (page 119 in the textbook) handle a packet that attempts to
open a connection to an FTP server? Explain.
6. For each of the following passwords, first state the kind of attack that would be necessary to
crack it. Justify your answer. Then say whether or not it is an adequate password, again giving
specific reasons.
a) password
b) Winter1
c) SpringBreaK
d) 2!T*d (00-10)
e) 9g&8tY7#?s+445=232+
Solution
Answers:
1)How do the viruses propagate between computers?
-> The browsers gain more features and functions more methods for virus propagation appear
directly to a computer across the network.
Web pages :
->Either via a download, web code, or a cross-site scripting (CSS) attack, web pages can be a
source of viral infection. As browsers gain more features and functions more methods for virus
propagation appear.
Messages:
-> E-mail messages as well as IM or chat messages can be a vector for viral infection.
-> Whether the virus is spread through links, embedded code, or an attachment depends on the
type of message and the client software.
Infected programs or Trojans :
-> Running an infected program that you copied from a CD or other media (or downloaded) is a
means of infection. Sometimes the virus is a Trojan, where it appears to be a useful program, but
is in fact a virus.
2. What is social engineering?
-> Social engineering is an attack vector that relies heavily on human interaction and often
involves tricking people into breaking normal security procedures.
-> Social engineering is the art of manipulating people so they give up confidential information.
-> A social engineer runs what used to be called a \"con game.
Popular types of social engineering attacks :
1)Phishing:
-> Phishing is when a malicious party sends a fraudulent email disguised as a legitimate email,
often purporting to be from a trusted source.
2) Pretexting:
->Pretexting is another form of social engineering where attackers focus on creating a good
pretext, or a fabricated scenario, that they can use to try and steal their victims’ personal
information.
3. How will an SPI firewall handle a packet containing a TCP segment which is an
acknowledgement?
-->Process it through the Access Control List (ACL)
-> Stateful Packet Inspection: Connections have distinct states or stages.
->Different states are subject to different attacks. Stateful firewalls use different filtering rules
for different states. Different because it uses filter rules in order to accept or deny traffic.
->An example would be windows firewall.
4. Revise the access control list (ACL) in Figure 3-23 (page 119 in.
Briefly describe the two competing theories regarding the origin of H.pdf
Briefly describe the two competing theories regarding the origin of Homo sapiens and its
distributional expansion from its origin in Africa. What does the genetic data say about these
two hypothesis, which genes are most useful for these analyses and why? Given a large section
of a genome, say a complete chromosome, how could you locate genes? Given a large section of
a gene, say a complete exon, how could you determine reading frame?
Solution
1. There are many competing therapies for the origin of Homo sapiens. Two such theories are
Multi-regional hypothesis and Out of Africa 2 model.
Multi-regional Hypothesis: The multi-regional hypothesis argues that our early hominid
ancestors, including Homo ergaster and Homo heidelbergensis, migrated out of Africa and thus
the evolution of modern humans took place in different parts of the world, a process termed
regional continuity. This theory places great emphasis on the notion of steady evolutionary
alterations or changes that happen in different regions and produce diverse variations of the
species. Evolution of this kind is kept at a regular rate due to an amalgamation of cultural
progress and ‘gene flow’ or interbreeding, thus keeping all lineages evolving at the same time.
This process is characterized as parallel evolution, which suggests a subtle morphological
resemblance between populations of species who are geographically separated.
Out of Africa 2 model: It is widely recognized among both archaeological and anthropological
academics that archaic Homo populations did leave Africa in an initial phase of globalisation,
called the Out of Africa 1 model. Following this, the population replacement hypothesis
maintains that modern humans evolved in Africa from the ancestral hominids that did not travel
out of this continent in the first stage of global colonisation. It is then argued in this model that
once evolved as anatomically modern humans, Homo sapiens travelled out of Africa to explore,
colonise and rapidly replace the archaic Homo population. This movement suggests a second
wave of hominid species emerging from Africa, thus the title: the Out of Africa 2 model..
Atlas Advertising Atlas Advertising is a regional advertising agency.pdf
Atlas Advertising Atlas Advertising is a regional advertising agency with oces in Boston, New
York, Providence, Washington, D.C., and Philadelphia. 1. Describe the types of networks you
think they would have (e.g., LANs, BNs, WANs) and where they are likely to be located. 2.
What types of standard proto-cols and technologies do you think they are using at each layer
(e.g., see Figure 1-5)?
Solution
1.
Within their offices,internally they will use local area network (LAN) to communicate among
themselves.In the same city ,if they have another office,they can use metropolitian area
network(MAN),which is used to communicate in a city or within a cities.For their offices in
other cities,they will use WAN(Wide area network)which is used for communication worldwide
.
2.
The standard protocols and technologies used are TCP-IP for the wired transmission and UDP
for wireless Transmission.
4. Show that Prefix(L)-x for some y, xy E L is regular if L is regul.pdf
4. Show that Prefix(L)-x: for some y, xy E L is regular if L is regular
Solution
As L is a regular language it will have a DFA. The new language prefix(L) should accept x if xy
(for some y) belongs to L. As we are given that y can be anything, this can be achieved by taking
all states which have a path from start and will reach a final state for some string, we make all
those states as final states. This DFA will accept a substring of any string accepted by L. Now as
prefix(L) has a DFA, it is a regular language..
26) Which is a function of an operating system O Browse the Internet.pdf
26) Which is a function of an operating system? O Browse the Internet O Edit images O Protect
against malware O Provide user interface Next Question
Solution
26.
Provide user interface
All other mentioned functionality can be achieved by other software
27.
Defender
28.
APIs
29.
device driver
Without device drivers you can use your hardware functionality optimally
30.
Application Software:
User can not directly interact with hardware directly,
they need some kind of application software to interact with them.
1.Declare a student structure that contains - Students first an.pdf
1.Declare a student structure that contains :
- Student\'s first and last name
-Student ID
2. Create the following functions:
-getStudentInfo(void)
Declares a single student, uses printf()/scanf() to get input
Returns the single student back
- printStudentInfo(student_t *st_ptr)
Takes the pointer to a student (to avoid making a copy)
-Prints out all of the student information
3. Declare an array of five students
4. Using a for loop and getStudentInfo() function, get input of all the students
5. Using a for loop and printStudentInfo() function, print all the output of all students.
Solution
getStudentInfo(void)
{
printStudentInfo(student_t *st_ptr)
{
str_ptr.getStudentInfo();
display(stud);
return 0;
}
int main()
{
printStudentInfo(str_ptr);
}
return 0;
}.
1. Describe how plant vasculature is optimized for water transport i.pdf
1. Describe how plant vasculature is optimized for water transport in the tallest trees
2. What property of water molecules allow water to stay together while being transported to
extreme heights? And how does temperature affect this?
3.Water and minerals enter the plant through _______, and are then transported through plant\'s
vasculature, via the _________. This flow is maintained through constant ________, where
water exists through __________ in the leaves. The sugars that are made during photosynthesis
in the ____________ are transported downwards through the vasculature via the _______.
4. Give an example of ways plant have evolved to cope with drought,
Solution
Water transport in tree occurs by cappilary action .Water is trasported by root to leaves through
xylem . xylem is made up by tube like cells tracheids and vessel . These join together from root
to leaves . Two forces are usful to flow water upward 1. root pressure and 2. transpiration . Root
pressure occure when water flow into root due to osmotic pressure . Transpiration occures due to
evoporation from leaf surface . The adhesive nature of water molecules to inner cappilary wall
help to keep water column against gravitational force .
2 Adhesive nature of water to inner cappilary wall . Temprature increases transpiration and low
osmatic pressure in leaves .
3. water and mineral entered the plant through ROOT and are then trasported through plant
vasculature , via the XYLEM These flow is maintained through constant TRANSPIRATION
where water exist through STOMETA in the leaves . Thesuger that are made during
photosynthesis in the leaf are transported downward through the vasculature via PHLOEM.
4 Plant have evolved to cope with drought as xerophytic plant have well develop root system ,
fleshy leaves and stometa sunk in fleshy leave to avoid tranpiration ..
A19730 Which, ofthe foliowrgconsidered a tusa.a.for restructuring ro.pdf
A19730 Which, ofthe foliowrgconsidered a tusa.a.for restructuring? road sector More road
engreers ) Increased donor funding Use of contractors V) None of the above (9) 12) What is the
function of a ) Planning for extensions 1o the rod network Providing contractiors for carrying out
road works i) Providing finance for road maintenance iv) Establishing road policy v) None of the
above (h) What other tem can be used for a supplier 12) i) Client i) Manager iv) Producer v)
None of the above (i) 121 Where is road ownership defined? i) Ministry procedures i) Ministerial
memo i) Legislation iv) Contract v) None of the above 0) How is official notice given that a road
has been designated? 121 i) Publication in the road administration asset plan ili) Letter circulated
to residents of the area served by the road iv) Road listed in a policy document v) None of the
above What action may be required by the road administration following road 12 designation?
(k) roviding length marker posts on the road ii) Provision of a maintenance depot on the road ili)
Guaranteed minimum level of maintenance funding for the road iv) Road will be paved v) None
of the above (1) Which is considered a functional classification? i) Secondary road ii) Collector
road ii) Class B road iv) Municipal road v) None of the above
Solution
Question number (f) Answer is (ii) Strenthened management.
Explanation: There are 3 bulding blocks of Road sector restructring :
1) continuous and stable finance.
2) Strenthened management.
3) Creating Ownership and Establishing Responsibility.
Question number (g) Answer is (i) Planning for extension to the road network.
Question number (h) Answer is (v) None of the above.
Question number (i) Answer is (iii) Legislation.
Question number (g) Answer is (v) None of the above.
Explanation when a road is designated a notice is published in a government gazette or other
official publication.
Question number (k) Answer is (i) Providing length marker posts on the road.
Question number (l) Answer is (ii) Collector road..
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