Explain the answers in sufficient details i.e show the work ace Wha.pdf
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Explain the answers in sufficient details i.e show the work ace? What is the probability of drawing a red ace? Are these events (\"ace\" and \"red\") mutually exclusive? Are they independent? e. List two events that are mutually exclusive for a single draw from a deck of cards. What is the probability of drawing a red f. king? What is the probability of drawing a face card in hearts? Are these two events mutually exclusive? Are they independent? 20. The human genome is composed of the four DNA nucleotides: A, T, G, and C. Some regions of the human genome are extremely G-C rich (i.e., a high proportion of the DNA nucleotides there are guanine and cytosine). Other regions are relatively A-T rich (i.e., a high proportion of the DNA nucleotides there are adenine and thymine). Imagine that you want to compare nucleotide sequences from two regions of the Solution a) Let we have two regions, Region1 and Region2. It is given that 60% nucleotides in 1st region is A&T (30% each) and remaining 40% is of G&C nucleotides(20% each). Probability of the nucleotides in Region1: Similarly for region2, Probability of the nucleotides in Region2: Probability of getting both are same nucleiods,we must multiply the individual probabilities. Probability of getting both has type A nucleoids=P(A1)*P(A2) Probability of getting both has type T nucleoids=P(T1)*P(T2) Probability of getting both has type G nucleoids=P(G1)*P(G2) Probability of getting both has type C nucleoids=P(C1)*P(C2) Required probability=P(A1)*P(A2)+ P(T1)*P(T2)+ P(G1)*P(G2)+ P(C1)*P(C2) =0.25 b) It is given that two regions are independent. To find probability to select three nucleoids sequence (triplets) from each of the two regions are same. All combinations made with the nucleoids A, T, G and C taking three at a time 4C3. The number of all combinations of 4 things taken 3 at a time is 4C3=4 Possible selections are ATG, TGC, GCA and CAT Suppose we get outcome as ATG from Region1, so Probability of getting it isNucleotidesP (N)A10.2T10.2G10.3C10.3.
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Recommended
On Some Measures of Genetic Distance Based on Rates of Nucleotide Substitution
An old class paper illustrating the use of the Laplace transform in population biology.
Recombination and LinkageA Three point test cross in Drosophil.docx
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distance more accurately
Today’s Goal: Construct a crude genetic map
Find the gene position
calculate c.o.c., interference
Perform X²(to determine if the map distance from the pooled class data are statistically same with the published map distance)
F2 Trihybrid Test Cross example
axp / +++ X axp / Y
gametes a x p ...
Large variance and fat tail of damage by natural disaster
In order to account for large variance and fat tail of damage by natural disaster, we study a simple model by combining distributions of disaster and population/property with their spatial correlation. We assume fat-tailed or power-law distributions for disaster and population/property exposed to the disaster, and a constant vulnerability for exposed population/property. Our model suggests that the fat tail property of damage can be determined either by that of disaster or by those of population/property depending on which tail is fatter. It is also found that the spatial correlations of population/property can enhance or reduce the variance of damage depending on how fat the tails of population/property are. In case of tornadoes in the United States, we show that the damage does have fat tail property. Our results support that the standard cost-benefit analysis would not be reliable for social investment in vulnerability reduction and disaster prevention.
http://ascelibrary.org/doi/abs/10.1061/9780784413609.277
http://arxiv.org/abs/1407.6209
Solution Manual Image Processing for Engineers by Yagle and Ulaby
This is a sample from "Sample for Solution Manual Image Processing for Engineers by Yagle and Ulaby".
Full Complete Solutions are available too.
I can send full complete Solution Manual for anyone who contact me on E. M ail.
Métodos computacionales para el estudio de modelos epidemiológicos con incer...
Métodos computacionales para el estudio de modelos epidemiológicos con incer...Facultad de Informática UCM
La introducción de la incertidumbre en modelos epidemiológicos es un área de incipiente actividad en la actualidad. En la mayor parte de los enfoques adoptados se asume un comportamiento gaussiano en la formulación de dichos modelos a través de la perturbación de los parámetros vía el proceso de Wiener o movimiento browiniano u otro proceso discretizado equivalente.
En esta conferencia se expone un método alternativo de introducir la incertidumbre en modelos de tipo epidemiológico que permite considerar patrones no necesariamente normales o gaussianos. Con el enfoque adoptado se determinará en contextos epidemiológicos que tienen un gran número de aplicaciones, la primera función de densidad de probabilidad del proceso estocástico solución. Esto permite la determinación exacta de la respuesta media y su variabilidad, así como la construcción de predicciones probabilísticas con intervalos de confianza sin necesidad de recurrir a aproximaciones asintóticas, a veces de difícil legitimación. El enfoque adoptado también permite determinar la distribución probabilística de parámetros que tienen gran importancia para los epidemiólogos, incluyendo la distribución del tiempo hasta que un cierto número de infectados permanecen en la población, lo cual, por ejemplo, permite tener información probabilística para declarar el estado de epidemia o pandemia de una determinada enfermedad. Finalmente, se expondrá algunos de los retos computacionales inmediatos a los que se enfrenta la técnica expuesta.
Recommended
On Some Measures of Genetic Distance Based on Rates of Nucleotide Substitution
An old class paper illustrating the use of the Laplace transform in population biology.
Recombination and LinkageA Three point test cross in Drosophil.docx
Recombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distance more accurately
Today’s Goal: Construct a crude genetic map
Find the gene position
calculate c.o.c., interference
Perform X²(to determine if the map distance from the pooled class data are statistically same with the published map distance)
F2 Trihybrid Test Cross example
axp / +++ X axp / Y
gametes a x p ...
Large variance and fat tail of damage by natural disaster
In order to account for large variance and fat tail of damage by natural disaster, we study a simple model by combining distributions of disaster and population/property with their spatial correlation. We assume fat-tailed or power-law distributions for disaster and population/property exposed to the disaster, and a constant vulnerability for exposed population/property. Our model suggests that the fat tail property of damage can be determined either by that of disaster or by those of population/property depending on which tail is fatter. It is also found that the spatial correlations of population/property can enhance or reduce the variance of damage depending on how fat the tails of population/property are. In case of tornadoes in the United States, we show that the damage does have fat tail property. Our results support that the standard cost-benefit analysis would not be reliable for social investment in vulnerability reduction and disaster prevention.
http://ascelibrary.org/doi/abs/10.1061/9780784413609.277
http://arxiv.org/abs/1407.6209
Solution Manual Image Processing for Engineers by Yagle and Ulaby
This is a sample from "Sample for Solution Manual Image Processing for Engineers by Yagle and Ulaby".
Full Complete Solutions are available too.
I can send full complete Solution Manual for anyone who contact me on E. M ail.
Métodos computacionales para el estudio de modelos epidemiológicos con incer...
Métodos computacionales para el estudio de modelos epidemiológicos con incer...Facultad de Informática UCM
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ECS 111March 2018These are suggested homework. Yo.docx
ECS 111 March 2018
These are suggested homework. You may also choose any of the articles introduced in class as a review topic. As discussed in class, you should not try and answer all of these. Choose one or two and do a complete essay on them. References should not be random blogs, but juried articles or news items from archived media such as NYT or the Economist.
1) Choose a wild animal and discuss its basic habitat and its role in this ecosystem or set of ecosystems. What is its current status relative to conservation concerns? What constitutes its primary food source?
2) Describe one of the major biomes found on earth. Include a discussion of its primary physical and biological attributes. In particular what are the climate, water and soils associated with this biome? What sort of plants and animals define it?
3) Compare and contrast two of the types of soil profiles discussed during the class. Discuss their distribution and the role they have in maintaining their associated ecosystems.
4) Discuss von Humboldt’s observation that the different biomes seen as one goes from the equator to the poles are also found with altitude in mountain systems. Why does this vary depending on which side of the mountain range is considered? Provide some examples.
5) Consider the following population equations.
Compute the steady states and discuss them. What type of system is this?
6) Choose a fishery and describe its attributes. What are the target species? Is this particular fishery sustainable? How does this fishery deal with issues such as markets, management techniques, and bycatch?
7) Discuss the immigration issues currently facing the European Union. How are these being handled by the various member states? Is this a threat to the EU itself? What are the consequences of these demographic shift in the Middle East?
8) Write an essay on the use of fire by human cultures. You might approach this broadly or focus on a particular culture such as the native Americans or the Bantu of Africa.
9) What factors are involved in the transition of a forest ecosystem into a grassland? Consider the anthropogenic place in such transitions. What are the evolutionary issues here?
10) Consider the tundra biome and its physical attributes. What uses of this biome by various countries have affected these regions in the past one hundred years?
1
ELEG-548: Low Power VLSI Circuit Design, Spring 2018, Homework #3
1.(15’) A 3-input NAND gate is shown in Figure 1. Assume “1”-probabilities of 3 inputs are:
p(A)=0.5, p(B)=0.5, p(C)=0.5.
1). Use equation to find out the “1”-probability of output node p(Out)=?
2). Use equation to find the switching probability of Psw(out) for output node.
3). Now assume there are following spatiotemporal correlations among input signals,
i). Every “0” applied to A is immediately followed by “1”,
ii). CB
Given the above spatiotemporal correlations, use switching table (as shown in ...
Reading Seminar (140515) Spectral Learning of L-PCFGs
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Wind/ Solar Power Forecasting
Those who wishes to research and build the system of Forecasting of Wind & Solar power generation, they need to solve the fundamental equation as discussed in the attached presentation. This was discussed at Technical Session 2 - Advanced Statistical Approach used in Wind/ Solar Power Forecasting NIWE - Workshop on theme"Current Practices in Wind & Solar Forecasting"
Bonding in Tranisiton Metal Compounds - Part 2
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LOS At the time of her death on September 4, 2015, Alicia held the .pdf
LOS At the time of her death on September 4, 2015, Alicia held the following assets. Bonds of
Emerald Tool Corporation Stock in Drab Corporation Insurance policy (face amount of
$400,000) Fair Market Value $ 900,000 1,100,000 on the life of her father, Mitch Traditional
IRAs *Cash surrender value. 80,000* 300,000 Alicia was also the life tenant of a trust (fair
market value of $2 million) created by her late husband Bert. (The executor of Bert\'s estate had
made a QTIP election.) In October, Alicia\'s estate received an interest payment of $11,500
($6,000 accrued before September 4, 2015) paid by Emerald and a cash dividend of $9,000 from
Drab. The Drab dividend was declared on August 19 and was payable to date of re- cord
shareholders on September 3, 2015. Although Mitch survives Alicia, she is the designated
beneficiary of the policy. The IRAs are distributed to Alicia\'s children. What amount is included
in Alicia\'s gross estate?
Solution
Alicia’s Gross Estate:-
Bonds of Emrald Tool Corporation - 900,000
Interest on Emrald Tool Bonds – 11,500…………..(accrued interest is to be included )
Stock in Drab Corporation - 1,100,000
Cash Dividend – 9,000
Insurance Policy – 80,000 ……….( Cash surrender value)
Traditional IRA – 300,000
Value of Trust – 2,000,000
Thus, Total asset – 4,400,500.
Match the following terms with the definitions given belowTerms Gi.pdf
Match the following terms with the definitions given below
Terms Given
a. Heap
b. Memory leak
c. Inaccessible object
d. Dangling pointer
e. Destructor
f. Dynamic data
Definitions given
i. The loss of available space that occurs when dynamic data is not deallocation properly.
ii. The area of memory used for allocation and deallocation of dynamic data.
iii. An object that has been allocated but has no pointer pointing to it.
iv. Variables created with the new operation.
v. A pointer that points to a dellocated object.
vi. A member function invoked when an object goes out of scope.
Solution
Heap - The area of memory used for allocation and deallocation of dynamic data.
Memory leak - The loss of available space that occurs when dynamic data is not deallocation
properly.
Inaccessible object - An object that has been allocated but has no pointer pointing to it.
Dangling pointer - A pointer that points to a dellocated object.
Destructor - vi. A member function invoked when an object goes out of scope.
Dynamic data - Variables created with the new operation..
JAVA...With N.E.T_B.E.A.N.S___________________________________.pdf
JAVA...W\'i\'t\'h
N.E.T_B.E.A.N.S_______________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
____________________________________________________
***********Product*********************
package rooster.nest;
import java.text.NumberFormat;
public class Product {
private long id;
private String code;
private String description;
private double price;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
public String getDescription() {
return description;
}
public void setDescription(String name) {
this.description = name;
}
public double getPrice() {
return price;
}
public void setPrice(double price) {
this.price = price;
}
public String getPriceFormatted() {
NumberFormat currencyFormatter =
NumberFormat.getCurrencyInstance();
return currencyFormatter.format(getPrice());
}
}
********DBException************
package crows.foot;
/*
* This is just a wrapper class so we can throw a common exception for
* the UI to catch without tightly coupling the UI to the database layer.
*/
public class DBException extends Exception {
DBException() {}
DBException(Exception e) {
super(e);
}
}
*******DBUtil************
package crows.foot;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;
public class DBUtil {
private static Connection connection;
private DBUtil() {}
public static synchronized Connection getConnection() throws DBException {
if (connection != null) {
return connection;
}
else {
try {
// set the db url, username, and password
String url = \"jdbc:mysql://localhost:3306/mma\";
String username = \"mma_user\";
String password = \"sesame\";
// get and return connection
connection = DriverManager.getConnection(
url, username, password);
return connection;
} catch (SQLException e) {
throw new DBException(e);
}
}
}
public static synchronized void closeConnection() throws DBException {
if (connection != null) {
try {
connection.close();
} catch (SQLException e) {
throw new DBException(e);
} finally {
connection = null;
}
}
}
}
*************ProductDB**************
package crows.foot;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;
import rooster.nest.Product;
public class ProductDB {
private static Product getProductFromRow(ResultSet rs) throws SQLException {
int productID = rs.getInt(1);
String code = rs.getString(2);
String description = rs.getString(3);
double price = rs.getDouble(4);
String note = rs.getString(5);
Product p = new Product();
p.setId(productID);
p.setCode(code);
p.setDescription(description);
p.setPrice(price);
return p;
}
public static List getAll() throws DBException {
String sql = \"SELECT * FROM Product.
Let us roll a die twice. The number from the first throw is denoted .pdf
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
Solution
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
We can use the following:
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36
sum of dice
y
p
y^2
y*p
y^2*p
2
9
1/36
81
0.25
2.25
3
4
1/18
16
0.222222
0.888889
4
1
1/12
1
0.083333
0.083333
5
0
1/9
0
0
0
6
1
5/36
1
0.138889
0.138889
7
4
1/6
16
0.666667
2.666667
8
9
5/36
81
1.25
11.25
9
16
1/9
256
1.777778
28.44444
10
25
1/12
625
2.083333
52.08333
11
36
1/18
1296
2
72
12
49
1/36
2401
1.361111
66.69444
sums:
1
9.833333
236.5
To get the y column, subtract 5 from the sum column and then square.
The second column and the third column of this table show the possible values of y and their
probabilities.
E(Y) and V(Y) are calculated using the sums in the last row of this table.
E(Y) = 9.8333
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36.
Is there relationship between having a high school education and kno.pdf
Is there relationship between having a high school education and knowledge of the orbit of the
earth, which one is independent variable and which one is dependent variable?
Solution
There is relationship between having a high school education and knowledge of the orbit of the
earth. While hving high school education is an independent variable, knowledge of the orbit of
earth is dependent upon the former information. If one has high school education then only he
can know about the orbit of the earth..
In a survey, 600 mothers and fathers were asked about the importance.pdf
In a survey, 600 mothers and fathers were asked about the importance of sports for boys and
girls. Of the parents interviewed, 70% said the genders are equal and should have equal
opportunities to participate in sports.
1. find the probability of getting at least 67% but no more than 73%. Use the exact binomial
calculation.
Solution
The probability is
P(0.67.
in html 1 A Header with font size ranging from h4 to h1 with at l.pdf
in html
1: A Header with font size ranging from h4 to h1 with at least your name and class
included. Follow the example on the printout I gave in class.
2. Two horizontal lines separating the head and body – spanning the screen and of size 5.
3. A background image chosen judiciously so that fonts will be displayed clearly. If you
wish not to use an image, you could use a different background color.
4. An image to the left and one to the right, aligned with each other and enough space
in between that you could add some written comments, description, etc. It should be
very similar to the example shown.
5. Keep adding more comments, etc. below the images that must now stretch across the
page. Somewhere in all of this writing, create a link that will take the user down
somewhere near the bottom of the page. After reading at the bottom of the page, there
must be another link that will take the user back to the point where he/she left.
6. Make all links blue and visited links red.
7. (a) Now create a bulleted list (unordered with square as the bulleted symbol).
(b) Inside of the bulleted list, you must have a sublist that is ordered (using numbers).
The font face of this sublist must be different than that of the rest of the
document.
(c) Inside of this sublist, create an unordered sublist with disc as the bulleted
symbols. Look at the example given.
8. Create a table with the following properties:
(a) A width of 80% and centered.
(b) 10 rows and 6 columns
(c) The background of the table must be different from rest of document.
(d) The first row must consist of an image not more than 2 inches in height.
(e) The label for the first column can be for example NAMES, centered, bold and take up two
rows
(f) The second row must span 5 columns and can have as its header TEST SCORES or WEEK
DAYS or ..., centered and bold.
(g) The third row must consist of 5 individual columns each with a header that is centered and
bold as shown in the example.
(h) Fill in the table with appropriate data as shown in the example making sure that the names
are left aligned and the scores are centered but not bold.
9. After the table, follow it up with some written material then include some sort of
quote as shown in the example. Make sure you use italics for the quote, different
color font and for the author of the quote also use different color font.
10. Use a horizontal line to separate this part of the document from what is to follow.
11. Create a link of some web site that you like. Make sure that the link is live i.e. I can
activate the link and be able to go to that web site.
12. Create a link to another page that you have created as demonstrated in the example.
(A) Only the written words are active.
(B) Both the image and the written words are active.
(C) Only the image is active.
13. Add your email address below. Again, make sure that it is active so that I can email
you. Also experiment by using an image for the email rather than words, as shown in
the example
Solut.
Identify these monomers (general classification). Are the monomers h.pdf
Identify these monomers (general classification). Are the monomers hydrophilic, or
amphipathic? Write your answer in the space above each monomer. How many functional
groups can you identify? Circle and name them. Show how you would make these monomers
into a polymer.
Solution
A and B. a.Cysteine (Cys, C) - Hydrophobic and aliphatic R group
b. Phenylalanine (Phe, F) - Hydrophobic and aromatic R group
C. Functional groups
Cysteine
- COOH - Carboxyl
- NH2 - Amino
- SH - Sulphydryl
Phenylalanine
- COOH - Carboxyl
- NH2- Amino
- Aromatic.
How is the Internet come about How was the World Wide Web ( WWW .pdf
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
Solution
It was in late 1960s the initial concepts of internet came into exostence, with ARPANET(
Advanced Research Projects Agency Networ), which was originally founded by US Department
of Defence. They introduced packet switching to communicate between systems over a single
network.Later on in 1970s the TCP/IP was got introduced, so the ARPANET adopted TCP/IP in
1983, and from there the network of network model came into existencce, which later known as
Internet.
WWW (World Wide Web) is an aspect of Internet, like email. Coomonly WWW is being
misinterpreted as the so called Internet. WWW is an information space in which one can locate
any kind of information through a URL(Uniform Resource Locator). It was in 1989, Tim
Berners-Lee invented the Web. He developed a web browser which he was an employee at
CERN in Switzerland. Lter on the commercial potential in web makes the growth of web in an
exponential paced manner..
Each figure is an isometry image of the figure at the right. Tell whe.pdf
Each figure is an isometry image of the figure at the right. Tell whether their orientations are the
same or opposite. Then classify the isometry.
Solution
27, 28, - Yes, they have same orientation
29 - Have opposite orientation
All images are isometric to the original image, its just that figure 29 is flipped..
E. None of the above.E. Actually, they differ in all of these ways.pdf
E. None of the above.
E. Actually, they differ in all of these ways.
E. Actually, all of these are accurate.
E. None of the above.1.Bacteria are good experimental organisms for genetic studies for all of
the following reasons EXCEPT:A. they have large numbers of progeny.B. they reproduce
rapidly.C. they are diploid.D. they reproduce asexually.E. Actually, all of these are reasons
bacteria are useful in genetic studies.
Solution
1. Bacteria are good experimental organisms for genetic studies for all of the following reasons
except they have a relatively small, diploid genome that is easier to characterize. So, the correct
answer is c.
2. A plasmid is a small, circular, double-stranded DNA molecule that is distinct from a cell\'s
chromosomal DNA. So, the correct answer is plasmid DNA is circular like bacterial DNA.
Option A.
3. Option B (Transference). Transference is the unconscious transfer to others of feelings, and
attitudes which were originally associated with important figures in one\'s early life.
4. Option C (minutes ) is the correct answer. It is a unit of meaure in baterial conjugation
experiment..
Describe the difference between a. The pulmonary and systemic circula.pdf
Describe the difference between a. The pulmonary and systemic circulation b. An artery and a
vein c. An atrium and a ventricle
Solution
a. Pulmonary and systemic circulation
The cardiovascular system comprise of two circulatory pathsways namely pulmonary and
sytemic circulation.The pulmonary circulation circuits through the lungs carrying the oxygenated
blood whereas the systemic circulation circuit through the rest of the body to provide oxygenated
blood to various body parts.
In the pulmonary circulation, blood travels through capillaries on the alveoli, air sacs in the lungs
allowing the gas exchange. Here, as blood flows, the size of the vessel decreases from artery to
arteriole, from vein to venuleand finally to capillaries,which serve as vessels for gas and nutrient
exchange.In the systemic circulation,the movement of blood takes place from the heart through
the body providing oxygen and nutrients to the tissues of the body while bringing back the
deoxygenated to the heart. Systemic and pulmonary circulation transition to the opposite to each
other. Systemic circulation is a much larger and higher pressure system than pulmonary
circulation.
Thus systemic circulation is the larger path that blood takes on, routing oxygenated blood around
the tissues of the body and deoxygenated blood from them back to the heart , while pulmonary
circulation is the smaller loop that blood takes in the lungs.
b. An artery and a Vein
Blood is circulated in the body through arteries, veins and capillaries in a closed circulatory
system.
All blood vessels leading from heart carrying oxygenated blood (except pulmonary artery) to
various organs of the body are called as arteries.
Whereas, vessels which carry deoxygenated blood (except pulmonary vein) from various organs
of the body to the heart are called as veins.
Arteries have thick elastic muscular walls and valves are absent, whereas, Veins have thin non-
elastic valves and valves are present to prevent bacl flow of blood.
c. An atrium and a Ventricle
There are 4 chambers in the heart; the atria refer to the upper chambers while the ventricles refer
to the lower chambers.
Atria (atrium) refers to the upper chambers of the heart that receive the impure blood from the
veins to send it to the ventricles.
Ventricles are small cavities or chambers that are present within the left chamber of the heart that
accepts blood from the arteries (left atrium) and then contracts to force into the aorta. Both right
and the left part of our heart has an atrium and one ventricle.
The walls of the ventricles are thicker, while that of the atria are thinner. They ventricles contain
valves to pump the blood in and out of the heart..
Cross a homozygous running, heterozygous black mouse with a waltzing.pdf
Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is
the probable phenotype ratio? Show punnet square and all work.
Solution
Dihybrid cross
Homozygous running & heterozygous black -mouse x heterozygous black & waltzing
mouse
Parental gametes --> RRBb x rrbb
RB Rb RB Rb
rb RrBb Rrbb RrBb Rabb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
F1-phenotypes
Running black mouse
Running brown mouse.
Describe some effects that cybertechnology has had so far for our se.pdf
Describe some effects that cybertechnology has had so far for our sense of self. Questions to
think about as you write: In the age of information how are we beginning to conceive ourselves
in relation to nature? How is cybertechnology changing the way he view ourselves and others?
Does cybertechnology bring us closer together or create a rift? Are humans becoming more
computer-like? How?
MUST BE AT LEAST 300 WORDS IN TOTAL.
Solution
Information technologies are in a stable level of change and break-through.We can see that
information technologies have brought great social changes compared to the past years before
they appeared.In the past days of disconnect,we plugged into each other\'s nervous system by
passing our books,on discussions with others and by sharing our thoughts and dreams.But
connectedness made one become others databases snd servers learning on eachother\'s
memories.The expanded technology increased the lenghts of people\'s thoughts with this
increasing world.It brought people\'s imaginations and thoughts into existence.
The effects of technology has to do with whether humans desire greater simplicity or complexity
as they progress. Many technology creators acclaim the value of convergence. While that might
severely reduce the number of ‘devices’ that we need and use, it also adds complexity to the task
of using a device.Alternately, people have written about how we desire purity, and will therefore
look for technologies that will do one thing, and do it extremely well. The challenge before
technology innovators, clearly, is to enable technologies to communicate with each other, and
sort problems out, the way humans do.Technology will enable diverse cultures to collaborate
more efficiently, in every sphere. It will bring people and organizations together, closer.
Technology will allow people to live multiple lives, assume and live out multiple
identities.Every day, as millions of people interact over the internet, they navigate worlds that are
simulated, they create virtual personalities, and forge online relationships. Fantasy and role-play,
both natural and important functions of child development, are fostered by adult chatrooms and
virtual reality. In a distributed, yet isolated world, people desire connections. In doing so, they
think not of identity but identity crises.
Technological world is a brilliant opportunity for global connections and it gives the exprience of
leading different lives and cultures.We must believe that internet brings us together as without
internet it is not so easy to communicate with eachother.But on the other hand, it does have some
negative effects on people, tearing them apart in way of face to face communication making such
relations undervalued.
Computers, or operating systems to be exact, are lacking in the the fundamental capabilities
possessed by all sentient systems: the ability to reason, to learn from observation and experience,
to grow intelectually over time. As of today, computers .
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9) What factors are involved in the transition of a forest ecosystem into a grassland? Consider the anthropogenic place in such transitions. What are the evolutionary issues here?
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ii). CB
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LOS At the time of her death on September 4, 2015, Alicia held the .pdf
LOS At the time of her death on September 4, 2015, Alicia held the following assets. Bonds of
Emerald Tool Corporation Stock in Drab Corporation Insurance policy (face amount of
$400,000) Fair Market Value $ 900,000 1,100,000 on the life of her father, Mitch Traditional
IRAs *Cash surrender value. 80,000* 300,000 Alicia was also the life tenant of a trust (fair
market value of $2 million) created by her late husband Bert. (The executor of Bert\'s estate had
made a QTIP election.) In October, Alicia\'s estate received an interest payment of $11,500
($6,000 accrued before September 4, 2015) paid by Emerald and a cash dividend of $9,000 from
Drab. The Drab dividend was declared on August 19 and was payable to date of re- cord
shareholders on September 3, 2015. Although Mitch survives Alicia, she is the designated
beneficiary of the policy. The IRAs are distributed to Alicia\'s children. What amount is included
in Alicia\'s gross estate?
Solution
Alicia’s Gross Estate:-
Bonds of Emrald Tool Corporation - 900,000
Interest on Emrald Tool Bonds – 11,500…………..(accrued interest is to be included )
Stock in Drab Corporation - 1,100,000
Cash Dividend – 9,000
Insurance Policy – 80,000 ……….( Cash surrender value)
Traditional IRA – 300,000
Value of Trust – 2,000,000
Thus, Total asset – 4,400,500.
Match the following terms with the definitions given belowTerms Gi.pdf
Match the following terms with the definitions given below
Terms Given
a. Heap
b. Memory leak
c. Inaccessible object
d. Dangling pointer
e. Destructor
f. Dynamic data
Definitions given
i. The loss of available space that occurs when dynamic data is not deallocation properly.
ii. The area of memory used for allocation and deallocation of dynamic data.
iii. An object that has been allocated but has no pointer pointing to it.
iv. Variables created with the new operation.
v. A pointer that points to a dellocated object.
vi. A member function invoked when an object goes out of scope.
Solution
Heap - The area of memory used for allocation and deallocation of dynamic data.
Memory leak - The loss of available space that occurs when dynamic data is not deallocation
properly.
Inaccessible object - An object that has been allocated but has no pointer pointing to it.
Dangling pointer - A pointer that points to a dellocated object.
Destructor - vi. A member function invoked when an object goes out of scope.
Dynamic data - Variables created with the new operation..
JAVA...With N.E.T_B.E.A.N.S___________________________________.pdf
JAVA...W\'i\'t\'h
N.E.T_B.E.A.N.S_______________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
____________________________________________________
***********Product*********************
package rooster.nest;
import java.text.NumberFormat;
public class Product {
private long id;
private String code;
private String description;
private double price;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
public String getDescription() {
return description;
}
public void setDescription(String name) {
this.description = name;
}
public double getPrice() {
return price;
}
public void setPrice(double price) {
this.price = price;
}
public String getPriceFormatted() {
NumberFormat currencyFormatter =
NumberFormat.getCurrencyInstance();
return currencyFormatter.format(getPrice());
}
}
********DBException************
package crows.foot;
/*
* This is just a wrapper class so we can throw a common exception for
* the UI to catch without tightly coupling the UI to the database layer.
*/
public class DBException extends Exception {
DBException() {}
DBException(Exception e) {
super(e);
}
}
*******DBUtil************
package crows.foot;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;
public class DBUtil {
private static Connection connection;
private DBUtil() {}
public static synchronized Connection getConnection() throws DBException {
if (connection != null) {
return connection;
}
else {
try {
// set the db url, username, and password
String url = \"jdbc:mysql://localhost:3306/mma\";
String username = \"mma_user\";
String password = \"sesame\";
// get and return connection
connection = DriverManager.getConnection(
url, username, password);
return connection;
} catch (SQLException e) {
throw new DBException(e);
}
}
}
public static synchronized void closeConnection() throws DBException {
if (connection != null) {
try {
connection.close();
} catch (SQLException e) {
throw new DBException(e);
} finally {
connection = null;
}
}
}
}
*************ProductDB**************
package crows.foot;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;
import rooster.nest.Product;
public class ProductDB {
private static Product getProductFromRow(ResultSet rs) throws SQLException {
int productID = rs.getInt(1);
String code = rs.getString(2);
String description = rs.getString(3);
double price = rs.getDouble(4);
String note = rs.getString(5);
Product p = new Product();
p.setId(productID);
p.setCode(code);
p.setDescription(description);
p.setPrice(price);
return p;
}
public static List getAll() throws DBException {
String sql = \"SELECT * FROM Product.
Let us roll a die twice. The number from the first throw is denoted .pdf
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
Solution
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
We can use the following:
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36
sum of dice
y
p
y^2
y*p
y^2*p
2
9
1/36
81
0.25
2.25
3
4
1/18
16
0.222222
0.888889
4
1
1/12
1
0.083333
0.083333
5
0
1/9
0
0
0
6
1
5/36
1
0.138889
0.138889
7
4
1/6
16
0.666667
2.666667
8
9
5/36
81
1.25
11.25
9
16
1/9
256
1.777778
28.44444
10
25
1/12
625
2.083333
52.08333
11
36
1/18
1296
2
72
12
49
1/36
2401
1.361111
66.69444
sums:
1
9.833333
236.5
To get the y column, subtract 5 from the sum column and then square.
The second column and the third column of this table show the possible values of y and their
probabilities.
E(Y) and V(Y) are calculated using the sums in the last row of this table.
E(Y) = 9.8333
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18
12
1/36
Total
36/36.
Is there relationship between having a high school education and kno.pdf
Is there relationship between having a high school education and knowledge of the orbit of the
earth, which one is independent variable and which one is dependent variable?
Solution
There is relationship between having a high school education and knowledge of the orbit of the
earth. While hving high school education is an independent variable, knowledge of the orbit of
earth is dependent upon the former information. If one has high school education then only he
can know about the orbit of the earth..
In a survey, 600 mothers and fathers were asked about the importance.pdf
In a survey, 600 mothers and fathers were asked about the importance of sports for boys and
girls. Of the parents interviewed, 70% said the genders are equal and should have equal
opportunities to participate in sports.
1. find the probability of getting at least 67% but no more than 73%. Use the exact binomial
calculation.
Solution
The probability is
P(0.67.
in html 1 A Header with font size ranging from h4 to h1 with at l.pdf
in html
1: A Header with font size ranging from h4 to h1 with at least your name and class
included. Follow the example on the printout I gave in class.
2. Two horizontal lines separating the head and body – spanning the screen and of size 5.
3. A background image chosen judiciously so that fonts will be displayed clearly. If you
wish not to use an image, you could use a different background color.
4. An image to the left and one to the right, aligned with each other and enough space
in between that you could add some written comments, description, etc. It should be
very similar to the example shown.
5. Keep adding more comments, etc. below the images that must now stretch across the
page. Somewhere in all of this writing, create a link that will take the user down
somewhere near the bottom of the page. After reading at the bottom of the page, there
must be another link that will take the user back to the point where he/she left.
6. Make all links blue and visited links red.
7. (a) Now create a bulleted list (unordered with square as the bulleted symbol).
(b) Inside of the bulleted list, you must have a sublist that is ordered (using numbers).
The font face of this sublist must be different than that of the rest of the
document.
(c) Inside of this sublist, create an unordered sublist with disc as the bulleted
symbols. Look at the example given.
8. Create a table with the following properties:
(a) A width of 80% and centered.
(b) 10 rows and 6 columns
(c) The background of the table must be different from rest of document.
(d) The first row must consist of an image not more than 2 inches in height.
(e) The label for the first column can be for example NAMES, centered, bold and take up two
rows
(f) The second row must span 5 columns and can have as its header TEST SCORES or WEEK
DAYS or ..., centered and bold.
(g) The third row must consist of 5 individual columns each with a header that is centered and
bold as shown in the example.
(h) Fill in the table with appropriate data as shown in the example making sure that the names
are left aligned and the scores are centered but not bold.
9. After the table, follow it up with some written material then include some sort of
quote as shown in the example. Make sure you use italics for the quote, different
color font and for the author of the quote also use different color font.
10. Use a horizontal line to separate this part of the document from what is to follow.
11. Create a link of some web site that you like. Make sure that the link is live i.e. I can
activate the link and be able to go to that web site.
12. Create a link to another page that you have created as demonstrated in the example.
(A) Only the written words are active.
(B) Both the image and the written words are active.
(C) Only the image is active.
13. Add your email address below. Again, make sure that it is active so that I can email
you. Also experiment by using an image for the email rather than words, as shown in
the example
Solut.
Identify these monomers (general classification). Are the monomers h.pdf
Identify these monomers (general classification). Are the monomers hydrophilic, or
amphipathic? Write your answer in the space above each monomer. How many functional
groups can you identify? Circle and name them. Show how you would make these monomers
into a polymer.
Solution
A and B. a.Cysteine (Cys, C) - Hydrophobic and aliphatic R group
b. Phenylalanine (Phe, F) - Hydrophobic and aromatic R group
C. Functional groups
Cysteine
- COOH - Carboxyl
- NH2 - Amino
- SH - Sulphydryl
Phenylalanine
- COOH - Carboxyl
- NH2- Amino
- Aromatic.
How is the Internet come about How was the World Wide Web ( WWW .pdf
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
How is the Internet come about ? How was the World Wide Web ( WWW ) developed ?
Solution
It was in late 1960s the initial concepts of internet came into exostence, with ARPANET(
Advanced Research Projects Agency Networ), which was originally founded by US Department
of Defence. They introduced packet switching to communicate between systems over a single
network.Later on in 1970s the TCP/IP was got introduced, so the ARPANET adopted TCP/IP in
1983, and from there the network of network model came into existencce, which later known as
Internet.
WWW (World Wide Web) is an aspect of Internet, like email. Coomonly WWW is being
misinterpreted as the so called Internet. WWW is an information space in which one can locate
any kind of information through a URL(Uniform Resource Locator). It was in 1989, Tim
Berners-Lee invented the Web. He developed a web browser which he was an employee at
CERN in Switzerland. Lter on the commercial potential in web makes the growth of web in an
exponential paced manner..
Each figure is an isometry image of the figure at the right. Tell whe.pdf
Each figure is an isometry image of the figure at the right. Tell whether their orientations are the
same or opposite. Then classify the isometry.
Solution
27, 28, - Yes, they have same orientation
29 - Have opposite orientation
All images are isometric to the original image, its just that figure 29 is flipped..
E. None of the above.E. Actually, they differ in all of these ways.pdf
E. None of the above.
E. Actually, they differ in all of these ways.
E. Actually, all of these are accurate.
E. None of the above.1.Bacteria are good experimental organisms for genetic studies for all of
the following reasons EXCEPT:A. they have large numbers of progeny.B. they reproduce
rapidly.C. they are diploid.D. they reproduce asexually.E. Actually, all of these are reasons
bacteria are useful in genetic studies.
Solution
1. Bacteria are good experimental organisms for genetic studies for all of the following reasons
except they have a relatively small, diploid genome that is easier to characterize. So, the correct
answer is c.
2. A plasmid is a small, circular, double-stranded DNA molecule that is distinct from a cell\'s
chromosomal DNA. So, the correct answer is plasmid DNA is circular like bacterial DNA.
Option A.
3. Option B (Transference). Transference is the unconscious transfer to others of feelings, and
attitudes which were originally associated with important figures in one\'s early life.
4. Option C (minutes ) is the correct answer. It is a unit of meaure in baterial conjugation
experiment..
Describe the difference between a. The pulmonary and systemic circula.pdf
Describe the difference between a. The pulmonary and systemic circulation b. An artery and a
vein c. An atrium and a ventricle
Solution
a. Pulmonary and systemic circulation
The cardiovascular system comprise of two circulatory pathsways namely pulmonary and
sytemic circulation.The pulmonary circulation circuits through the lungs carrying the oxygenated
blood whereas the systemic circulation circuit through the rest of the body to provide oxygenated
blood to various body parts.
In the pulmonary circulation, blood travels through capillaries on the alveoli, air sacs in the lungs
allowing the gas exchange. Here, as blood flows, the size of the vessel decreases from artery to
arteriole, from vein to venuleand finally to capillaries,which serve as vessels for gas and nutrient
exchange.In the systemic circulation,the movement of blood takes place from the heart through
the body providing oxygen and nutrients to the tissues of the body while bringing back the
deoxygenated to the heart. Systemic and pulmonary circulation transition to the opposite to each
other. Systemic circulation is a much larger and higher pressure system than pulmonary
circulation.
Thus systemic circulation is the larger path that blood takes on, routing oxygenated blood around
the tissues of the body and deoxygenated blood from them back to the heart , while pulmonary
circulation is the smaller loop that blood takes in the lungs.
b. An artery and a Vein
Blood is circulated in the body through arteries, veins and capillaries in a closed circulatory
system.
All blood vessels leading from heart carrying oxygenated blood (except pulmonary artery) to
various organs of the body are called as arteries.
Whereas, vessels which carry deoxygenated blood (except pulmonary vein) from various organs
of the body to the heart are called as veins.
Arteries have thick elastic muscular walls and valves are absent, whereas, Veins have thin non-
elastic valves and valves are present to prevent bacl flow of blood.
c. An atrium and a Ventricle
There are 4 chambers in the heart; the atria refer to the upper chambers while the ventricles refer
to the lower chambers.
Atria (atrium) refers to the upper chambers of the heart that receive the impure blood from the
veins to send it to the ventricles.
Ventricles are small cavities or chambers that are present within the left chamber of the heart that
accepts blood from the arteries (left atrium) and then contracts to force into the aorta. Both right
and the left part of our heart has an atrium and one ventricle.
The walls of the ventricles are thicker, while that of the atria are thinner. They ventricles contain
valves to pump the blood in and out of the heart..
Cross a homozygous running, heterozygous black mouse with a waltzing.pdf
Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is
the probable phenotype ratio? Show punnet square and all work.
Solution
Dihybrid cross
Homozygous running & heterozygous black -mouse x heterozygous black & waltzing
mouse
Parental gametes --> RRBb x rrbb
RB Rb RB Rb
rb RrBb Rrbb RrBb Rabb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
rb RrBb Rrbb RrBb Rrbb
F1-phenotypes
Running black mouse
Running brown mouse.
Describe some effects that cybertechnology has had so far for our se.pdf
Describe some effects that cybertechnology has had so far for our sense of self. Questions to
think about as you write: In the age of information how are we beginning to conceive ourselves
in relation to nature? How is cybertechnology changing the way he view ourselves and others?
Does cybertechnology bring us closer together or create a rift? Are humans becoming more
computer-like? How?
MUST BE AT LEAST 300 WORDS IN TOTAL.
Solution
Information technologies are in a stable level of change and break-through.We can see that
information technologies have brought great social changes compared to the past years before
they appeared.In the past days of disconnect,we plugged into each other\'s nervous system by
passing our books,on discussions with others and by sharing our thoughts and dreams.But
connectedness made one become others databases snd servers learning on eachother\'s
memories.The expanded technology increased the lenghts of people\'s thoughts with this
increasing world.It brought people\'s imaginations and thoughts into existence.
The effects of technology has to do with whether humans desire greater simplicity or complexity
as they progress. Many technology creators acclaim the value of convergence. While that might
severely reduce the number of ‘devices’ that we need and use, it also adds complexity to the task
of using a device.Alternately, people have written about how we desire purity, and will therefore
look for technologies that will do one thing, and do it extremely well. The challenge before
technology innovators, clearly, is to enable technologies to communicate with each other, and
sort problems out, the way humans do.Technology will enable diverse cultures to collaborate
more efficiently, in every sphere. It will bring people and organizations together, closer.
Technology will allow people to live multiple lives, assume and live out multiple
identities.Every day, as millions of people interact over the internet, they navigate worlds that are
simulated, they create virtual personalities, and forge online relationships. Fantasy and role-play,
both natural and important functions of child development, are fostered by adult chatrooms and
virtual reality. In a distributed, yet isolated world, people desire connections. In doing so, they
think not of identity but identity crises.
Technological world is a brilliant opportunity for global connections and it gives the exprience of
leading different lives and cultures.We must believe that internet brings us together as without
internet it is not so easy to communicate with eachother.But on the other hand, it does have some
negative effects on people, tearing them apart in way of face to face communication making such
relations undervalued.
Computers, or operating systems to be exact, are lacking in the the fundamental capabilities
possessed by all sentient systems: the ability to reason, to learn from observation and experience,
to grow intelectually over time. As of today, computers .
Choose a deployment platform that allows for the implementation o.pdf
Choose a deployment platform that allows for the implementation of SOA. Describe the
architectural layers that are included in the platform and analyze the flexibility of the platform
for supporting different types of services. Determine whether this platform supports loose
coupling or if it is strongly implementation dependent.
Choose a deployment platform that allows for the implementation of SOA. Describe the
architectural layers that are included in the platform and analyze the flexibility of the platform
for supporting different types of services. Determine whether this platform supports loose
coupling or if it is strongly implementation dependent.
Solution
Service-oriented architectures (SOA) are based on the software services, which are high-level
software components that include web services like (Rest , Soap). Implementation of an SOA
requires tools and run-time infrastructure software. This is collectively referred to as a service-
oriented architecture implementation framework or (SOAIF).
Application development and deployment :
In the traditional software development process, translating the requirements into working
distributed systems is time-consuming and difficult, requiring several stages of manual
development and then its deployment. This complex, error-friendlytask can be effectively done
using a higher-level, component-based SOAIF. The SOAIF incorporates tools that let processes
that are developed, using standards such as Business Process Execution Language (BPEL), can
be easily translated into distributed, high-level services, which are easy to develop, manipulate,
and debug and hence more effecient . These services are easily composed of implementation-
level data flows without the user or developer having to track complex middleware concepts,
such as topics or queues. Further, the implementation-level services can run on any machine
across the network due to the built-in dynamic deployment support SOAIF provides.
Architectural Layers
Enterprise service bus :core infrastructure of an SOAIF is typically provided by an enterprise
service bus (ESB), which points out the challenges in composing, deploying, and managing
distributed, service-based enterprise applications. The ESB incorporates systems connectivity
using web services, Java EE, .NET Framework, and other standards also
ESB also do not provide any guaranty of loose coupling to meet evolving Service-oriented needs
. Implementing ESB to meet SOA requirements require the addition of extra functionality to
compose for business Services and provide secure Service interactions.
Business process management :
Business process management (BPM) considers systems and IT assets as activities or tasks that
participate in well-coordinated and centrally organised Business processes. Traditionally, the
challenge of BPM is that while it is possible to construct any processes that achieve integration
goals, enterprises typically use BPM tools only at the time of designing , modeling.
Abstract Base Class (C++ Program)Create an abstract base class cal.pdf
Abstract Base Class (C++ Program)
Create an abstract base class called Animal which has the following pure virtual member
functions: talk( ), move( ) and the following attribute: animalType (which is a pointer to a
dynamically allocated string.)
Then create the following derived classes (from the animal class): reptile, bird, and mammal.
Here are the concrete classes:
From the reptile class derive lizard and snake.
From bird derive eagle and chicken.
From mammal derive bear, hyena and lion.
Each concrete class needs to override the talk() and move() functions. Also each concrete class
will have the operator<<( ) defined. The output of cout << smokey; (assuming smokey is a bear)
will be the animalType , the sound he makes (talk( ) ) and how he moves ( move( )). Here is an
example:
int main()
{
bear smokey; // using the concrete class
cout << smokey // prints out: bear, growl, walk
animal *aptr = new bear; // I could have used an animal pointer
cout << *aptr; // prints the same results: bear, growl, walk
return 0;
}
Create an array of animal pointers called zoo. Make it 10 in size. Here is what the menu should
look like:
----Zoo Cages----
1 [empty]
2 [empty]
3 [empty]
4 [snake]
5 [eagle]
6 [bear]
7 [empty]
8 [empty]
9 [lion]
10 [hyena]
----Menu----
1) Add
2) Remove
3) Display all animals
4) Display animal count
5) Exit
If the user selects 1) then ask for a cage number and the animal he wishes to put in that cage. Be
sure to check the cage for an existing animal first. The user cannot put an animal in a cage that is
already occupied.
If the user selects 2) then ask for the cage number and remove the animal from that cage (use the
destructor). If there is no animal in the cage, issue an error to the user.
If the user selects 3) then display the contents of each cage. The display function will show each
cage number as well as the animalType of the animal, the sound it makes and how it moves, in
other words, use the operator<<( ). All this information must be on a single screen. The output
screen is up to you.
If the user selects 4) then display the count of the number of animals in your zoo. In the above
example, you will have 5. You should use a static data member in the animal class to keep track
of the number of animals. You should increase that count each time the animal constructor is
called and decrease the count when the animal destructor is called.
***Use Virtual Functions and the animal base class MUST be abstract
Solution
main.cpp
#include //input output
#include //strings
#include //Randoms
#include //Randoms
#include //For input
#include \"lizard.h\"
#include \"snake.h\"
#include \"chicken.h\"
#include \"eagle.h\"
#include \"hyena.h\"
#include \"bear.h\"
#include \"lion.h\"
using namespace std;
//declare our static variable from the animal class
int animal::numAnimals;
int main()
{
//array of animal pointers
animal *zoo[10] =
{ NULL };
//initialize our static variable to zero
animal::numAnimals = 0;
//To test if cage is empty, it is nu.
A range of values estimated to have a high probably of containing th.pdf
A range of values estimated to have a high probably of containing the population value is called:
Point estimate
Standard error of the mean
Confidence interval
Standard deviation
Point estimate
Standard error of the mean
Confidence interval
Standard deviation
Solution
It is
OPTION C: CONFIDENCE INTERVAL. [ANSWER]
***********
The confidence interval is the range of values in which the true population parameter value is
there..
A thrown lump of clay hits a block in mid air and sticks to it. what.pdf
A thrown lump of clay hits a block in mid air and sticks to it. what happens to the total kinetic
energy of the ball+clay system during this inetraction, ignoring gravity?
Solution
As the lump of clay sticks to th block, it is an example of inelasti collision. During the inelastic
collision, the momentum will be conserved for ball+clay system but Kinetic energy will not.
Total Kinetic Energy of ball+clay system will decrease during the interaction..
Answer die following questions with short answers Explain the d.pdf
Answer die following questions with short answers: Explain the difference between stream
ciphers and block cipher. Explain the difference between passive and active attacks. Describe
the difference between authentication and non-repudiation Explain the difference between a
MAC and message digest. Give one advantage and one disadvantage of public-key cryptography
over symmetric-key cryptogra Why is CBC mode considered preferable to ECB mode?
Solution
a)
stream cipher and block cipher differ in whitch the plain text is encoded and decoded.
the basic difference is how the logic used for the encryption and decryption. the idea of block
cipher is partition the plain text into relatively larger blocks and further each block will be
encoded.
and same key is used for both encryption and decryption.example DES,IDEA
the idea of stream cipher is plain text is encrypted and transfered bit by bit.but key property of
stream cipher is different key is generated for each block or bit.example -ISAAC,FISH
b)
passive attack is indirect attack.the attacked host is completely unaware about the attack.
hence it is called passive attack like some body listens the packet transformation or observing
the host in network.
active attack is direct attack.its nothing but the attacked one gets aware of the attack.
like somebody listens the id and password and stop the services(denial of service) or logic
bomb.
both are harmful.
c)
authentication and non repudiation are two different things.
authentication verifies the who you are(user-id).where the non-repudiation verifies the what you
did(messages).
authentication is done through a central instance where non repudiation can done by anyone
d)
message digest is a hash of a message.its output of a cryptographic hash function appilied to
input data it refered to as message.
mac is a piece of information that proves the integrity of a message.
message digest is take only single input and produces a message digest(aka hash) if the message
change it produce different hash code.
mac (message authentication code) is take input and security code to check the integrity of
message. HMAC is one of this kind.
e)
advantage of public key cryptography over the symmetric key cryptography is increased
security and convenience.
and no need of key distribution(transmitted to anyone).and it provide digital signatures that
cannot be repudiated.
the disadvantage of public key cryptography over symmetric cryptography is speed.they are
many symmetic key methods runs faster than the public key cryptography.
f)
the reason why cbc is considered better than ecb has nothing to do with situations involving
an attacker with partial cipher text.
problem with ECB is that it leaks the information like when two blocks have same plain text
then the corresponding cipher blocks also be same where in case
of the CBC it randomly independent of plain text to produce the cipher text..
A single breeding pair of rabbits is introduced to Australia in 1900.pdf
A single breeding pair of rabbits is introduced to Australia in 1900. Assume (for simplicity) that
each pair can produce 4 offspring each year (2 male, 2 female). Also assume, that all births
happen in the last hour of the last day each year and the parents die immediately after offspring
are born. a) How many rabbits are there after 10 years? b) How many rabbits are there after 25
years? If they have 20 offspring each year, c) How many rabbits are there after 10 years? d) How
many rabbits are there after 25 years?
Solution
Each pair can produce 4 offsprings each year and parents die after giving birth to offsprings. So,
offsprings makes the population and parents don\'t add to population.
After first year total population is 4. Next year these 4 rabbits make 2 pairs and give birth to 4 off
springs each. Second year population becomes 8, multiplies by 2 from last year. This continues
till 10th year and then till 25th year.
At the end of 10th year, the population of rabbits will be 2048 and at the end of 25th year the
population of rabbits will be 33554432.
Now when each year 20 offsprings are produced, after the end of first year total population is 20.
At the end of second year, 20 rabbits make 10 pairs and give birth to 20 rabbits each taking total
population to 10*20=200. At the end of 3rd year 200 rabbits make 100 pairs and give birth to 20
each taking total population to 100*20=2000.
In this way at the end of each year population gets multiplied by 10.Therefore, at the end of 10
year total population will be 2*1010 and at the end of 25 years total rabbit population is will be
2*1025..
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