D=10 log (P1P2) Where the values of P1 and P2 are expressed in the .pdf
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D=10 log (P1/P2) Where the values of P1 and P2 are expressed in the same units, such as watts (w). If P2 = 75 w, find the value of P1 at which D=0.7 Solution D = 10 log ( P1 / P2 ) P2 = 75 D = 0.7 0.7 = 10 log ( P1 / 75 ) dividing both sides by 10 0.7 / 10 = log ( P1 / 75 ) 0.07 = log ( P1 / 75 ) 10^0.07 = P1 /75 P1 = 10^0.07* 75 P1 = 88.12.
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Blossom Corporation had 120,000 common shares outstanding on Decembe.pdf
Blossom Corporation had 120,000 common shares outstanding on December 31, 2016. During
2017, the company issued 14,500 shares on March 1, retired 4,100 shares on July 1, issued a
20% stock dividend on October 1, and issued 21,600 shares on December 1. For 2017, the
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(net of tax). The company issued a 2-for-1 stock split on February 1, 2018, and the company’s
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Calculate earnings per share for 2017 as it should be reported to shareholders.
Solution
Share Outstanding As of December 31,2016 120000 Share Issued as of March 1, 2017 14500
Total Share Oustanding as of March 1, 2017 134500 Retired Shares as of July 1, 2017 4100
Total Share Oustanding as of July 1, 2017 130400 20% Stock Dividend ( 130400*20%) as of
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Dec 1 ) 480000/178080 EPS= 2.70.
A system experiences a phase change from liquid to solid. Is this exo.pdf
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Solution
If heat is released then the system is exothermic
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Write an equation of a tangent function with the following characters.pdf
Write an equation of a tangent function with the following characters Period = 1 Vertical
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Solution
comparing with y =tan(Bx+C) +D
period /B=1
=>B=
midline y =-1
=>D=-1
vertical asymptotes when x =1/2,(Bx+C)=/2
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Why are the 1960s relatively turbulentSolutionAnswer1960s .pdf
Why are the 1960s relatively turbulent?
Solution
Answer:
1960s was a most different year for the world in general and US in particular. In 1960s
particularly US accepted the fact that they are loosing from the world . It was 20/01/1961 , John
F. became the president of the US . In his governance , US came out from the problems . Since
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When acting as a neurotransmitter NO binds to the iron atom in vario.pdf
When acting as a neurotransmitter NO binds to the iron atom in various heme containing
proteins. Is the N or the O atom more likely to be directly attached to the Fe? Explain.
Solution
Most iron enzymes engage iron either as heme or as a special arrangement of iron with sulphur
groups referred to as iron–sulfur centers (FeS) .Iron in heme bears a striking resemblance to
magnesium ion in chlorophyll. Heme basically is a porphyrin ring system with iron positioned in
the center, is the most common form of iron in biological proteins. In cytochrome c, a common
heme protein in the mitochondria, the axial ligands to the iron are occupied by histidine and
methionine from the protein. Heme enzymes include calalase and peroxidase. As components of
iron–sulphur centers, iron enters into multiple cluster arrangements with cysteine residues on
enzymes that offer a more direct contact with the protein. These centers differ in their complexity
from the simple 2Fe–2S to the more elaborate 4Fe–4S . Iron in these centers binds substrates as
well as transfer electrons and takes part in reactions involving dehydrations and rearrangements.
Enzymes with iron–sulphur centers include xanthine oxidase, succinate dehydrogenase,
aconitase, and nitrogenase
As a neuromodulator or neurotransmitter, NO is released from the synthesis place
diffuses through the plasma membranes due to its concentration gradients. NO can
directly interact with the cysteinyl side chains of peptides and proteins, iron-
containing heme groups, superoxide anion and oxygen. Since the life-span of NO‚
is Only few seconds it is easily inactivated. It has been suggested that some intermediate
compounds, for example S-nitrosothiols may act as a carrier system to stabilize, target
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there is a good chance for the reaction to occur. There is indeed evidence that GSNO
is synthetized in the CNS.
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Which of the following are reasons that Drosophila are an ideal exper.pdf
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Solution
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comparing with y =tan(Bx+C) +D
period /B=1
=>B=
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vertical asymptotes when x =1/2,(Bx+C)=/2
((/2)+C)=/2
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Why is a mixture of alcohol and water used, rather than simply water itself, to saponify the fat or
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Why are the 1960s relatively turbulentSolutionAnswer1960s .pdf
Why are the 1960s relatively turbulent?
Solution
Answer:
1960s was a most different year for the world in general and US in particular. In 1960s
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F. became the president of the US . In his governance , US came out from the problems . Since
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turned to be the Radical 60s..
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When acting as a neurotransmitter NO binds to the iron atom in various heme containing
proteins. Is the N or the O atom more likely to be directly attached to the Fe? Explain.
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Most iron enzymes engage iron either as heme or as a special arrangement of iron with sulphur
groups referred to as iron–sulfur centers (FeS) .Iron in heme bears a striking resemblance to
magnesium ion in chlorophyll. Heme basically is a porphyrin ring system with iron positioned in
the center, is the most common form of iron in biological proteins. In cytochrome c, a common
heme protein in the mitochondria, the axial ligands to the iron are occupied by histidine and
methionine from the protein. Heme enzymes include calalase and peroxidase. As components of
iron–sulphur centers, iron enters into multiple cluster arrangements with cysteine residues on
enzymes that offer a more direct contact with the protein. These centers differ in their complexity
from the simple 2Fe–2S to the more elaborate 4Fe–4S . Iron in these centers binds substrates as
well as transfer electrons and takes part in reactions involving dehydrations and rearrangements.
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aconitase, and nitrogenase
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diffuses through the plasma membranes due to its concentration gradients. NO can
directly interact with the cysteinyl side chains of peptides and proteins, iron-
containing heme groups, superoxide anion and oxygen. Since the life-span of NO‚
is Only few seconds it is easily inactivated. It has been suggested that some intermediate
compounds, for example S-nitrosothiols may act as a carrier system to stabilize, target
and to preserve its biological activity. S-nitrosothiols are readily formed from
glutathione and NO. Since neurons and glial cells contain GSH at high concentrations,
there is a good chance for the reaction to occur. There is indeed evidence that GSNO
is synthetized in the CNS.
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platelet aggregation, bronchodilation and vasodilation..
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that apply. r They are easy to grow and maintain in a lab. n Afemale has very few offspring each
reproductive cycle. r They are relatively cheap to maintain in a lab r Their life cycle is not well
understood. r They have a small number of well characterized chromosomes. They do not have
very many easily observable phenotypes. They have a long generation time.
Solution
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have a small number of well characterized chromosomes..
What Michael acceptor is needed for the conjugate addition Beta - .pdf
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step process: Michel reaction followed by intramolecular aidol reaction. What Michael acceptor
is needed for the conjugate addition?
Solution.
What is the role of intuition in decision-makingSolutionIntui.pdf
What is the role of intuition in decision-making?
Solution
Intuition is ability to acquire the knowledge, information and skills without any forceful activity;
intuition is something which is come from inner sense that provoked itself.
Decision- making process always influence by the intuition because decision making is a kind of
a practices which is not founded in every one but mostly founded in leaders ,managers and
businessmen.
Intuition always linked with emotions, intuition linked to knowledge, intuition always fast and
processing of information.
Role of intuition in decision-making as follows-.
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What is the difference between a HTML element’s id attribute and name attribute?
Solution
The form input element id has nothing to do with the data contained within the element. Input
element ids are for hooking the element with JavaScript and CSS. The name attribute is used in
the HTTP request sent by your browser to the server as a variable name associated with the data
contained in the value attribute.
In other words, By using id attribute we can validate and manipulate the value of an element at
client side. By using name attribut, we can validate and manipulate the value of an element at
server side.
In java script, by using id attribute we can get the value of input element.
document.getElementById(\"id\").value;
In server side, by using name attribute we can get the value of input element.
request.getParameter(\"name\");.
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What happens when you\'re Running Scripts without a console in Linux/UNIX.
Solution
The script will run as a background process that is daemon process. These daemon processes
doesnot interact with the user. Console is used to send text output and receive user input and is
under the direct control of the user. However daemon process is run in the background by
forking a child process and the parent is exited immediately..
TrueFalse A virtual private network is a way to use the Internet to.pdf
True/False A virtual private network is a way to use the Internet to create a connection between
a remote user and a central location. IPsec encrypts the packet data but not the header
information. L2TP uses IPsec for its encryption. Kerberos is an authentication protocol used for
logging into Windows and uses a ticket granting system that sends an encrypted ticket to the
user\'s machine. A digital signature is used to guarantee who sent a message. This is referred to
as authenticity. Public key encryption is fast becoming the most widely used type of encryption
because there are no issues to deal with concerning distribution of keys. PGP involves private
key encryption. The most widely used symmetric key algorithm is Advanced Encryption
Standard. Cipher text is encrypted text. Binary numbers are made up of 0s and 1s.
Solution
Answer:
1) True
explanation:
=> A VPN, or Virtual Private Network, allows you to create a secure connection to another
network over the Internet.
2)true
explanation:
=>IPsec includes Encapsulating Security Payload (ESP) to encrypt the data and provide
confidentiality
3) TRUE
explanation:
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5) True
explanation:
=> Digital signatures are the public-key primitives of message authentication.
=> A valid digital signature gives a recipient reason to believe that the message was created by a
known sender.
6) True
explanation:
=> Public key encryption in turn provides a solution to key distribution and data transmission
issues. Used together, performance and key distribution are improved without any sacrifice in
security.
7) False.
8) True
explanation:
=> Encryption algorithms that use the same key for encrypting and for decrypting information
are called symmetric- key algorithms.
=>The symmetric key is also called a secret key because it is kept as a shared secret between the
sender and receiver of information.
9) true
explanation:
=>An encryption function takes cleartext and a key as input and returns ciphertext.
10) TRUE
explanation:
=>A binary code represents text or computer processor instructions using the binary number
system\'s two binary digits, 0 and 1..
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The \"total physical units accounted for\" is the sum of the units completed and transferred out
during the period plus the True ? False
Solution
Total physical units accounted for = Units of completed and transferred out+Ending work in
process inventory Units
So above statement that the total physical units accounted for is the sum of the units completed
and transferred out during the period plus the units in beginning wip inventory is False
So above statement is False.
The problem can be found below. I would like someone to run this. I .pdf
The problem can be found below. I would like someone to run this. I don’t get it because there’s
an error. Please, fix this because I don’t know what is happening. What version of python should
I use?
Please, if you don’t know what is happening let someone else to do it. Bp Project untitled
PycharmProjects/untitled 3 import sys ServerPort 80 new server Socket socket (AF INET, S0CK
STREAM) Illi External Libraries 6 #Prepare a sever socket 7 #Fill in start server Socket.bind
server Port)) server Socket.listen (1) print the web server is up on port server Port 10 #Fill in end
11 while True: Establish the connection 13 print \"Ready to serve... 14 connectionSocket add J
server Socket. accept() 15 16 try: message connectionSocket.recv(1024) 17 print message
message split 01, message. split 1] 18 filename 3 message. split [1] print filename \'II filename
[1:] 20 open (filename (1:1) 21 output data f.read 22 print outputdata 23 Send one HTTP header
line into socket Fill in start 25 connection Socket.send(\' InHTTP/1.1 200 0KNnNn 26
connectionSocket.send(outputdata) 27 Fill in end 28 Send the content of the requested file to the
client 29 for i in range (0, len(outputdata)): 30 connection Socket.send(outputdata [i]) 31
connection Socket.close 32 33 except IOError w Send response message for file not found 34
Fill in start 35 connectionSocket.send NnHTTP/1.1 404 Not Found nAn 36
connectionSocket.send \'InHTTP/1.1 404 Not Found n\ \') 37 38 39
/Library/Frameworks/Python, framework/Versions/2.7/bin/python2.7 /Users. File User S untitled
new py\", line 33 except IOError: SyntaxError: invalid syntax Process finished with exit code 1
Solution
Update your code as below :
#!/bin/python
import sys
import socket
serverPort=80
serverSocket = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
serverSocket.bind((\'\',serverPort))
serverSocket.listen(1)
while True :
connectionSocket, addr = serverSocket.accept()
try :
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filename = message.split()[1]
f = open(filename[1:])
outputdata = f.read()
connectionSocket.send(\'\ HTTP/1.1 200 OK\ \ \')
connectionSocket.send(outputdata)
for i in range(0,len(outputdata)):
connectionSocket.send(outputdata[i])
connectionSocket.close()
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serverSocket.close();.
The probability that a teacher will give an unannounced test duri.pdf
The probability that a teacher will give an unannounced test during any class is 1/5. If a student
is absent twice ,then probability that misses atleast one test is ?
Solution
Given the probability of teacher will give an unannounced test during any class is 1/5.
The probability of not missing test 1-1/5=4/5
probability of missing test in only one day 2(4/25)=8/25
probability of missing test on both the day (1/5)*(1/5)=1/25
probability that misses atleast one test is (8/25)+(1/25)=9/25
probability that misses atleast one test is 9/25.
The Orlando MedicalOrlando Medical Corporation financial statements.pdf
The Orlando MedicalOrlando Medical Corporation financial statements? follow: LOADING...
?(Click the icon to view the consolidated balance? sheets.) LOADING... ?(Click the icon to view
the consolidated income? statements.) Orlando MedicalOrlando Medical?\'s inventory turnover
during fiscal year 20142014 was A. 8484 times. B. ?$35 comma 16435,164. C. very slow. D.
138.0138.0 times. Orlando Medical Corporation Consolidated Statements of Financial Position
December 31, (In Millions) 2014 2013 Assets: Current assets Cash and cash equivalents $4,344
$4,237 Accounts and notes receivable 3,402 2,409 Short-term investments 855 504 Inventories,
at cost 435 404 Prepaid expense and other current assets 1,638 1,220 Total current assets 10,674
8,774 Property and equipment, net 1,560 933 Investments 6,804 5,300 Other non-current assets
301 160 Total assets $19,339 $15,167 Liabilities and stockholder\'s equity: Current liabilities
Accounts payable $7,703 $6,002 Accrued and other liabilities 3,787 3,011 Total current
liabilities 11,490 9,013 Long-term debt 305 306 Other non-current liabilities 1,687 1,177 Total
liabilities 13,482 10,496 Stockholders’ equity Preferred stock and capital in excess of $0.02 par
value; shares issued and outstanding: none - - Common stock and capital in excess of $0.05 par
value; shares authorized: 6,000; shares issued: 2,112 and 1,852, respectively 7,804 7,001
Treasury stock, at cost; 184 and 124 shares, respectively (6,333) (4,401) Retained earnings 4,516
2,162 Other comprehensive loss (98) (49) Other (32) (42) Total stockholders’ equity 5,857 4,671
Total liabilities and stockholders’ equity $19,339 $15,167 PrintDone Orlando Medical
Corporation Consolidated Statements of Income Year ended December 31, (In Millions, Except
per Share Amounts) 2014 2013 2012 Net revenue $42,667 $35,613 $31,124 Cost of goods sold
35,164 29,270 25,478 Gross profit 7,503 6,343 5,646 Operating expenses: Selling, general, and
administrative 3,747 3,275 2,600 Research, development, and engineering 584 560 534 Special
charges - - 503 Total operating expenses 4,331 3,835 3,637 Operating income 3,172 2,508 2,009
Investment and other income (loss), net 195 194 (78) Income before income taxes 3,367 2,702
1,931 Income tax expense 1,100 912 475 Net income $2,267 $1,790 $1,456 Earnings per
common share: Basic $1.24 $0.94 $0.41
Solution
Answer : A. 84 times
=> Inventory t/o = COGS/average inventoryYear20142014Cost of Goods
Sold3516435164Average Inventory(435+404)/2419.5Inventory Turnover83.82.
The IP network is a virtual network and must rely on a link layer ne.pdf
The IP network is a virtual network and must rely on a link layer network to transport a frame.
What elements support getting an IP frame over an Ethernet LAN?
Solution
The data link layer (L2) is the second layer of the seven-layer OSI model of computer
networking. This layer is the protocol layer that transfers data between adjacent network nodes in
a wide area network (WAN) or between nodes on the same local area network (LAN) segment.
The data link layer, also sometimes termed as the link layer, is where many wired and wireless
local area networking (LAN) technologies primarily function.
Ethernet is the most widely used local area network (LAN) in metropolitan area networks
(MAN). Ethernet is a link layer protocol in the TCP/IP stack. It describes how networked devices
can format data for transmission to other network devices on the same network segment, and
how to put that data out on the network connection. It was commercially introduced in 1980 and
first standardized in 1983 as IEEE 802.3, and has since been refined to support higher bit rates
and longer link distances.
The data link layer is divided into two sublayers: logical link control (LLC) and media access
control (MAC).By separating LLC and MAC functions, interoperability of different network
technologies is made easier.
Data Link Layer Functions: Given below are the key tasks performed at the data link layer:
Logical Link Control (LLC): Logical link control refers to the functions required for the
establishment and control of logical links between local devices on a network. It provides
services to the network layer above it and hides the rest of the details of the data link layer to
allow different technologies to work seamlessly with the higher layers. Most local area
networking technologies use the IEEE 802.2 LLC protocol.
Media Access Control (MAC): This refers to the procedures used by devices to control access to
the network medium. Since many networks use a shared medium, it is necessary to have rules for
managing the medium to avoid conflicts. Ethernet uses the CSMA/CD method of media access
control, while Token Ring uses token passing.
Data Framing: The data link layer is responsible for the final encapsulation of higher-level
messages into frames that are sent over the network at the physical layer.
Addressing: The data link layer is the lowest layer in the OSI model that is concerned with
addressing: labeling information with a particular destination location. Each device on a network
has a unique hardware address or MAC address imbined with the hardware i.e present in network
interface card (NIC) to uniquely identify the device.
Error Detection and Handling: The data link layer handles errors that occur at the lower levels of
the network stack. For example, a cyclic redundancy check (CRC) field is often employed to
allow the station receiving data to detect if it was received correctly.
The hardware address is used for the MAC sublayer addressing. Ethernet uses the.
Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Ma.pdf
Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Machines 1. Based on
Table 1, according to the Heckscher-Ohlin Theorem, Canada exports should be goods that A)
intensively use labor input. B) intensively use capital input. C) use capital and labor in about
equal proportions D) use either labor or capital input, depending on the good. 2. Suppose that
Brazil is capital abundant and Chile is natural resource abundant. If timber is natural resource
intensive and computers are capital intensive, then according to the Stolper Samuelson Theorem,
the incomes of the owners of trade with Brazil begins A) capital B) labor C) natural resources D)
It is impossible to determine which will be favored. are likely to rise in Chile after
Solution
1) According to Heckscher Ohlin Theorem, A country will export the commodity in which it
uses its intensively endowed factor. Since Canada has more machine per worker compared to
US. Thus Canada is more capital intensive and will export the good that intensively uses capital
input. ans is B.
2) Since Brazil is capital abundant, and computer are capital intensive then Brazil will export
computer and Chile will export Timber. Thus income of natural resources are likely to rise in
Chile
Thus ans is C.
State if you agree or disagree with the statement made and whyLes.pdf
State if you agree or disagree with the statement made and why?
Lesson #1 - Discussion Forum – Linkage between Public Administration and Information
Technology
The article “Computing and Public Organizations” by Kraemer and King is an interesting look
back at the prevailing questions and concerns thirty years ago about the potential impact of the
use of computers in public organizations. The main point of the authors\' discussion is whether a
wide-spread organizational use of computers would affect the centralized or decentralized nature
of an organization. While the authors conclude that any move, either way, would involve more
factors than just the use of computers, they do seem to come down on the side of computer use
increasing the centralization of organizational decision making. When considering the impact of
computer use on the centralization of decision making in public organizations as it stands today,
thirty years after the article was written, it is still difficult to make a definitive statement on
whether organizations have become more or less centralized due to the use of information
technology. In some cases, data gathering both with and without IT is so decentralized
throughout an agency that a desired level of centralization is never achieved. A case in point is
the Veterans Administration, where data from field offices were either inefficiently or never sent
to regional offices that could make service-related decisions, an outcome of decentralization that
had very negative consequences for the veterans that are supposed to be the focus of the agency.
Other government agencies also suffer from having so many points of entry to a centralized
system that the public is uncertain how to access the information they need. An example would
be the military’s health insurance program, which has three or four portals that each accomplish
a different task within the overall program.
A strong and well-organized linkage between public agencies and information technology is
a deciding factor in how beneficial the use of technology is to public organizations.
Solution
I disagree with the fact the IT helps in the centralization of decision making as well as with the
point that it is difficult to determine whether IT helps in decentralization or centralization. I
believe IT has helped in the decentralization of decision making especially in public
administration. The banking sector is a very good example.
Centralization is where the distance- between where the problems arise and where the solution is
decided- is greater. In decentralization, this distance is smaller and decisions are solved closer to
the place where the problems arise.
Decision making depends upon many factors- like access to information, knowledge, experience,
skills, abilities, talent etc. With the inflow of technology into the hands of everyone, data
collection, gathering, assortment and analysis has become easier. Data is converted into
meaningful information and then into.
subject.....project management.Tools and Processes Based on the pr.pdf
subject.....project management.
Tools and Processes Based on the progressive case study (PCS) project, you will need to develop
a list of tools and processes that you would use as the contractor of record for opening the new
division office.
•Will you use subcontractors for different parts of the project?
•How will you maintain contact communication? •How will you track progress, milestones, and
setbacks?
•How will you monitor resources and budget?
•What reporting systems will you use?
You are not expected to address all of these areas, though this is not an exhaustive list. Choose
one or two components and describe the tools and processes you will use, then explain the
rationale for using these tools. You may choose to use a certain tool as that is where your
experience is, or it is more cost effective for example.
not less than 150 words
A project team requires individuals from the cross-functional area who can come and work
together on different aspects of the project. In this regard, a project based organizational structure
will be developed that will be flat in nature. The first parameter of the selection of the team
member is the ability to work in the respective areas of the project and or the past experience of
working on similar projects. The second parameter is the ability to manage a dual line of
command (line and project) while working on the project. The third parameter is the propensity
to work in a team and complement each other so that project work runs smoothly. The fourth
parameter is the ability to take decisions with the given authority so that project work is not
delayed due to the decision-making process. Besides, there should be guidelines related to the
scope, objectives, time-based milestones as well as closure of the project. It should be based on
the ethical principles.
Regarding the challenges, the first challenge is the procurement of resources to build capacity as
no any department or function wants to release the best employees. The second challenge is the
building of the clarity in command so that any ambiguity is eliminated. The third challenge is the
timely availability of resources and funds other than HR to start the project. The next challenge
is to build the cooperation based team culture in the project team
Solution
TRACKING PROGRESS, MILESTONES, AND SETBACKS:
A contractor needs to keep a close check on the progress of the project on which he is working.
Project management is a difficult and very important task and requires skills and expertise of the
contractor so as to make sure that the project is working as planned for.
Keeping a track on the progress of the project is must from time to time and making sure that the
milestones are achieved properly and if in case of any setbacks, recording why it happened.
Following are the list of the tools that can be used to track the progress, milestones and setback
of the project:
MONITORING RESOURCES AND BUDGET:
For successful completion of every project it is .
Read and discuss the following three articles 1. ACAs Perform.pdf
Read and discuss the following three articles:
1. ACAs Performance Based Healthcare
Standards ACAsPerformanceBasedHealthCareStandards.pdf
2. Road to Accreditation RoadToAccreditation.pdf
3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO
accreditation and quality of care for acute myocardial infarction.pdf
Have an open discussion about these articles. Share your thoughts
For example, here are some questions to answer and discuss:
Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is
the value of accreditation? Do quality concerns initiate changes in staff behavior? Should
accreditation be based on results?
Read and discuss the following three articles:
1. ACAs Performance Based Healthcare
Standards ACAsPerformanceBasedHealthCareStandards.pdf
2. Road to Accreditation RoadToAccreditation.pdf
3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO
accreditation and quality of care for acute myocardial infarction.pdf
Have an open discussion about these articles. Share your thoughts
For example, here are some questions to answer and discuss:
Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is
the value of accreditation? Do quality concerns initiate changes in staff behavior? Should
accreditation be based on results?
Solution
ACAs Performance Based Healthcare Standards-
This booklet is intended to assist anyone dealing with or affected by Health and Employment
issues. It is one of a series of booklets and handbooks designed to give impartial advice on
employment matters to employers, employees and their representatives. Legal information is
provided for guidance only and should not be regarded as an authoritative statement of the law,
which can only be made by reference to the particular circumstances which apply. It may,
therefore, be wise to seek legal advice.
Acas is committed to building better relationships in the workplace and offers training to suit
you. From a two-hour session on the key points of new legislation or employing people to
courses specially designed for people in your organisation. Click here to find out about training
sessions in your area. We also offer hands-on practical help and support to tackleissues in your
business with you. This might be through one of our wellknown problem-solving services or a
programme we have worked out together to put your business firmly on track for effective
employment
relations.
road to accrediation -
Accreditation is usually a voluntary program, sponsored by a non-governmental organization
(NGO), in which trained external peer reviewers evaluate a healthcare organization\'s
compliance and compare it with pre-established performance standard
Accreditation is a quality assurance process designed to ensure an educational program meets a
national standard. It serves to support and encourage program responsiveness to the ra pidly
changing environmental .
Question 1. Retrieve the first and last names and department numbe.pdf
Question 1. Retrieve the first and last names and department
number and name of all employees directly supervised by James Borg.
Show results in ascending alpha order (by last name and then first
name).
Column Headings: FNAME LNAME DNUMBER DNAME
Solution
select FNAME,LNAME,DNUMBER,DNAME
from TABLE_NAME where DNAME = \'James Borg\'
order by LNAME,FNAME;.
Polysaccharides, lipids, and proteins are similar in that they A) ar.pdf
Polysaccharides, lipids, and proteins are similar in that they A) are synthesized from monomers
by dehydration reactions. B) are synthesized as a result of peptide bond formation between
monomers. C) are decomposed into their subunits by dehydration reactions. D) are synthesized
from monomers by the process of hydrolysis. E) all contain nitrogen in their monomer building
blocks.
Solution
The answer is A. Are synthesized from monomers by dehydration reactions.
Peptide bond is present in polypeptides only hence the option b is wrong.
They are formed by dehydration decomposed by hrdration hence options c and d are wrong.
N molecule is not present in all polysac hence option E is wrong..
Please make the complete program, Distinguish between header files a.pdf
Please make the complete program, Distinguish between header files and class files. I give three
of the classes below, but I do not know if they are complete.I already posted this question before,
but the answer was incorreect. ask any questions if you need any help.
---------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------
-----------------------------------
Section 1: Project Objectives
By given a class hierarchy, focus on Object-Oriented-Design concept, design and write multiple
classes.
Be able to apply the concepts of inheritance, polymorphism, abstract classes, pure virtual
functions, function overriding, etc.
Be familiar with common functions in vector class in C++.
Section 2: Background Introduction
Video game programmers often have to maintain a collection of figures that are simultaneously
moving in various directions on the screen. In this project, we will design a solution to a
simplified version of the problem.
We will maintain a collection of geometric shapes and simultaneously animate those shapes. The
functions used to directly access the screen and manage the timer are peculiar to Microsoft
Windows, but the principles used are very general and are applicable to all operating systems.
Section 3: Project Description
We begin with a class that represents a shape that is able to move in any of eight different
directions, with each direction being specified by a pair of integer (X, Y) coordinates. Upward or
downward motion is indicated by a Y component of ±1. A value of 0 for an X or Y coordinate
indicates lack of motion in that direction. Thus, a value of (0,-1) for (X, Y) indicates motion
straight up, a value of (-1, 0) indicates motion to the left, and (1, 1) indicates motion that is
simultaneously downward and to the right.
In this project, you will need to design five classes, namely Shape, ComplexShape, SimpleShape,
Box and Tent. These classes form the inheritance hierarchy as showing in the following figure.
Shape
ComplexShape
SimpleShape
Box
Tent
Figure 1
1
Section 3.1: Classes Description
The Shape class has a move() function that is pure virtual. This is because a shape is moved by
erasing it at its current position and redrawing it at new position and it is not possible to know
how to draw a shape without knowing what type of shape it is.
The SimpleShape class represents objects that can be drawn at a given position in a specified
color. Accordingly, it has member variables for representing position and color and member
functions for setting and accessing those values. The SimpleShape class inherits from Shape,
notice that the SimpleShape class is still abstract because if provides no implementation for the
draw() method. The class does implement the move() method, though. The is because the move()
method works the same way for all subclasses of.
Part CYou are a graduate student in the lab of a famous fly geneti.pdf
Part C
You are a graduate student in the lab of a famous fly geneticist. You need to analyze a batch of
mutant flies that were recently created in her laboratory to identify the single gene that is most
likely mutated in each of the flies.
Drag the mutations on the left to the appropriate blanks on the right to complete the sentences.
Help
Reset
knirps (a gap gene)
hunchback (a gap gene)
hedgehog (a segment-polarity gene)
wingless (a segment-polarity gene)
Kruppel (a gap gene)
even-skipped (a pair-rule gene)
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in .
2. The wings are missing in mutant fly L, which is likely the result of a mutation in .
3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely
responsible for this mutation is .
4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely
candidate gene is .
5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation
is .
6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior
regions of each segment. The gene that most likely caused this mutation is .
Help
Reset
knirps (a gap gene)
hunchback (a gap gene)
hedgehog (a segment-polarity gene)
wingless (a segment-polarity gene)
Kruppel (a gap gene)
even-skipped (a pair-rule gene)
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in .
2. The wings are missing in mutant fly L, which is likely the result of a mutation in .
3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely
responsible for this mutation is .
4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely
candidate gene is .
5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation
is .
6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior
regions of each segment. The gene that most likely caused this mutation is .
Solution
Answer:
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in knirps (a gap gene). Explaination: The knirps gene is a type of gap gene (which is
responsible for the regulation of transcription for the product resulting to the development of
embryo). The mutation in the gap gene results to the gaps in the embryo, resulting to the missing
of many adjacent segments in phenotype. And the mutation in knirps gene resulting to the lack of
posterior segments in the fly phenotype.
2. The wings are missing in mutant fly L, which is likely the result of a mutation in wingless (a
segment-polarity gene).
Explaination: The wingless gene is type of Segment polarity gene, which is responsible for the
regulation of transcription for the product resulting to the development of anterior or posterior
regions of each segment. And the mutation in wingless g.
Part A Calculations for Solution Preparation d. Calculate the mass o.pdf
Part A: Calculations for
Solution
Preparation d. Calculate the mass of solid malonic acid (CH(COOH)2) necessary to prepare 50.0
mL of a 0.15 M (CH2(COOH)2) solution. b. Calculate the mass of solid manganese sulfate
monohydrate (MnSO4 H2O) necessary to prepare 50.0 mL of a 0.020 M manganese ion (Mn *)..
In java.Write a class called GenDoubleLinkedList which is a generi.pdf
In java.
Write a class called GenDoubleLinkedList which is a generic double linked list. This link list is
similar to the single linked list that was shown in class except that each node in addition to
having data and nextLink (which was originally called link) now has prevLink.
The class GenDoubleLinkedList needs to have the following:
Internal class ListNode which hasInstance Variables
data of type T
nextLink of type ListNode
prevLink of type ListNode
Constructors
Default
Parameterized
Instance Variables
head of type ListNode which always points to the beginning of the linked list
current of type ListNode which moves around pointing to one of the nodes
Constructor
A default constructor that initializes head to an empty ListNode and sets current to point at the
head.
Methods
goToNext – This moves the current node forward in the list by one node. It doesn’t move
forward if that node is null
goToPrev – This moves the current node backwards in the list by one node. It doesn’t move
backwards if that node is null.
getDataAtCurrent – returns the data at the current node as long as the current isn’t null
setDataAtCurrent – takes in a parameter and sets the data at the current node to that value as long
as current is not null
insertNodeAfterCurrent – creates a new node based on data that is passed in by a parameter and
puts that node after the current position
deleteCurrentNode – removes the current node from the list by resetting the links
showList – prints out the contents of the list line-by-line
inList – returns a true or false value based on whether or not the data passed in via a parameter is
in the list
Solution
package com.quad.generic;
public class GenDoubleLinkedList {
private ListNode head;
private ListNode tail;
private int size;
public int getSize() {
return size;
}
/**
*
*/
public GenDoubleLinkedList() {
this.head = null;
this.size = 0;
}
@SuppressWarnings(\"hiding\")
class ListNode{
private T data;
private ListNode nextLink;
private ListNode prevLink;
/**
*
*/
public ListNode() {
nextLink=null;
prevLink=null;
data=null;
}
/**
* @param data
* @param nextLink
* @param prevLink
*/
public ListNode(T data, ListNode nextLink,
ListNode prevLink) {
super();
this.data = data;
this.nextLink = nextLink;
this.prevLink = prevLink;
}
public T getData() {
return data;
}
public void setData(T data) {
this.data = data;
}
public ListNode getNextLink() {
return nextLink;
}
public void setNextLink(ListNode nextLink) {
this.nextLink = nextLink;
}
public ListNode getPrevLink() {
return prevLink;
}
public void setPrevLink(ListNode prevLink) {
this.prevLink = prevLink;
}
}
public void goToNext(){
ListNode temp = head;
while(temp != null){
temp = temp.getNextLink();
}
}
public void goToPrev() {
ListNode temporary = tail;
while(temporary != null){
temporary = temporary.getPrevLink();
}
}
public T getDataAtCurrent (int index) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
} else {
ListNode current = head;
for (int i = 0; i < index; i.
Let G and H be disjoint events in some sample space . The probabilit.pdf
Let G and H be disjoint events in some sample space . The probability that G occurs is P(G) and
H occurs is P(H).
a. Describe the event G U H.
b. What is P(G U H) in terms of P(G) and P(H)?
c. Describe the event G intersection H.
d. What is the probability of the event G intersection H?
Solution
G U H can be denoted by G +H as G and H are disjoint events P(G U H) = P(G) + P(H) G H ={
€ S : € G and € H} P(G H) = (null set).
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
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What is the role of intuition in decision-makingSolutionIntui.pdf
What is the role of intuition in decision-making?
Solution
Intuition is ability to acquire the knowledge, information and skills without any forceful activity;
intuition is something which is come from inner sense that provoked itself.
Decision- making process always influence by the intuition because decision making is a kind of
a practices which is not founded in every one but mostly founded in leaders ,managers and
businessmen.
Intuition always linked with emotions, intuition linked to knowledge, intuition always fast and
processing of information.
Role of intuition in decision-making as follows-.
What is the difference between a HTML element’s id attribute and nam.pdf
What is the difference between a HTML element’s id attribute and name attribute?
Solution
The form input element id has nothing to do with the data contained within the element. Input
element ids are for hooking the element with JavaScript and CSS. The name attribute is used in
the HTTP request sent by your browser to the server as a variable name associated with the data
contained in the value attribute.
In other words, By using id attribute we can validate and manipulate the value of an element at
client side. By using name attribut, we can validate and manipulate the value of an element at
server side.
In java script, by using id attribute we can get the value of input element.
document.getElementById(\"id\").value;
In server side, by using name attribute we can get the value of input element.
request.getParameter(\"name\");.
What happens when youre Running Scripts without a console in Linux.pdf
What happens when you\'re Running Scripts without a console in Linux/UNIX.
Solution
The script will run as a background process that is daemon process. These daemon processes
doesnot interact with the user. Console is used to send text output and receive user input and is
under the direct control of the user. However daemon process is run in the background by
forking a child process and the parent is exited immediately..
TrueFalse A virtual private network is a way to use the Internet to.pdf
True/False A virtual private network is a way to use the Internet to create a connection between
a remote user and a central location. IPsec encrypts the packet data but not the header
information. L2TP uses IPsec for its encryption. Kerberos is an authentication protocol used for
logging into Windows and uses a ticket granting system that sends an encrypted ticket to the
user\'s machine. A digital signature is used to guarantee who sent a message. This is referred to
as authenticity. Public key encryption is fast becoming the most widely used type of encryption
because there are no issues to deal with concerning distribution of keys. PGP involves private
key encryption. The most widely used symmetric key algorithm is Advanced Encryption
Standard. Cipher text is encrypted text. Binary numbers are made up of 0s and 1s.
Solution
Answer:
1) True
explanation:
=> A VPN, or Virtual Private Network, allows you to create a secure connection to another
network over the Internet.
2)true
explanation:
=>IPsec includes Encapsulating Security Payload (ESP) to encrypt the data and provide
confidentiality
3) TRUE
explanation:
=> L2TP does not encrypt the tunnel itself but often uses IPsec for encryption.
5) True
explanation:
=> Digital signatures are the public-key primitives of message authentication.
=> A valid digital signature gives a recipient reason to believe that the message was created by a
known sender.
6) True
explanation:
=> Public key encryption in turn provides a solution to key distribution and data transmission
issues. Used together, performance and key distribution are improved without any sacrifice in
security.
7) False.
8) True
explanation:
=> Encryption algorithms that use the same key for encrypting and for decrypting information
are called symmetric- key algorithms.
=>The symmetric key is also called a secret key because it is kept as a shared secret between the
sender and receiver of information.
9) true
explanation:
=>An encryption function takes cleartext and a key as input and returns ciphertext.
10) TRUE
explanation:
=>A binary code represents text or computer processor instructions using the binary number
system\'s two binary digits, 0 and 1..
The total physical units accounted for is the sum of the units co.pdf
The \"total physical units accounted for\" is the sum of the units completed and transferred out
during the period plus the True ? False
Solution
Total physical units accounted for = Units of completed and transferred out+Ending work in
process inventory Units
So above statement that the total physical units accounted for is the sum of the units completed
and transferred out during the period plus the units in beginning wip inventory is False
So above statement is False.
The problem can be found below. I would like someone to run this. I .pdf
The problem can be found below. I would like someone to run this. I don’t get it because there’s
an error. Please, fix this because I don’t know what is happening. What version of python should
I use?
Please, if you don’t know what is happening let someone else to do it. Bp Project untitled
PycharmProjects/untitled 3 import sys ServerPort 80 new server Socket socket (AF INET, S0CK
STREAM) Illi External Libraries 6 #Prepare a sever socket 7 #Fill in start server Socket.bind
server Port)) server Socket.listen (1) print the web server is up on port server Port 10 #Fill in end
11 while True: Establish the connection 13 print \"Ready to serve... 14 connectionSocket add J
server Socket. accept() 15 16 try: message connectionSocket.recv(1024) 17 print message
message split 01, message. split 1] 18 filename 3 message. split [1] print filename \'II filename
[1:] 20 open (filename (1:1) 21 output data f.read 22 print outputdata 23 Send one HTTP header
line into socket Fill in start 25 connection Socket.send(\' InHTTP/1.1 200 0KNnNn 26
connectionSocket.send(outputdata) 27 Fill in end 28 Send the content of the requested file to the
client 29 for i in range (0, len(outputdata)): 30 connection Socket.send(outputdata [i]) 31
connection Socket.close 32 33 except IOError w Send response message for file not found 34
Fill in start 35 connectionSocket.send NnHTTP/1.1 404 Not Found nAn 36
connectionSocket.send \'InHTTP/1.1 404 Not Found n\ \') 37 38 39
/Library/Frameworks/Python, framework/Versions/2.7/bin/python2.7 /Users. File User S untitled
new py\", line 33 except IOError: SyntaxError: invalid syntax Process finished with exit code 1
Solution
Update your code as below :
#!/bin/python
import sys
import socket
serverPort=80
serverSocket = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
serverSocket.bind((\'\',serverPort))
serverSocket.listen(1)
while True :
connectionSocket, addr = serverSocket.accept()
try :
message = connectionSocket.recv[1024]
filename = message.split()[1]
f = open(filename[1:])
outputdata = f.read()
connectionSocket.send(\'\ HTTP/1.1 200 OK\ \ \')
connectionSocket.send(outputdata)
for i in range(0,len(outputdata)):
connectionSocket.send(outputdata[i])
connectionSocket.close()
except IOError : #indentation is not correct here
connectionSocket.send(\'\ HTTP/1.1 404 Not Found\ \ \')
serverSocket.close();.
The probability that a teacher will give an unannounced test duri.pdf
The probability that a teacher will give an unannounced test during any class is 1/5. If a student
is absent twice ,then probability that misses atleast one test is ?
Solution
Given the probability of teacher will give an unannounced test during any class is 1/5.
The probability of not missing test 1-1/5=4/5
probability of missing test in only one day 2(4/25)=8/25
probability of missing test on both the day (1/5)*(1/5)=1/25
probability that misses atleast one test is (8/25)+(1/25)=9/25
probability that misses atleast one test is 9/25.
The Orlando MedicalOrlando Medical Corporation financial statements.pdf
The Orlando MedicalOrlando Medical Corporation financial statements? follow: LOADING...
?(Click the icon to view the consolidated balance? sheets.) LOADING... ?(Click the icon to view
the consolidated income? statements.) Orlando MedicalOrlando Medical?\'s inventory turnover
during fiscal year 20142014 was A. 8484 times. B. ?$35 comma 16435,164. C. very slow. D.
138.0138.0 times. Orlando Medical Corporation Consolidated Statements of Financial Position
December 31, (In Millions) 2014 2013 Assets: Current assets Cash and cash equivalents $4,344
$4,237 Accounts and notes receivable 3,402 2,409 Short-term investments 855 504 Inventories,
at cost 435 404 Prepaid expense and other current assets 1,638 1,220 Total current assets 10,674
8,774 Property and equipment, net 1,560 933 Investments 6,804 5,300 Other non-current assets
301 160 Total assets $19,339 $15,167 Liabilities and stockholder\'s equity: Current liabilities
Accounts payable $7,703 $6,002 Accrued and other liabilities 3,787 3,011 Total current
liabilities 11,490 9,013 Long-term debt 305 306 Other non-current liabilities 1,687 1,177 Total
liabilities 13,482 10,496 Stockholders’ equity Preferred stock and capital in excess of $0.02 par
value; shares issued and outstanding: none - - Common stock and capital in excess of $0.05 par
value; shares authorized: 6,000; shares issued: 2,112 and 1,852, respectively 7,804 7,001
Treasury stock, at cost; 184 and 124 shares, respectively (6,333) (4,401) Retained earnings 4,516
2,162 Other comprehensive loss (98) (49) Other (32) (42) Total stockholders’ equity 5,857 4,671
Total liabilities and stockholders’ equity $19,339 $15,167 PrintDone Orlando Medical
Corporation Consolidated Statements of Income Year ended December 31, (In Millions, Except
per Share Amounts) 2014 2013 2012 Net revenue $42,667 $35,613 $31,124 Cost of goods sold
35,164 29,270 25,478 Gross profit 7,503 6,343 5,646 Operating expenses: Selling, general, and
administrative 3,747 3,275 2,600 Research, development, and engineering 584 560 534 Special
charges - - 503 Total operating expenses 4,331 3,835 3,637 Operating income 3,172 2,508 2,009
Investment and other income (loss), net 195 194 (78) Income before income taxes 3,367 2,702
1,931 Income tax expense 1,100 912 475 Net income $2,267 $1,790 $1,456 Earnings per
common share: Basic $1.24 $0.94 $0.41
Solution
Answer : A. 84 times
=> Inventory t/o = COGS/average inventoryYear20142014Cost of Goods
Sold3516435164Average Inventory(435+404)/2419.5Inventory Turnover83.82.
The IP network is a virtual network and must rely on a link layer ne.pdf
The IP network is a virtual network and must rely on a link layer network to transport a frame.
What elements support getting an IP frame over an Ethernet LAN?
Solution
The data link layer (L2) is the second layer of the seven-layer OSI model of computer
networking. This layer is the protocol layer that transfers data between adjacent network nodes in
a wide area network (WAN) or between nodes on the same local area network (LAN) segment.
The data link layer, also sometimes termed as the link layer, is where many wired and wireless
local area networking (LAN) technologies primarily function.
Ethernet is the most widely used local area network (LAN) in metropolitan area networks
(MAN). Ethernet is a link layer protocol in the TCP/IP stack. It describes how networked devices
can format data for transmission to other network devices on the same network segment, and
how to put that data out on the network connection. It was commercially introduced in 1980 and
first standardized in 1983 as IEEE 802.3, and has since been refined to support higher bit rates
and longer link distances.
The data link layer is divided into two sublayers: logical link control (LLC) and media access
control (MAC).By separating LLC and MAC functions, interoperability of different network
technologies is made easier.
Data Link Layer Functions: Given below are the key tasks performed at the data link layer:
Logical Link Control (LLC): Logical link control refers to the functions required for the
establishment and control of logical links between local devices on a network. It provides
services to the network layer above it and hides the rest of the details of the data link layer to
allow different technologies to work seamlessly with the higher layers. Most local area
networking technologies use the IEEE 802.2 LLC protocol.
Media Access Control (MAC): This refers to the procedures used by devices to control access to
the network medium. Since many networks use a shared medium, it is necessary to have rules for
managing the medium to avoid conflicts. Ethernet uses the CSMA/CD method of media access
control, while Token Ring uses token passing.
Data Framing: The data link layer is responsible for the final encapsulation of higher-level
messages into frames that are sent over the network at the physical layer.
Addressing: The data link layer is the lowest layer in the OSI model that is concerned with
addressing: labeling information with a particular destination location. Each device on a network
has a unique hardware address or MAC address imbined with the hardware i.e present in network
interface card (NIC) to uniquely identify the device.
Error Detection and Handling: The data link layer handles errors that occur at the lower levels of
the network stack. For example, a cyclic redundancy check (CRC) field is often employed to
allow the station receiving data to detect if it was received correctly.
The hardware address is used for the MAC sublayer addressing. Ethernet uses the.
Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Ma.pdf
Table 1 Endowment of Labor and Capital US 100 20 Canada 10 Workers Machines 1. Based on
Table 1, according to the Heckscher-Ohlin Theorem, Canada exports should be goods that A)
intensively use labor input. B) intensively use capital input. C) use capital and labor in about
equal proportions D) use either labor or capital input, depending on the good. 2. Suppose that
Brazil is capital abundant and Chile is natural resource abundant. If timber is natural resource
intensive and computers are capital intensive, then according to the Stolper Samuelson Theorem,
the incomes of the owners of trade with Brazil begins A) capital B) labor C) natural resources D)
It is impossible to determine which will be favored. are likely to rise in Chile after
Solution
1) According to Heckscher Ohlin Theorem, A country will export the commodity in which it
uses its intensively endowed factor. Since Canada has more machine per worker compared to
US. Thus Canada is more capital intensive and will export the good that intensively uses capital
input. ans is B.
2) Since Brazil is capital abundant, and computer are capital intensive then Brazil will export
computer and Chile will export Timber. Thus income of natural resources are likely to rise in
Chile
Thus ans is C.
State if you agree or disagree with the statement made and whyLes.pdf
State if you agree or disagree with the statement made and why?
Lesson #1 - Discussion Forum – Linkage between Public Administration and Information
Technology
The article “Computing and Public Organizations” by Kraemer and King is an interesting look
back at the prevailing questions and concerns thirty years ago about the potential impact of the
use of computers in public organizations. The main point of the authors\' discussion is whether a
wide-spread organizational use of computers would affect the centralized or decentralized nature
of an organization. While the authors conclude that any move, either way, would involve more
factors than just the use of computers, they do seem to come down on the side of computer use
increasing the centralization of organizational decision making. When considering the impact of
computer use on the centralization of decision making in public organizations as it stands today,
thirty years after the article was written, it is still difficult to make a definitive statement on
whether organizations have become more or less centralized due to the use of information
technology. In some cases, data gathering both with and without IT is so decentralized
throughout an agency that a desired level of centralization is never achieved. A case in point is
the Veterans Administration, where data from field offices were either inefficiently or never sent
to regional offices that could make service-related decisions, an outcome of decentralization that
had very negative consequences for the veterans that are supposed to be the focus of the agency.
Other government agencies also suffer from having so many points of entry to a centralized
system that the public is uncertain how to access the information they need. An example would
be the military’s health insurance program, which has three or four portals that each accomplish
a different task within the overall program.
A strong and well-organized linkage between public agencies and information technology is
a deciding factor in how beneficial the use of technology is to public organizations.
Solution
I disagree with the fact the IT helps in the centralization of decision making as well as with the
point that it is difficult to determine whether IT helps in decentralization or centralization. I
believe IT has helped in the decentralization of decision making especially in public
administration. The banking sector is a very good example.
Centralization is where the distance- between where the problems arise and where the solution is
decided- is greater. In decentralization, this distance is smaller and decisions are solved closer to
the place where the problems arise.
Decision making depends upon many factors- like access to information, knowledge, experience,
skills, abilities, talent etc. With the inflow of technology into the hands of everyone, data
collection, gathering, assortment and analysis has become easier. Data is converted into
meaningful information and then into.
subject.....project management.Tools and Processes Based on the pr.pdf
subject.....project management.
Tools and Processes Based on the progressive case study (PCS) project, you will need to develop
a list of tools and processes that you would use as the contractor of record for opening the new
division office.
•Will you use subcontractors for different parts of the project?
•How will you maintain contact communication? •How will you track progress, milestones, and
setbacks?
•How will you monitor resources and budget?
•What reporting systems will you use?
You are not expected to address all of these areas, though this is not an exhaustive list. Choose
one or two components and describe the tools and processes you will use, then explain the
rationale for using these tools. You may choose to use a certain tool as that is where your
experience is, or it is more cost effective for example.
not less than 150 words
A project team requires individuals from the cross-functional area who can come and work
together on different aspects of the project. In this regard, a project based organizational structure
will be developed that will be flat in nature. The first parameter of the selection of the team
member is the ability to work in the respective areas of the project and or the past experience of
working on similar projects. The second parameter is the ability to manage a dual line of
command (line and project) while working on the project. The third parameter is the propensity
to work in a team and complement each other so that project work runs smoothly. The fourth
parameter is the ability to take decisions with the given authority so that project work is not
delayed due to the decision-making process. Besides, there should be guidelines related to the
scope, objectives, time-based milestones as well as closure of the project. It should be based on
the ethical principles.
Regarding the challenges, the first challenge is the procurement of resources to build capacity as
no any department or function wants to release the best employees. The second challenge is the
building of the clarity in command so that any ambiguity is eliminated. The third challenge is the
timely availability of resources and funds other than HR to start the project. The next challenge
is to build the cooperation based team culture in the project team
Solution
TRACKING PROGRESS, MILESTONES, AND SETBACKS:
A contractor needs to keep a close check on the progress of the project on which he is working.
Project management is a difficult and very important task and requires skills and expertise of the
contractor so as to make sure that the project is working as planned for.
Keeping a track on the progress of the project is must from time to time and making sure that the
milestones are achieved properly and if in case of any setbacks, recording why it happened.
Following are the list of the tools that can be used to track the progress, milestones and setback
of the project:
MONITORING RESOURCES AND BUDGET:
For successful completion of every project it is .
Read and discuss the following three articles 1. ACAs Perform.pdf
Read and discuss the following three articles:
1. ACAs Performance Based Healthcare
Standards ACAsPerformanceBasedHealthCareStandards.pdf
2. Road to Accreditation RoadToAccreditation.pdf
3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO
accreditation and quality of care for acute myocardial infarction.pdf
Have an open discussion about these articles. Share your thoughts
For example, here are some questions to answer and discuss:
Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is
the value of accreditation? Do quality concerns initiate changes in staff behavior? Should
accreditation be based on results?
Read and discuss the following three articles:
1. ACAs Performance Based Healthcare
Standards ACAsPerformanceBasedHealthCareStandards.pdf
2. Road to Accreditation RoadToAccreditation.pdf
3. JCAHO Accreditation and Quality of Care for Acute Myocardial Infarction JCAHO
accreditation and quality of care for acute myocardial infarction.pdf
Have an open discussion about these articles. Share your thoughts
For example, here are some questions to answer and discuss:
Does accreditation impact quality? Are there less errors in hospitals that are accredited? What is
the value of accreditation? Do quality concerns initiate changes in staff behavior? Should
accreditation be based on results?
Solution
ACAs Performance Based Healthcare Standards-
This booklet is intended to assist anyone dealing with or affected by Health and Employment
issues. It is one of a series of booklets and handbooks designed to give impartial advice on
employment matters to employers, employees and their representatives. Legal information is
provided for guidance only and should not be regarded as an authoritative statement of the law,
which can only be made by reference to the particular circumstances which apply. It may,
therefore, be wise to seek legal advice.
Acas is committed to building better relationships in the workplace and offers training to suit
you. From a two-hour session on the key points of new legislation or employing people to
courses specially designed for people in your organisation. Click here to find out about training
sessions in your area. We also offer hands-on practical help and support to tackleissues in your
business with you. This might be through one of our wellknown problem-solving services or a
programme we have worked out together to put your business firmly on track for effective
employment
relations.
road to accrediation -
Accreditation is usually a voluntary program, sponsored by a non-governmental organization
(NGO), in which trained external peer reviewers evaluate a healthcare organization\'s
compliance and compare it with pre-established performance standard
Accreditation is a quality assurance process designed to ensure an educational program meets a
national standard. It serves to support and encourage program responsiveness to the ra pidly
changing environmental .
Question 1. Retrieve the first and last names and department numbe.pdf
Question 1. Retrieve the first and last names and department
number and name of all employees directly supervised by James Borg.
Show results in ascending alpha order (by last name and then first
name).
Column Headings: FNAME LNAME DNUMBER DNAME
Solution
select FNAME,LNAME,DNUMBER,DNAME
from TABLE_NAME where DNAME = \'James Borg\'
order by LNAME,FNAME;.
Polysaccharides, lipids, and proteins are similar in that they A) ar.pdf
Polysaccharides, lipids, and proteins are similar in that they A) are synthesized from monomers
by dehydration reactions. B) are synthesized as a result of peptide bond formation between
monomers. C) are decomposed into their subunits by dehydration reactions. D) are synthesized
from monomers by the process of hydrolysis. E) all contain nitrogen in their monomer building
blocks.
Solution
The answer is A. Are synthesized from monomers by dehydration reactions.
Peptide bond is present in polypeptides only hence the option b is wrong.
They are formed by dehydration decomposed by hrdration hence options c and d are wrong.
N molecule is not present in all polysac hence option E is wrong..
Please make the complete program, Distinguish between header files a.pdf
Please make the complete program, Distinguish between header files and class files. I give three
of the classes below, but I do not know if they are complete.I already posted this question before,
but the answer was incorreect. ask any questions if you need any help.
---------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------
-----------------------------------
Section 1: Project Objectives
By given a class hierarchy, focus on Object-Oriented-Design concept, design and write multiple
classes.
Be able to apply the concepts of inheritance, polymorphism, abstract classes, pure virtual
functions, function overriding, etc.
Be familiar with common functions in vector class in C++.
Section 2: Background Introduction
Video game programmers often have to maintain a collection of figures that are simultaneously
moving in various directions on the screen. In this project, we will design a solution to a
simplified version of the problem.
We will maintain a collection of geometric shapes and simultaneously animate those shapes. The
functions used to directly access the screen and manage the timer are peculiar to Microsoft
Windows, but the principles used are very general and are applicable to all operating systems.
Section 3: Project Description
We begin with a class that represents a shape that is able to move in any of eight different
directions, with each direction being specified by a pair of integer (X, Y) coordinates. Upward or
downward motion is indicated by a Y component of ±1. A value of 0 for an X or Y coordinate
indicates lack of motion in that direction. Thus, a value of (0,-1) for (X, Y) indicates motion
straight up, a value of (-1, 0) indicates motion to the left, and (1, 1) indicates motion that is
simultaneously downward and to the right.
In this project, you will need to design five classes, namely Shape, ComplexShape, SimpleShape,
Box and Tent. These classes form the inheritance hierarchy as showing in the following figure.
Shape
ComplexShape
SimpleShape
Box
Tent
Figure 1
1
Section 3.1: Classes Description
The Shape class has a move() function that is pure virtual. This is because a shape is moved by
erasing it at its current position and redrawing it at new position and it is not possible to know
how to draw a shape without knowing what type of shape it is.
The SimpleShape class represents objects that can be drawn at a given position in a specified
color. Accordingly, it has member variables for representing position and color and member
functions for setting and accessing those values. The SimpleShape class inherits from Shape,
notice that the SimpleShape class is still abstract because if provides no implementation for the
draw() method. The class does implement the move() method, though. The is because the move()
method works the same way for all subclasses of.
Part CYou are a graduate student in the lab of a famous fly geneti.pdf
Part C
You are a graduate student in the lab of a famous fly geneticist. You need to analyze a batch of
mutant flies that were recently created in her laboratory to identify the single gene that is most
likely mutated in each of the flies.
Drag the mutations on the left to the appropriate blanks on the right to complete the sentences.
Help
Reset
knirps (a gap gene)
hunchback (a gap gene)
hedgehog (a segment-polarity gene)
wingless (a segment-polarity gene)
Kruppel (a gap gene)
even-skipped (a pair-rule gene)
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in .
2. The wings are missing in mutant fly L, which is likely the result of a mutation in .
3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely
responsible for this mutation is .
4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely
candidate gene is .
5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation
is .
6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior
regions of each segment. The gene that most likely caused this mutation is .
Help
Reset
knirps (a gap gene)
hunchback (a gap gene)
hedgehog (a segment-polarity gene)
wingless (a segment-polarity gene)
Kruppel (a gap gene)
even-skipped (a pair-rule gene)
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in .
2. The wings are missing in mutant fly L, which is likely the result of a mutation in .
3. Thoracic and abdominal structures are missing in mutant fly Q. The gene most likely
responsible for this mutation is .
4. Segment-sized sections of every other segment are missing in mutant fly Z. One likely
candidate gene is .
5. Mutant fly X has lost the head and thorax. The gene most likely responsible for this mutation
is .
6. The mutation observed in mutant fly O resulted in defects within the anterior or posterior
regions of each segment. The gene that most likely caused this mutation is .
Solution
Answer:
1. Mutant fly F has predominantly lost abdominal structures, which is likely the result of a
mutation in knirps (a gap gene). Explaination: The knirps gene is a type of gap gene (which is
responsible for the regulation of transcription for the product resulting to the development of
embryo). The mutation in the gap gene results to the gaps in the embryo, resulting to the missing
of many adjacent segments in phenotype. And the mutation in knirps gene resulting to the lack of
posterior segments in the fly phenotype.
2. The wings are missing in mutant fly L, which is likely the result of a mutation in wingless (a
segment-polarity gene).
Explaination: The wingless gene is type of Segment polarity gene, which is responsible for the
regulation of transcription for the product resulting to the development of anterior or posterior
regions of each segment. And the mutation in wingless g.
Part A Calculations for Solution Preparation d. Calculate the mass o.pdf
Part A: Calculations for
Solution
Preparation d. Calculate the mass of solid malonic acid (CH(COOH)2) necessary to prepare 50.0
mL of a 0.15 M (CH2(COOH)2) solution. b. Calculate the mass of solid manganese sulfate
monohydrate (MnSO4 H2O) necessary to prepare 50.0 mL of a 0.020 M manganese ion (Mn *)..
In java.Write a class called GenDoubleLinkedList which is a generi.pdf
In java.
Write a class called GenDoubleLinkedList which is a generic double linked list. This link list is
similar to the single linked list that was shown in class except that each node in addition to
having data and nextLink (which was originally called link) now has prevLink.
The class GenDoubleLinkedList needs to have the following:
Internal class ListNode which hasInstance Variables
data of type T
nextLink of type ListNode
prevLink of type ListNode
Constructors
Default
Parameterized
Instance Variables
head of type ListNode which always points to the beginning of the linked list
current of type ListNode which moves around pointing to one of the nodes
Constructor
A default constructor that initializes head to an empty ListNode and sets current to point at the
head.
Methods
goToNext – This moves the current node forward in the list by one node. It doesn’t move
forward if that node is null
goToPrev – This moves the current node backwards in the list by one node. It doesn’t move
backwards if that node is null.
getDataAtCurrent – returns the data at the current node as long as the current isn’t null
setDataAtCurrent – takes in a parameter and sets the data at the current node to that value as long
as current is not null
insertNodeAfterCurrent – creates a new node based on data that is passed in by a parameter and
puts that node after the current position
deleteCurrentNode – removes the current node from the list by resetting the links
showList – prints out the contents of the list line-by-line
inList – returns a true or false value based on whether or not the data passed in via a parameter is
in the list
Solution
package com.quad.generic;
public class GenDoubleLinkedList {
private ListNode head;
private ListNode tail;
private int size;
public int getSize() {
return size;
}
/**
*
*/
public GenDoubleLinkedList() {
this.head = null;
this.size = 0;
}
@SuppressWarnings(\"hiding\")
class ListNode{
private T data;
private ListNode nextLink;
private ListNode prevLink;
/**
*
*/
public ListNode() {
nextLink=null;
prevLink=null;
data=null;
}
/**
* @param data
* @param nextLink
* @param prevLink
*/
public ListNode(T data, ListNode nextLink,
ListNode prevLink) {
super();
this.data = data;
this.nextLink = nextLink;
this.prevLink = prevLink;
}
public T getData() {
return data;
}
public void setData(T data) {
this.data = data;
}
public ListNode getNextLink() {
return nextLink;
}
public void setNextLink(ListNode nextLink) {
this.nextLink = nextLink;
}
public ListNode getPrevLink() {
return prevLink;
}
public void setPrevLink(ListNode prevLink) {
this.prevLink = prevLink;
}
}
public void goToNext(){
ListNode temp = head;
while(temp != null){
temp = temp.getNextLink();
}
}
public void goToPrev() {
ListNode temporary = tail;
while(temporary != null){
temporary = temporary.getPrevLink();
}
}
public T getDataAtCurrent (int index) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
} else {
ListNode current = head;
for (int i = 0; i < index; i.
Let G and H be disjoint events in some sample space . The probabilit.pdf
Let G and H be disjoint events in some sample space . The probability that G occurs is P(G) and
H occurs is P(H).
a. Describe the event G U H.
b. What is P(G U H) in terms of P(G) and P(H)?
c. Describe the event G intersection H.
d. What is the probability of the event G intersection H?
Solution
G U H can be denoted by G +H as G and H are disjoint events P(G U H) = P(G) + P(H) G H ={
€ S : € G and € H} P(G H) = (null set).
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