man ABO blood groups are controlled by a single gene with three all.pdf
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MACROECONOMICS IN THE GLOBAL ENVIRONMENT ECONOMIC GROWTH Write a 400 word note on the prospects for economic growth in New Zealand for the next twenty years. Be sure to highlight the main policy choices that the government faces in encouraging higher rates of growth, including any trade-offs. Solution on the prospects for economic growth in New Zealand for the next twenty years. be sure to highlight the main policy choices that the government faces in encouraging higher rates of growth, including any trade-offs. Asset price bubble is a serious threat as it would push inflation to high and over evaluate the assets which would burst later because of unable to generate the income as expected. So some of the measures it needs to adopt are: The monetary policies need to control the inflation The government investments to be designed so that it would create more jobs Create conducive environment for exports.
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easylearningwithned.blogspot.com-BLOOD GROUP.pdf
This document discusses blood groups and blood typing. It explains that there are four main blood groups - A, B, AB, and O - which are determined by the presence or absence of antigens on red blood cells and antibodies in plasma. Group O blood is the universal donor as it lacks antigens, while Group AB blood is the universal recipient as it possesses both A and B antigens but lacks antibodies. The document also covers Rh factor blood typing and discusses issues that can arise during pregnancy when the mother's and baby's Rh factors are incompatible. It describes how blood grouping is performed using the agglutination of red blood cells and specific antisera to identify antigens.
1 BIOL 102 Lab 9 Simulated ABO and Rh Blood Typing.docx
1
BIOL 102: Lab 9
Simulated ABO and Rh Blood Typing
Objectives:
After completing this laboratory assignment, students will be able to:
• explain the biology of blood typing systems ABO and Rh
• explain the genetics of blood types
• determine the blood types of several patients
Introduction:
Before Karl Landsteiner discovered the ABO human blood groups in 1901, it was thought that all blood was the
same. This misunderstanding led to fatal blood transfusions. Later, in 1940, Landsteiner was part of a team
who discovered another blood group, the Rh blood group system. There are many blood group systems known
today, but the ABO and the Rh blood groups are the most important ones used for blood transfusions. The
designation Rh is derived from the Rhesus monkey in which the existence of the Rh blood group was
discovered.
Although all blood is made of the same basic elements, not all blood is alike. In fact, there are eight different
common blood types, which are determined by the presence or absence of certain antigens – substances that
can trigger an immune response if they are foreign to the body – on the surface of the red blood cells (RBCs
also known as erythrocytes).
ABO System:
The antigens on RBCs are agglutinating antigens or agglutinogens. They have been designated as A and B.
Antibodies against antigens A and B begin to build up in the blood plasma shortly after birth. A person
normally produces antibodies (agglutinins) against those antigens that are not present on his/her erythrocytes
but does not produce antibodies against those antigens that are present on his/her erythrocytes.
• A person who is blood type A will have A antigens on the surface of her/his RBCs and will have
antibodies against B antigens (anti-B antibodies). See picture below.
• A person with blood type B will have B antigens on the surface of her/his RBCs and will have antibodies
against antigen A (anti-A antibodies).
• A person with blood type O will have neither A nor B antigens on the surface of her/his RBCs and has
BOTH anti-A and anti-B antibodies.
• A person with blood type AB will have both A and B antigens on the surface of her/his RBCs and has
neither anti-A nor anti-B antibodies.
The individual’s blood type is based on the antigens (not the antibodies) he/she has. The four blood groups
are known as types A, B, AB, and O. Blood type O, characterized by an absence of A and B agglutinogens, is
the most common in the United States (45% of the population). Type A is the next in frequency, found in 39%
of the population. The incidences of types B and AB are 12% and 4%, respectively.
2
Table 1: The ABO System
Blood
Type
Antigens on
RBCs
Antibodies
in the Blood
Can GIVE Blood
to Groups:
Can RECEIVE
Blood from Groups:
A A Anti-B A, AB O, A
B B Anti-A B, AB O, B
AB A and B
Neither anti-A
nor anti-B
AB O, A, B, AB
O
Neither A nor
B
Both anti-A.
Blood grouping and crosss matching
This document discusses blood grouping and Rh typing. It begins by explaining that blood grouping is based on the antigens present on red blood cells, with ABO and Rh being clinically most important. It then covers the history and discovery of blood grouping, the various blood group systems, and Landsteiner's law regarding antigens and antibodies. The rest of the document discusses methods for blood grouping and Rh typing, including slide and tube methods. It covers interpretation of results, universal donor/recipient groups, and rare Bombay blood group. The importance and techniques of cross-matching for blood transfusions are also summarized.
Abo blood groups
The ABO blood group system was discovered in 1900 by Karl Landsteiner, who identified three main blood types: A, B, and O. The fourth type, AB, was discovered later. People have antigens on their red blood cells and corresponding antibodies in their plasma. Blood type is inherited and determines compatibility for transfusions. Type O negative blood can be donated to all recipients, while type AB positive can receive from all donors.
1.blood group
types of blood group bettre Rh factor cross match Rh incompatibility mis matc hazaeds nlood transfusion
Blood Groups & Rh antigens.pptx
This document provides information about blood groups and blood typing. It discusses the key blood group systems, including ABO and Rh, which are most important for blood transfusions. The ABO system divides blood into four main types - A, B, AB, and O - based on the presence or absence of A and B antigens on red blood cells and antibodies in the plasma. The Rh system indicates if the blood is positive or negative for the Rh antigen. Proper matching of blood types is essential to prevent dangerous transfusion reactions from the immune system attacking foreign blood antigens.
Recommended
easylearningwithned.blogspot.com-BLOOD GROUP.pdf
This document discusses blood groups and blood typing. It explains that there are four main blood groups - A, B, AB, and O - which are determined by the presence or absence of antigens on red blood cells and antibodies in plasma. Group O blood is the universal donor as it lacks antigens, while Group AB blood is the universal recipient as it possesses both A and B antigens but lacks antibodies. The document also covers Rh factor blood typing and discusses issues that can arise during pregnancy when the mother's and baby's Rh factors are incompatible. It describes how blood grouping is performed using the agglutination of red blood cells and specific antisera to identify antigens.
1 BIOL 102 Lab 9 Simulated ABO and Rh Blood Typing.docx
1
BIOL 102: Lab 9
Simulated ABO and Rh Blood Typing
Objectives:
After completing this laboratory assignment, students will be able to:
• explain the biology of blood typing systems ABO and Rh
• explain the genetics of blood types
• determine the blood types of several patients
Introduction:
Before Karl Landsteiner discovered the ABO human blood groups in 1901, it was thought that all blood was the
same. This misunderstanding led to fatal blood transfusions. Later, in 1940, Landsteiner was part of a team
who discovered another blood group, the Rh blood group system. There are many blood group systems known
today, but the ABO and the Rh blood groups are the most important ones used for blood transfusions. The
designation Rh is derived from the Rhesus monkey in which the existence of the Rh blood group was
discovered.
Although all blood is made of the same basic elements, not all blood is alike. In fact, there are eight different
common blood types, which are determined by the presence or absence of certain antigens – substances that
can trigger an immune response if they are foreign to the body – on the surface of the red blood cells (RBCs
also known as erythrocytes).
ABO System:
The antigens on RBCs are agglutinating antigens or agglutinogens. They have been designated as A and B.
Antibodies against antigens A and B begin to build up in the blood plasma shortly after birth. A person
normally produces antibodies (agglutinins) against those antigens that are not present on his/her erythrocytes
but does not produce antibodies against those antigens that are present on his/her erythrocytes.
• A person who is blood type A will have A antigens on the surface of her/his RBCs and will have
antibodies against B antigens (anti-B antibodies). See picture below.
• A person with blood type B will have B antigens on the surface of her/his RBCs and will have antibodies
against antigen A (anti-A antibodies).
• A person with blood type O will have neither A nor B antigens on the surface of her/his RBCs and has
BOTH anti-A and anti-B antibodies.
• A person with blood type AB will have both A and B antigens on the surface of her/his RBCs and has
neither anti-A nor anti-B antibodies.
The individual’s blood type is based on the antigens (not the antibodies) he/she has. The four blood groups
are known as types A, B, AB, and O. Blood type O, characterized by an absence of A and B agglutinogens, is
the most common in the United States (45% of the population). Type A is the next in frequency, found in 39%
of the population. The incidences of types B and AB are 12% and 4%, respectively.
2
Table 1: The ABO System
Blood
Type
Antigens on
RBCs
Antibodies
in the Blood
Can GIVE Blood
to Groups:
Can RECEIVE
Blood from Groups:
A A Anti-B A, AB O, A
B B Anti-A B, AB O, B
AB A and B
Neither anti-A
nor anti-B
AB O, A, B, AB
O
Neither A nor
B
Both anti-A.
Blood grouping and crosss matching
This document discusses blood grouping and Rh typing. It begins by explaining that blood grouping is based on the antigens present on red blood cells, with ABO and Rh being clinically most important. It then covers the history and discovery of blood grouping, the various blood group systems, and Landsteiner's law regarding antigens and antibodies. The rest of the document discusses methods for blood grouping and Rh typing, including slide and tube methods. It covers interpretation of results, universal donor/recipient groups, and rare Bombay blood group. The importance and techniques of cross-matching for blood transfusions are also summarized.
Abo blood groups
The ABO blood group system was discovered in 1900 by Karl Landsteiner, who identified three main blood types: A, B, and O. The fourth type, AB, was discovered later. People have antigens on their red blood cells and corresponding antibodies in their plasma. Blood type is inherited and determines compatibility for transfusions. Type O negative blood can be donated to all recipients, while type AB positive can receive from all donors.
1.blood group
types of blood group bettre Rh factor cross match Rh incompatibility mis matc hazaeds nlood transfusion
Blood Groups & Rh antigens.pptx
This document provides information about blood groups and blood typing. It discusses the key blood group systems, including ABO and Rh, which are most important for blood transfusions. The ABO system divides blood into four main types - A, B, AB, and O - based on the presence or absence of A and B antigens on red blood cells and antibodies in the plasma. The Rh system indicates if the blood is positive or negative for the Rh antigen. Proper matching of blood types is essential to prevent dangerous transfusion reactions from the immune system attacking foreign blood antigens.
Find the domain of the function. (Enter your answer using interval no.pdf
Find an equation of the line through (-6, 10), parallel to the line with equation 3x - 7y = 14.
Write the new equation in point-slope form. Convert the equation of the new line found in
problem #14 to standard form, Ax + By = C, where A, B, and C are integers.
Solution
14) Given line is 3x-7y =14
7y = 3x-14
y = (3/7)x - 2
It is of the form, y = mx + c
Hence, slope = 3/7
We know that two parallel lines have same slope.
Hence,
The equation parallel to 3x-7y =14 has slope equal to 3/7.
Hence, the required line passing through (-6,10) and parallel to 3x-7y=14 in slope-point form is
slope point formula is y-y1= m (x-x1)
y-10 = (3/7) (x+6)
15) From 14) , y-10 = (3/7) (x+6)
7y-70 = 3x+18
3x-7y = -88
It is of the form Ax + By = C.
factors that are significant in the development of disease includ.pdf
Explain why the vapor pressure of a solution made with a nonvolatile solution must be less than
the vapor pressure of the pure solvent?
Solution
Escaping of solvent molecules from the surface of liquid to space is called vaporization. The
vapors are then condensed. When there is an equilibrium between vaporization and
condensation, the pressure exerted by the vapors of the liquid over the liquid is called vapor
pressure.
In a pure liquid, the entire surface is occupied by the molecules of the liquid whereas in the case
of solution, the surface is occupied by the liquid molecuels and nonvolatile solute molecules. So,
the number of solvent molecules escaping from the surface is reduced compare to the pure
solvent, thus, the vapor pressure of solution is reduced.
Hence, the vapor pressure of a solution of nonvolatile solution is less than the vapor pressure of
the pure solvent..
economic institutions 8. According to Thomas Piketty, what is the si.pdf
ECO 355 Game Theory I need a right answer of it. QUESTION 9 Player II Player A 6,3 2,6 B
4,3 8,1 Suppose that in the game above Player 2 plays strategy C with probability q-0.3. The
value for El1(A,q) is: O a. 4.6 b. 3.2 O c. 8.3 O d. 3.9
Solution
Suppose that in the game above Player 2 plays strategy C with probability q=0.3, the value for
E(A,q)
The Correct Answer is:- D:- 3.9
Reason:- Value E(A,q) = 6*0.3 + 3*0.7 = 3.9.
Discuss bio films , formation location etc Discuss bio films , .pdf
Discuss antecedents of target marketing in a case scenario.
GEL-1.2: Demonstrate college-level communication through the composition of original
materials in Standard English.
GEL-6.2: Apply a professionally recognized method of formatting and documentation.
In this Assessment you will discuss antecedents of target marketing (i.e. what steps come before
target marketing).
Using the information on target market antecedents you learned from reading Chapters 6 and 8 in
your text, original research on the Internet, and the various Learning Activities provided, read the
scenario and respond to the checklist items in a business memo format.
Scenario: The product is a portable rolled computer made by the start-up company UclidTouch.
The screen and accompanying keyboard roll up for travelers or those who want to easily
transport their computer. The unrolled computer measures 8 inches by 16 inches. When rolled
up, the computer is 1 3/4 inches in diameter and weighs 1 pound. It fits in a lightweight
aluminum canister that has an optional mini-combination lock. It is available in 7 different colors
and can be hung around the customer’s neck on a lanyard which also is available in 7 colors. The
price is comparable to the pricing for popular tablets already on the market.
Business Memo: You are a marketing assistant at UclidTouch tasked with providing a business
memo (i.e., an internal company document) to the U.S. A. executive management team of
UclidTouch concerning the checklist items provided below.
Write a minimum 500-word original business memo. In a concise manner, describe to executive
management the data you will analyze to determine a potential market for this product. State
your purpose in the opening paragraph and provide sufficient information to explain the required
items. Finally, provide a concluding paragraph to summarize your suggestions.
Address the following items in your Assessment submission.
1. Identify a market or product category for review.
2. Describe bases for segmenting the market.
3. Identify segmentation descriptors.
4. Explain potential segments based on your reading in Chapter 8 “Steps in Segmenting a
Market.”
Size of segment
Expected growth
Purchase frequency
Brand loyalty
Sales potential
Profit potential
Chapter Reading Includes:
Chapter 6: “Consumer Decision Making”
Chapter 6 explains why consumer behavior is essential to the marketer who endeavors to satisfy
the needs and wants of customers and wants to communicate effectively with them. When
buying products, particularly new or expensive items, consumers generally follow the five step
consumer decision-making process shown in Exhibit 6.1 in the Lamb et al. (2014) textbook:
Needs recognition
Information search
Evaluation of alternatives
Purchase
Post-purchase behavior
Each of these steps in the consumer decision process is discussed thoroughly.
What is a brand and brand loyalty?
A brand can be a company name, symbol, or design that distinguishes one company’s services .
Discuss the importance of documentation in the medical record.So.pdf
Discuss bio films , formation location etc
Discuss bio films , formation location etc
Solution
biofilm is a thin mat of microorganism covered with mucilage or slime screted by them. biofilms
usually occur in nutrition rich surfaces, where abundant growth factors are available for biofilm
producers. the mucilage or extracellular polymeric substance contains cell debris, amino acids
and other byproducts of bacterial metabolism.
biofilm formation: it forms at surfaces which are easy to attach and provides suitable
environment for supporting the growth of attached organism. microorganism binds to surface via
suitable receptors or adhesins and starts utilizing the available material, as they start growing
they secrete etracellular mucilage which contains many attractants(molecules that attracts other
microorganisms) such as signailing molecules which attracts other species towards the growth of
mat thus, a layer of cells develops.
location: can be found on any surfaces which renders the bacteria to grow. commonly observed
surfaces are: water pipes, tooth surfaces(dental caries), urinary catheters, surgical sites,
aquarium. all these surfaces provide abundant food material and optimum growth fascilities.
functions: it prevents organisms from antibiotics and thus make them resistant, ample growth
oppurtunities, acts as a infection/virulence factor
example: dental caries is a good example of bofilm production by Streptococcus mutans,
Catheter associated UTI caused by many gram negtive is also due to the biofilm production..
Determine the molecular geometry of CO32â».Solution .pdf
Design a word guess game. Ask player 1 to input a word. Then player 2 guesses the word. If the
guessed word is not correct, program says \"Your guess is not correct!\". Player 2 can guess
unlimited times until the answer is correct, or type \"@exit\" to quit the program. When player 2
types \"@hint\", print out the first and last letter of the word. When players 2 types \"@hint\"
again, randomly tells one of the other unknown letter. After this, every time player 2 input
\"@hint\", randomly tells one of the other unknown letter until no more unknown letters. If the
guess is correct, print out \"Yes! ! ! ! !\"
Solution
from random import randint
word=raw_input(\"Player 1:Enter word\ \")
guess=\"\"
flag=0
while guess!=word:
guess=raw_input(\"Player 2:Enter guess word\ \")
if guess==\"exit\":
break
elif guess==\"hint\" and flag==1:
print word[randint(0,len(word))]
elif guess==\"hint\" and flag==0:
print word[0],word[len(word)-1]
flag=1
elif guess==word:
print \"Yes\ \"
else:
print \"Your guess is not correct\ \"
======================================================
Output:
akshay@akshay-Inspiron-3537:~/Chegg$ python wordguess.py
Player 1:Enter word
hello
Player 2:Enter guess word
hint
h o
Player 2:Enter guess word
hello
Yes.
Define each class of incident.types of incidents classifcationsr.pdf
DATA SET G Mileage and Vehicle Weight
Obtain the regression equation. (Negative amount should be indicated by a minus sign. Round
your answers to 4 decimal places.)
Calculate R2. (Round your answer to 4 decimal places.)
Calculate the degrees of freedom and t-critical. (Round your answers to 2 decimal places.)Using
the data set G, answer the questions given below.
Solution
Here dependent variable y is City MPG and indepedent variable x is weight.
All the parts we can done by using MINITAB.
Enter the data.
STAT --> Regression --> Regression --> Response : y --> Predictors : x --> Results : select
second option --> ok --> ok
output is :
Regression Analysis: y versus x
The regression equation is
y = 36.6 - 0.00456 x
Predictor Coef SE Coef T P
Constant 36.634 1.879 19.49 0.000
x -0.0045564 0.0004871 -9.35 0.000
S = 2.49891 R-Sq = 68.1% R-Sq(adj) = 67.3%
Analysis of Variance
Source DF SS MS F P
Regression 1 546.44 546.44 87.51 0.000
Residual Error 41 256.03 6.24
Total 42 802.47
y = -0.00456 x + 36.6
R-sq is 68.1% or 0.681
Degrees of freedom is n-1.
where n is number of data pairs.
df = n - 1 = 43 - 1 = 42
And critical value by using EXCEL is,
syntax :
=TINV(probability, deg_freedom)
where probability = alpha
deg_freedom = df = 42
t-critical = 2.0181
We see that | t | = | -9.35 | = 9.35
t > t-critical
Reject H0 at 5% level of significance.
Conclusion : Population slope for x is differ than 0..
Complete a force field analysis for an organization with which you a.pdf
Compare the golf equation of h(x) = -16x^2 +120x and the baseball equation of h(x) = -16x^2 +
60x + 4. What do you see as a major difference between these 2 formats? How does this
difference directly relate to hitting a golf ball or a baseball? Rewrite the golf equation of h(x) = -
16x^2 +120x to show Marc hitting the ball while standing on top of a platform that is 10 feet off
the ground. Using our quadratic equation knowledge match the following scenarios. Hitting a
golf ball on the moon Hitting a golf ball on Earth from the top of a tall building Hitting a
baseball on Jupiter Hitting a baseball on Earth by a bionic man Hitting a baseball on Earth by a
10 foot tall person h(x) = -16x^2 + 500x + 4 h(x) = -50x^2 + 50x + 4 h(x) = -8x^2 +100x h(x)
= -16x^2 +75x + 8 h(x) = -16x^2 + 150x + 300
Solution
h(x) = -16x^2 + 120x ( golf)
h(x) = -16x^2 + 60x + 4 ( base ball)
a) Difference between the two is the presence of constant 4 in equation of base ball) .
b) This means that base ball is thrown from a height of 4 ft
Golf ball is hit from the ground.
c) h(x) = -16x^2 + 120x
Ball is hit from the platform :
h(x) = -16x^2 + 120x + 10.
Write CC++ a program that inputs a weighted undirected graph and fi.pdf
Write a shell command to search all the files in the working directory with names containing one
character extension for the lines that does not start with a character A or character B.
Solution
Answer:
#!/bin/bash
# to search all the files in the working directory with names containing one character extension
#for the lines that does not start with a character A or character B
find . -type f -name \"*.?\" | -not -name \'[ac]*\' -type f.
Why was the iodide ion successful for the SN2 reactions What if you.pdf
Why is it helpful to approach the marketing research process from a \"backward” approach?
How can this backward approach lead to more useful research findings?
Solution
Backward approach in marketing design helps to analyse each stage properly and determines
what comes after it. This will also give more control over realistic market picture rather than just
historical data analysis. This gives more sensitive information from the end users and can work
backward to integrate the necessary things more precisely.The backward approach will help the
researchers to design the outcome more better so that they can define the problem statement
more precisely. The backward approach avoids the usual glitches in research as it has more
control over the end outcomes and can guide the researcher with more rational support. This will
also reduce the overall time taken for data interpretation and give better control over
classification of relevenat information and creating value added insights.Thys backward
approach will help to guide the research forward and it is all about what it is required and what
they need from the study so that routemap will be clear for all data collection sampling and
processing..
What were the triumphs and limitations of democracy in Periclean Ath.pdf
What process causes plant ion-uptake to acidify soil? What soil pH range generally promotes the
greatest availability of cations?
Solution
The process that causes plant ion-uptake to acidify soil is known as soil acidification. It is a
process by which the soil pH decreases over time without producing any visible signs. Slow soil
acidification occurs under natural conditions over thousands of years, with high rainfall areas
most affected, whereas, rapid acidification occurs over a few years under intensive agricultural
practices. Soil acidification affects the surface and the subsoil simultaneously. However, areas
that remain unaffected by industrial pollution, there soil acidification is mainly caused by the
release of protons (H+) during the oxidation of carbon (C), sulphur (S) and nitrogen (N)
compounds in soils. During nitrogen cycling, the major processes leading to acidification
consists of the cation imbalance over anion during uptake in the rhizosphere of plants, during
nitrification from fixation or from the fertilizers, and during the removal of plant and animal
products by leaching. The uptake of excess cations over anions by plants results in the
acidification of the rhizosphere which is a “localized” effect and can be balanced by the release
of hydroxyl (OH-) ions during subsequent plant decomposition.
Soils are considered acidic below a pH of 5, and alkaline above a pH of 7.5 and very alkaline
above a pH of 8. It has been observed that the “ideal” soil pH is close to neutral, which fall
within a range from a slightly acidic pH of 6.5 to slightly alkaline pH of 7.5. It has been
determined that most plant nutrients are optimally available to plants within this range helpful for
plant root growth. However, Nitrogen (N), Potassium (K), and Sulfur (S) are major plant
nutrients that enhance plant growth and development. Most of the plant nutrients (cations) tend
to be less available when soil pH is above 7.5, and are optimally available at a slightly acidic pH
in the range of 6.5 to 6.8..
What process causes plant ion-uptake to acidify soil What soil pH r.pdf
What postulates that the state and activity of a particular region of chromatin depends upon the
specific modifications, or combinations of modifications, to the histone tails in that region? It
suggests that the pattern of modifications adorning the tails of the core histones contains encoded
information governing the properties of the nucleosomes containing them. the DaVinci code the
chemiosmotic hypothesis the histone code the genetic code
Solution
Answer is the histone code.
Modifications to the histone proteins which bind to the dna, make the fragment of dna accessible
to various enzymes..
what is the name and stereochemistry of the compoundSolution.pdf
What is the maximum order of any element in Azo
Solution
Here is A20.
So, First break 20 into different sums.
20 = 1+2+3+4+10
LCM (1,2,3,4,10) = 60
20 = 1+2+3+5+9
LCM (1,2,3,5,9) = 90
20 = 1+2+3+6+8
LCM (1,2,3,6,8) = 24
20 = 1+2+3+7+7
LCM (1,2,3,7,7) = 42
20 = 1+2+3+14
LCM (1,2,3,14) = 42
20 = 1+2+4+13
LCM (1,2,4,13) = 52
20 = 5+7+8
LCM (5,7,8) = 280
Hence 280 is maximum order..
What are some of the architectural decisions that you need to make w.pdf
What are lichens? What are foliose lichens?
Solution
Answers :
Lichens : Lichens are the symbiotic association between algae (phycobiont) and fungi
(mycobiont). In this association, fungus makes available the moisture and minerals to the algae
which prepares food by photosynthesis. in nature, neither of the two can grow alone
independently..
What are lichens What are foliose lichensSolutionAnswers .pdf
We learned the various design reviews during the application life cycle. What is the role of DBA
in each review stage? Who, other than DBA, should be involved at each stage and why? Do you
think some of the design reviews may need to be reviewed again as a result of application
development? Why or why not?
Solution
Design reviews are an important factor of the system development lifecycle for database
applications. It is during the design review that all aspects of the database and application code
are reviewed for efficiency, effectiveness, and accuracy. It is imperative that all database
applications, regardless of their size, are reviewed to assure that the application was design
properly, efficient coding techniques were used, and the database is accessed and modified
correctly and efficiently. The design review is an important process for checking the validity of
design decisions and correcting errors before applications and databases are promoted to
production status.
Multiple design reviews should be conducted throughout the application’s life. There are seven
basic design review phases for a database application:
There are many aspects of the development process and its resulting application that are
addressed by design reviews. Some of the areas that can be addressed by a design review
include:
The DBA usually must act as the center point for organizing and conducting design reviews. If
the DBA does not organize design reviews it is unlikely that any design review will be
conducted. And if design reviews are not conducted, the application is more apt to suffer
performance and availability problems in a production environment. A group consisting of
subject matter experts and the developer’s peers and co-workers should conduct each design
review.
Formal design review roles should include the leader, the scribe, the mediator, and the
participants. The leader can change from one design review to the next, but within the scope of a
single design review a single leader must be assigned. The primary role of the leader is to act as a
master of ceremonies to keep the review process moving along. Other duties include creating an
agenda and ensuring that the agenda is followed, engaging all participants of the design review to
solicit everyone’s input, ensuring that all participants maintain proper decorum, working with the
proper participants before the meeting to ensure that all required documentation will be made
available for the design review, and any other tasks necessary to ensure a successful design
review. Though it is not mandatory, the DBA typically acts as the leader of design reviews for
applications using a database.
Thank you..
These are questions to be answered in a paper about the Chernobyl in.pdf
The wire fencing costs $3.25 per linear foot. The privacy fencing costs $5.45 per linear foot. The
homeowner wants a total garden area of 600 square feet. a. write a model for the cost of the
constructing the fence, in terms of a single input variable. b. calculate the dimensions that will
result in minimum cost. c. How much of each type of fencing will be used?
Solution
w = length of wire fence
p = length of privacy fence
cost = w*3.25 + p*5.45
w*p = 600
a. cost = w*3.25 + (600/w)*5.45
cost = 3.25*w + 3270/w
d(cost)/dw = 3.25 - 3270/w^2 = 0
3.25 = 3270/w^2
w^2 = 1006.15
b. w = 31.72 feet.
The following is the (incomplete) header file for the class Fracti.pdf
The following comparative income statement (in thousands of dollars) for the two recent fiscal
years was adapted from the annual report of Speedway Motorsports, Inc., owner and operator of
several major motor speedways, such as the Atlanta, Texas, and Las Vegas Motor Speedways.
Current Year Previous Year 2 Revenues: 3 Admissions 4Event-related revenue $104,312.00
$116,644.00 131,164.00 170,368.00 65,824.00 $472,000.00 $484,000.00 42,544.00 180,304.00
44,840.00 NASCAR broadcasting revenue Other operating revenue 7 Total revenue 8 Expenses
and other: 9 Direct expense of events 10 NASCAR purse and sanction fees 91,568.00
$101,640.00 117,612.00 20,328.00 220,220.00 424800.00 $459,800.00 $24,200.00 118,000.00
19,352.00 195,880.00 11 Other direct expenses 2 General and administrative 13 Total expenses
and other Income from continuing operations $47,200.00 14 Required A. Prepare a comparative
income statement for these two years in vertical form, stating each item as a percent of revenues.
Enter all amounts as positive numbers. Rounding instructions
Solution
VERTICAL ANALYSIS OF INCOME STATEMENT Particulars Current Year
Amount Current Year Percentage Previous Year Amount Previous Year Percentage
Admissions $ 1,04,312.00 22.10% $ 1,16,644.00 24.10% Event Related revenue $ 1,42,544.00
30.20% $ 1,31,164.00 27.10% NASCAR Broadcasting Revenue $ 1,80,304.00 38.20% $
1,70,368.00 35.20% Other Operating Income $ 44,840.00 9.50% $ 65,824.00 13.60%
Total Revenue $ 4,72,000.00 100.00% $ 4,84,000.00 100.00% Expenses and others
Direct expenses of events $ 91,568.00 19.40% $ 1,01,640.00 21.00% NASCAR purse and
sanction fees $ 1,18,000.00 25.00% $ 1,17,612.00 24.30% Other direct expenses $ 19,352.00
4.10% $ 20,328.00 4.20% General and adminstrative $ 1,95,880.00 41.50% $ 2,20,220.00
45.50% Total Expenses and others $ 4,24,800.00 90.00% $ 4,59,800.00 95.00% Income
from continuing operation $ 47,200.00 10.00% $ 24,200.00 5.00%.
The adjusted trial balance for Tybalt Construction as of December 31.pdf
The accounting records for Eisner Manufacturing Company included the following cost
information relating to its first year of operations: Direct materials Direct labor Fixed
manufacturing overhead Variable manufacturing overhead $ 60,000 $ 80,000 100,000 $ 20,000
Assume the company produced 10,000 units of inventory and sold 6,000 of these units for
$192,000. What amount of finished goods will be reported on the balance sheet at the end of the
year under absorption costing? Under variable costing? A. $104,000, $64,000 B. $64,000,
$104,000 C. $130,667, $64,000 D. $64,000, $130,667
Solution
Calculate unit product cost and finished goods :
So answer is a) $104000 ; $64000AbsorptionvariableDirect material66Direct labour88Variable
manufacturing overhead22Fixed manufacturing overhead100Unit product cost2616Finished
goods inventory40004000Finished goods inventory cost10400064000.
Summarize the case Surgery Iturralde v. Hilo Medical Center USA , i.pdf
Steel can be strengthened by all but which of the following Annealing Quenching and
tempering Age or precipitation hardening Work hardening
Solution
Sol: option A
Steel can be strengthened by quenching or tempering, Precipitation hardening, work hardening.
Steel can not be strengthened by Anneling process.
1) Anneling is a heat treatment that alters the physical and sometimes chemical properties of a
material to increase its ductility and reduce its hardness, making it more workable. It involves
heating a material to above its recrystallization temperature, maintaining a suitable temperature,
and then cooling. In annealing, atoms migrate in the crystal lattice and the number of
dislocations decreases, leading to the change in ductility and hardness.
2) Quench hardening is a mechanical process in which steel and cast iron alloys are strengthened
and hardened. These metals consist of ferrous metals and alloys. This is done by heating the
material to a certain temperature, depending on the material
3) Precipitation hardening, also called age hardening, is a heat treatment technique used to
increase the yield strength of malleable materials, including most structural alloys of aluminium,
magnesium, nickel, titanium, and some steels and stainless steels
4) Work hardening also known as strain hardening or cold working, is the strengthening of a
metal by plastic deformation. This strengthening occurs because of dislocation movements and
dislocation generation within the crystal structure of the material..
Suppose we draw 2 cards out of a deck of 52. Let A = “the first card.pdf
Suppose we are searching for some key r in a BST. As we perform the search, we compare z to
the keys on the path from the root to the leaf in the BST. Which sequence below cannot be the
sequence of keys examined during the search? Explain why not. Do not assume that the search
below is performed on the same BST 85, 76, 30, 15, 6, 1 13, 98, 28, 94, 33, 85, 48, 77, 51, 62,
55 99, 78, 44, 87, 57, 52 16, 33, 77, 54, 73, 60 67, 24, 74, 55, 48
Solution
(e) is the option.
In BST ,First we search for the root. If the key is greater than root then we recur in the right sub
tree or we recur in the left sub tree.
In (e) the root is 67. The key is not 67 that\'s why we recurred to left sub tree that is 24 less than
67. Now in the sequence it cannot be 74 because this number is greater that the root 67. The
number in the sequence can be greater than 24 but strictly less than 67..
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
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Find the domain of the function. (Enter your answer using interval no.pdf
Find an equation of the line through (-6, 10), parallel to the line with equation 3x - 7y = 14.
Write the new equation in point-slope form. Convert the equation of the new line found in
problem #14 to standard form, Ax + By = C, where A, B, and C are integers.
Solution
14) Given line is 3x-7y =14
7y = 3x-14
y = (3/7)x - 2
It is of the form, y = mx + c
Hence, slope = 3/7
We know that two parallel lines have same slope.
Hence,
The equation parallel to 3x-7y =14 has slope equal to 3/7.
Hence, the required line passing through (-6,10) and parallel to 3x-7y=14 in slope-point form is
slope point formula is y-y1= m (x-x1)
y-10 = (3/7) (x+6)
15) From 14) , y-10 = (3/7) (x+6)
7y-70 = 3x+18
3x-7y = -88
It is of the form Ax + By = C.
factors that are significant in the development of disease includ.pdf
Explain why the vapor pressure of a solution made with a nonvolatile solution must be less than
the vapor pressure of the pure solvent?
Solution
Escaping of solvent molecules from the surface of liquid to space is called vaporization. The
vapors are then condensed. When there is an equilibrium between vaporization and
condensation, the pressure exerted by the vapors of the liquid over the liquid is called vapor
pressure.
In a pure liquid, the entire surface is occupied by the molecules of the liquid whereas in the case
of solution, the surface is occupied by the liquid molecuels and nonvolatile solute molecules. So,
the number of solvent molecules escaping from the surface is reduced compare to the pure
solvent, thus, the vapor pressure of solution is reduced.
Hence, the vapor pressure of a solution of nonvolatile solution is less than the vapor pressure of
the pure solvent..
economic institutions 8. According to Thomas Piketty, what is the si.pdf
ECO 355 Game Theory I need a right answer of it. QUESTION 9 Player II Player A 6,3 2,6 B
4,3 8,1 Suppose that in the game above Player 2 plays strategy C with probability q-0.3. The
value for El1(A,q) is: O a. 4.6 b. 3.2 O c. 8.3 O d. 3.9
Solution
Suppose that in the game above Player 2 plays strategy C with probability q=0.3, the value for
E(A,q)
The Correct Answer is:- D:- 3.9
Reason:- Value E(A,q) = 6*0.3 + 3*0.7 = 3.9.
Discuss bio films , formation location etc Discuss bio films , .pdf
Discuss antecedents of target marketing in a case scenario.
GEL-1.2: Demonstrate college-level communication through the composition of original
materials in Standard English.
GEL-6.2: Apply a professionally recognized method of formatting and documentation.
In this Assessment you will discuss antecedents of target marketing (i.e. what steps come before
target marketing).
Using the information on target market antecedents you learned from reading Chapters 6 and 8 in
your text, original research on the Internet, and the various Learning Activities provided, read the
scenario and respond to the checklist items in a business memo format.
Scenario: The product is a portable rolled computer made by the start-up company UclidTouch.
The screen and accompanying keyboard roll up for travelers or those who want to easily
transport their computer. The unrolled computer measures 8 inches by 16 inches. When rolled
up, the computer is 1 3/4 inches in diameter and weighs 1 pound. It fits in a lightweight
aluminum canister that has an optional mini-combination lock. It is available in 7 different colors
and can be hung around the customer’s neck on a lanyard which also is available in 7 colors. The
price is comparable to the pricing for popular tablets already on the market.
Business Memo: You are a marketing assistant at UclidTouch tasked with providing a business
memo (i.e., an internal company document) to the U.S. A. executive management team of
UclidTouch concerning the checklist items provided below.
Write a minimum 500-word original business memo. In a concise manner, describe to executive
management the data you will analyze to determine a potential market for this product. State
your purpose in the opening paragraph and provide sufficient information to explain the required
items. Finally, provide a concluding paragraph to summarize your suggestions.
Address the following items in your Assessment submission.
1. Identify a market or product category for review.
2. Describe bases for segmenting the market.
3. Identify segmentation descriptors.
4. Explain potential segments based on your reading in Chapter 8 “Steps in Segmenting a
Market.”
Size of segment
Expected growth
Purchase frequency
Brand loyalty
Sales potential
Profit potential
Chapter Reading Includes:
Chapter 6: “Consumer Decision Making”
Chapter 6 explains why consumer behavior is essential to the marketer who endeavors to satisfy
the needs and wants of customers and wants to communicate effectively with them. When
buying products, particularly new or expensive items, consumers generally follow the five step
consumer decision-making process shown in Exhibit 6.1 in the Lamb et al. (2014) textbook:
Needs recognition
Information search
Evaluation of alternatives
Purchase
Post-purchase behavior
Each of these steps in the consumer decision process is discussed thoroughly.
What is a brand and brand loyalty?
A brand can be a company name, symbol, or design that distinguishes one company’s services .
Discuss the importance of documentation in the medical record.So.pdf
Discuss bio films , formation location etc
Discuss bio films , formation location etc
Solution
biofilm is a thin mat of microorganism covered with mucilage or slime screted by them. biofilms
usually occur in nutrition rich surfaces, where abundant growth factors are available for biofilm
producers. the mucilage or extracellular polymeric substance contains cell debris, amino acids
and other byproducts of bacterial metabolism.
biofilm formation: it forms at surfaces which are easy to attach and provides suitable
environment for supporting the growth of attached organism. microorganism binds to surface via
suitable receptors or adhesins and starts utilizing the available material, as they start growing
they secrete etracellular mucilage which contains many attractants(molecules that attracts other
microorganisms) such as signailing molecules which attracts other species towards the growth of
mat thus, a layer of cells develops.
location: can be found on any surfaces which renders the bacteria to grow. commonly observed
surfaces are: water pipes, tooth surfaces(dental caries), urinary catheters, surgical sites,
aquarium. all these surfaces provide abundant food material and optimum growth fascilities.
functions: it prevents organisms from antibiotics and thus make them resistant, ample growth
oppurtunities, acts as a infection/virulence factor
example: dental caries is a good example of bofilm production by Streptococcus mutans,
Catheter associated UTI caused by many gram negtive is also due to the biofilm production..
Determine the molecular geometry of CO32â».Solution .pdf
Design a word guess game. Ask player 1 to input a word. Then player 2 guesses the word. If the
guessed word is not correct, program says \"Your guess is not correct!\". Player 2 can guess
unlimited times until the answer is correct, or type \"@exit\" to quit the program. When player 2
types \"@hint\", print out the first and last letter of the word. When players 2 types \"@hint\"
again, randomly tells one of the other unknown letter. After this, every time player 2 input
\"@hint\", randomly tells one of the other unknown letter until no more unknown letters. If the
guess is correct, print out \"Yes! ! ! ! !\"
Solution
from random import randint
word=raw_input(\"Player 1:Enter word\ \")
guess=\"\"
flag=0
while guess!=word:
guess=raw_input(\"Player 2:Enter guess word\ \")
if guess==\"exit\":
break
elif guess==\"hint\" and flag==1:
print word[randint(0,len(word))]
elif guess==\"hint\" and flag==0:
print word[0],word[len(word)-1]
flag=1
elif guess==word:
print \"Yes\ \"
else:
print \"Your guess is not correct\ \"
======================================================
Output:
akshay@akshay-Inspiron-3537:~/Chegg$ python wordguess.py
Player 1:Enter word
hello
Player 2:Enter guess word
hint
h o
Player 2:Enter guess word
hello
Yes.
Define each class of incident.types of incidents classifcationsr.pdf
DATA SET G Mileage and Vehicle Weight
Obtain the regression equation. (Negative amount should be indicated by a minus sign. Round
your answers to 4 decimal places.)
Calculate R2. (Round your answer to 4 decimal places.)
Calculate the degrees of freedom and t-critical. (Round your answers to 2 decimal places.)Using
the data set G, answer the questions given below.
Solution
Here dependent variable y is City MPG and indepedent variable x is weight.
All the parts we can done by using MINITAB.
Enter the data.
STAT --> Regression --> Regression --> Response : y --> Predictors : x --> Results : select
second option --> ok --> ok
output is :
Regression Analysis: y versus x
The regression equation is
y = 36.6 - 0.00456 x
Predictor Coef SE Coef T P
Constant 36.634 1.879 19.49 0.000
x -0.0045564 0.0004871 -9.35 0.000
S = 2.49891 R-Sq = 68.1% R-Sq(adj) = 67.3%
Analysis of Variance
Source DF SS MS F P
Regression 1 546.44 546.44 87.51 0.000
Residual Error 41 256.03 6.24
Total 42 802.47
y = -0.00456 x + 36.6
R-sq is 68.1% or 0.681
Degrees of freedom is n-1.
where n is number of data pairs.
df = n - 1 = 43 - 1 = 42
And critical value by using EXCEL is,
syntax :
=TINV(probability, deg_freedom)
where probability = alpha
deg_freedom = df = 42
t-critical = 2.0181
We see that | t | = | -9.35 | = 9.35
t > t-critical
Reject H0 at 5% level of significance.
Conclusion : Population slope for x is differ than 0..
Complete a force field analysis for an organization with which you a.pdf
Compare the golf equation of h(x) = -16x^2 +120x and the baseball equation of h(x) = -16x^2 +
60x + 4. What do you see as a major difference between these 2 formats? How does this
difference directly relate to hitting a golf ball or a baseball? Rewrite the golf equation of h(x) = -
16x^2 +120x to show Marc hitting the ball while standing on top of a platform that is 10 feet off
the ground. Using our quadratic equation knowledge match the following scenarios. Hitting a
golf ball on the moon Hitting a golf ball on Earth from the top of a tall building Hitting a
baseball on Jupiter Hitting a baseball on Earth by a bionic man Hitting a baseball on Earth by a
10 foot tall person h(x) = -16x^2 + 500x + 4 h(x) = -50x^2 + 50x + 4 h(x) = -8x^2 +100x h(x)
= -16x^2 +75x + 8 h(x) = -16x^2 + 150x + 300
Solution
h(x) = -16x^2 + 120x ( golf)
h(x) = -16x^2 + 60x + 4 ( base ball)
a) Difference between the two is the presence of constant 4 in equation of base ball) .
b) This means that base ball is thrown from a height of 4 ft
Golf ball is hit from the ground.
c) h(x) = -16x^2 + 120x
Ball is hit from the platform :
h(x) = -16x^2 + 120x + 10.
Write CC++ a program that inputs a weighted undirected graph and fi.pdf
Write a shell command to search all the files in the working directory with names containing one
character extension for the lines that does not start with a character A or character B.
Solution
Answer:
#!/bin/bash
# to search all the files in the working directory with names containing one character extension
#for the lines that does not start with a character A or character B
find . -type f -name \"*.?\" | -not -name \'[ac]*\' -type f.
Why was the iodide ion successful for the SN2 reactions What if you.pdf
Why is it helpful to approach the marketing research process from a \"backward” approach?
How can this backward approach lead to more useful research findings?
Solution
Backward approach in marketing design helps to analyse each stage properly and determines
what comes after it. This will also give more control over realistic market picture rather than just
historical data analysis. This gives more sensitive information from the end users and can work
backward to integrate the necessary things more precisely.The backward approach will help the
researchers to design the outcome more better so that they can define the problem statement
more precisely. The backward approach avoids the usual glitches in research as it has more
control over the end outcomes and can guide the researcher with more rational support. This will
also reduce the overall time taken for data interpretation and give better control over
classification of relevenat information and creating value added insights.Thys backward
approach will help to guide the research forward and it is all about what it is required and what
they need from the study so that routemap will be clear for all data collection sampling and
processing..
What were the triumphs and limitations of democracy in Periclean Ath.pdf
What process causes plant ion-uptake to acidify soil? What soil pH range generally promotes the
greatest availability of cations?
Solution
The process that causes plant ion-uptake to acidify soil is known as soil acidification. It is a
process by which the soil pH decreases over time without producing any visible signs. Slow soil
acidification occurs under natural conditions over thousands of years, with high rainfall areas
most affected, whereas, rapid acidification occurs over a few years under intensive agricultural
practices. Soil acidification affects the surface and the subsoil simultaneously. However, areas
that remain unaffected by industrial pollution, there soil acidification is mainly caused by the
release of protons (H+) during the oxidation of carbon (C), sulphur (S) and nitrogen (N)
compounds in soils. During nitrogen cycling, the major processes leading to acidification
consists of the cation imbalance over anion during uptake in the rhizosphere of plants, during
nitrification from fixation or from the fertilizers, and during the removal of plant and animal
products by leaching. The uptake of excess cations over anions by plants results in the
acidification of the rhizosphere which is a “localized” effect and can be balanced by the release
of hydroxyl (OH-) ions during subsequent plant decomposition.
Soils are considered acidic below a pH of 5, and alkaline above a pH of 7.5 and very alkaline
above a pH of 8. It has been observed that the “ideal” soil pH is close to neutral, which fall
within a range from a slightly acidic pH of 6.5 to slightly alkaline pH of 7.5. It has been
determined that most plant nutrients are optimally available to plants within this range helpful for
plant root growth. However, Nitrogen (N), Potassium (K), and Sulfur (S) are major plant
nutrients that enhance plant growth and development. Most of the plant nutrients (cations) tend
to be less available when soil pH is above 7.5, and are optimally available at a slightly acidic pH
in the range of 6.5 to 6.8..
What process causes plant ion-uptake to acidify soil What soil pH r.pdf
What postulates that the state and activity of a particular region of chromatin depends upon the
specific modifications, or combinations of modifications, to the histone tails in that region? It
suggests that the pattern of modifications adorning the tails of the core histones contains encoded
information governing the properties of the nucleosomes containing them. the DaVinci code the
chemiosmotic hypothesis the histone code the genetic code
Solution
Answer is the histone code.
Modifications to the histone proteins which bind to the dna, make the fragment of dna accessible
to various enzymes..
what is the name and stereochemistry of the compoundSolution.pdf
What is the maximum order of any element in Azo
Solution
Here is A20.
So, First break 20 into different sums.
20 = 1+2+3+4+10
LCM (1,2,3,4,10) = 60
20 = 1+2+3+5+9
LCM (1,2,3,5,9) = 90
20 = 1+2+3+6+8
LCM (1,2,3,6,8) = 24
20 = 1+2+3+7+7
LCM (1,2,3,7,7) = 42
20 = 1+2+3+14
LCM (1,2,3,14) = 42
20 = 1+2+4+13
LCM (1,2,4,13) = 52
20 = 5+7+8
LCM (5,7,8) = 280
Hence 280 is maximum order..
What are some of the architectural decisions that you need to make w.pdf
What are lichens? What are foliose lichens?
Solution
Answers :
Lichens : Lichens are the symbiotic association between algae (phycobiont) and fungi
(mycobiont). In this association, fungus makes available the moisture and minerals to the algae
which prepares food by photosynthesis. in nature, neither of the two can grow alone
independently..
What are lichens What are foliose lichensSolutionAnswers .pdf
We learned the various design reviews during the application life cycle. What is the role of DBA
in each review stage? Who, other than DBA, should be involved at each stage and why? Do you
think some of the design reviews may need to be reviewed again as a result of application
development? Why or why not?
Solution
Design reviews are an important factor of the system development lifecycle for database
applications. It is during the design review that all aspects of the database and application code
are reviewed for efficiency, effectiveness, and accuracy. It is imperative that all database
applications, regardless of their size, are reviewed to assure that the application was design
properly, efficient coding techniques were used, and the database is accessed and modified
correctly and efficiently. The design review is an important process for checking the validity of
design decisions and correcting errors before applications and databases are promoted to
production status.
Multiple design reviews should be conducted throughout the application’s life. There are seven
basic design review phases for a database application:
There are many aspects of the development process and its resulting application that are
addressed by design reviews. Some of the areas that can be addressed by a design review
include:
The DBA usually must act as the center point for organizing and conducting design reviews. If
the DBA does not organize design reviews it is unlikely that any design review will be
conducted. And if design reviews are not conducted, the application is more apt to suffer
performance and availability problems in a production environment. A group consisting of
subject matter experts and the developer’s peers and co-workers should conduct each design
review.
Formal design review roles should include the leader, the scribe, the mediator, and the
participants. The leader can change from one design review to the next, but within the scope of a
single design review a single leader must be assigned. The primary role of the leader is to act as a
master of ceremonies to keep the review process moving along. Other duties include creating an
agenda and ensuring that the agenda is followed, engaging all participants of the design review to
solicit everyone’s input, ensuring that all participants maintain proper decorum, working with the
proper participants before the meeting to ensure that all required documentation will be made
available for the design review, and any other tasks necessary to ensure a successful design
review. Though it is not mandatory, the DBA typically acts as the leader of design reviews for
applications using a database.
Thank you..
These are questions to be answered in a paper about the Chernobyl in.pdf
The wire fencing costs $3.25 per linear foot. The privacy fencing costs $5.45 per linear foot. The
homeowner wants a total garden area of 600 square feet. a. write a model for the cost of the
constructing the fence, in terms of a single input variable. b. calculate the dimensions that will
result in minimum cost. c. How much of each type of fencing will be used?
Solution
w = length of wire fence
p = length of privacy fence
cost = w*3.25 + p*5.45
w*p = 600
a. cost = w*3.25 + (600/w)*5.45
cost = 3.25*w + 3270/w
d(cost)/dw = 3.25 - 3270/w^2 = 0
3.25 = 3270/w^2
w^2 = 1006.15
b. w = 31.72 feet.
The following is the (incomplete) header file for the class Fracti.pdf
The following comparative income statement (in thousands of dollars) for the two recent fiscal
years was adapted from the annual report of Speedway Motorsports, Inc., owner and operator of
several major motor speedways, such as the Atlanta, Texas, and Las Vegas Motor Speedways.
Current Year Previous Year 2 Revenues: 3 Admissions 4Event-related revenue $104,312.00
$116,644.00 131,164.00 170,368.00 65,824.00 $472,000.00 $484,000.00 42,544.00 180,304.00
44,840.00 NASCAR broadcasting revenue Other operating revenue 7 Total revenue 8 Expenses
and other: 9 Direct expense of events 10 NASCAR purse and sanction fees 91,568.00
$101,640.00 117,612.00 20,328.00 220,220.00 424800.00 $459,800.00 $24,200.00 118,000.00
19,352.00 195,880.00 11 Other direct expenses 2 General and administrative 13 Total expenses
and other Income from continuing operations $47,200.00 14 Required A. Prepare a comparative
income statement for these two years in vertical form, stating each item as a percent of revenues.
Enter all amounts as positive numbers. Rounding instructions
Solution
VERTICAL ANALYSIS OF INCOME STATEMENT Particulars Current Year
Amount Current Year Percentage Previous Year Amount Previous Year Percentage
Admissions $ 1,04,312.00 22.10% $ 1,16,644.00 24.10% Event Related revenue $ 1,42,544.00
30.20% $ 1,31,164.00 27.10% NASCAR Broadcasting Revenue $ 1,80,304.00 38.20% $
1,70,368.00 35.20% Other Operating Income $ 44,840.00 9.50% $ 65,824.00 13.60%
Total Revenue $ 4,72,000.00 100.00% $ 4,84,000.00 100.00% Expenses and others
Direct expenses of events $ 91,568.00 19.40% $ 1,01,640.00 21.00% NASCAR purse and
sanction fees $ 1,18,000.00 25.00% $ 1,17,612.00 24.30% Other direct expenses $ 19,352.00
4.10% $ 20,328.00 4.20% General and adminstrative $ 1,95,880.00 41.50% $ 2,20,220.00
45.50% Total Expenses and others $ 4,24,800.00 90.00% $ 4,59,800.00 95.00% Income
from continuing operation $ 47,200.00 10.00% $ 24,200.00 5.00%.
The adjusted trial balance for Tybalt Construction as of December 31.pdf
The accounting records for Eisner Manufacturing Company included the following cost
information relating to its first year of operations: Direct materials Direct labor Fixed
manufacturing overhead Variable manufacturing overhead $ 60,000 $ 80,000 100,000 $ 20,000
Assume the company produced 10,000 units of inventory and sold 6,000 of these units for
$192,000. What amount of finished goods will be reported on the balance sheet at the end of the
year under absorption costing? Under variable costing? A. $104,000, $64,000 B. $64,000,
$104,000 C. $130,667, $64,000 D. $64,000, $130,667
Solution
Calculate unit product cost and finished goods :
So answer is a) $104000 ; $64000AbsorptionvariableDirect material66Direct labour88Variable
manufacturing overhead22Fixed manufacturing overhead100Unit product cost2616Finished
goods inventory40004000Finished goods inventory cost10400064000.
Summarize the case Surgery Iturralde v. Hilo Medical Center USA , i.pdf
Steel can be strengthened by all but which of the following Annealing Quenching and
tempering Age or precipitation hardening Work hardening
Solution
Sol: option A
Steel can be strengthened by quenching or tempering, Precipitation hardening, work hardening.
Steel can not be strengthened by Anneling process.
1) Anneling is a heat treatment that alters the physical and sometimes chemical properties of a
material to increase its ductility and reduce its hardness, making it more workable. It involves
heating a material to above its recrystallization temperature, maintaining a suitable temperature,
and then cooling. In annealing, atoms migrate in the crystal lattice and the number of
dislocations decreases, leading to the change in ductility and hardness.
2) Quench hardening is a mechanical process in which steel and cast iron alloys are strengthened
and hardened. These metals consist of ferrous metals and alloys. This is done by heating the
material to a certain temperature, depending on the material
3) Precipitation hardening, also called age hardening, is a heat treatment technique used to
increase the yield strength of malleable materials, including most structural alloys of aluminium,
magnesium, nickel, titanium, and some steels and stainless steels
4) Work hardening also known as strain hardening or cold working, is the strengthening of a
metal by plastic deformation. This strengthening occurs because of dislocation movements and
dislocation generation within the crystal structure of the material..
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Suppose we are searching for some key r in a BST. As we perform the search, we compare z to
the keys on the path from the root to the leaf in the BST. Which sequence below cannot be the
sequence of keys examined during the search? Explain why not. Do not assume that the search
below is performed on the same BST 85, 76, 30, 15, 6, 1 13, 98, 28, 94, 33, 85, 48, 77, 51, 62,
55 99, 78, 44, 87, 57, 52 16, 33, 77, 54, 73, 60 67, 24, 74, 55, 48
Solution
(e) is the option.
In BST ,First we search for the root. If the key is greater than root then we recur in the right sub
tree or we recur in the left sub tree.
In (e) the root is 67. The key is not 67 that\'s why we recurred to left sub tree that is 24 less than
67. Now in the sequence it cannot be 74 because this number is greater that the root 67. The
number in the sequence can be greater than 24 but strictly less than 67..
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