Which of the following is an economic system where the major means o.pdf
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Which of the following is an economic system where the major means of production are privately owned Mercantilism Monopoly Normative system engineering Capitalism Utilitarianism Mercantilism Monopoly Normative system engineering Capitalism Utilitarianism Solution Capitalism is a structure where mostly firms are owned and controlled by private individuals and there is no interference by the government. The motive of the firms is to earn more and more profits. Whereas in a socialist structure the firms are owned and controlled by the government and the main aim is welfare of the society and not to earn profits..
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A health program planning committee wants to address various health .pdf
A health program planning committee wants to address various health problems by developing
programs for each health problem at all four levels of the public health pyramid. What would be
some implications of this perspective on setting priorities and on the nature of the subsequent
community health assessment?
Solution
The bottom tire consists of Infrastructure development: The most essential and set with high
priority. As it affects a larger population .This includes development of basic infrastructures like
electricity, sanitation, clean drinking water, hospitals, transportation and above all education.
The second tire from bottom tire consists of Population based services: It includes public
awareness programmes through print and television media for control of disease and prevention
of population from getting affected. It is a less costly but has a greater impact on mass
population. Healthier food and clean drinking water should be made available to all. It holds the
second highest priority as it will decrease the chance of illness.
The third tire from bottom consists of enabling services. It provides health education and
promotion of health care products among a community. Immunization programmes and
increasing awareness among population regarding the impact of hazardous chemicals on their
health. Enforcement of laws if required for betterment of public health like avoid alcohol while
driving, no smoking zones etc. It holds the third highest priority as prevention is better than cure.
The top tire consists of direct health care: It includes availability clinical care for individuals
affected and their proper treatment in due course of time. This applies to a smaller population but
is also very essential for communities..
Which one of the following significantly contributes to the specifici.pdf
Which one of the following significantly contributes to the specificity of base pairing in double-
stranded DNA, i.e., limits base pairing to cytosine-guanine (C:G) or adenine-thymine (A:T)? the
right handedness of double-stranded DNA hydrogen bonding between bases base-stacking
interactions the phosphodiester bonds between phosphate groups and deoxyribose sugar
molecules
Solution
Answer:
Base-stacking interactions
Explanation:
The purine and pyrimidine bases are hydrophobic and relatively insoluble in water at the near-
neutral pH of the cell. At acidic or alkaline pH the bases become charged and their solubility in
water increases. Hydrophobic stacking interactions in which two or more bases are positioned
with the planes of their rings parallel (like a stack of coins) are one of two important modes of
interaction between bases in nucleic acids. The stacking also involves a combination of van der
Waals and dipole-dipole interactions between the bases. Base stacking helps to minimize contact
of the bases with water, and base-stacking interactions are very important in stabilizing the three-
dimensional structure of nucleic acids..
Which of the following is true of a mainframe, or central, computer t.pdf
Which of the following is true of a mainframe, or central, computer that you may find in an
industrial application? It interfaces with manufacturing processes directly. It eliminates any need
for system firmware. It does away with most applications software. It receives information from
one or more workstations. Multiplexing is a means of in order to channel data. organizing the
transmission of signals from many lines onto one line transmitting signals by changing their
amplitude, frequency, or phase mixing and transmitting both synchronous and asynchronous
signals simultaneously mixing and transmitting both analog and digital signals simultaneously
The device shown isolates digital inputs and outputs from machinery grounds by emitting to a
phototransistor. visible light X rays infrared sound The primary purpose of interface hardware in
an industrial control system is to reduce the processor\'s overhead runtime. provide compatibility
between synchronous and asynchronous transmissions. provide a communications link between
the MPU and peripheral devices. connect the MPU to its internal registers.
Solution
1) Which of the following is true of a mainframe, or central, computer that you may
find in an industrial application?
A. It inter,faces with manufacturing processes directly.
B. It eliminates any need for system firmware.
C. It does away with most applications software.
D. It receives information from one or more workstations.
Answer:
A) It inter,faces with manufacturing processes directly.
2. Multiplexing is a means of----- in order to channel data.
A. organizing the transmission of signals from many lines onto one line
B. transmitting signals by changing their amplitude, frequency, or phase
C. mixing and transmitting both synchronous and asynchronous signals simultaneously
D. mixing and transmitting both analog and digital signals simultaneously
Answer:
A) organizing the transmission of signals from many lines onto one line.
4. The primary purpose of interface hardware in an industrial control system is to
A. reduce the processor\'s overhead runtime.
B. provide compatibility between synchronous and asynchronous transmissions.
C. provide a communications link between the MPU and peripheral devices.
D. connect the MPU to its internal registers.
Answer:
C) provide a communications link between the MPU and peripheral devices..
What type of security is there for a computer serverSolutionA.pdf
What type of security is there for a computer server?
Solution
Answer:
Security is the main concern to every IT organisation as it should be because everything is been
sent through the internet. It is responsibility of IT department manager to provide security to a
computer server. Here are the few types of security for a computer server :.
Which of the following are affine properties (Which are preserved u.pdf
Which of the following are affine properties? (Which are preserved under affine
transformations?)
Distance, circularity, collinearity, magnitude of angle, midpoint of line segment.
Solution
An affine transformation preserves the collinearity relation between points such that points
which lie on the same line (called collinear points) continue to be collinear after the
transformation..
What is a muno and what is a scintillation What is a muno and .pdf
What is a muno and what is a scintillation?
What is a muno and what is a scintillation?
Solution
I suppose you were to ask about Muon and not Muno.
To put it simply, Muons are sub-atomic particles, and do not consitute any other known particle,
that is, they are fundamental particles. Muons are similar to electrons but heavier, in fact, 207
times heavier than electrons.
Muons are fundamental particles, still very unfamiliar for the fact that they are radioactive and
decay with an afterlife of 1.5 millionth of a second. Muons decay into an electron and two
neutrinos as per:
µ ---> e+ e + µ [muon ---> electron + electron antineutrino + muon neutrino]
Scintillation refers to the flashes that appear on a scintillation counter whenever a particle passes
through them. Scintillation counters exist for gamma, x-ray, beta, and alpha radiation. By the
amount of Scintillation energy produced, we can predict the properties of the particles..
A psychologist wishes to conduct a study on the effects of music dep.pdf
A psychologist wishes to conduct a study on the effects of music deprivation on high school
students. A high school class consists of the 30 students numbered in the list below. The
researcher establishes a treatment group of 15 students who will have their portable music
players replaced by experimental players that present the sound of water running. The control
group of 15 students will all get regular portable music players, stuffed full of their favorite
songs.Use the line of random numbers below to select the first 5 students to receive the
treatment. What is the name of the fourth student selected? 59784 44312 15954 09233 00046
74318 02610 57396 16843 38454
00 Aaron 01 Buffy 02 Chandler 03 Cindy 04 Drusilla 05 Eric 06 Fallon 07 Graham 08 heather
09 Hsin-Chi 10 Ismail 11 Jasmine 12 Kiefer 13 Lucia 14 Monte 15 Naomi 16 Otis 17 Polly 18
Quincy 19 Rachael 20 Sarah 21 Stacy 22 Tasha 23 Tuan 24 Ukiah 25 Valerie 26 Wahib 27
Xavier 28 Yolanda 29 Zachary
Solution
The given random can be arranged in two digits numbers as follows:
59, 78, 44, 43, 12, 15, 95, 40, 92, 33, 00, 04, 67, 43, 18, 02, 61, 05, 73, 96, 16, 84, 33, 84, 54
Out of these we have to select the numbers that are between 00 and 29. Selected numbers are as
follows:
12, 15, 00, 04, 18, 02, 05, 16,
So the five first selected students are as follows:
12- Kiefer
15- Naomi
00- Aaron
04- Drusilla
18- Quincy
------------
The name of the fourth student selected is Drusilla..
This week you surveyed many different groups of living things. What D.pdf
This week you surveyed many different groups of living things. What Domain of living things is
thought to have been the first form of life on Earth? Name one piece of evidence that exists for
the claim that these were the earliest living things?
Solution
Archaea are found in the harshest environments on Earth, and are the oldest known organisms on
Earth, appearing in the fossil record over 3.6 BILLION years ago
Archaea display unique capacities, such as methanogenesis and survival at temperatures higher
than 90 °C, that make them crucial for understanding the nature of the biota of early Earth..
Recommended
A health program planning committee wants to address various health .pdf
A health program planning committee wants to address various health problems by developing
programs for each health problem at all four levels of the public health pyramid. What would be
some implications of this perspective on setting priorities and on the nature of the subsequent
community health assessment?
Solution
The bottom tire consists of Infrastructure development: The most essential and set with high
priority. As it affects a larger population .This includes development of basic infrastructures like
electricity, sanitation, clean drinking water, hospitals, transportation and above all education.
The second tire from bottom tire consists of Population based services: It includes public
awareness programmes through print and television media for control of disease and prevention
of population from getting affected. It is a less costly but has a greater impact on mass
population. Healthier food and clean drinking water should be made available to all. It holds the
second highest priority as it will decrease the chance of illness.
The third tire from bottom consists of enabling services. It provides health education and
promotion of health care products among a community. Immunization programmes and
increasing awareness among population regarding the impact of hazardous chemicals on their
health. Enforcement of laws if required for betterment of public health like avoid alcohol while
driving, no smoking zones etc. It holds the third highest priority as prevention is better than cure.
The top tire consists of direct health care: It includes availability clinical care for individuals
affected and their proper treatment in due course of time. This applies to a smaller population but
is also very essential for communities..
Which one of the following significantly contributes to the specifici.pdf
Which one of the following significantly contributes to the specificity of base pairing in double-
stranded DNA, i.e., limits base pairing to cytosine-guanine (C:G) or adenine-thymine (A:T)? the
right handedness of double-stranded DNA hydrogen bonding between bases base-stacking
interactions the phosphodiester bonds between phosphate groups and deoxyribose sugar
molecules
Solution
Answer:
Base-stacking interactions
Explanation:
The purine and pyrimidine bases are hydrophobic and relatively insoluble in water at the near-
neutral pH of the cell. At acidic or alkaline pH the bases become charged and their solubility in
water increases. Hydrophobic stacking interactions in which two or more bases are positioned
with the planes of their rings parallel (like a stack of coins) are one of two important modes of
interaction between bases in nucleic acids. The stacking also involves a combination of van der
Waals and dipole-dipole interactions between the bases. Base stacking helps to minimize contact
of the bases with water, and base-stacking interactions are very important in stabilizing the three-
dimensional structure of nucleic acids..
Which of the following is true of a mainframe, or central, computer t.pdf
Which of the following is true of a mainframe, or central, computer that you may find in an
industrial application? It interfaces with manufacturing processes directly. It eliminates any need
for system firmware. It does away with most applications software. It receives information from
one or more workstations. Multiplexing is a means of in order to channel data. organizing the
transmission of signals from many lines onto one line transmitting signals by changing their
amplitude, frequency, or phase mixing and transmitting both synchronous and asynchronous
signals simultaneously mixing and transmitting both analog and digital signals simultaneously
The device shown isolates digital inputs and outputs from machinery grounds by emitting to a
phototransistor. visible light X rays infrared sound The primary purpose of interface hardware in
an industrial control system is to reduce the processor\'s overhead runtime. provide compatibility
between synchronous and asynchronous transmissions. provide a communications link between
the MPU and peripheral devices. connect the MPU to its internal registers.
Solution
1) Which of the following is true of a mainframe, or central, computer that you may
find in an industrial application?
A. It inter,faces with manufacturing processes directly.
B. It eliminates any need for system firmware.
C. It does away with most applications software.
D. It receives information from one or more workstations.
Answer:
A) It inter,faces with manufacturing processes directly.
2. Multiplexing is a means of----- in order to channel data.
A. organizing the transmission of signals from many lines onto one line
B. transmitting signals by changing their amplitude, frequency, or phase
C. mixing and transmitting both synchronous and asynchronous signals simultaneously
D. mixing and transmitting both analog and digital signals simultaneously
Answer:
A) organizing the transmission of signals from many lines onto one line.
4. The primary purpose of interface hardware in an industrial control system is to
A. reduce the processor\'s overhead runtime.
B. provide compatibility between synchronous and asynchronous transmissions.
C. provide a communications link between the MPU and peripheral devices.
D. connect the MPU to its internal registers.
Answer:
C) provide a communications link between the MPU and peripheral devices..
What type of security is there for a computer serverSolutionA.pdf
What type of security is there for a computer server?
Solution
Answer:
Security is the main concern to every IT organisation as it should be because everything is been
sent through the internet. It is responsibility of IT department manager to provide security to a
computer server. Here are the few types of security for a computer server :.
Which of the following are affine properties (Which are preserved u.pdf
Which of the following are affine properties? (Which are preserved under affine
transformations?)
Distance, circularity, collinearity, magnitude of angle, midpoint of line segment.
Solution
An affine transformation preserves the collinearity relation between points such that points
which lie on the same line (called collinear points) continue to be collinear after the
transformation..
What is a muno and what is a scintillation What is a muno and .pdf
What is a muno and what is a scintillation?
What is a muno and what is a scintillation?
Solution
I suppose you were to ask about Muon and not Muno.
To put it simply, Muons are sub-atomic particles, and do not consitute any other known particle,
that is, they are fundamental particles. Muons are similar to electrons but heavier, in fact, 207
times heavier than electrons.
Muons are fundamental particles, still very unfamiliar for the fact that they are radioactive and
decay with an afterlife of 1.5 millionth of a second. Muons decay into an electron and two
neutrinos as per:
µ ---> e+ e + µ [muon ---> electron + electron antineutrino + muon neutrino]
Scintillation refers to the flashes that appear on a scintillation counter whenever a particle passes
through them. Scintillation counters exist for gamma, x-ray, beta, and alpha radiation. By the
amount of Scintillation energy produced, we can predict the properties of the particles..
A psychologist wishes to conduct a study on the effects of music dep.pdf
A psychologist wishes to conduct a study on the effects of music deprivation on high school
students. A high school class consists of the 30 students numbered in the list below. The
researcher establishes a treatment group of 15 students who will have their portable music
players replaced by experimental players that present the sound of water running. The control
group of 15 students will all get regular portable music players, stuffed full of their favorite
songs.Use the line of random numbers below to select the first 5 students to receive the
treatment. What is the name of the fourth student selected? 59784 44312 15954 09233 00046
74318 02610 57396 16843 38454
00 Aaron 01 Buffy 02 Chandler 03 Cindy 04 Drusilla 05 Eric 06 Fallon 07 Graham 08 heather
09 Hsin-Chi 10 Ismail 11 Jasmine 12 Kiefer 13 Lucia 14 Monte 15 Naomi 16 Otis 17 Polly 18
Quincy 19 Rachael 20 Sarah 21 Stacy 22 Tasha 23 Tuan 24 Ukiah 25 Valerie 26 Wahib 27
Xavier 28 Yolanda 29 Zachary
Solution
The given random can be arranged in two digits numbers as follows:
59, 78, 44, 43, 12, 15, 95, 40, 92, 33, 00, 04, 67, 43, 18, 02, 61, 05, 73, 96, 16, 84, 33, 84, 54
Out of these we have to select the numbers that are between 00 and 29. Selected numbers are as
follows:
12, 15, 00, 04, 18, 02, 05, 16,
So the five first selected students are as follows:
12- Kiefer
15- Naomi
00- Aaron
04- Drusilla
18- Quincy
------------
The name of the fourth student selected is Drusilla..
This week you surveyed many different groups of living things. What D.pdf
This week you surveyed many different groups of living things. What Domain of living things is
thought to have been the first form of life on Earth? Name one piece of evidence that exists for
the claim that these were the earliest living things?
Solution
Archaea are found in the harshest environments on Earth, and are the oldest known organisms on
Earth, appearing in the fossil record over 3.6 BILLION years ago
Archaea display unique capacities, such as methanogenesis and survival at temperatures higher
than 90 °C, that make them crucial for understanding the nature of the biota of early Earth..
Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains.pdf
Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains 10 yellow marbles and 6
brown marbles. Urn III contains 7 yellow marbles and 11 brown marbles. The probability that
Urn I will be selected is 0.42, th e probability that Urn II will be selected is 0.45, and the
probability that Urn III will be selected i s 0.13. Find each probability: D. The probability that
the marble selected was from Urn II, if it is known that the marble was yellow.
Solution.
The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequenc.pdf
The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequence. On average, how often
would this enzyme cut a random sequence of DNA? How many times would the enzyme cut the
E. coli genome? (The E. coli genome contains 4.6 x 106 bp).
Solution
There are 4 types of bases so the chances of each base to cut in the sequence is
(1/4) (1/4) (1/4) (1/4) (1/4) (1/4)= (1/4)6 base pairs=1/4096. Now as the DNA sequence is
completely random and we can expect that the enzyme may cut the DNA once every 4096 bases.
Now the E.coli genome contains 4.6 x 106 base pairs of sequence.
Then the cutting site will be in the whole E.coli genome is (4.6*106 /4096)= 1123 times..
The vertical motion of a mass attached to a spring is described by t.pdf
The vertical motion of a mass attached to a spring is described by the IVP
Solution
the associated equation is 1/4x^2+x+1=0 or x^2+4x+4=0 (x+2)^2=0 so the solution
is of the form x=Ae^(-2t)+Bte^(-2t) using x(0)=4 we get A=4 x\'(0)=-2A+B=2 B=10 x=4e^(-
2t)+10te^(-2t) x\'=-8e^(-2t)+10e^(-2t)-16te^(-2t)=0 2-16t=0 t=1/8 x\'>0 before 1/8 and x\'>0
after 1/8 So the maximum vertical is when t=1/8.
There are several types of bacterial toxins that are categorized as .pdf
There are several types of bacterial toxins that are categorized as spreading factors. One of those
is leukocidins. Do the 3 species of staphylococci you\'ve described produce leukocidins? What is
their function? How do they work?
Solution
Leukocidins are toxins secreted by Staphylococcus aureus, Streptococcus pyogenes, and S.
pneumoniae
Among the three important Staphylococci, viz., S. aureus, S.epidermidis, and S.saprophiticus,
leukocidins play a major role in the pathogenesis of S. aureus.
Leukocidins are toxins that kill leukocytes (lit., Leuko: for leukocytes and caedo: killing) with or
without lysis.
In Streptococcus aureus, they attacks neutrophils, disrupt their membrane that leads to the release
of the lysosomal contents, thereby, killing them..
the life span of the general population. is the distribution uni.pdf
the life span of the general population. is the distribution uniform, unimodal or bimodal?
the life span of the general population. is the distribution uniform, unimodal or bimodal?
the life span of the general population. is the distribution uniform, unimodal or bimodal?
Solution
the life span of the general population follows exponential distribution.It is unimodel..
1.2.3.4.SolutionThe number of observations per cell was 3To.pdf
1.2.3.4.
Solution
The number of observations per cell was: 3
Total df=23
Total observation = 24
Interaction Df=3
Df for amplifier and position are (1 and 3) any order
Number of factors for amplifier and position = (2 and 4) any order
Observations per cell = 24/(2*4) =3
Two-factor ANOVA with replication
The two factors are PVC and Temperature.
Each cell having 3 observations or replications..
A volume of 0.0125 m3 at a temperature of 1300C and pressure at 90kP.pdf
A volume of 0.0125 m3 at a temperature of 1300C and pressure at 90kPa has the temperature
increased by 150oF.
What is the new volume?
Solution
increase in temperature = 150 F = 65.556 C
As process is not mentioned, considering isobaric process (p=Constant)
New volume = Vf
= (Tf*Vi/Ti)
= 0.0125 * (1300+65.556)/1300
= 0.01313 m3.
Please help!!!I can change a link layer protocol and the applicati.pdf
Please help!!!
I can change a link layer protocol and the application layer protocol does not need to know or be
made aware of the change. Explain why?
Solution
Answer :-
The protocols are like a pile of building blocks stacked one upon another. Because of this
appearance, the structure is often called a stack or protocol stack.
A layer does not define a single protocol it defines a data communications function that may be
performed by any number of protocols. Therefore, each layer may contain multiple protocols,
each providing a service suitable to the function of that layer. For example, a file transfer
protocol and an electronic mail protocol both provide user services, and both are part of the
Application Layer.
1) Application Layer :-
The Application Layer is the level of the protocol hierarchy where user-accessed network
processes reside. In this text, a TCP/IP application is any network process that occurs above the
Transport Layer. This includes all of the processes that users directly interact with as well as
other processes at this level that users are not necessarily aware of.
2) Presentation Layer :-
For cooperating applications to exchange data, they must agree about how data is represented. In
OSI, the Presentation Layer provides standard data presentation routines. This function is
frequently handled within the applications in TCP/IP, though TCP/IP protocols such as XDR and
MIME also perform this function.
3) Session Layer :-
As with the Presentation Layer, the Session Layer is not identifiable as a separate layer in the
TCP/IP protocol hierarchy. The OSI Session Layer manages the sessions (connections) between
cooperating applications. In TCP/IP, this function largely occurs in the Transport Layer, and the
term “session” is not used; instead, the terms “socket” and “port” are used to describe the path
over which cooperating applications communicate.
4) Transport Layer:
Much of our discussion of TCP/IP is directed to the protocols that occur in the Transport Layer.
The Transport Layer in the OSI reference model guarantees that the receiver gets the data exactly
as it was sent. In TCP/IP, this function is performed by the Transmission Control Protocol
(TCP). However, TCP/IP offers a second Transport Layer service, User Datagram Protocol
(UDP), that does not perform the end-to-end reliability checks.
5) Network Layer :-
The Network Layer manages connections across the network and isolates the upper layer
protocols from the details of the underlying network. The Internet Protocol (IP), which isolates
the upper layers from the underlying network and handles the addressing and delivery of data, is
usually described as TCP/IP’s Network Layer.
6) Data Link Layer :-
The reliable delivery of data across the underlying physical network is handled by the Data Link
Layer. TCP/IP rarely creates protocols in the Data Link Layer. Most RFCs that relate to the Data
Link Layer discuss how IP can make use of existing data link protocols.
7) Physi.
4. (a) Negate the statement Sarah has a spaceship and has three f.pdf
4. (a) Negate the statement \'\'Sarah has a spaceship and has three fingers and is from Venus\'\'
and convert it to formal logic. (b) Simplify the statement from (a) as much as possible using
equivalence rules.
Solution
a) sarah neither have spaceship nor has three fingers nor is from venus.
b)s: sarah
S: spaceship
t: three fingers
v:venus
s-->S ^ t ^ v.
Match the bones in column A with the features in column B. Place the .pdf
Match the bones in column A with the features in column B. Place the letter of your choice in
the space provided. (Some answers are used more than once.) a. Ethmoid bone b. Frontal bone
c. Occipital bone d. Parietal bone e. Sphenoid bone f. Temporal bone
Solution
correct answers
a) Ethmoid bone - 2) Cribriform plate
b) Frontal lobe - 8) middle nosal concha
c) Occipital bone - 5) foramen magnum
d) Parietal bone - 1) form sagittal, coronal , squamous and lambdiod sutures
e) Sphenoid bone - 10) Sella turcica
f) Temporal bone - 11) Styloid process.
1. Prepare and analize the common statement for An.pdf
1.
Prepare and analize the common statement for Anandam Manufacturing Company. Show all
calculations.
2.
Prepare and analize the cash flow statement for Anandam Manufacturing Company. Show all
calculations.
3.
Calculate the ratios based on case Exhibit 3. Show all calculations. Based on financial analysis
of financial statements, would a loan officer grant a loan to Anandam financial company?
EXHIBIT 3: INDUSTRY AVERAGE OF KEY RATIos Sector Average Ratio 2:301 Current
ratio Acid test ratio (quick ratio) 120 Receivable tunover ratioimes turnover 52 days Receivable
days Inventory turmover ratio Inventory days Long-term detttototal debt Debt-to-equity ratio
Gross profit rato Net profit ratio Retun on equity Retum on total assets 4.85 times 75 days 24%
35% 40% 18% 22% 10% Total asset turnover ratio Ti Fixed asset turnover ratio Current asset
tunover ratio3 interest coverage ratio (tmes 10 nterest earned Warking captal smover ratio Retu
on ed sset
Solution
1….Common -size Income statement 2012-13 % to Total sales 2013-14 % to Total sales
2014-15 % to Total sales Analysis of % to Total sales proportion Sales Cash 200 10.00%
480 10.00% 800 10.00% Credit 1800 90.00% 4320 90.00% 7200 90.00% Total Sales 2000
100.00% 4800 100.00% 8000 100.00% COGS 1240 62.00% 2832 59.00% 4800 60.00%
COGS has reduced in the last 2 yrs. Gross profit 760 38.00% 1968 41.00% 3200 40.00% so,
G/P has increased Operating Expenses: Gen.adm.& sell.exp. 80 4.00% 450 9.38% 1000
12.50% % to sales has doubled in 2013-14 & increased by more than 25% in 2014-15
Depreciation 100 5.00% 400 8.33% 660 8.25% Has increased by 3% in the last 2 yrs.
Int.exp.(on borrowings) 60 3.00% 158 3.29% 340 4.25% Has slightly increased in 2014-15
Profit before tax(PBT) 520 26.00% 960 20.00% 1200 15.00% PBT% has reduced gradually in
both the yrs.due to increase in all operating expenses esp.gen admn.& sell exp. Tax at 30% 156
7.80% 288 6.00% 360 4.50% Decrease % due to decrease in profit % Profit after tax(PAT) 364
18.20% 672 14.00% 840 10.50% PAT% has reduced gradually in both the yrs.due to increase in
all operating expenses esp.gen admn.& sell exp. Common-size Balance
Sheet Assets % to Total % to Total % to Total Fixed assets(net of dep) 1900 74.22%
2500 44.64% 4700 51.33% Assets have decreased to the total Current assets Cash & Cash
equivalents 40 1.56% 100 1.79% 106 1.16% Slight variation in ratio to total Accounts
Receivables 300 11.72% 1500 26.79% 2100 22.94% Has increased in 2013-14 & again
decreased in 2014-15 Inventories 320 12.50% 1500 26.79% 2250 24.57% Has increased in
2013-14 & again decreased in 2014-15 Total 2560 100.00% 5600 100.00% 9156 100.00%
Equity & Liabilities Equity share capital($ 10) 1200 46.88% 1600 28.57% 2000 21.84%
Decreased steadily in all the 2 yrs. Reserves & surplus 364 14.22% 1036 18.50% 1876 20.49%
Increased steadily in both 2 yrs.due to increase in $ net income in those yrs. Long-term
borrowings 736 28.75% 1236 22.07% 2500 27.30% Decreased in 2013-14 & again increased i.
Julia primarily uses her minivan transport her grandchildren, Vera, C.pdf
Julia primarily uses her minivan transport her grandchildren, Vera, Chuck, and Dave, to and
from school. She, 25/l00 bodily injury$50, 000 property damage coverage. Julia has a $50
deductible. Her driver-rating factor is 1.75. Her is classified as B, 12. What is Julia\'s annual
premium?
Solution
you have deducted 50 twice.That is mistake as she has either collision deductible or
comprehensive deductible.so it is to be summed only once..
I have an interface Peron that has firstnamestring, lastnamestring.pdf
I have an interface Peron that has firstname:string, lastname:string. I have a Class Professor
implements Person has salary:Dollar, and Class Student implements Person has major:string. I
need to write a program that compiles. I cannot upload the UML diagram.
Solution
import java.util.*;
public interface persom {
String Fname=\"\";
String Lname=\"\";
public void getint();
}
class proffesor implements persom{
String Fname;
String Lname;
int salary;
public void getint(){
Scanner sc=new Scanner(System.in);
System.out.println(\"enter Fist name:\");
Fname=sc.next();
System.out.println(\"enter Last name:\");
Lname=sc.next();
System.out.println(\"enter salaruy:\");
salary=sc.nextInt();
System.out.println(\"Fname is\"+Fname);
System.out.println(\"Lname is\"+Lname);
System.out.println(\"major is\"+salary);
}
}
class student implements persom{
String Fname;
String Lname;
String major;
public void getint(){
Scanner sc=new Scanner(System.in);
System.out.println(\"enter Fist name:\");
Fname=sc.next();
System.out.println(\"enter Last name:\");
Lname=sc.next();
System.out.println(\"enter major:\");
major=sc.next();
System.out.println(\"Fname is\"+Fname);
System.out.println(\"Lname is\"+Lname);
System.out.println(\"major is\"+major);
}
}
public class mymainever {
public static void main(String[] args) {
student s=new student();
//s.getint();
persom p;
p=s;
p.getint();
}
}.
How do condition variables and semaphores differ when sending a sign.pdf
How do condition variables and semaphores differ when sending a signal?
Solution
conditional varaibles:
1) these are the variable which are combined with mutexes and used for implementing
primitives
which are synchronized
2)these variable allows the thread for wait for completion by using wait after the completion of
job then it invoke other by using signal method
semaphore:
1)here the shared resource can accesible by only one thread at a time
2)here at first n threads can access a resource concurrently and then it provide signal mechanism
by using of wait and signal methods.
Function of yolk sac in human embryos Forms blood cells shock absorb.pdf
Function of yolk sac in human embryos? Forms blood cells shock absorber forms the placenta
Nutrition. Which cells allow embryos to implant in the Uterus? Syncytriophoblast trophoblast E-
cadherin\'s goner cell mass
Solution
Option D. Yolk sac is responsible for the process of viteline circulation in which it provides
nutrition to the developing embryo through the primitive aorta.
2. Option B. As trophoblasts are the cells which make the placenta. Placenta connects the foetus
to the uterine wall which helps in transfer of nutrients etc to the developing foetus..
Explain the basic characteristics of a partnership.SolutionThe.pdf
Explain the basic characteristics of a partnership.
Solution
There must be at least two or more persons to form Partnership.
There must be agreement between partners and such agreement can be Oral or Written. The
agreement forms the basis of Mutual rights and duties of Partners.
Partnership can be formed for the purpose of carrying on some business with the intension of
earning some profit.
The Agreement must include how the profit or loss to be shared. If the persons joins firm for
some Charity then it will not be called partnership and if the persons is deprived of his right to
share the profit then he/she will not be called partner. And it is not necessary that all partner
share loss also, it may be agreed between partners that some oo one of the partners will not be
liable for losses.
All the persons have Unlimited liability i.e. if the assets of the company is not sufficient to pay
the liabilities then the Creditors can come after the personal property of the Partners.
In the eyes of Law, A partner and Partnership firm are same, i.e. Partnership firm doesn’t have
its own Legal identity.
The business must be Legal..
Explain the functions of the different types of meristems found in pl.pdf
Explain the functions of the different types of meristems found in plants. Be able to distinguish
between secretory tissues.
Solution
In plants, meristems are resent broadly of three types namely:
1. Apical meristem
2. intercalary Meristem
3. Lateral meristem
1. APICAL MERISTEM
This is present at the root and shoot apex of the plant and is actively dividing zone. Shoot apcal
meristem causes plant to grow up and root apical meristem causes thr roots to grow down in the
soil.
Apical meristems keep dividing and result in the primary growth of plant and the tissues formed
as their division as named as Protoderm and Procambium.
2. INTERCALARY MERISTEM
Intercalary meristem is found between the mature tisues like nodes and internodes. It functions to
regulate the regenration of Tissues when the leaves and grasses are grazed by herbivores. This
meristem is found mainly in monocot plants.
3. LATERAL MERISTEM
This type of meristems is developed late in the plant life cycle and thus it is also known as
secondary meristem. It functions in the secondary growth of plant where plant grows in girth not
in height. Cork cambium fascisular cambium and interfascicular cambium are types of lateral
meristems..
Crossing-over occurs during prophase I of meiosis.True or False.pdf
Crossing-over occurs during prophase I of meiosis.
True or False
Solution
It is true answer.
Because crossing over occurs in the prophase I of meiotic cell division the substage of prophase I
which is pachytene where the non sister chromatids exchanges their arms and that leads to
variation in organisms..
A single card is drawn from a standard deck of cards. What is the pr.pdf
A single card is drawn from a standard deck of cards. What is the probability that the card is a
spade given that it is not a heart?
Solution
P(spade|not heart)= P(spade) because spade and heart can\'t happen in the same card.
They are independent event.
let A be the event of getting a spade.
total cards = n(s) = 52
total spade = n(A) = 13
P(A) = n(A)/ n(S) = 13/52 = 1/4..
Consider the security of a personal computer. List a few of the atta.pdf
Consider the security of a personal computer. List a few of the attackers who might break into
the computer, their incentives, and the associated costs and risks to the attacker.
Solution
A top-down approach to identifying threats and countermeasures, CISOs should consider a threat
modeling technique allows the target application to be decomposed to reveal its attack surface
and subsequently its relevant threats, associated countermeasures, and finally, its security control
gaps and design flaws. applications are also a source of potential risks especially when new or
different technologies are integrated within applications. As applications evolve by offering new
services to citizens, clients, customers and employees, it is also necessary to plan for mitigation
of new vulnerabilities introduced by the adoption and implementation of new technologies such
as mobile devices, web 2.0 and new services such as cloud computing. Adopting a risk
framework to evaluate the risks introduced by new technologies is essential to determine which
countermeasures to adopt to mitigate these new risks. threat agents seek financial gain such as by
attacking applications to compromise users\' sensitive data and company’s proprietary
information for financial gain, fraud as well as for competitive advantage the risks posed by
these threat agents, it is necessary to determine the risk exposure and factor the probability and
the impact of these threats as well as to identify the type of application vulnerabilities that can be
exploited by these threat agents. The exploit of some of these application vulnerabilities might
severely and negatively impact the organization and personal computer.
data from legal in relation to law suits and regulatory fines and fraud data that includes amount
of money losses incurred because of online fraud. All this type of information is essential to
determine the overall impact. In absence of this data, the best the CISO can do is to use data
breach incident data from public sources and data breach incident reports. In part I of this guide,
we provided some examples of how this data can be used to estimate impact. We documented
what are the critical factors to estimate impacts of data breaches: these as the value of the data
assets liability for the organization in case these assets are lost. Once the potential business
impact of a data breach is estimated, the next step is to determine how much should be spent to
mitigate the risk. At high level, this is a risk strategy decision that depends on the organization
risk culture and the organization priorities for mitigating risks..
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
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Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains.pdf
Urn I contains 9 yellow marbles and 7 brown marbles. Urn II contains 10 yellow marbles and 6
brown marbles. Urn III contains 7 yellow marbles and 11 brown marbles. The probability that
Urn I will be selected is 0.42, th e probability that Urn II will be selected is 0.45, and the
probability that Urn III will be selected i s 0.13. Find each probability: D. The probability that
the marble selected was from Urn II, if it is known that the marble was yellow.
Solution.
The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequenc.pdf
The restriction enzyme EcoRI recognizes a 6 bp (palindromic) sequence. On average, how often
would this enzyme cut a random sequence of DNA? How many times would the enzyme cut the
E. coli genome? (The E. coli genome contains 4.6 x 106 bp).
Solution
There are 4 types of bases so the chances of each base to cut in the sequence is
(1/4) (1/4) (1/4) (1/4) (1/4) (1/4)= (1/4)6 base pairs=1/4096. Now as the DNA sequence is
completely random and we can expect that the enzyme may cut the DNA once every 4096 bases.
Now the E.coli genome contains 4.6 x 106 base pairs of sequence.
Then the cutting site will be in the whole E.coli genome is (4.6*106 /4096)= 1123 times..
The vertical motion of a mass attached to a spring is described by t.pdf
The vertical motion of a mass attached to a spring is described by the IVP
Solution
the associated equation is 1/4x^2+x+1=0 or x^2+4x+4=0 (x+2)^2=0 so the solution
is of the form x=Ae^(-2t)+Bte^(-2t) using x(0)=4 we get A=4 x\'(0)=-2A+B=2 B=10 x=4e^(-
2t)+10te^(-2t) x\'=-8e^(-2t)+10e^(-2t)-16te^(-2t)=0 2-16t=0 t=1/8 x\'>0 before 1/8 and x\'>0
after 1/8 So the maximum vertical is when t=1/8.
There are several types of bacterial toxins that are categorized as .pdf
There are several types of bacterial toxins that are categorized as spreading factors. One of those
is leukocidins. Do the 3 species of staphylococci you\'ve described produce leukocidins? What is
their function? How do they work?
Solution
Leukocidins are toxins secreted by Staphylococcus aureus, Streptococcus pyogenes, and S.
pneumoniae
Among the three important Staphylococci, viz., S. aureus, S.epidermidis, and S.saprophiticus,
leukocidins play a major role in the pathogenesis of S. aureus.
Leukocidins are toxins that kill leukocytes (lit., Leuko: for leukocytes and caedo: killing) with or
without lysis.
In Streptococcus aureus, they attacks neutrophils, disrupt their membrane that leads to the release
of the lysosomal contents, thereby, killing them..
the life span of the general population. is the distribution uni.pdf
the life span of the general population. is the distribution uniform, unimodal or bimodal?
the life span of the general population. is the distribution uniform, unimodal or bimodal?
the life span of the general population. is the distribution uniform, unimodal or bimodal?
Solution
the life span of the general population follows exponential distribution.It is unimodel..
1.2.3.4.SolutionThe number of observations per cell was 3To.pdf
1.2.3.4.
Solution
The number of observations per cell was: 3
Total df=23
Total observation = 24
Interaction Df=3
Df for amplifier and position are (1 and 3) any order
Number of factors for amplifier and position = (2 and 4) any order
Observations per cell = 24/(2*4) =3
Two-factor ANOVA with replication
The two factors are PVC and Temperature.
Each cell having 3 observations or replications..
A volume of 0.0125 m3 at a temperature of 1300C and pressure at 90kP.pdf
A volume of 0.0125 m3 at a temperature of 1300C and pressure at 90kPa has the temperature
increased by 150oF.
What is the new volume?
Solution
increase in temperature = 150 F = 65.556 C
As process is not mentioned, considering isobaric process (p=Constant)
New volume = Vf
= (Tf*Vi/Ti)
= 0.0125 * (1300+65.556)/1300
= 0.01313 m3.
Please help!!!I can change a link layer protocol and the applicati.pdf
Please help!!!
I can change a link layer protocol and the application layer protocol does not need to know or be
made aware of the change. Explain why?
Solution
Answer :-
The protocols are like a pile of building blocks stacked one upon another. Because of this
appearance, the structure is often called a stack or protocol stack.
A layer does not define a single protocol it defines a data communications function that may be
performed by any number of protocols. Therefore, each layer may contain multiple protocols,
each providing a service suitable to the function of that layer. For example, a file transfer
protocol and an electronic mail protocol both provide user services, and both are part of the
Application Layer.
1) Application Layer :-
The Application Layer is the level of the protocol hierarchy where user-accessed network
processes reside. In this text, a TCP/IP application is any network process that occurs above the
Transport Layer. This includes all of the processes that users directly interact with as well as
other processes at this level that users are not necessarily aware of.
2) Presentation Layer :-
For cooperating applications to exchange data, they must agree about how data is represented. In
OSI, the Presentation Layer provides standard data presentation routines. This function is
frequently handled within the applications in TCP/IP, though TCP/IP protocols such as XDR and
MIME also perform this function.
3) Session Layer :-
As with the Presentation Layer, the Session Layer is not identifiable as a separate layer in the
TCP/IP protocol hierarchy. The OSI Session Layer manages the sessions (connections) between
cooperating applications. In TCP/IP, this function largely occurs in the Transport Layer, and the
term “session” is not used; instead, the terms “socket” and “port” are used to describe the path
over which cooperating applications communicate.
4) Transport Layer:
Much of our discussion of TCP/IP is directed to the protocols that occur in the Transport Layer.
The Transport Layer in the OSI reference model guarantees that the receiver gets the data exactly
as it was sent. In TCP/IP, this function is performed by the Transmission Control Protocol
(TCP). However, TCP/IP offers a second Transport Layer service, User Datagram Protocol
(UDP), that does not perform the end-to-end reliability checks.
5) Network Layer :-
The Network Layer manages connections across the network and isolates the upper layer
protocols from the details of the underlying network. The Internet Protocol (IP), which isolates
the upper layers from the underlying network and handles the addressing and delivery of data, is
usually described as TCP/IP’s Network Layer.
6) Data Link Layer :-
The reliable delivery of data across the underlying physical network is handled by the Data Link
Layer. TCP/IP rarely creates protocols in the Data Link Layer. Most RFCs that relate to the Data
Link Layer discuss how IP can make use of existing data link protocols.
7) Physi.
4. (a) Negate the statement Sarah has a spaceship and has three f.pdf
4. (a) Negate the statement \'\'Sarah has a spaceship and has three fingers and is from Venus\'\'
and convert it to formal logic. (b) Simplify the statement from (a) as much as possible using
equivalence rules.
Solution
a) sarah neither have spaceship nor has three fingers nor is from venus.
b)s: sarah
S: spaceship
t: three fingers
v:venus
s-->S ^ t ^ v.
Match the bones in column A with the features in column B. Place the .pdf
Match the bones in column A with the features in column B. Place the letter of your choice in
the space provided. (Some answers are used more than once.) a. Ethmoid bone b. Frontal bone
c. Occipital bone d. Parietal bone e. Sphenoid bone f. Temporal bone
Solution
correct answers
a) Ethmoid bone - 2) Cribriform plate
b) Frontal lobe - 8) middle nosal concha
c) Occipital bone - 5) foramen magnum
d) Parietal bone - 1) form sagittal, coronal , squamous and lambdiod sutures
e) Sphenoid bone - 10) Sella turcica
f) Temporal bone - 11) Styloid process.
1. Prepare and analize the common statement for An.pdf
1.
Prepare and analize the common statement for Anandam Manufacturing Company. Show all
calculations.
2.
Prepare and analize the cash flow statement for Anandam Manufacturing Company. Show all
calculations.
3.
Calculate the ratios based on case Exhibit 3. Show all calculations. Based on financial analysis
of financial statements, would a loan officer grant a loan to Anandam financial company?
EXHIBIT 3: INDUSTRY AVERAGE OF KEY RATIos Sector Average Ratio 2:301 Current
ratio Acid test ratio (quick ratio) 120 Receivable tunover ratioimes turnover 52 days Receivable
days Inventory turmover ratio Inventory days Long-term detttototal debt Debt-to-equity ratio
Gross profit rato Net profit ratio Retun on equity Retum on total assets 4.85 times 75 days 24%
35% 40% 18% 22% 10% Total asset turnover ratio Ti Fixed asset turnover ratio Current asset
tunover ratio3 interest coverage ratio (tmes 10 nterest earned Warking captal smover ratio Retu
on ed sset
Solution
1….Common -size Income statement 2012-13 % to Total sales 2013-14 % to Total sales
2014-15 % to Total sales Analysis of % to Total sales proportion Sales Cash 200 10.00%
480 10.00% 800 10.00% Credit 1800 90.00% 4320 90.00% 7200 90.00% Total Sales 2000
100.00% 4800 100.00% 8000 100.00% COGS 1240 62.00% 2832 59.00% 4800 60.00%
COGS has reduced in the last 2 yrs. Gross profit 760 38.00% 1968 41.00% 3200 40.00% so,
G/P has increased Operating Expenses: Gen.adm.& sell.exp. 80 4.00% 450 9.38% 1000
12.50% % to sales has doubled in 2013-14 & increased by more than 25% in 2014-15
Depreciation 100 5.00% 400 8.33% 660 8.25% Has increased by 3% in the last 2 yrs.
Int.exp.(on borrowings) 60 3.00% 158 3.29% 340 4.25% Has slightly increased in 2014-15
Profit before tax(PBT) 520 26.00% 960 20.00% 1200 15.00% PBT% has reduced gradually in
both the yrs.due to increase in all operating expenses esp.gen admn.& sell exp. Tax at 30% 156
7.80% 288 6.00% 360 4.50% Decrease % due to decrease in profit % Profit after tax(PAT) 364
18.20% 672 14.00% 840 10.50% PAT% has reduced gradually in both the yrs.due to increase in
all operating expenses esp.gen admn.& sell exp. Common-size Balance
Sheet Assets % to Total % to Total % to Total Fixed assets(net of dep) 1900 74.22%
2500 44.64% 4700 51.33% Assets have decreased to the total Current assets Cash & Cash
equivalents 40 1.56% 100 1.79% 106 1.16% Slight variation in ratio to total Accounts
Receivables 300 11.72% 1500 26.79% 2100 22.94% Has increased in 2013-14 & again
decreased in 2014-15 Inventories 320 12.50% 1500 26.79% 2250 24.57% Has increased in
2013-14 & again decreased in 2014-15 Total 2560 100.00% 5600 100.00% 9156 100.00%
Equity & Liabilities Equity share capital($ 10) 1200 46.88% 1600 28.57% 2000 21.84%
Decreased steadily in all the 2 yrs. Reserves & surplus 364 14.22% 1036 18.50% 1876 20.49%
Increased steadily in both 2 yrs.due to increase in $ net income in those yrs. Long-term
borrowings 736 28.75% 1236 22.07% 2500 27.30% Decreased in 2013-14 & again increased i.
Julia primarily uses her minivan transport her grandchildren, Vera, C.pdf
Julia primarily uses her minivan transport her grandchildren, Vera, Chuck, and Dave, to and
from school. She, 25/l00 bodily injury$50, 000 property damage coverage. Julia has a $50
deductible. Her driver-rating factor is 1.75. Her is classified as B, 12. What is Julia\'s annual
premium?
Solution
you have deducted 50 twice.That is mistake as she has either collision deductible or
comprehensive deductible.so it is to be summed only once..
I have an interface Peron that has firstnamestring, lastnamestring.pdf
I have an interface Peron that has firstname:string, lastname:string. I have a Class Professor
implements Person has salary:Dollar, and Class Student implements Person has major:string. I
need to write a program that compiles. I cannot upload the UML diagram.
Solution
import java.util.*;
public interface persom {
String Fname=\"\";
String Lname=\"\";
public void getint();
}
class proffesor implements persom{
String Fname;
String Lname;
int salary;
public void getint(){
Scanner sc=new Scanner(System.in);
System.out.println(\"enter Fist name:\");
Fname=sc.next();
System.out.println(\"enter Last name:\");
Lname=sc.next();
System.out.println(\"enter salaruy:\");
salary=sc.nextInt();
System.out.println(\"Fname is\"+Fname);
System.out.println(\"Lname is\"+Lname);
System.out.println(\"major is\"+salary);
}
}
class student implements persom{
String Fname;
String Lname;
String major;
public void getint(){
Scanner sc=new Scanner(System.in);
System.out.println(\"enter Fist name:\");
Fname=sc.next();
System.out.println(\"enter Last name:\");
Lname=sc.next();
System.out.println(\"enter major:\");
major=sc.next();
System.out.println(\"Fname is\"+Fname);
System.out.println(\"Lname is\"+Lname);
System.out.println(\"major is\"+major);
}
}
public class mymainever {
public static void main(String[] args) {
student s=new student();
//s.getint();
persom p;
p=s;
p.getint();
}
}.
How do condition variables and semaphores differ when sending a sign.pdf
How do condition variables and semaphores differ when sending a signal?
Solution
conditional varaibles:
1) these are the variable which are combined with mutexes and used for implementing
primitives
which are synchronized
2)these variable allows the thread for wait for completion by using wait after the completion of
job then it invoke other by using signal method
semaphore:
1)here the shared resource can accesible by only one thread at a time
2)here at first n threads can access a resource concurrently and then it provide signal mechanism
by using of wait and signal methods.
Function of yolk sac in human embryos Forms blood cells shock absorb.pdf
Function of yolk sac in human embryos? Forms blood cells shock absorber forms the placenta
Nutrition. Which cells allow embryos to implant in the Uterus? Syncytriophoblast trophoblast E-
cadherin\'s goner cell mass
Solution
Option D. Yolk sac is responsible for the process of viteline circulation in which it provides
nutrition to the developing embryo through the primitive aorta.
2. Option B. As trophoblasts are the cells which make the placenta. Placenta connects the foetus
to the uterine wall which helps in transfer of nutrients etc to the developing foetus..
Explain the basic characteristics of a partnership.SolutionThe.pdf
Explain the basic characteristics of a partnership.
Solution
There must be at least two or more persons to form Partnership.
There must be agreement between partners and such agreement can be Oral or Written. The
agreement forms the basis of Mutual rights and duties of Partners.
Partnership can be formed for the purpose of carrying on some business with the intension of
earning some profit.
The Agreement must include how the profit or loss to be shared. If the persons joins firm for
some Charity then it will not be called partnership and if the persons is deprived of his right to
share the profit then he/she will not be called partner. And it is not necessary that all partner
share loss also, it may be agreed between partners that some oo one of the partners will not be
liable for losses.
All the persons have Unlimited liability i.e. if the assets of the company is not sufficient to pay
the liabilities then the Creditors can come after the personal property of the Partners.
In the eyes of Law, A partner and Partnership firm are same, i.e. Partnership firm doesn’t have
its own Legal identity.
The business must be Legal..
Explain the functions of the different types of meristems found in pl.pdf
Explain the functions of the different types of meristems found in plants. Be able to distinguish
between secretory tissues.
Solution
In plants, meristems are resent broadly of three types namely:
1. Apical meristem
2. intercalary Meristem
3. Lateral meristem
1. APICAL MERISTEM
This is present at the root and shoot apex of the plant and is actively dividing zone. Shoot apcal
meristem causes plant to grow up and root apical meristem causes thr roots to grow down in the
soil.
Apical meristems keep dividing and result in the primary growth of plant and the tissues formed
as their division as named as Protoderm and Procambium.
2. INTERCALARY MERISTEM
Intercalary meristem is found between the mature tisues like nodes and internodes. It functions to
regulate the regenration of Tissues when the leaves and grasses are grazed by herbivores. This
meristem is found mainly in monocot plants.
3. LATERAL MERISTEM
This type of meristems is developed late in the plant life cycle and thus it is also known as
secondary meristem. It functions in the secondary growth of plant where plant grows in girth not
in height. Cork cambium fascisular cambium and interfascicular cambium are types of lateral
meristems..
Crossing-over occurs during prophase I of meiosis.True or False.pdf
Crossing-over occurs during prophase I of meiosis.
True or False
Solution
It is true answer.
Because crossing over occurs in the prophase I of meiotic cell division the substage of prophase I
which is pachytene where the non sister chromatids exchanges their arms and that leads to
variation in organisms..
A single card is drawn from a standard deck of cards. What is the pr.pdf
A single card is drawn from a standard deck of cards. What is the probability that the card is a
spade given that it is not a heart?
Solution
P(spade|not heart)= P(spade) because spade and heart can\'t happen in the same card.
They are independent event.
let A be the event of getting a spade.
total cards = n(s) = 52
total spade = n(A) = 13
P(A) = n(A)/ n(S) = 13/52 = 1/4..
Consider the security of a personal computer. List a few of the atta.pdf
Consider the security of a personal computer. List a few of the attackers who might break into
the computer, their incentives, and the associated costs and risks to the attacker.
Solution
A top-down approach to identifying threats and countermeasures, CISOs should consider a threat
modeling technique allows the target application to be decomposed to reveal its attack surface
and subsequently its relevant threats, associated countermeasures, and finally, its security control
gaps and design flaws. applications are also a source of potential risks especially when new or
different technologies are integrated within applications. As applications evolve by offering new
services to citizens, clients, customers and employees, it is also necessary to plan for mitigation
of new vulnerabilities introduced by the adoption and implementation of new technologies such
as mobile devices, web 2.0 and new services such as cloud computing. Adopting a risk
framework to evaluate the risks introduced by new technologies is essential to determine which
countermeasures to adopt to mitigate these new risks. threat agents seek financial gain such as by
attacking applications to compromise users\' sensitive data and company’s proprietary
information for financial gain, fraud as well as for competitive advantage the risks posed by
these threat agents, it is necessary to determine the risk exposure and factor the probability and
the impact of these threats as well as to identify the type of application vulnerabilities that can be
exploited by these threat agents. The exploit of some of these application vulnerabilities might
severely and negatively impact the organization and personal computer.
data from legal in relation to law suits and regulatory fines and fraud data that includes amount
of money losses incurred because of online fraud. All this type of information is essential to
determine the overall impact. In absence of this data, the best the CISO can do is to use data
breach incident data from public sources and data breach incident reports. In part I of this guide,
we provided some examples of how this data can be used to estimate impact. We documented
what are the critical factors to estimate impacts of data breaches: these as the value of the data
assets liability for the organization in case these assets are lost. Once the potential business
impact of a data breach is estimated, the next step is to determine how much should be spent to
mitigate the risk. At high level, this is a risk strategy decision that depends on the organization
risk culture and the organization priorities for mitigating risks..
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