This type of deformation is irreversible. However.pdf
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This type of deformation is irreversible. However, an object in the plastic deformation range will first have undergone elastic deformation, which is reversible, so the object will return part way to its original shape Solution This type of deformation is irreversible. However, an object in the plastic deformation range will first have undergone elastic deformation, which is reversible, so the object will return part way to its original shape.
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X has +2 charge It has low first and second ioniz.pdf
X has +2 charge It has low first and second ionization energies. Hence it can give 2
electrons easily. Hence it will have +2 charge Y has larger ionization energy than X Hence atom
of Y is smaller than that of X(Same group)
Solution
X has +2 charge It has low first and second ionization energies. Hence it can give 2
electrons easily. Hence it will have +2 charge Y has larger ionization energy than X Hence atom
of Y is smaller than that of X(Same group).
Valence Bond Theory According to valence band the.pdf
Valence Bond Theory According to valence band theory a covalent bond is formed
between the two atoms by the overlap of half filled valence atomic orbitals of each atom
containing one unpaired electron. Valence bond theory considers that the overlapping atomic
orbitals of the participating atoms form a chemical bond. Because of the overlapping, it is most
probable that electrons should be in the bond region. Valence bond theory views bonds as
weakly coupled orbitals (small overlap). Valence bond theory is typically easier to employ in
ground state molecules. Hybridization It is the concept of mixing atomic orbitals to form new
hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised
orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is
an integral part of valence bond theory. Molecular orbital theory molecular orbital (MO) theory
is a method for determining molecular structure in which electrons are not assigned to individual
bonds between atoms, but are treated as moving under the influence of the nuclei in the whole
molecule. In this theory, each molecule has a set of molecular orbitals, in which it is assumed
that the molecular orbital wave function ?f may be written as a simple weighted sum of the n
constituent atomic orbitals ?i Metallic Bonding Metallic bonding is the electrostatic attractive
forces between the delocalized electrons, called conduction electrons, gathered in an \"electron
sea\", and the positively charged metal ions.
Solution
Valence Bond Theory According to valence band theory a covalent bond is formed
between the two atoms by the overlap of half filled valence atomic orbitals of each atom
containing one unpaired electron. Valence bond theory considers that the overlapping atomic
orbitals of the participating atoms form a chemical bond. Because of the overlapping, it is most
probable that electrons should be in the bond region. Valence bond theory views bonds as
weakly coupled orbitals (small overlap). Valence bond theory is typically easier to employ in
ground state molecules. Hybridization It is the concept of mixing atomic orbitals to form new
hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised
orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is
an integral part of valence bond theory. Molecular orbital theory molecular orbital (MO) theory
is a method for determining molecular structure in which electrons are not assigned to individual
bonds between atoms, but are treated as moving under the influence of the nuclei in the whole
molecule. In this theory, each molecule has a set of molecular orbitals, in which it is assumed
that the molecular orbital wave function ?f may be written as a simple weighted sum of the n
constituent atomic orbitals ?i Metallic Bonding Metallic bonding is the electrostatic attractive
forces between the delocalized e.
total volume =25+25=50 ml total millimoles of KI=.pdf
total volume =25+25=50 ml total millimoles of KI=25*0.2=5
conce=millimoles/volume in ml =5/50=0.1 similarly for other compound 25*0.2/50=0.1
Solution
total volume =25+25=50 ml total millimoles of KI=25*0.2=5
conce=millimoles/volume in ml =5/50=0.1 similarly for other compound 25*0.2/50=0.1.
Some elements receive their respective symbol bas.pdf
Some elements receive their respective symbol based on their latin equivalence...
Ex: iron(Fe) = ferrous, lead(Pb) = plumbum, mercury(Hg) = hydrargyrum While others are
receive their symbol for where they are discovered or to the discoverer\'s home country... Ex:
Germanium (Germany), Californium (University of California) While others are named after
their founders... Ex: Einsteinium (Albert Einstein), Mendelevium (Dmitri Ivanovich Mendeleev)
When these elements are combined into compounds, there are a couple of tricks to help with
remembering them... Ex: 1. -ous compounds will generally have a lower charge than -ic
compounds 2. Hydro- compounds will have an addition of an H+ proton making the compound
an acid
Solution
Some elements receive their respective symbol based on their latin equivalence...
Ex: iron(Fe) = ferrous, lead(Pb) = plumbum, mercury(Hg) = hydrargyrum While others are
receive their symbol for where they are discovered or to the discoverer\'s home country... Ex:
Germanium (Germany), Californium (University of California) While others are named after
their founders... Ex: Einsteinium (Albert Einstein), Mendelevium (Dmitri Ivanovich Mendeleev)
When these elements are combined into compounds, there are a couple of tricks to help with
remembering them... Ex: 1. -ous compounds will generally have a lower charge than -ic
compounds 2. Hydro- compounds will have an addition of an H+ proton making the compound
an acid.
phosphonium ylide is produced when phosphonium ha.pdf
phosphonium ylide is produced when phosphonium halide salt reacts with a strong
base such as nBuLi.
Solution
phosphonium ylide is produced when phosphonium halide salt reacts with a strong
base such as nBuLi..
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1..pdf
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1.06 M
Solution
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1.06 M.
In chemistry, an amphoteric species is a molecule.pdf
In chemistry, an amphoteric species is a molecule or ion that can react as an acid as
well as a base. Zinc oxide (ZnO) reacts with both acids and with bases: In acid: ZnO + 2H+ ?
Zn2+ + H2O In base: ZnO + H2O + 2OH- ? [Zn(OH)4]2-
Solution
In chemistry, an amphoteric species is a molecule or ion that can react as an acid as
well as a base. Zinc oxide (ZnO) reacts with both acids and with bases: In acid: ZnO + 2H+ ?
Zn2+ + H2O In base: ZnO + H2O + 2OH- ? [Zn(OH)4]2-.
Recommended
X has +2 charge It has low first and second ioniz.pdf
X has +2 charge It has low first and second ionization energies. Hence it can give 2
electrons easily. Hence it will have +2 charge Y has larger ionization energy than X Hence atom
of Y is smaller than that of X(Same group)
Solution
X has +2 charge It has low first and second ionization energies. Hence it can give 2
electrons easily. Hence it will have +2 charge Y has larger ionization energy than X Hence atom
of Y is smaller than that of X(Same group).
Valence Bond Theory According to valence band the.pdf
Valence Bond Theory According to valence band theory a covalent bond is formed
between the two atoms by the overlap of half filled valence atomic orbitals of each atom
containing one unpaired electron. Valence bond theory considers that the overlapping atomic
orbitals of the participating atoms form a chemical bond. Because of the overlapping, it is most
probable that electrons should be in the bond region. Valence bond theory views bonds as
weakly coupled orbitals (small overlap). Valence bond theory is typically easier to employ in
ground state molecules. Hybridization It is the concept of mixing atomic orbitals to form new
hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised
orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is
an integral part of valence bond theory. Molecular orbital theory molecular orbital (MO) theory
is a method for determining molecular structure in which electrons are not assigned to individual
bonds between atoms, but are treated as moving under the influence of the nuclei in the whole
molecule. In this theory, each molecule has a set of molecular orbitals, in which it is assumed
that the molecular orbital wave function ?f may be written as a simple weighted sum of the n
constituent atomic orbitals ?i Metallic Bonding Metallic bonding is the electrostatic attractive
forces between the delocalized electrons, called conduction electrons, gathered in an \"electron
sea\", and the positively charged metal ions.
Solution
Valence Bond Theory According to valence band theory a covalent bond is formed
between the two atoms by the overlap of half filled valence atomic orbitals of each atom
containing one unpaired electron. Valence bond theory considers that the overlapping atomic
orbitals of the participating atoms form a chemical bond. Because of the overlapping, it is most
probable that electrons should be in the bond region. Valence bond theory views bonds as
weakly coupled orbitals (small overlap). Valence bond theory is typically easier to employ in
ground state molecules. Hybridization It is the concept of mixing atomic orbitals to form new
hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised
orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is
an integral part of valence bond theory. Molecular orbital theory molecular orbital (MO) theory
is a method for determining molecular structure in which electrons are not assigned to individual
bonds between atoms, but are treated as moving under the influence of the nuclei in the whole
molecule. In this theory, each molecule has a set of molecular orbitals, in which it is assumed
that the molecular orbital wave function ?f may be written as a simple weighted sum of the n
constituent atomic orbitals ?i Metallic Bonding Metallic bonding is the electrostatic attractive
forces between the delocalized e.
total volume =25+25=50 ml total millimoles of KI=.pdf
total volume =25+25=50 ml total millimoles of KI=25*0.2=5
conce=millimoles/volume in ml =5/50=0.1 similarly for other compound 25*0.2/50=0.1
Solution
total volume =25+25=50 ml total millimoles of KI=25*0.2=5
conce=millimoles/volume in ml =5/50=0.1 similarly for other compound 25*0.2/50=0.1.
Some elements receive their respective symbol bas.pdf
Some elements receive their respective symbol based on their latin equivalence...
Ex: iron(Fe) = ferrous, lead(Pb) = plumbum, mercury(Hg) = hydrargyrum While others are
receive their symbol for where they are discovered or to the discoverer\'s home country... Ex:
Germanium (Germany), Californium (University of California) While others are named after
their founders... Ex: Einsteinium (Albert Einstein), Mendelevium (Dmitri Ivanovich Mendeleev)
When these elements are combined into compounds, there are a couple of tricks to help with
remembering them... Ex: 1. -ous compounds will generally have a lower charge than -ic
compounds 2. Hydro- compounds will have an addition of an H+ proton making the compound
an acid
Solution
Some elements receive their respective symbol based on their latin equivalence...
Ex: iron(Fe) = ferrous, lead(Pb) = plumbum, mercury(Hg) = hydrargyrum While others are
receive their symbol for where they are discovered or to the discoverer\'s home country... Ex:
Germanium (Germany), Californium (University of California) While others are named after
their founders... Ex: Einsteinium (Albert Einstein), Mendelevium (Dmitri Ivanovich Mendeleev)
When these elements are combined into compounds, there are a couple of tricks to help with
remembering them... Ex: 1. -ous compounds will generally have a lower charge than -ic
compounds 2. Hydro- compounds will have an addition of an H+ proton making the compound
an acid.
phosphonium ylide is produced when phosphonium ha.pdf
phosphonium ylide is produced when phosphonium halide salt reacts with a strong
base such as nBuLi.
Solution
phosphonium ylide is produced when phosphonium halide salt reacts with a strong
base such as nBuLi..
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1..pdf
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1.06 M
Solution
N1 V1 = N2 V2 50 x M1 = 1.50 x 2 x 17.7 M M1 = 1.06 M.
In chemistry, an amphoteric species is a molecule.pdf
In chemistry, an amphoteric species is a molecule or ion that can react as an acid as
well as a base. Zinc oxide (ZnO) reacts with both acids and with bases: In acid: ZnO + 2H+ ?
Zn2+ + H2O In base: ZnO + H2O + 2OH- ? [Zn(OH)4]2-
Solution
In chemistry, an amphoteric species is a molecule or ion that can react as an acid as
well as a base. Zinc oxide (ZnO) reacts with both acids and with bases: In acid: ZnO + 2H+ ?
Zn2+ + H2O In base: ZnO + H2O + 2OH- ? [Zn(OH)4]2-.
Chemistry is the science of matter, especially it.pdf
Chemistry is the science of matter, especially its chemical reactions, but also its
composition, structure and properties.[1][2] Chemistry is concerned with atoms and their
interactions with other atoms, and particularly with the properties of chemical bonds.
Solution
Chemistry is the science of matter, especially its chemical reactions, but also its
composition, structure and properties.[1][2] Chemistry is concerned with atoms and their
interactions with other atoms, and particularly with the properties of chemical bonds..
Before this confederates began their general advance, General Lee pl.pdf
Before this confederates began their general advance, General Lee planned to weaken any union
opposition with
(c) an artillery barrage
Solution
Before this confederates began their general advance, General Lee planned to weaken any union
opposition with
(c) an artillery barrage.
An understanding of periodic trends is important .pdf
An understanding of periodic trends is important because the trends:- 4. relate to
properties of elements and how they may react.
Solution
An understanding of periodic trends is important because the trends:- 4. relate to
properties of elements and how they may react..
Acid contains hydrogen ions. Thus, iodate will re.pdf
Acid contains hydrogen ions. Thus, iodate will react with hydrogen ion to form
iodine and water. 2IO3- + 12H+ + 10e- -> I2 + 6H2O Iodine and water will be further reduced
to form iodide ions. Thus, this will affect the titration process, thus it should be titrated
immediately. IO3- + 3H2O + 6e- -> I- + 6OH- Hope this helps!
Solution
Acid contains hydrogen ions. Thus, iodate will react with hydrogen ion to form
iodine and water. 2IO3- + 12H+ + 10e- -> I2 + 6H2O Iodine and water will be further reduced
to form iodide ions. Thus, this will affect the titration process, thus it should be titrated
immediately. IO3- + 3H2O + 6e- -> I- + 6OH- Hope this helps!.
There is a difference between to connect to a network and to communi.pdf
There is a difference between to connect to a network and to communicate over a network.
Connecting to a network requires that special hardware be installed in your computer. This
hardware is called connection devices. There are two common types of connection devices: 1-
Modem – short for modulator / demodulator - which connects computers using a standard
telephone line 2- Network interface card (NIC) – which connects computers using a special type
of network cabling Note that, NICs are not synonymous with Ethernet cards. An Ethernet card is
a particular type of NIC, and is the most popular in personal computers (PCs). Most of the time,
these devices are internal hardware, although they are available as an external peripherals. 6 In
order to communicate with another computer over a network, you must do two things: 1. Use the
set of rules governing communication over the network, called a protocol. Your computer will
generally handle this. 2. Know the address of the computer you want to communicate with.
There are two types of network addresses: a. Medium access control (MAC) address which is
used inside a single network b. Internet protocol (IP) address which is used on the Internet World
Wide Web The World Wide Web (web for short or www) is a collection of interlinked
multimedia documents that are stored on the Internet and accessed using a common protocol
(HTTP). Form this definition we can say that the web is part of the system that is primarily
include software components (text files, graphics files, sound files, video files…). So, what is the
relation between the internet and the World Wide Web? World Wide Web (WWW) is an Internet
based software application. Other Internet applications are: Email ftp (file transfer protocol)
Messenger 7 More than 90 percent of the internet applications are World Wide Web so it is valid
to use Internet and World Wide Web terms interchangeably. World Wide Web is tied up to
several terms such as web server, web browser and HTML or Hypertext Markup Language. In
the following paragraph those terms are briefly explained.
Solution
There is a difference between to connect to a network and to communicate over a network.
Connecting to a network requires that special hardware be installed in your computer. This
hardware is called connection devices. There are two common types of connection devices: 1-
Modem – short for modulator / demodulator - which connects computers using a standard
telephone line 2- Network interface card (NIC) – which connects computers using a special type
of network cabling Note that, NICs are not synonymous with Ethernet cards. An Ethernet card is
a particular type of NIC, and is the most popular in personal computers (PCs). Most of the time,
these devices are internal hardware, although they are available as an external peripherals. 6 In
order to communicate with another computer over a network, you must do two things: 1. Use the
set of rules governing communication over the network, calle.
The picture is not uploaded properly it is corrupted. please upload .pdf
The picture is not uploaded properly it is corrupted. please upload it properly for the answer
Thank you.
Solution
The picture is not uploaded properly it is corrupted. please upload it properly for the answer
Thank you..
No clarity in question . Not visibleSolutionNo clarity in ques.pdf
No clarity in question . Not visible
Solution
No clarity in question . Not visible.
Date Account title Debit credit jan 30 cash 45000 common sto.pdf
Date Account title Debit credit jan 30 cash 45000 common stock 45000 [being common
stock issued] feb1 Prepaid rent 24000 cash 24000 [being rent paid in advance for
2years] Apr 10 supplies 3200 Accounts payable 3200 [being supplies purchased on
account] july 1 cash 24000 Unearned service revenue 24000 [being fees received in
advance] july 20 Accounts payable 1500 cash 1500 [Being accounts payable paid]
Aug 15 Accounts receivable 18000 Service revenue 18000 [being services provided]
sep 15 cash 8400 service revenue 8400 [being services provided] oct 1 Salaries 12000
cash 12000 [salary paid ] oct cash 15000 Accounts receivable 15000 [being
receivables due collected] nov 16 Accounts receivable 42000 service revenue 42000
[being services provided] dec 1 Dividend paid 15000 cash 15000 [being dividend
paid] dec 31 Unearned service revenue 12000 service revenue 12000 [being services
provided during the year recorded] 24000 *6/12 [july - dec = 6 months ] dec 31 Salary
expense 3600 salary payable 3600 [being salary due but not paid ] dec 31 Rent
expense 11000 prepaid rent 11000 [being rent for the year recorded (24000/2 = 12000per
year or 12000*11/12 = 11000 ] [1 feb - 31 dec = 11 months] dec 31 supplies expense
2920 supplies 2920 [being supplies used recorded (3200 -280 ]
Solution
Date Account title Debit credit jan 30 cash 45000 common stock 45000 [being common
stock issued] feb1 Prepaid rent 24000 cash 24000 [being rent paid in advance for
2years] Apr 10 supplies 3200 Accounts payable 3200 [being supplies purchased on
account] july 1 cash 24000 Unearned service revenue 24000 [being fees received in
advance] july 20 Accounts payable 1500 cash 1500 [Being accounts payable paid]
Aug 15 Accounts receivable 18000 Service revenue 18000 [being services provided]
sep 15 cash 8400 service revenue 8400 [being services provided] oct 1 Salaries 12000
cash 12000 [salary paid ] oct cash 15000 Accounts receivable 15000 [being
receivables due collected] nov 16 Accounts receivable 42000 service revenue 42000
[being services provided] dec 1 Dividend paid 15000 cash 15000 [being dividend
paid] dec 31 Unearned service revenue 12000 service revenue 12000 [being services
provided during the year recorded] 24000 *6/12 [july - dec = 6 months ] dec 31 Salary
expense 3600 salary payable 3600 [being salary due but not paid ] dec 31 Rent
expense 11000 prepaid rent 11000 [being rent for the year recorded (24000/2 = 12000per
year or 12000*11/12 = 11000 ] [1 feb - 31 dec = 11 months] dec 31 supplies expense
2920 supplies 2920 [being supplies used recorded (3200 -280 ].
Initial Value=2.5Drift rate per annum=0.4Variance rate per annum.pdf
Initial Value=2.5
Drift rate per annum=0.4
Variance rate per annum=0.5
Standard deviation=(0.5^0.5)= 0.707107
At the end of second year:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Double of initial value=2*2.5=5
Z value=(5-mean)/Standard deviation=(5-3.3)/1=1.7
Cumulative area Under the Standard Normal Distribution Table
For z value=1.7
Cumulative area=N(d)=0.9554
The probability that cash position will be more than double(greater than 5):
(1-0.9554)=0.0446
The probability that cash position will be more than double(greater than 5)=4.46%
Cash position negative:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Cash position zero=0
Z value=(0-mean)/Standard deviation=(0-3.3)/1=-3.3
Cumulative area Under the Standard Normal Distribution Table
For z value=-3.3
Cumulative area=N(d)=0.0013 {Cumulative area N(d) given in table for -3, below -3 is not
mentioned in the table. The z value is rounded to -3}
The probability that cash position will be negative:
=0.0013
The probability that cash position negative=0.13%
The probability that cash position will be more than double=4.46%
The probability that cash position will be negative=0.13%
Solution
Initial Value=2.5
Drift rate per annum=0.4
Variance rate per annum=0.5
Standard deviation=(0.5^0.5)= 0.707107
At the end of second year:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Double of initial value=2*2.5=5
Z value=(5-mean)/Standard deviation=(5-3.3)/1=1.7
Cumulative area Under the Standard Normal Distribution Table
For z value=1.7
Cumulative area=N(d)=0.9554
The probability that cash position will be more than double(greater than 5):
(1-0.9554)=0.0446
The probability that cash position will be more than double(greater than 5)=4.46%
Cash position negative:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Cash position zero=0
Z value=(0-mean)/Standard deviation=(0-3.3)/1=-3.3
Cumulative area Under the Standard Normal Distribution Table
For z value=-3.3
Cumulative area=N(d)=0.0013 {Cumulative area N(d) given in table for -3, below -3 is not
mentioned in the table. The z value is rounded to -3}
The probability that cash position will be negative:
=0.0013
The probability that cash position negative=0.13%
The probability that cash position will be more than double=4.46%
The probability that cash position will be negative=0.13%.
IntroductionDarwin argused that living organisms evolved from one .pdf
Introduction
Darwin argused that living organisms evolved from one or a few simple forms of live through
the process of natural selection. The essence of this comprise a population differ in the numper
of viable offspring the produce during their lifetimes.
Give that sexual reproduction is the means by which genes are propagated into future
generations, sexual selection plays a large role in human evolution. Human sexuality then is of
interest to evolutionary psychologists who aim to investigate evolved mechanisms to attract and
secure mates.
When an environment stress is constant and lasts for many generation , successful adaptation
may develop through biological evolution. The human body raedily responds to changing
environment stresses in a variety and cultural ways. We can acclimatize to a wide range of
temperature and humidity. When traveling to hige altitudes, our bodies adjust so, that our cells
still receive sufficient oxygen. We also are constantly responding in physiological ways to
internal and external stresses such as, bactrial and viral infections aie, water pollution, dietary
imbalance, and overcrowding.
Humans normally respond to environmentalstresses in four ways;
1. genetic change
2. developmental adjustment
3. acclimatization
4. cutural practices and technology
Human sexuality refers to people sexual interst in and attraction to others it is the capacity to
have erotic or sexual feling and experiences.
Human sexuality can be understood as part of the social life of humans, governed by social
norms,m implied rules of behavior, and the status quo. Socity views on sexuality have changed
throughout history and are continuously evolved.
Each society has different norma about premarital sex, the age of sexual consent, homosexuality,
mastrubation, and other sextual behaviours. Individuals are socialized to these from an early age
by their family, education system, peers, media, and religion.
With the advent of patriarchal societies, gender roles around sexality became much more
stringent, and sexual norma began focusing on sexual possessiveness and the control of female
sexuality.
The United States is relatively restrictive compared to other industrialized nation when it comes
to its citizens general attitudes about sex when.
The messages that children are taught about sex play an important role in how they will grow
into their sexual selves and express their sexual motivations. How, what, when, and by whom
children should be taught about sex is a matter oa great debate in sex education.
Solution
Introduction
Darwin argused that living organisms evolved from one or a few simple forms of live through
the process of natural selection. The essence of this comprise a population differ in the numper
of viable offspring the produce during their lifetimes.
Give that sexual reproduction is the means by which genes are propagated into future
generations, sexual selection plays a large role in human evolution. Human sexuality then is of
int.
import java.awt.LinearGradientPaint; public class searchComparis.pdf
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}
Solution
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}.
Distance VectorLink StateIt Sends the entire routing tableIt S.pdf
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Solution
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS.
Evolution of gymnospersm or plants without fruit which have a naked .pdf
Evolution of gymnospersm or plants without fruit which have a naked seed vs other seedless
vascular plants such as ferns, mosses etc. Though both of these types of plants survive and
dominate their respective ecological successions highly significantly and possess differential and
discrete adaptive features according to their respective environments, their evolutionary
advances remain highly different in nature. Many aspects of their evolutionary differences can be
discussed but survival and longevity of the seed remains the most important and vital feature.
This can be discussed as below:
Dispersal and longevity of the seed life: Seedless vascular plants follow a medium-mediated
dispersal of seeds in nature. They do not possess bright and eye-catching flowers which can
attract the insects to aid pollinatation. Their seeds are directly localized onto the upper/lower
surface of the leaves itself. After fertilization, the seed is produced and released in the
environment. On the other hand, the gymnosperms do not possess bright flowers to attract insects
which can aid pollination. Hence, after fertilization, the seeds are dispersed in highly discrete
manner or remain dorman as such for very long times. This also decreases the net density of
gymnosperms in nature.
This property of low fertilization and dispersal of seeds has made the gymnosperms more sturdy
enough to survive under very harsh environments as well, such as high heat, low oxygen (at high
altitudes), high salinity etc. Thus, the gymosperms have naturally evolved with high endurance
against environmental stress/stimuli which makes thier life-time more long as compared to other
seedless vascular plants which can be directly reflected into their mode of pollination and seed-
dispersal.
Thus, this explains the evolutionary advances in gymnospers as compared to other plants.
Solution
Evolution of gymnospersm or plants without fruit which have a naked seed vs other seedless
vascular plants such as ferns, mosses etc. Though both of these types of plants survive and
dominate their respective ecological successions highly significantly and possess differential and
discrete adaptive features according to their respective environments, their evolutionary
advances remain highly different in nature. Many aspects of their evolutionary differences can be
discussed but survival and longevity of the seed remains the most important and vital feature.
This can be discussed as below:
Dispersal and longevity of the seed life: Seedless vascular plants follow a medium-mediated
dispersal of seeds in nature. They do not possess bright and eye-catching flowers which can
attract the insects to aid pollinatation. Their seeds are directly localized onto the upper/lower
surface of the leaves itself. After fertilization, the seed is produced and released in the
environment. On the other hand, the gymnosperms do not possess bright flowers to attract insects
which can aid pollination. Hence, after fertilization, the.
ANSimport javax.swing.;import java.awt.; import java.awt.ev.pdf
ANS:
import javax.swing.*;
import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
public class Frame extends JFrame {
private double tn = 0;//here declare the variables
private double tn2 = 0;
private byte nom = -1;
private JTextField jtf;
public Frame() {
JButton[] bn = new JButton[10];
for ( int i = 9; i >= 0; i--) {
bn[i] = new JButton(Integer.toString(i));
}
//here declare the buttons
JButton eb= new JButton(\"e1\");
JButton cb= new JButton(\"C\");
JButton ab= new JButton(\"+\");
JButton sb= new JButton(\"-\");
JButton mb= new JButton(\"*\");
JButton db= new JButton(\"/\");
JButton ab= new JButton(\"+\");
JButton sb= new JButton(\"-\");
//here to set the size,bg,dimensionals
jtf = new JTextField();
jtf.setPreferredSize(new Dimension(180, 30));
jtf.setBackground(Color.BLUE);
jtf.setEnabled(false);
jtf.setHorizontalAlignment(4);
jtf.setDisabledTextColor(Color.GREEN);
//here set the layouts
JPanel PM = new JPanel();
PM.setLayout(new BoxLayout(PM, BoxLayout.Y_AXIS));
JPanel pt = new JPanel();
pt.setPreferredSize(new Dimension(160, 20));
pt.add(jtf);
JPanel pnb = new JPanel();
pnb.setPreferredSize(new Dimension(160, 100));
for(int i = 9; i>=0; i--) {
pnb.add(bn[i]);
}
//here buttons add to panels
JPanel pb_nom = new JPanel();
pb_nom.setPreferredSize(new Dimension(160, 35));
pb_nom.add(eb);
pb_nom.add(cb);
pb_nom.add(mb);
pb_nom.add(db);
pb_nom.add(ab);
pb_nom.add(sb);
umo_ji[] pb_action = new umo_ji[10];
for ( int i = 0; i < 10; i++ ) {
pb_action[i] = new umo_ji(bn[i]);
bn[i].addActionListener(pb_action[i]);
}
//here perform action buttons
eb e1 = new eb();
eb.addActionListener(e1);
cb c = new cb();
cb.addActionListener(c);
mb e2 = new mb();
mb.addActionListener(e2);
db e3 = new db();
db.addActionListener(e3);
ab add = new ab();
ab.addActionListener(add);
Sub_But subtract = new Sub_But();
sb.addActionListener(subtract);
PM.add(pt);
PM.add(pnb);
PM.add(pb_nom);
add(PM);
//To set the title and size
setTitle(\"Button_Test\");
setSize(180, 290);
setLocationByPlatform(true);
setDefaultCloseOperation(WindowConstants.EXIT_ON_CLOSE);
setVisible(true);
}
private class umo_ji implements ActionListener {
//here implements actionlistener
private String c;
public umo_ji(JButton a) {
this.c = a.text_get();
}
public void actionPerformed(ActionEvent e) {
if (!jtf.text_get().equals(\"0.0\")) {
jtf.text_set(jtf.text_get() + c);
} else {
jtf.text_set(\"\");
actionPerformed(e);
}
}
}
private class eb implements ActionListener {
@Override//overide buttons
public void actionPerformed(ActionEvent e) {
tn2 = Double.parseDouble(jtf.text_get());
if (nom == 0) {
jtf.text_set(Double.toString((Math.round((tn / tn2) * 100)) / 100));
} else if (nom == 1) {
jtf.text_set(Double.toString(tn * tn2));
} else if (nom == 2) {
jtf.text_set(Double.toString(tn2 + tn));
} else if (nom == 3) {
jtf.text_set(Double.toString(tn - tn2));
} else {
jtf.text_set(String.valueOf(tn));
}
tn = Double.parseDouble(jtf.text_get());
}
}
private class cb implements.
a. Total asset turnover = Return on assets Profit margin= 0.145 .pdf
a. Total asset turnover = Return on assets / Profit margin
= 0.145 / 0.055
= 2.64 times
b. Return on equity = Return on assets / (1 - Debt-to-total-assets ratio)
= 0.145 / (1 - 0.65)
= 0.145 / 0.35
= 0.4143 or 41.43%
Solution
a. Total asset turnover = Return on assets / Profit margin
= 0.145 / 0.055
= 2.64 times
b. Return on equity = Return on assets / (1 - Debt-to-total-assets ratio)
= 0.145 / (1 - 0.65)
= 0.145 / 0.35
= 0.4143 or 41.43%.
d) All but Iv is the right answer.Solutiond) All but Iv is the.pdf
d) All but Iv is the right answer.
Solution
d) All but Iv is the right answer..
2. b) fed on molecules with a chemical structure to pesticides.The.pdf
2. b) fed on molecules with a chemical structure to pesticides.
They are also called as biopesticides.
Solution
2. b) fed on molecules with a chemical structure to pesticides.
They are also called as biopesticides..
1.Management information systemMany of the different types of Inf.pdf
1.Management information system:
Many of the different types of Information Systems found in commercial organizations are
referred to as \"Management Information Systems\". Management Information Systems are
management-level systems that are used by middle managers to help ensure the smooth running
of the organization in the short to medium term. The highly structured information provided by
these systems allows managers to evaluate an organization\'s performance by comparing current
with previous outputs.
The functions of Management Information Systems is
Internal transactions sorting and maintain a summery reports.
Structured data summerizing and maintain detailed reports.
2. Executive information system :
Executive Information Systems are strategic-level information systems. Executive Information
System help executives and senior managers analyze the environment in which the organization
operates, to identify long-term trends, and to plan appropriate courses of action. The information
in such systems is often weakly structured and comes from both internal and external sources.
Executive Information System are designed to be operated directly by executives without the
need for intermediaries.
Executive Information System organizes and presents data and information from both external
data sources and internal Management Information System.
Functions
Summerizing the external data and maintain summery reports.
Drilling down the predefined models i.e graphs
Solution
1.Management information system:
Many of the different types of Information Systems found in commercial organizations are
referred to as \"Management Information Systems\". Management Information Systems are
management-level systems that are used by middle managers to help ensure the smooth running
of the organization in the short to medium term. The highly structured information provided by
these systems allows managers to evaluate an organization\'s performance by comparing current
with previous outputs.
The functions of Management Information Systems is
Internal transactions sorting and maintain a summery reports.
Structured data summerizing and maintain detailed reports.
2. Executive information system :
Executive Information Systems are strategic-level information systems. Executive Information
System help executives and senior managers analyze the environment in which the organization
operates, to identify long-term trends, and to plan appropriate courses of action. The information
in such systems is often weakly structured and comes from both internal and external sources.
Executive Information System are designed to be operated directly by executives without the
need for intermediaries.
Executive Information System organizes and presents data and information from both external
data sources and internal Management Information System.
Functions
Summerizing the external data and maintain summery reports.
Drilling down the predefined models i.e graphs.
1. MnO4- is purple in solution. When the reaction is complete the so.pdf
1. MnO4- is purple in solution. When the reaction is complete the solution turns colorless as
MnO4- is reduced to Mn2+. The end point can thus be determined using the color change of the
reactant itself.
2. The reaction needs to occur in acidic medium for MnO4- to be fully reduced to Mn2+. In
neutral or alkaine media, the MnO4- is only partially reduced to MnO2.
3. The initial reaction has a high activation energy and requires heating to increase the reaction
rate so that the titration can be completed within a reasonable time.
Solution
1. MnO4- is purple in solution. When the reaction is complete the solution turns colorless as
MnO4- is reduced to Mn2+. The end point can thus be determined using the color change of the
reactant itself.
2. The reaction needs to occur in acidic medium for MnO4- to be fully reduced to Mn2+. In
neutral or alkaine media, the MnO4- is only partially reduced to MnO2.
3. The initial reaction has a high activation energy and requires heating to increase the reaction
rate so that the titration can be completed within a reasonable time..
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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Chemistry is the science of matter, especially it.pdf
Chemistry is the science of matter, especially its chemical reactions, but also its
composition, structure and properties.[1][2] Chemistry is concerned with atoms and their
interactions with other atoms, and particularly with the properties of chemical bonds.
Solution
Chemistry is the science of matter, especially its chemical reactions, but also its
composition, structure and properties.[1][2] Chemistry is concerned with atoms and their
interactions with other atoms, and particularly with the properties of chemical bonds..
Before this confederates began their general advance, General Lee pl.pdf
Before this confederates began their general advance, General Lee planned to weaken any union
opposition with
(c) an artillery barrage
Solution
Before this confederates began their general advance, General Lee planned to weaken any union
opposition with
(c) an artillery barrage.
An understanding of periodic trends is important .pdf
An understanding of periodic trends is important because the trends:- 4. relate to
properties of elements and how they may react.
Solution
An understanding of periodic trends is important because the trends:- 4. relate to
properties of elements and how they may react..
Acid contains hydrogen ions. Thus, iodate will re.pdf
Acid contains hydrogen ions. Thus, iodate will react with hydrogen ion to form
iodine and water. 2IO3- + 12H+ + 10e- -> I2 + 6H2O Iodine and water will be further reduced
to form iodide ions. Thus, this will affect the titration process, thus it should be titrated
immediately. IO3- + 3H2O + 6e- -> I- + 6OH- Hope this helps!
Solution
Acid contains hydrogen ions. Thus, iodate will react with hydrogen ion to form
iodine and water. 2IO3- + 12H+ + 10e- -> I2 + 6H2O Iodine and water will be further reduced
to form iodide ions. Thus, this will affect the titration process, thus it should be titrated
immediately. IO3- + 3H2O + 6e- -> I- + 6OH- Hope this helps!.
There is a difference between to connect to a network and to communi.pdf
There is a difference between to connect to a network and to communicate over a network.
Connecting to a network requires that special hardware be installed in your computer. This
hardware is called connection devices. There are two common types of connection devices: 1-
Modem – short for modulator / demodulator - which connects computers using a standard
telephone line 2- Network interface card (NIC) – which connects computers using a special type
of network cabling Note that, NICs are not synonymous with Ethernet cards. An Ethernet card is
a particular type of NIC, and is the most popular in personal computers (PCs). Most of the time,
these devices are internal hardware, although they are available as an external peripherals. 6 In
order to communicate with another computer over a network, you must do two things: 1. Use the
set of rules governing communication over the network, called a protocol. Your computer will
generally handle this. 2. Know the address of the computer you want to communicate with.
There are two types of network addresses: a. Medium access control (MAC) address which is
used inside a single network b. Internet protocol (IP) address which is used on the Internet World
Wide Web The World Wide Web (web for short or www) is a collection of interlinked
multimedia documents that are stored on the Internet and accessed using a common protocol
(HTTP). Form this definition we can say that the web is part of the system that is primarily
include software components (text files, graphics files, sound files, video files…). So, what is the
relation between the internet and the World Wide Web? World Wide Web (WWW) is an Internet
based software application. Other Internet applications are: Email ftp (file transfer protocol)
Messenger 7 More than 90 percent of the internet applications are World Wide Web so it is valid
to use Internet and World Wide Web terms interchangeably. World Wide Web is tied up to
several terms such as web server, web browser and HTML or Hypertext Markup Language. In
the following paragraph those terms are briefly explained.
Solution
There is a difference between to connect to a network and to communicate over a network.
Connecting to a network requires that special hardware be installed in your computer. This
hardware is called connection devices. There are two common types of connection devices: 1-
Modem – short for modulator / demodulator - which connects computers using a standard
telephone line 2- Network interface card (NIC) – which connects computers using a special type
of network cabling Note that, NICs are not synonymous with Ethernet cards. An Ethernet card is
a particular type of NIC, and is the most popular in personal computers (PCs). Most of the time,
these devices are internal hardware, although they are available as an external peripherals. 6 In
order to communicate with another computer over a network, you must do two things: 1. Use the
set of rules governing communication over the network, calle.
The picture is not uploaded properly it is corrupted. please upload .pdf
The picture is not uploaded properly it is corrupted. please upload it properly for the answer
Thank you.
Solution
The picture is not uploaded properly it is corrupted. please upload it properly for the answer
Thank you..
No clarity in question . Not visibleSolutionNo clarity in ques.pdf
No clarity in question . Not visible
Solution
No clarity in question . Not visible.
Date Account title Debit credit jan 30 cash 45000 common sto.pdf
Date Account title Debit credit jan 30 cash 45000 common stock 45000 [being common
stock issued] feb1 Prepaid rent 24000 cash 24000 [being rent paid in advance for
2years] Apr 10 supplies 3200 Accounts payable 3200 [being supplies purchased on
account] july 1 cash 24000 Unearned service revenue 24000 [being fees received in
advance] july 20 Accounts payable 1500 cash 1500 [Being accounts payable paid]
Aug 15 Accounts receivable 18000 Service revenue 18000 [being services provided]
sep 15 cash 8400 service revenue 8400 [being services provided] oct 1 Salaries 12000
cash 12000 [salary paid ] oct cash 15000 Accounts receivable 15000 [being
receivables due collected] nov 16 Accounts receivable 42000 service revenue 42000
[being services provided] dec 1 Dividend paid 15000 cash 15000 [being dividend
paid] dec 31 Unearned service revenue 12000 service revenue 12000 [being services
provided during the year recorded] 24000 *6/12 [july - dec = 6 months ] dec 31 Salary
expense 3600 salary payable 3600 [being salary due but not paid ] dec 31 Rent
expense 11000 prepaid rent 11000 [being rent for the year recorded (24000/2 = 12000per
year or 12000*11/12 = 11000 ] [1 feb - 31 dec = 11 months] dec 31 supplies expense
2920 supplies 2920 [being supplies used recorded (3200 -280 ]
Solution
Date Account title Debit credit jan 30 cash 45000 common stock 45000 [being common
stock issued] feb1 Prepaid rent 24000 cash 24000 [being rent paid in advance for
2years] Apr 10 supplies 3200 Accounts payable 3200 [being supplies purchased on
account] july 1 cash 24000 Unearned service revenue 24000 [being fees received in
advance] july 20 Accounts payable 1500 cash 1500 [Being accounts payable paid]
Aug 15 Accounts receivable 18000 Service revenue 18000 [being services provided]
sep 15 cash 8400 service revenue 8400 [being services provided] oct 1 Salaries 12000
cash 12000 [salary paid ] oct cash 15000 Accounts receivable 15000 [being
receivables due collected] nov 16 Accounts receivable 42000 service revenue 42000
[being services provided] dec 1 Dividend paid 15000 cash 15000 [being dividend
paid] dec 31 Unearned service revenue 12000 service revenue 12000 [being services
provided during the year recorded] 24000 *6/12 [july - dec = 6 months ] dec 31 Salary
expense 3600 salary payable 3600 [being salary due but not paid ] dec 31 Rent
expense 11000 prepaid rent 11000 [being rent for the year recorded (24000/2 = 12000per
year or 12000*11/12 = 11000 ] [1 feb - 31 dec = 11 months] dec 31 supplies expense
2920 supplies 2920 [being supplies used recorded (3200 -280 ].
Initial Value=2.5Drift rate per annum=0.4Variance rate per annum.pdf
Initial Value=2.5
Drift rate per annum=0.4
Variance rate per annum=0.5
Standard deviation=(0.5^0.5)= 0.707107
At the end of second year:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Double of initial value=2*2.5=5
Z value=(5-mean)/Standard deviation=(5-3.3)/1=1.7
Cumulative area Under the Standard Normal Distribution Table
For z value=1.7
Cumulative area=N(d)=0.9554
The probability that cash position will be more than double(greater than 5):
(1-0.9554)=0.0446
The probability that cash position will be more than double(greater than 5)=4.46%
Cash position negative:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Cash position zero=0
Z value=(0-mean)/Standard deviation=(0-3.3)/1=-3.3
Cumulative area Under the Standard Normal Distribution Table
For z value=-3.3
Cumulative area=N(d)=0.0013 {Cumulative area N(d) given in table for -3, below -3 is not
mentioned in the table. The z value is rounded to -3}
The probability that cash position will be negative:
=0.0013
The probability that cash position negative=0.13%
The probability that cash position will be more than double=4.46%
The probability that cash position will be negative=0.13%
Solution
Initial Value=2.5
Drift rate per annum=0.4
Variance rate per annum=0.5
Standard deviation=(0.5^0.5)= 0.707107
At the end of second year:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Double of initial value=2*2.5=5
Z value=(5-mean)/Standard deviation=(5-3.3)/1=1.7
Cumulative area Under the Standard Normal Distribution Table
For z value=1.7
Cumulative area=N(d)=0.9554
The probability that cash position will be more than double(greater than 5):
(1-0.9554)=0.0446
The probability that cash position will be more than double(greater than 5)=4.46%
Cash position negative:
Mean=2.5+(2*0.4)=3.3
Standard deviation=0.707107*(2^0.5)=1
Cash position zero=0
Z value=(0-mean)/Standard deviation=(0-3.3)/1=-3.3
Cumulative area Under the Standard Normal Distribution Table
For z value=-3.3
Cumulative area=N(d)=0.0013 {Cumulative area N(d) given in table for -3, below -3 is not
mentioned in the table. The z value is rounded to -3}
The probability that cash position will be negative:
=0.0013
The probability that cash position negative=0.13%
The probability that cash position will be more than double=4.46%
The probability that cash position will be negative=0.13%.
IntroductionDarwin argused that living organisms evolved from one .pdf
Introduction
Darwin argused that living organisms evolved from one or a few simple forms of live through
the process of natural selection. The essence of this comprise a population differ in the numper
of viable offspring the produce during their lifetimes.
Give that sexual reproduction is the means by which genes are propagated into future
generations, sexual selection plays a large role in human evolution. Human sexuality then is of
interest to evolutionary psychologists who aim to investigate evolved mechanisms to attract and
secure mates.
When an environment stress is constant and lasts for many generation , successful adaptation
may develop through biological evolution. The human body raedily responds to changing
environment stresses in a variety and cultural ways. We can acclimatize to a wide range of
temperature and humidity. When traveling to hige altitudes, our bodies adjust so, that our cells
still receive sufficient oxygen. We also are constantly responding in physiological ways to
internal and external stresses such as, bactrial and viral infections aie, water pollution, dietary
imbalance, and overcrowding.
Humans normally respond to environmentalstresses in four ways;
1. genetic change
2. developmental adjustment
3. acclimatization
4. cutural practices and technology
Human sexuality refers to people sexual interst in and attraction to others it is the capacity to
have erotic or sexual feling and experiences.
Human sexuality can be understood as part of the social life of humans, governed by social
norms,m implied rules of behavior, and the status quo. Socity views on sexuality have changed
throughout history and are continuously evolved.
Each society has different norma about premarital sex, the age of sexual consent, homosexuality,
mastrubation, and other sextual behaviours. Individuals are socialized to these from an early age
by their family, education system, peers, media, and religion.
With the advent of patriarchal societies, gender roles around sexality became much more
stringent, and sexual norma began focusing on sexual possessiveness and the control of female
sexuality.
The United States is relatively restrictive compared to other industrialized nation when it comes
to its citizens general attitudes about sex when.
The messages that children are taught about sex play an important role in how they will grow
into their sexual selves and express their sexual motivations. How, what, when, and by whom
children should be taught about sex is a matter oa great debate in sex education.
Solution
Introduction
Darwin argused that living organisms evolved from one or a few simple forms of live through
the process of natural selection. The essence of this comprise a population differ in the numper
of viable offspring the produce during their lifetimes.
Give that sexual reproduction is the means by which genes are propagated into future
generations, sexual selection plays a large role in human evolution. Human sexuality then is of
int.
import java.awt.LinearGradientPaint; public class searchComparis.pdf
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}
Solution
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}.
Distance VectorLink StateIt Sends the entire routing tableIt S.pdf
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Solution
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS
Distance Vector
Link State
It Sends the entire routing table
It Sends only link state information
Distance Vector is slow convergence
Link State is fast convergence
Distance Vector susceptible to routing loops
Link State less susceptible to routing loops
The updates are sometimes sent using broadcast
Link State always uses multicast for the routing updates
It doesn’t know the network topology
It knows the entire network topology
It is Simpler to configure
It can be harder to configure
Ex : RIP, IGRP
Ex : OSPF, IS-IS.
Evolution of gymnospersm or plants without fruit which have a naked .pdf
Evolution of gymnospersm or plants without fruit which have a naked seed vs other seedless
vascular plants such as ferns, mosses etc. Though both of these types of plants survive and
dominate their respective ecological successions highly significantly and possess differential and
discrete adaptive features according to their respective environments, their evolutionary
advances remain highly different in nature. Many aspects of their evolutionary differences can be
discussed but survival and longevity of the seed remains the most important and vital feature.
This can be discussed as below:
Dispersal and longevity of the seed life: Seedless vascular plants follow a medium-mediated
dispersal of seeds in nature. They do not possess bright and eye-catching flowers which can
attract the insects to aid pollinatation. Their seeds are directly localized onto the upper/lower
surface of the leaves itself. After fertilization, the seed is produced and released in the
environment. On the other hand, the gymnosperms do not possess bright flowers to attract insects
which can aid pollination. Hence, after fertilization, the seeds are dispersed in highly discrete
manner or remain dorman as such for very long times. This also decreases the net density of
gymnosperms in nature.
This property of low fertilization and dispersal of seeds has made the gymnosperms more sturdy
enough to survive under very harsh environments as well, such as high heat, low oxygen (at high
altitudes), high salinity etc. Thus, the gymosperms have naturally evolved with high endurance
against environmental stress/stimuli which makes thier life-time more long as compared to other
seedless vascular plants which can be directly reflected into their mode of pollination and seed-
dispersal.
Thus, this explains the evolutionary advances in gymnospers as compared to other plants.
Solution
Evolution of gymnospersm or plants without fruit which have a naked seed vs other seedless
vascular plants such as ferns, mosses etc. Though both of these types of plants survive and
dominate their respective ecological successions highly significantly and possess differential and
discrete adaptive features according to their respective environments, their evolutionary
advances remain highly different in nature. Many aspects of their evolutionary differences can be
discussed but survival and longevity of the seed remains the most important and vital feature.
This can be discussed as below:
Dispersal and longevity of the seed life: Seedless vascular plants follow a medium-mediated
dispersal of seeds in nature. They do not possess bright and eye-catching flowers which can
attract the insects to aid pollinatation. Their seeds are directly localized onto the upper/lower
surface of the leaves itself. After fertilization, the seed is produced and released in the
environment. On the other hand, the gymnosperms do not possess bright flowers to attract insects
which can aid pollination. Hence, after fertilization, the.
ANSimport javax.swing.;import java.awt.; import java.awt.ev.pdf
ANS:
import javax.swing.*;
import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
public class Frame extends JFrame {
private double tn = 0;//here declare the variables
private double tn2 = 0;
private byte nom = -1;
private JTextField jtf;
public Frame() {
JButton[] bn = new JButton[10];
for ( int i = 9; i >= 0; i--) {
bn[i] = new JButton(Integer.toString(i));
}
//here declare the buttons
JButton eb= new JButton(\"e1\");
JButton cb= new JButton(\"C\");
JButton ab= new JButton(\"+\");
JButton sb= new JButton(\"-\");
JButton mb= new JButton(\"*\");
JButton db= new JButton(\"/\");
JButton ab= new JButton(\"+\");
JButton sb= new JButton(\"-\");
//here to set the size,bg,dimensionals
jtf = new JTextField();
jtf.setPreferredSize(new Dimension(180, 30));
jtf.setBackground(Color.BLUE);
jtf.setEnabled(false);
jtf.setHorizontalAlignment(4);
jtf.setDisabledTextColor(Color.GREEN);
//here set the layouts
JPanel PM = new JPanel();
PM.setLayout(new BoxLayout(PM, BoxLayout.Y_AXIS));
JPanel pt = new JPanel();
pt.setPreferredSize(new Dimension(160, 20));
pt.add(jtf);
JPanel pnb = new JPanel();
pnb.setPreferredSize(new Dimension(160, 100));
for(int i = 9; i>=0; i--) {
pnb.add(bn[i]);
}
//here buttons add to panels
JPanel pb_nom = new JPanel();
pb_nom.setPreferredSize(new Dimension(160, 35));
pb_nom.add(eb);
pb_nom.add(cb);
pb_nom.add(mb);
pb_nom.add(db);
pb_nom.add(ab);
pb_nom.add(sb);
umo_ji[] pb_action = new umo_ji[10];
for ( int i = 0; i < 10; i++ ) {
pb_action[i] = new umo_ji(bn[i]);
bn[i].addActionListener(pb_action[i]);
}
//here perform action buttons
eb e1 = new eb();
eb.addActionListener(e1);
cb c = new cb();
cb.addActionListener(c);
mb e2 = new mb();
mb.addActionListener(e2);
db e3 = new db();
db.addActionListener(e3);
ab add = new ab();
ab.addActionListener(add);
Sub_But subtract = new Sub_But();
sb.addActionListener(subtract);
PM.add(pt);
PM.add(pnb);
PM.add(pb_nom);
add(PM);
//To set the title and size
setTitle(\"Button_Test\");
setSize(180, 290);
setLocationByPlatform(true);
setDefaultCloseOperation(WindowConstants.EXIT_ON_CLOSE);
setVisible(true);
}
private class umo_ji implements ActionListener {
//here implements actionlistener
private String c;
public umo_ji(JButton a) {
this.c = a.text_get();
}
public void actionPerformed(ActionEvent e) {
if (!jtf.text_get().equals(\"0.0\")) {
jtf.text_set(jtf.text_get() + c);
} else {
jtf.text_set(\"\");
actionPerformed(e);
}
}
}
private class eb implements ActionListener {
@Override//overide buttons
public void actionPerformed(ActionEvent e) {
tn2 = Double.parseDouble(jtf.text_get());
if (nom == 0) {
jtf.text_set(Double.toString((Math.round((tn / tn2) * 100)) / 100));
} else if (nom == 1) {
jtf.text_set(Double.toString(tn * tn2));
} else if (nom == 2) {
jtf.text_set(Double.toString(tn2 + tn));
} else if (nom == 3) {
jtf.text_set(Double.toString(tn - tn2));
} else {
jtf.text_set(String.valueOf(tn));
}
tn = Double.parseDouble(jtf.text_get());
}
}
private class cb implements.
a. Total asset turnover = Return on assets Profit margin= 0.145 .pdf
a. Total asset turnover = Return on assets / Profit margin
= 0.145 / 0.055
= 2.64 times
b. Return on equity = Return on assets / (1 - Debt-to-total-assets ratio)
= 0.145 / (1 - 0.65)
= 0.145 / 0.35
= 0.4143 or 41.43%
Solution
a. Total asset turnover = Return on assets / Profit margin
= 0.145 / 0.055
= 2.64 times
b. Return on equity = Return on assets / (1 - Debt-to-total-assets ratio)
= 0.145 / (1 - 0.65)
= 0.145 / 0.35
= 0.4143 or 41.43%.
d) All but Iv is the right answer.Solutiond) All but Iv is the.pdf
d) All but Iv is the right answer.
Solution
d) All but Iv is the right answer..
2. b) fed on molecules with a chemical structure to pesticides.The.pdf
2. b) fed on molecules with a chemical structure to pesticides.
They are also called as biopesticides.
Solution
2. b) fed on molecules with a chemical structure to pesticides.
They are also called as biopesticides..
1.Management information systemMany of the different types of Inf.pdf
1.Management information system:
Many of the different types of Information Systems found in commercial organizations are
referred to as \"Management Information Systems\". Management Information Systems are
management-level systems that are used by middle managers to help ensure the smooth running
of the organization in the short to medium term. The highly structured information provided by
these systems allows managers to evaluate an organization\'s performance by comparing current
with previous outputs.
The functions of Management Information Systems is
Internal transactions sorting and maintain a summery reports.
Structured data summerizing and maintain detailed reports.
2. Executive information system :
Executive Information Systems are strategic-level information systems. Executive Information
System help executives and senior managers analyze the environment in which the organization
operates, to identify long-term trends, and to plan appropriate courses of action. The information
in such systems is often weakly structured and comes from both internal and external sources.
Executive Information System are designed to be operated directly by executives without the
need for intermediaries.
Executive Information System organizes and presents data and information from both external
data sources and internal Management Information System.
Functions
Summerizing the external data and maintain summery reports.
Drilling down the predefined models i.e graphs
Solution
1.Management information system:
Many of the different types of Information Systems found in commercial organizations are
referred to as \"Management Information Systems\". Management Information Systems are
management-level systems that are used by middle managers to help ensure the smooth running
of the organization in the short to medium term. The highly structured information provided by
these systems allows managers to evaluate an organization\'s performance by comparing current
with previous outputs.
The functions of Management Information Systems is
Internal transactions sorting and maintain a summery reports.
Structured data summerizing and maintain detailed reports.
2. Executive information system :
Executive Information Systems are strategic-level information systems. Executive Information
System help executives and senior managers analyze the environment in which the organization
operates, to identify long-term trends, and to plan appropriate courses of action. The information
in such systems is often weakly structured and comes from both internal and external sources.
Executive Information System are designed to be operated directly by executives without the
need for intermediaries.
Executive Information System organizes and presents data and information from both external
data sources and internal Management Information System.
Functions
Summerizing the external data and maintain summery reports.
Drilling down the predefined models i.e graphs.
1. MnO4- is purple in solution. When the reaction is complete the so.pdf
1. MnO4- is purple in solution. When the reaction is complete the solution turns colorless as
MnO4- is reduced to Mn2+. The end point can thus be determined using the color change of the
reactant itself.
2. The reaction needs to occur in acidic medium for MnO4- to be fully reduced to Mn2+. In
neutral or alkaine media, the MnO4- is only partially reduced to MnO2.
3. The initial reaction has a high activation energy and requires heating to increase the reaction
rate so that the titration can be completed within a reasonable time.
Solution
1. MnO4- is purple in solution. When the reaction is complete the solution turns colorless as
MnO4- is reduced to Mn2+. The end point can thus be determined using the color change of the
reactant itself.
2. The reaction needs to occur in acidic medium for MnO4- to be fully reduced to Mn2+. In
neutral or alkaine media, the MnO4- is only partially reduced to MnO2.
3. The initial reaction has a high activation energy and requires heating to increase the reaction
rate so that the titration can be completed within a reasonable time..
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