Is this a reduction oxidation reaction If so, wh.pdf
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Is this a reduction oxidation reaction? If so, what you use is the periodic table. Hydrogen will always have a +1 charge associated with it and Br and F are both halogens which have a -1 charge associated with them. For your reaction, Hydrogen is staying constant at +1 while Br is going from -1 to 0 and F is going from 0 to -1. When atoms are bound to other atoms, they display a charge. When they are not bound or in their natural state, the associated charge is 0. Since Br and F are both diatomic (2 atoms=natural state), their charges are 0. So hydrogen is staying constant, Br is being oxidized and F is being reduced. Hope this helps! Solution Is this a reduction oxidation reaction? If so, what you use is the periodic table. Hydrogen will always have a +1 charge associated with it and Br and F are both halogens which have a -1 charge associated with them. For your reaction, Hydrogen is staying constant at +1 while Br is going from -1 to 0 and F is going from 0 to -1. When atoms are bound to other atoms, they display a charge. When they are not bound or in their natural state, the associated charge is 0. Since Br and F are both diatomic (2 atoms=natural state), their charges are 0. So hydrogen is staying constant, Br is being oxidized and F is being reduced. Hope this helps!.
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Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^.pdf![Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^.pdf](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^-3 =2.14302x10^-3 Step2
Equivalents of acid = .225/E ; E be the equivalent eight of acid Step3 .225/E = 2.14302x10^-3
or E=105
Solution
Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^-3 =2.14302x10^-3 Step2
Equivalents of acid = .225/E ; E be the equivalent eight of acid Step3 .225/E = 2.14302x10^-3
or E=105.
step 1 Br2 under UV light step 2 NMe3 Solu.pdf
step 1: Br2 under UV light step 2: NMe3
Solution
step 1: Br2 under UV light step 2: NMe3.
Since column chromatography substances are separa.pdf
Since column chromatography substances are separated on the basis of their
movement. As impurity atoms are less adsorbed in solution so their movement is less and we get
separated solution form impurity
Solution
Since column chromatography substances are separated on the basis of their
movement. As impurity atoms are less adsorbed in solution so their movement is less and we get
separated solution form impurity.
Oh group is removed by Br yes ther can be two ty.pdf
Oh group is removed by Br yes ther can be two types of isomers Br can be above
the plane or below the plane of ring
Solution
Oh group is removed by Br yes ther can be two types of isomers Br can be above
the plane or below the plane of ring.
Moles of Na2O2 = 2578=0.32 Moles of H2O = 12.11.pdf
Moles of Na2O2 = 25/78=0.32 Moles of H2O = 12.1/18=0.6722 Since Na2O2 is
the limiting reagent Heat released = 0.32*126 =40.32 kj
Solution
Moles of Na2O2 = 25/78=0.32 Moles of H2O = 12.1/18=0.6722 Since Na2O2 is
the limiting reagent Heat released = 0.32*126 =40.32 kj.
HNO3========= H+ (aq) + NO3-(aq) [H+] = 0.0578 .pdf
HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^-
14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = -
log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76
Solution
HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^-
14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = -
log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76.
Recommended
Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^.pdf
Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^-3 =2.14302x10^-3 Step2
Equivalents of acid = .225/E ; E be the equivalent eight of acid Step3 .225/E = 2.14302x10^-3
or E=105
Solution
Step1 Equivalents of NaOH= .1146 x[19.25-.55]x10^-3 =2.14302x10^-3 Step2
Equivalents of acid = .225/E ; E be the equivalent eight of acid Step3 .225/E = 2.14302x10^-3
or E=105.
step 1 Br2 under UV light step 2 NMe3 Solu.pdf
step 1: Br2 under UV light step 2: NMe3
Solution
step 1: Br2 under UV light step 2: NMe3.
Since column chromatography substances are separa.pdf
Since column chromatography substances are separated on the basis of their
movement. As impurity atoms are less adsorbed in solution so their movement is less and we get
separated solution form impurity
Solution
Since column chromatography substances are separated on the basis of their
movement. As impurity atoms are less adsorbed in solution so their movement is less and we get
separated solution form impurity.
Oh group is removed by Br yes ther can be two ty.pdf
Oh group is removed by Br yes ther can be two types of isomers Br can be above
the plane or below the plane of ring
Solution
Oh group is removed by Br yes ther can be two types of isomers Br can be above
the plane or below the plane of ring.
Moles of Na2O2 = 2578=0.32 Moles of H2O = 12.11.pdf
Moles of Na2O2 = 25/78=0.32 Moles of H2O = 12.1/18=0.6722 Since Na2O2 is
the limiting reagent Heat released = 0.32*126 =40.32 kj
Solution
Moles of Na2O2 = 25/78=0.32 Moles of H2O = 12.1/18=0.6722 Since Na2O2 is
the limiting reagent Heat released = 0.32*126 =40.32 kj.
HNO3========= H+ (aq) + NO3-(aq) [H+] = 0.0578 .pdf
HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^-
14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = -
log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76
Solution
HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^-
14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = -
log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76.
Valley fever is the initial form of coccidioidomycosis infection. Th.pdf
Valley fever is the initial form of coccidioidomycosis infection. This initial, acute illness can
develop into a more serious disease, including chronic and disseminated coccidioidomycosis.
Microrganism is Coccidioides immitis
Acute coccidioidomycosis (valley fever)
The initial, or acute, form of coccidioidomycosis is often mild, with few, if any, symptoms.
When signs and symptoms do occur, they appear one to three weeks after exposure. They tend to
resemble those of the flu, and can range from minor to severe, including:
The rash that sometimes accompanies valley fever is made up of painful red bumps that may
later turn brown. The rash mainly appears on your lower legs, but sometimes on your chest, arms
and back. Others may have a raised red rash with blisters or eruptions that look like pimples.
If you don\'t become ill from valley fever, you may only find out you\'ve been infected when you
later have a positive skin or blood test or when small areas of residual infection (nodules) in the
lungs show up on a routine chest X-ray. Although the nodules typically don\'t cause problems,
they can look like cancer on X-rays.
If you do develop symptoms, especially severe ones, the course of the disease is highly variable.
It can take months to fully recover, and fatigue and joint aches can last even longer. The severity
of the disease depends on several factors, including your overall health and the number of fungus
spores you inhale.
Chronic coccidioidomycosis
If the initial coccidioidomycosis infection doesn\'t completely resolve, it may progress to a
chronic form of pneumonia. This complication is most common in people with weakened
immune systems.
Signs and symptoms include:
Disseminated coccidioidomycosis
The most serious form of the disease, disseminated coccidioidomycosis, occurs when the
infection spreads (disseminates) beyond the lungs to other parts of the body. Most often these
parts include the skin, bones, liver, brain, heart, and the membranes that protect the brain and
spinal cord (meninges).
The signs and symptoms of disseminated disease depend on which parts of your body are
affected and may include:
Solution
Valley fever is the initial form of coccidioidomycosis infection. This initial, acute illness can
develop into a more serious disease, including chronic and disseminated coccidioidomycosis.
Microrganism is Coccidioides immitis
Acute coccidioidomycosis (valley fever)
The initial, or acute, form of coccidioidomycosis is often mild, with few, if any, symptoms.
When signs and symptoms do occur, they appear one to three weeks after exposure. They tend to
resemble those of the flu, and can range from minor to severe, including:
The rash that sometimes accompanies valley fever is made up of painful red bumps that may
later turn brown. The rash mainly appears on your lower legs, but sometimes on your chest, arms
and back. Others may have a raised red rash with blisters or eruptions that look like pimples.
If you don\'t become ill from .
Yes,The interthalamic adhesion (also known as the mass intermedia or.pdf
Yes,The interthalamic adhesion (also known as the mass intermedia or middle commissure) is a
flattened band of tissue that connects both parts of the thalamus at their medial surfaces. The
medial surfaces form the upper part of the lateral wall to the third ventricle.
Solution
Yes,The interthalamic adhesion (also known as the mass intermedia or middle commissure) is a
flattened band of tissue that connects both parts of the thalamus at their medial surfaces. The
medial surfaces form the upper part of the lateral wall to the third ventricle..
Think of the X and Y as totally different events. The expected value.pdf
Think of the X and Y as totally different events. The expected value for X will be the middle of
the interval because all the points are equally weighted.
E(X) = 0.5
The same goes for Y.
E(Y) = 0.5
The triangle (0, 0.5) (0.5, 0) (0,0) has base=0.5 and height = 0.5
A = (1/2)(BH) = (1/2)(0.5)(0.5) = 0.125 = 1/8
Solution
Think of the X and Y as totally different events. The expected value for X will be the middle of
the interval because all the points are equally weighted.
E(X) = 0.5
The same goes for Y.
E(Y) = 0.5
The triangle (0, 0.5) (0.5, 0) (0,0) has base=0.5 and height = 0.5
A = (1/2)(BH) = (1/2)(0.5)(0.5) = 0.125 = 1/8.
There are two source terms in Maxwell’s equations, the static charge.pdf
There are two source terms in Maxwell’s equations, the static charge density and the dynamic
current density J.
Solution
There are two source terms in Maxwell’s equations, the static charge density and the dynamic
current density J..
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to .pdf
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to give perxentate (XeO64-)
2HXeO4- + 2OH- = XeO64- + Xe + O2 + 2H2O
Solution
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to give perxentate (XeO64-)
2HXeO4- + 2OH- = XeO64- + Xe + O2 + 2H2O.
The strengths of the Indian Parliamentary System of governanace1..pdf
The strengths of the Indian Parliamentary System of governanace:
1.Smooth functioning of the government through the legislature and administration when the
party enjoys majority in the Parliament.
2. Decision making is quick - Majority of the ruling party in the Parliament facilitates speedy
decision making and implementation of policies.
3. Flexible system - can quickly adapt to changing situations. During grave situations the
leadership can be changed without harassment like during the \"emergency\" period.
4. Transparent Administration - Executive remains ever vigilant to secure its future electoral
prospects and the Parliament controls the executive through the requirement of the party have the
confidence of the house and through the financial powers vested in the Parliament.This controls
the executive from misuse of power and funds.
5. A nation with so many varied groups get an opportunity for representation.
6. The executive and legislature are not totally separated and thus while the executive wing has
the freedom to act on policies and yet it is accountable to the Parliament for its actions.
Disadvantages:
1. Absolute majority in the Parliament brings about a tendency of the ruling party to be
dictatorial and engenders the possibilities of corruption.
2. Politicization of the administration as decision making takes into account the political
ramifications more than the merits of the decision.
3. Unsuitable for multi-party system as coalition governments bring about logjams and much
delayed decision making in the Parliament. The bureaucracy too lacks direction and leadership in
those times and the policy making and implemntation thereof \'crawls\' during such periods.
4. Unsuitable for managing crisis situations rapidly as the PM and his Council of Ministers have
the political ramifications weighing over them more so than the criticality of the situation that
they have to respond to.
The above is from data from the Net and personal experience.
Solution
The strengths of the Indian Parliamentary System of governanace:
1.Smooth functioning of the government through the legislature and administration when the
party enjoys majority in the Parliament.
2. Decision making is quick - Majority of the ruling party in the Parliament facilitates speedy
decision making and implementation of policies.
3. Flexible system - can quickly adapt to changing situations. During grave situations the
leadership can be changed without harassment like during the \"emergency\" period.
4. Transparent Administration - Executive remains ever vigilant to secure its future electoral
prospects and the Parliament controls the executive through the requirement of the party have the
confidence of the house and through the financial powers vested in the Parliament.This controls
the executive from misuse of power and funds.
5. A nation with so many varied groups get an opportunity for representation.
6. The executive and legislature are not totally separated and thus while .
Pulse code modulationPulse code modulation is a digital sign of a.pdf
Pulse code modulation:
Pulse code modulation is a digital sign of an analog signal that takes samples of the amplitude of
the analog signal at usual intervals. The sampled analog data is changed and then represented in
the form of binary data
Solution
Pulse code modulation:
Pulse code modulation is a digital sign of an analog signal that takes samples of the amplitude of
the analog signal at usual intervals. The sampled analog data is changed and then represented in
the form of binary data.
PFA JAVA programimport java.io.; import java.util.; import .pdf
PFA JAVA program:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Bus {
static int[] applyPulse(String registers, String pulses)
{
int[] register=new int[4];
int[] pulse=new int[3];
int i=0;
StringTokenizer st = new StringTokenizer(registers,\" \");
while (st.hasMoreElements()) {
register[i++]=Integer.parseInt(st.nextElement().toString());
}
i=0;
st = new StringTokenizer(pulses,\" \");
while (st.hasMoreElements()) {
pulse[i++]=Integer.parseInt(st.nextElement().toString());
}
//Now ewe have, register [0] is A, register [1] is AB,register [3] is C and register [4] is D
if(pulse[0]==1) // if L=1
{
register[1]=register[0];
register[2]=register[0];
}
if(pulse[0]==1) // if M=1
{
register[0]=register[3];
}
if(pulse[0]==1) // if N=1
{
register[1]=register[2];
register[3]=register[2];
}
return register;
}
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
System.out.println(\"Enter initial values of four regisers A B C D seperated by spaces\");
String registers=sc.nextLine();
System.out.println(\"Ener value of Clock Pulse L M N seperated by spaces\");
String pulses=sc.nextLine();
System.out.println(\"Values of registers A B C D respectively after applying pulse:\");
int[] arr=applyPulse(registers,pulses);
for(int i=0;i<4;i++)
{
System.out.print(arr[i]+\" \");
}
}
}
Solution
PFA JAVA program:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Bus {
static int[] applyPulse(String registers, String pulses)
{
int[] register=new int[4];
int[] pulse=new int[3];
int i=0;
StringTokenizer st = new StringTokenizer(registers,\" \");
while (st.hasMoreElements()) {
register[i++]=Integer.parseInt(st.nextElement().toString());
}
i=0;
st = new StringTokenizer(pulses,\" \");
while (st.hasMoreElements()) {
pulse[i++]=Integer.parseInt(st.nextElement().toString());
}
//Now ewe have, register [0] is A, register [1] is AB,register [3] is C and register [4] is D
if(pulse[0]==1) // if L=1
{
register[1]=register[0];
register[2]=register[0];
}
if(pulse[0]==1) // if M=1
{
register[0]=register[3];
}
if(pulse[0]==1) // if N=1
{
register[1]=register[2];
register[3]=register[2];
}
return register;
}
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
System.out.println(\"Enter initial values of four regisers A B C D seperated by spaces\");
String registers=sc.nextLine();
System.out.println(\"Ener value of Clock Pulse L M N seperated by spaces\");
String pulses=sc.nextLine();
System.out.println(\"Values of registers A B C D respectively after applying pulse:\");
int[] arr=applyPulse(registers,pulses);
for(int i=0;i<4;i++)
{
System.out.print(arr[i]+\" \");
}
}
}.
D) four s bonds Pi bonds are double bonds as carb.pdf
D) four s bonds Pi bonds are double bonds as carbon cannot support them...four
valence electrons,
Solution
D) four s bonds Pi bonds are double bonds as carbon cannot support them...four
valence electrons,.
out of Africa hypothesis out of Africa hypothesis is about .pdf
\"out of Africa hypothesis\": \"out of Africa hypothesis\" is about the origin of human beings
from africa. this theory is highly acceptedby various researchers because of the strong evidence
of
Solution
\"out of Africa hypothesis\": \"out of Africa hypothesis\" is about the origin of human beings
from africa. this theory is highly acceptedby various researchers because of the strong evidence
of.
Leukocytes or WBC They are of two typesGranulocytes Neutrophils.pdf
Leukocytes or WBC: They are of two types:
Granulocytes: Neutrophils, eosinohils, and basophils
Agranulocytes: Monocytes and lymphocytes
Neutrophils: MOst numberous of all the WBC, make 60 - 80% of WBC. Upon Giemsa staining,
the granules take up both eosin red and methylene blue, thus appear purple color. The cytoplasm
stains blue with multilobed nucleus. Their number increases in pathological conditions like
infectionss.
Eosinophils: upto 6% of total WBC, their granules take up Eosin red and appear red to pink
colored, due to eosinophilic cationic protein that binds the dye. Cytoplasm stains blue, bilobed
nucleus. The number increases with conditions like parasitic manifestations and viral infections.
Basophils: Generally absent or rarely found in healthy people. Their granules take up methylene
blue due to negatively charged molecles like heparin present in them. They are responsible for
certian types of hypersensitivities. Their number increaes during allergies.
Monocytes: They have a slightly indented nucleus that occupies about half of the cell. They are
larger than other cells. The cytoplasm shows granules but are not conspicuously stained. They
are cells that have the function of phagocytosis, next to neutrophils.
Lymphocytes: Large nucleus that occupies most of the cell leaving a thin rim of cytoplasm. They
produce antibodies.Nucleus stains purple and cytoplasm blue. Their number increases in
conditions like leukemia.
Solution
Leukocytes or WBC: They are of two types:
Granulocytes: Neutrophils, eosinohils, and basophils
Agranulocytes: Monocytes and lymphocytes
Neutrophils: MOst numberous of all the WBC, make 60 - 80% of WBC. Upon Giemsa staining,
the granules take up both eosin red and methylene blue, thus appear purple color. The cytoplasm
stains blue with multilobed nucleus. Their number increases in pathological conditions like
infectionss.
Eosinophils: upto 6% of total WBC, their granules take up Eosin red and appear red to pink
colored, due to eosinophilic cationic protein that binds the dye. Cytoplasm stains blue, bilobed
nucleus. The number increases with conditions like parasitic manifestations and viral infections.
Basophils: Generally absent or rarely found in healthy people. Their granules take up methylene
blue due to negatively charged molecles like heparin present in them. They are responsible for
certian types of hypersensitivities. Their number increaes during allergies.
Monocytes: They have a slightly indented nucleus that occupies about half of the cell. They are
larger than other cells. The cytoplasm shows granules but are not conspicuously stained. They
are cells that have the function of phagocytosis, next to neutrophils.
Lymphocytes: Large nucleus that occupies most of the cell leaving a thin rim of cytoplasm. They
produce antibodies.Nucleus stains purple and cytoplasm blue. Their number increases in
conditions like leukemia..
In User Interface, Menu Architecture contains the following types.pdf
In User Interface, Menu Architecture contains the following types:
Answer: c.Single - Linear sequences
Solution
In User Interface, Menu Architecture contains the following types:
Answer: c.Single - Linear sequences.
import static org.junit.Assert.;import java.util.Arrays; import.pdf
import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;
import org.junit.runners.Parameterized.Parameter;
import org.junit.runners.Parameterized.Parameters;
import com.att.osd.best.module.registration.bean.Min;
@RunWith(Parameterized.class)
public class DataDrivenMinTest {
private List list;
private String expected;
public DataDrivenMinTest(List list, String expected) {
this.list = list;
this.expected = expected;
}
@Parameterized.Parameters
public static Iterable data() {
return Arrays.asList(
new Object[][]{
{Arrays.asList(new Object[]{\"A\",\"B\",\"C\"}),\"A\"}
}
);
}
@Test
public void test_min() {
assertEquals(expected,
Min.min(list));
}
}
Solution
import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;
import org.junit.runners.Parameterized.Parameter;
import org.junit.runners.Parameterized.Parameters;
import com.att.osd.best.module.registration.bean.Min;
@RunWith(Parameterized.class)
public class DataDrivenMinTest {
private List list;
private String expected;
public DataDrivenMinTest(List list, String expected) {
this.list = list;
this.expected = expected;
}
@Parameterized.Parameters
public static Iterable data() {
return Arrays.asList(
new Object[][]{
{Arrays.asList(new Object[]{\"A\",\"B\",\"C\"}),\"A\"}
}
);
}
@Test
public void test_min() {
assertEquals(expected,
Min.min(list));
}
}.
Fresh cut vegitation a non-generic name, are any of 47 species of.pdf
Fresh cut vegitation:
a non-generic name, are any of 47 species of leaf-chewing ants belonging to the two genera Atta
and Acromyrmex. These species of tropical, fungus-growing ants are all endemic to South and
Central America, Mexico, and parts of the southern United States. Leafcutter ants can carry more
than 5000 times their body weight and cut and process fresh vegetation (leaves, flowers, and
grasses) to serve as the nutritional substrate for their fungal cultivars.
Acromyrmex and Atta ants have much in common anatomically; however, the two can be
identified by their external differences. Atta ants have three pairs of spines and a smooth
exoskeleton on the upper surface of the thorax, while Acromyrmex ants have four pairs and a
rough exoskeleton.
Ants:
Ants are eusocial insects of the family Formicidae and, along with the related wasps and bees,
belong to the order Hymenoptera.Ants form colonies that range in size from a few dozen
predatory individuals living in small natural cavities to highly organised colonies that may
occupy large territories and consist of millions of individuals. Larger colonies consist mostly of
sterile, wingless females forming castes of \"workers\", \"soldiers\", or other specialised groups.
Nearly all ant colonies also have some fertile males called \"drones\" and one or more fertile
females called \"queens\". The colonies are described as superorganisms because the ants appear
to operate as a unified entity, collectively working together to support the colony. Ants have
colonised almost every landmass on Earth. The only places lacking indigenous ants are
Antarctica and a few remote or inhospitable islands.
Actinobacter:
Actinobacteria is a phylum of gram-positive bacteria. They can be terrestrial or aquatic. They
are of great economic importance to humans because agriculture and forests depend on their
contributions to soil systems. In soil, they behave much like fungi, helping to decompose the
organic matter of dead organisms so that the molecules can be taken up anew by plants. In this
role the colonies often grow extensive mycelia, like a fungus would, and the name of an
important order of the phylum, Actinomycetales
Fertilizers:
Bio-fertilizers provide \"eco-friendly\" organic agro-input. Bio-fertilizers such as Rhizobium,
Azotobacter, Azospirilium and blue green algae (BGA) have been in use a long time.
Rhizobiuminoculant is used for leguminous crops. Azotobacter can be used with crops like
wheat, maize, mustard, cotton, potato and other vegetable crops. Azospirillum inoculations are
recommended mainly for sorghum, millets, maize, sugarcane and wheat. Blue green algae
belonging to a general cyanobacteria genus, Nostoc or Anabaena or Tolypothrix or Aulosira, fix
atmospheric nitrogen and are used as inoculations for paddy crop grown both under upland and
low-land conditions. Anabaena in association with water fern Azolla contributes nitrogen up to
60 kg/ha/season and also enriches soils with organic matter
A.
answer is E its a oxydising agent finally leading.pdf
answer is E its a oxydising agent finally leading to bond break for forming double
oxygen bond
Solution
answer is E its a oxydising agent finally leading to bond break for forming double
oxygen bond.
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths
Solution
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths.
CSF flows from the 3rd ventricle through the cerebral aqueduct into .pdf
CSF flows from the 3rd ventricle through the cerebral aqueduct into the 4th ventricle.
CSF is produced by the choroid plexus in the ventricles.
CSF flows through the paired lateral apertures or the single medial aperture and into the central
canal of the spinal cord.
CSF flows into the arachnoid villi and drains into the dural venous sinuses.
CSF flows through the subarachnoid space and removes waste products.
Solution
CSF flows from the 3rd ventricle through the cerebral aqueduct into the 4th ventricle.
CSF is produced by the choroid plexus in the ventricles.
CSF flows through the paired lateral apertures or the single medial aperture and into the central
canal of the spinal cord.
CSF flows into the arachnoid villi and drains into the dural venous sinuses.
CSF flows through the subarachnoid space and removes waste products..
c.) it reduces random influences that would otherwise add to the ove.pdf
c.) it reduces random influences that would otherwise add to the overall variance
Solution
c.) it reduces random influences that would otherwise add to the overall variance.
Ans The comorbid means a morbidity or disease which is present alon.pdf
Ans: The comorbid means a morbidity or disease which is present along with the new or current
disease.
For example diabetes mellitus, hypertension, hypothyroidism.
Example of childhood comorbidity is dyslexia.
Solution
Ans: The comorbid means a morbidity or disease which is present along with the new or current
disease.
For example diabetes mellitus, hypertension, hypothyroidism.
Example of childhood comorbidity is dyslexia..
Answer ½Colorblindness is an X linked recessive inheritance. Affe.pdf
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2
Solution
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
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.
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Valley fever is the initial form of coccidioidomycosis infection. Th.pdf
Valley fever is the initial form of coccidioidomycosis infection. This initial, acute illness can
develop into a more serious disease, including chronic and disseminated coccidioidomycosis.
Microrganism is Coccidioides immitis
Acute coccidioidomycosis (valley fever)
The initial, or acute, form of coccidioidomycosis is often mild, with few, if any, symptoms.
When signs and symptoms do occur, they appear one to three weeks after exposure. They tend to
resemble those of the flu, and can range from minor to severe, including:
The rash that sometimes accompanies valley fever is made up of painful red bumps that may
later turn brown. The rash mainly appears on your lower legs, but sometimes on your chest, arms
and back. Others may have a raised red rash with blisters or eruptions that look like pimples.
If you don\'t become ill from valley fever, you may only find out you\'ve been infected when you
later have a positive skin or blood test or when small areas of residual infection (nodules) in the
lungs show up on a routine chest X-ray. Although the nodules typically don\'t cause problems,
they can look like cancer on X-rays.
If you do develop symptoms, especially severe ones, the course of the disease is highly variable.
It can take months to fully recover, and fatigue and joint aches can last even longer. The severity
of the disease depends on several factors, including your overall health and the number of fungus
spores you inhale.
Chronic coccidioidomycosis
If the initial coccidioidomycosis infection doesn\'t completely resolve, it may progress to a
chronic form of pneumonia. This complication is most common in people with weakened
immune systems.
Signs and symptoms include:
Disseminated coccidioidomycosis
The most serious form of the disease, disseminated coccidioidomycosis, occurs when the
infection spreads (disseminates) beyond the lungs to other parts of the body. Most often these
parts include the skin, bones, liver, brain, heart, and the membranes that protect the brain and
spinal cord (meninges).
The signs and symptoms of disseminated disease depend on which parts of your body are
affected and may include:
Solution
Valley fever is the initial form of coccidioidomycosis infection. This initial, acute illness can
develop into a more serious disease, including chronic and disseminated coccidioidomycosis.
Microrganism is Coccidioides immitis
Acute coccidioidomycosis (valley fever)
The initial, or acute, form of coccidioidomycosis is often mild, with few, if any, symptoms.
When signs and symptoms do occur, they appear one to three weeks after exposure. They tend to
resemble those of the flu, and can range from minor to severe, including:
The rash that sometimes accompanies valley fever is made up of painful red bumps that may
later turn brown. The rash mainly appears on your lower legs, but sometimes on your chest, arms
and back. Others may have a raised red rash with blisters or eruptions that look like pimples.
If you don\'t become ill from .
Yes,The interthalamic adhesion (also known as the mass intermedia or.pdf
Yes,The interthalamic adhesion (also known as the mass intermedia or middle commissure) is a
flattened band of tissue that connects both parts of the thalamus at their medial surfaces. The
medial surfaces form the upper part of the lateral wall to the third ventricle.
Solution
Yes,The interthalamic adhesion (also known as the mass intermedia or middle commissure) is a
flattened band of tissue that connects both parts of the thalamus at their medial surfaces. The
medial surfaces form the upper part of the lateral wall to the third ventricle..
Think of the X and Y as totally different events. The expected value.pdf
Think of the X and Y as totally different events. The expected value for X will be the middle of
the interval because all the points are equally weighted.
E(X) = 0.5
The same goes for Y.
E(Y) = 0.5
The triangle (0, 0.5) (0.5, 0) (0,0) has base=0.5 and height = 0.5
A = (1/2)(BH) = (1/2)(0.5)(0.5) = 0.125 = 1/8
Solution
Think of the X and Y as totally different events. The expected value for X will be the middle of
the interval because all the points are equally weighted.
E(X) = 0.5
The same goes for Y.
E(Y) = 0.5
The triangle (0, 0.5) (0.5, 0) (0,0) has base=0.5 and height = 0.5
A = (1/2)(BH) = (1/2)(0.5)(0.5) = 0.125 = 1/8.
There are two source terms in Maxwell’s equations, the static charge.pdf
There are two source terms in Maxwell’s equations, the static charge density and the dynamic
current density J.
Solution
There are two source terms in Maxwell’s equations, the static charge density and the dynamic
current density J..
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to .pdf
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to give perxentate (XeO64-)
2HXeO4- + 2OH- = XeO64- + Xe + O2 + 2H2O
Solution
The xenate ion, [HXeO4]-, slowly decomposes in alkaline solution to give perxentate (XeO64-)
2HXeO4- + 2OH- = XeO64- + Xe + O2 + 2H2O.
The strengths of the Indian Parliamentary System of governanace1..pdf
The strengths of the Indian Parliamentary System of governanace:
1.Smooth functioning of the government through the legislature and administration when the
party enjoys majority in the Parliament.
2. Decision making is quick - Majority of the ruling party in the Parliament facilitates speedy
decision making and implementation of policies.
3. Flexible system - can quickly adapt to changing situations. During grave situations the
leadership can be changed without harassment like during the \"emergency\" period.
4. Transparent Administration - Executive remains ever vigilant to secure its future electoral
prospects and the Parliament controls the executive through the requirement of the party have the
confidence of the house and through the financial powers vested in the Parliament.This controls
the executive from misuse of power and funds.
5. A nation with so many varied groups get an opportunity for representation.
6. The executive and legislature are not totally separated and thus while the executive wing has
the freedom to act on policies and yet it is accountable to the Parliament for its actions.
Disadvantages:
1. Absolute majority in the Parliament brings about a tendency of the ruling party to be
dictatorial and engenders the possibilities of corruption.
2. Politicization of the administration as decision making takes into account the political
ramifications more than the merits of the decision.
3. Unsuitable for multi-party system as coalition governments bring about logjams and much
delayed decision making in the Parliament. The bureaucracy too lacks direction and leadership in
those times and the policy making and implemntation thereof \'crawls\' during such periods.
4. Unsuitable for managing crisis situations rapidly as the PM and his Council of Ministers have
the political ramifications weighing over them more so than the criticality of the situation that
they have to respond to.
The above is from data from the Net and personal experience.
Solution
The strengths of the Indian Parliamentary System of governanace:
1.Smooth functioning of the government through the legislature and administration when the
party enjoys majority in the Parliament.
2. Decision making is quick - Majority of the ruling party in the Parliament facilitates speedy
decision making and implementation of policies.
3. Flexible system - can quickly adapt to changing situations. During grave situations the
leadership can be changed without harassment like during the \"emergency\" period.
4. Transparent Administration - Executive remains ever vigilant to secure its future electoral
prospects and the Parliament controls the executive through the requirement of the party have the
confidence of the house and through the financial powers vested in the Parliament.This controls
the executive from misuse of power and funds.
5. A nation with so many varied groups get an opportunity for representation.
6. The executive and legislature are not totally separated and thus while .
Pulse code modulationPulse code modulation is a digital sign of a.pdf
Pulse code modulation:
Pulse code modulation is a digital sign of an analog signal that takes samples of the amplitude of
the analog signal at usual intervals. The sampled analog data is changed and then represented in
the form of binary data
Solution
Pulse code modulation:
Pulse code modulation is a digital sign of an analog signal that takes samples of the amplitude of
the analog signal at usual intervals. The sampled analog data is changed and then represented in
the form of binary data.
PFA JAVA programimport java.io.; import java.util.; import .pdf
PFA JAVA program:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Bus {
static int[] applyPulse(String registers, String pulses)
{
int[] register=new int[4];
int[] pulse=new int[3];
int i=0;
StringTokenizer st = new StringTokenizer(registers,\" \");
while (st.hasMoreElements()) {
register[i++]=Integer.parseInt(st.nextElement().toString());
}
i=0;
st = new StringTokenizer(pulses,\" \");
while (st.hasMoreElements()) {
pulse[i++]=Integer.parseInt(st.nextElement().toString());
}
//Now ewe have, register [0] is A, register [1] is AB,register [3] is C and register [4] is D
if(pulse[0]==1) // if L=1
{
register[1]=register[0];
register[2]=register[0];
}
if(pulse[0]==1) // if M=1
{
register[0]=register[3];
}
if(pulse[0]==1) // if N=1
{
register[1]=register[2];
register[3]=register[2];
}
return register;
}
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
System.out.println(\"Enter initial values of four regisers A B C D seperated by spaces\");
String registers=sc.nextLine();
System.out.println(\"Ener value of Clock Pulse L M N seperated by spaces\");
String pulses=sc.nextLine();
System.out.println(\"Values of registers A B C D respectively after applying pulse:\");
int[] arr=applyPulse(registers,pulses);
for(int i=0;i<4;i++)
{
System.out.print(arr[i]+\" \");
}
}
}
Solution
PFA JAVA program:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Bus {
static int[] applyPulse(String registers, String pulses)
{
int[] register=new int[4];
int[] pulse=new int[3];
int i=0;
StringTokenizer st = new StringTokenizer(registers,\" \");
while (st.hasMoreElements()) {
register[i++]=Integer.parseInt(st.nextElement().toString());
}
i=0;
st = new StringTokenizer(pulses,\" \");
while (st.hasMoreElements()) {
pulse[i++]=Integer.parseInt(st.nextElement().toString());
}
//Now ewe have, register [0] is A, register [1] is AB,register [3] is C and register [4] is D
if(pulse[0]==1) // if L=1
{
register[1]=register[0];
register[2]=register[0];
}
if(pulse[0]==1) // if M=1
{
register[0]=register[3];
}
if(pulse[0]==1) // if N=1
{
register[1]=register[2];
register[3]=register[2];
}
return register;
}
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
System.out.println(\"Enter initial values of four regisers A B C D seperated by spaces\");
String registers=sc.nextLine();
System.out.println(\"Ener value of Clock Pulse L M N seperated by spaces\");
String pulses=sc.nextLine();
System.out.println(\"Values of registers A B C D respectively after applying pulse:\");
int[] arr=applyPulse(registers,pulses);
for(int i=0;i<4;i++)
{
System.out.print(arr[i]+\" \");
}
}
}.
D) four s bonds Pi bonds are double bonds as carb.pdf
D) four s bonds Pi bonds are double bonds as carbon cannot support them...four
valence electrons,
Solution
D) four s bonds Pi bonds are double bonds as carbon cannot support them...four
valence electrons,.
out of Africa hypothesis out of Africa hypothesis is about .pdf
\"out of Africa hypothesis\": \"out of Africa hypothesis\" is about the origin of human beings
from africa. this theory is highly acceptedby various researchers because of the strong evidence
of
Solution
\"out of Africa hypothesis\": \"out of Africa hypothesis\" is about the origin of human beings
from africa. this theory is highly acceptedby various researchers because of the strong evidence
of.
Leukocytes or WBC They are of two typesGranulocytes Neutrophils.pdf
Leukocytes or WBC: They are of two types:
Granulocytes: Neutrophils, eosinohils, and basophils
Agranulocytes: Monocytes and lymphocytes
Neutrophils: MOst numberous of all the WBC, make 60 - 80% of WBC. Upon Giemsa staining,
the granules take up both eosin red and methylene blue, thus appear purple color. The cytoplasm
stains blue with multilobed nucleus. Their number increases in pathological conditions like
infectionss.
Eosinophils: upto 6% of total WBC, their granules take up Eosin red and appear red to pink
colored, due to eosinophilic cationic protein that binds the dye. Cytoplasm stains blue, bilobed
nucleus. The number increases with conditions like parasitic manifestations and viral infections.
Basophils: Generally absent or rarely found in healthy people. Their granules take up methylene
blue due to negatively charged molecles like heparin present in them. They are responsible for
certian types of hypersensitivities. Their number increaes during allergies.
Monocytes: They have a slightly indented nucleus that occupies about half of the cell. They are
larger than other cells. The cytoplasm shows granules but are not conspicuously stained. They
are cells that have the function of phagocytosis, next to neutrophils.
Lymphocytes: Large nucleus that occupies most of the cell leaving a thin rim of cytoplasm. They
produce antibodies.Nucleus stains purple and cytoplasm blue. Their number increases in
conditions like leukemia.
Solution
Leukocytes or WBC: They are of two types:
Granulocytes: Neutrophils, eosinohils, and basophils
Agranulocytes: Monocytes and lymphocytes
Neutrophils: MOst numberous of all the WBC, make 60 - 80% of WBC. Upon Giemsa staining,
the granules take up both eosin red and methylene blue, thus appear purple color. The cytoplasm
stains blue with multilobed nucleus. Their number increases in pathological conditions like
infectionss.
Eosinophils: upto 6% of total WBC, their granules take up Eosin red and appear red to pink
colored, due to eosinophilic cationic protein that binds the dye. Cytoplasm stains blue, bilobed
nucleus. The number increases with conditions like parasitic manifestations and viral infections.
Basophils: Generally absent or rarely found in healthy people. Their granules take up methylene
blue due to negatively charged molecles like heparin present in them. They are responsible for
certian types of hypersensitivities. Their number increaes during allergies.
Monocytes: They have a slightly indented nucleus that occupies about half of the cell. They are
larger than other cells. The cytoplasm shows granules but are not conspicuously stained. They
are cells that have the function of phagocytosis, next to neutrophils.
Lymphocytes: Large nucleus that occupies most of the cell leaving a thin rim of cytoplasm. They
produce antibodies.Nucleus stains purple and cytoplasm blue. Their number increases in
conditions like leukemia..
In User Interface, Menu Architecture contains the following types.pdf
In User Interface, Menu Architecture contains the following types:
Answer: c.Single - Linear sequences
Solution
In User Interface, Menu Architecture contains the following types:
Answer: c.Single - Linear sequences.
import static org.junit.Assert.;import java.util.Arrays; import.pdf
import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;
import org.junit.runners.Parameterized.Parameter;
import org.junit.runners.Parameterized.Parameters;
import com.att.osd.best.module.registration.bean.Min;
@RunWith(Parameterized.class)
public class DataDrivenMinTest {
private List list;
private String expected;
public DataDrivenMinTest(List list, String expected) {
this.list = list;
this.expected = expected;
}
@Parameterized.Parameters
public static Iterable data() {
return Arrays.asList(
new Object[][]{
{Arrays.asList(new Object[]{\"A\",\"B\",\"C\"}),\"A\"}
}
);
}
@Test
public void test_min() {
assertEquals(expected,
Min.min(list));
}
}
Solution
import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;
import org.junit.runners.Parameterized.Parameter;
import org.junit.runners.Parameterized.Parameters;
import com.att.osd.best.module.registration.bean.Min;
@RunWith(Parameterized.class)
public class DataDrivenMinTest {
private List list;
private String expected;
public DataDrivenMinTest(List list, String expected) {
this.list = list;
this.expected = expected;
}
@Parameterized.Parameters
public static Iterable data() {
return Arrays.asList(
new Object[][]{
{Arrays.asList(new Object[]{\"A\",\"B\",\"C\"}),\"A\"}
}
);
}
@Test
public void test_min() {
assertEquals(expected,
Min.min(list));
}
}.
Fresh cut vegitation a non-generic name, are any of 47 species of.pdf
Fresh cut vegitation:
a non-generic name, are any of 47 species of leaf-chewing ants belonging to the two genera Atta
and Acromyrmex. These species of tropical, fungus-growing ants are all endemic to South and
Central America, Mexico, and parts of the southern United States. Leafcutter ants can carry more
than 5000 times their body weight and cut and process fresh vegetation (leaves, flowers, and
grasses) to serve as the nutritional substrate for their fungal cultivars.
Acromyrmex and Atta ants have much in common anatomically; however, the two can be
identified by their external differences. Atta ants have three pairs of spines and a smooth
exoskeleton on the upper surface of the thorax, while Acromyrmex ants have four pairs and a
rough exoskeleton.
Ants:
Ants are eusocial insects of the family Formicidae and, along with the related wasps and bees,
belong to the order Hymenoptera.Ants form colonies that range in size from a few dozen
predatory individuals living in small natural cavities to highly organised colonies that may
occupy large territories and consist of millions of individuals. Larger colonies consist mostly of
sterile, wingless females forming castes of \"workers\", \"soldiers\", or other specialised groups.
Nearly all ant colonies also have some fertile males called \"drones\" and one or more fertile
females called \"queens\". The colonies are described as superorganisms because the ants appear
to operate as a unified entity, collectively working together to support the colony. Ants have
colonised almost every landmass on Earth. The only places lacking indigenous ants are
Antarctica and a few remote or inhospitable islands.
Actinobacter:
Actinobacteria is a phylum of gram-positive bacteria. They can be terrestrial or aquatic. They
are of great economic importance to humans because agriculture and forests depend on their
contributions to soil systems. In soil, they behave much like fungi, helping to decompose the
organic matter of dead organisms so that the molecules can be taken up anew by plants. In this
role the colonies often grow extensive mycelia, like a fungus would, and the name of an
important order of the phylum, Actinomycetales
Fertilizers:
Bio-fertilizers provide \"eco-friendly\" organic agro-input. Bio-fertilizers such as Rhizobium,
Azotobacter, Azospirilium and blue green algae (BGA) have been in use a long time.
Rhizobiuminoculant is used for leguminous crops. Azotobacter can be used with crops like
wheat, maize, mustard, cotton, potato and other vegetable crops. Azospirillum inoculations are
recommended mainly for sorghum, millets, maize, sugarcane and wheat. Blue green algae
belonging to a general cyanobacteria genus, Nostoc or Anabaena or Tolypothrix or Aulosira, fix
atmospheric nitrogen and are used as inoculations for paddy crop grown both under upland and
low-land conditions. Anabaena in association with water fern Azolla contributes nitrogen up to
60 kg/ha/season and also enriches soils with organic matter
A.
answer is E its a oxydising agent finally leading.pdf
answer is E its a oxydising agent finally leading to bond break for forming double
oxygen bond
Solution
answer is E its a oxydising agent finally leading to bond break for forming double
oxygen bond.
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths
Solution
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths.
CSF flows from the 3rd ventricle through the cerebral aqueduct into .pdf
CSF flows from the 3rd ventricle through the cerebral aqueduct into the 4th ventricle.
CSF is produced by the choroid plexus in the ventricles.
CSF flows through the paired lateral apertures or the single medial aperture and into the central
canal of the spinal cord.
CSF flows into the arachnoid villi and drains into the dural venous sinuses.
CSF flows through the subarachnoid space and removes waste products.
Solution
CSF flows from the 3rd ventricle through the cerebral aqueduct into the 4th ventricle.
CSF is produced by the choroid plexus in the ventricles.
CSF flows through the paired lateral apertures or the single medial aperture and into the central
canal of the spinal cord.
CSF flows into the arachnoid villi and drains into the dural venous sinuses.
CSF flows through the subarachnoid space and removes waste products..
c.) it reduces random influences that would otherwise add to the ove.pdf
c.) it reduces random influences that would otherwise add to the overall variance
Solution
c.) it reduces random influences that would otherwise add to the overall variance.
Ans The comorbid means a morbidity or disease which is present alon.pdf
Ans: The comorbid means a morbidity or disease which is present along with the new or current
disease.
For example diabetes mellitus, hypertension, hypothyroidism.
Example of childhood comorbidity is dyslexia.
Solution
Ans: The comorbid means a morbidity or disease which is present along with the new or current
disease.
For example diabetes mellitus, hypertension, hypothyroidism.
Example of childhood comorbidity is dyslexia..
Answer ½Colorblindness is an X linked recessive inheritance. Affe.pdf
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2
Solution
Answer: ½
Colorblindness is an X linked recessive inheritance. Affected father passes on the chromosome
to all of his daughters.
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
GrandFather: XcY
XcY (father) x XcX (mother)
Son: XcY or XY
Hence: 50% probability or 1/2.
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