import java.util.LinkedList; import java.util.Random; import jav.pdf

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import java.util.LinkedList; import java.util.Random; import java.util.Scanner; /** * author * */ public class LinkedListTest { /** * param args */ public static void main(String[] args) { int M, N, Lower, Upper; Scanner scanner = null; try { LinkedList list1 = new LinkedList(); LinkedList list2 = new LinkedList(); LinkedList list3 = new LinkedList(); Random random = new Random(); scanner = new Scanner(System.in); System.out.print(\"Enter the Lower and Upper bounds:\"); Lower = scanner.nextInt(); Upper = scanner.nextInt(); System.out.print(\"Enter the N value to Insert list1:\"); N = scanner.nextInt(); System.out.print(\"Enter the M value to Insert list2:\"); M = scanner.nextInt(); for (int i = 1; i <= N; i++) { list1.add(random.nextInt((Upper - Lower) + 1) + Lower); } for (int i = 1; i <= M; i++) { list2.add(random.nextInt((Upper - Lower) + 1) + Lower); } System.out.println(\"list1: \" + list1); System.out.println(\"list2: \" + list2); System.out.println(\"Minimum elelment in list1 :\" + getMin(list1)); System.out.println(\"Minimum elelment in list2 :\" + getMin(list2)); System.out .println(\"Count Divisibles by 3 and 5 elelment in list1 :\" + countDivisibles(list1)); System.out .println(\"Count Divisibles by 3 and 5 elelment in list2 :\" + countDivisibles(list2)); System.out.println(\"Common Elements From list1 and list2:\" + getCommonList(list1, list2)); } catch (Exception e) { // TODO: handle exception } } /** * param list * return */ public static int getMin(LinkedList list) { int min = list.get(0); for (Integer integer : list) { if (min > integer) { min = integer; } } return min; } /** * param list * return */ public static int countDivisibles(LinkedList list) { int count = 0; for (Integer integer : list) { if (integer % 3 == 0 && integer % 5 == 0) { count++; } } return count; } /** * param list1 * param list2 * return */ public static LinkedList getCommonList(LinkedList list1, LinkedList list2) { LinkedList list = new LinkedList(); LinkedList listCommon = new LinkedList(); list = (list1.size() > list2.size()) ? list1 : list2; for (int i = 0; i < list.size(); i++) { Integer number = list.get(i); if (list2.contains(number) && list1.contains(number)) { listCommon.add(number); } } return listCommon; } } OUTPUT: Enter the Lower and Upper bounds:1 10 Enter the N value to Insert list1:10 Enter the M value to Insert list2:20 list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5] list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2] Minimum elelment in list1 :1 Minimum elelment in list2 :1 Count Divisibles by 3 and 5 elelment in list1 :0 Count Divisibles by 3 and 5 elelment in list2 :0 Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6] Solution import java.util.LinkedList; import java.util.Random; import java.util.Scanner; /** * author * */ public class LinkedListTest { /** * param args */ public static void main(String[] args) { int M, N, Lower, Upper; Scanner scanner = null; try { LinkedList list1 = new LinkedList(); L.

Education

System.out.println("Minimum elelment in list1 :" + getMin(list1));
System.out.println("Minimum elelment in list2 :" + getMin(list2));
System.out
.println("Count Divisibles by 3 and 5 elelment in list1 :"
+ countDivisibles(list1));
System.out
.println("Count Divisibles by 3 and 5 elelment in list2 :"
+ countDivisibles(list2));
System.out.println("Common Elements From list1 and list2:"
+ getCommonList(list1, list2));
} catch (Exception e) {
// TODO: handle exception
}
}
/**
* param list
* return
*/
public static int getMin(LinkedList list) {
int min = list.get(0);
for (Integer integer : list) {
if (min > integer) {
min = integer;
}
}
return min;
}
/**
* param list
* return
*/
public static int countDivisibles(LinkedList list) {
int count = 0;
for (Integer integer : list) {
if (integer % 3 == 0 && integer % 5 == 0) {
count++;

}
}
return count;
}
/**
* param list1
* param list2
* return
*/
public static LinkedList getCommonList(LinkedList list1,
LinkedList list2) {
LinkedList list = new LinkedList();
LinkedList listCommon = new LinkedList();
list = (list1.size() > list2.size()) ? list1 : list2;
for (int i = 0; i < list.size(); i++) {
Integer number = list.get(i);
if (list2.contains(number) && list1.contains(number)) {
listCommon.add(number);
}
}
return listCommon;
}
}
OUTPUT:
Enter the Lower and Upper bounds:1 10
Enter the N value to Insert list1:10
Enter the M value to Insert list2:20
list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5]
list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2]
Minimum elelment in list1 :1
Minimum elelment in list2 :1
Count Divisibles by 3 and 5 elelment in list1 :0
Count Divisibles by 3 and 5 elelment in list2 :0
Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6]
Solution

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Module-2.pptx

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1. How new modern materials prompted changes in architecture in the late nineteenth century in reference to the construction of the Crystal Palace or Eiffel Tower. After the Baroque faded slowly away, eighteenth-century architecture consisted primarily of revivals of previous periods. This time was to be the calm before the storm, for the approaching Industrial Revolution was to change everything about the world as it was then, including architecture. Previously, building materials had been restricted to a few manmade materials along with those available in nature: timber, stone, timber, lime mortar, and concrete. Metals were not available in sufficient quantity or consistent quality to be used as anything more than ornamentation. Structure was limited by the capabilities of natural materials. The Industrial Revolution changed this situation dramatically. In 1800, the worldwide tonnage of iron produced was 825,000 tons. By 1900, with the Industrial Revolution in full swing, worldwide production stood at 40 million tons, almost 50 times as much. Iron was available in three forms. The least processed form, cast iron, was brittle due to a high percentage of impurities. It still displayed impressive compressive strength, however. Wrought iron was a more refined form of iron, malleable, though with low tensile strength. Steel was the strongest, most versatile form of iron. Through a conversion process, all of the impurities were burned out of the iron ore, then precise amounts of carbon were added for hardness. Steel had tensile and compressive strength greater than any material previously available, and its capabilities would revolutionize architecture. This change did not happen over night. Prior to the introduction of bulk iron, architecture relied on compressive strength to hold buildings up. Even great structures like the Chartres Cathedral or the Parthenon were essentially orderly piles of stone. Architects were accustomed to thinking of certain ways of creating structure, and though they glimpsed some of the possibilities of the new materials, the first applications were made using the old ideas. The explosion in the development of iron and steel structures was driven initially by the advance of the railroads. Bridges were required to span gorges and rivers. In 1779, the first iron bridge was built across the Severn River in Coalescence, England. It was not an iron bridge as we might conceive of it today, but rather a traditional arch made of iron instead of stone. The compressive strength of limestone is 20 tons per square foot. The compressive strength of cast iron is 10 tons per square inch, 72 times as high, permitting significantly larger spans. Later, the truss, long used in timber roofs, became the primary element of bridge building. A triangle is the strongest structural element known, and applied force only makes it more stable. When a diagonal is added to a square, the form can be viewed as two triangles sharing a side, the fundam.

1. How new modern materials prompted changes in architecture in the .pdf

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// c++ code for a Median of Integer Stream from Text File program #include #include #include #include #include #include #include #include #include #include using namespace std; #include double streamMedian(vector v, int size) { double* s = new double[size]; for (int i = 0; i < size; ++i) { s[i] = v[i]; } for (int i = size - 1; i > 0; --i) { for (int j = 0; j < i; ++j) { if (s[j] > s[j+1]) { double t = s[j]; s[j] = s[j+1]; s[j+1] = t; } } } double median = 0.0; if ((size % 2) == 0) { median = (s[size/2] + s[(size/2) - 1])/2.0; } else { median = s[size/2]; } return median; } int main() { int number; double median; std::vector v; int count = 1; ifstream myfile (\"input.txt\"); if (myfile.is_open()) { while ( myfile >> number ) { cout << \"After reading \" << count << \" element of stream - \"; v.push_back(number); for (int i = 0; i < v.size(); ++i) { cout << v[i] << \" \"; } cout << \"-> \"; sort(v.begin(), v.end()); cout << \"median - \" << streamMedian(v,v.size()) << endl; } myfile.close(); } return 0; } /* input.txt 5 15 1 3 output: After reading 1 element of stream - 5 -> median - 5 After reading 1 element of stream - 5 15 -> median - 10 After reading 1 element of stream - 5 15 1 -> median - 5 After reading 1 element of stream - 1 5 15 3 -> median - 4 */ Solution // c++ code for a Median of Integer Stream from Text File program #include #include #include #include #include #include #include #include #include #include using namespace std; #include double streamMedian(vector v, int size) { double* s = new double[size]; for (int i = 0; i < size; ++i) { s[i] = v[i]; } for (int i = size - 1; i > 0; --i) { for (int j = 0; j < i; ++j) { if (s[j] > s[j+1]) { double t = s[j]; s[j] = s[j+1]; s[j+1] = t; } } } double median = 0.0; if ((size % 2) == 0) { median = (s[size/2] + s[(size/2) - 1])/2.0; } else { median = s[size/2]; } return median; } int main() { int number; double median; std::vector v; int count = 1; ifstream myfile (\"input.txt\"); if (myfile.is_open()) { while ( myfile >> number ) { cout << \"After reading \" << count << \" element of stream - \"; v.push_back(number); for (int i = 0; i < v.size(); ++i) { cout << v[i] << \" \"; } cout << \"-> \"; sort(v.begin(), v.end()); cout << \"median - \" << streamMedian(v,v.size()) << endl; } myfile.close(); } return 0; } /* input.txt 5 15 1 3 output: After reading 1 element of stream - 5 -> median - 5 After reading 1 element of stream - 5 15 -> median - 10 After reading 1 element of stream - 5 15 1 -> median - 5 After reading 1 element of stream - 1 5 15 3 -> median - 4 */.

c++ code for a Median of Integer Stream from Text File program#.pdf

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1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled. 2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains waters of crystallization. Solution 1) cobalt iodide has an intense color, thus this colorless salt cannot be CoCl2. It is mislabeled. 2) No. usually hydrate salts are crystalline, but not chalky looking powder, since it contains waters of crystallization..

1) cobalt iodide has an intense color, thus this colorless salt cann.pdf

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the hybridisation is sp2 since the electron cloud.pdf

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Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes, granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide is used in many industries, mostly as a strong chemical base in the manufacture of pulp and paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide production in 2004 was approximately 60 million tonnes, while demand was 51 million tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original sample. Solution Sodium hydroxide, also known as lye and caustic soda,[1][2] with the molecular formula NaOH is a caustic metallic base which is a white solid available in pellets, flakes, granules, and as a 50% saturated solution. It is soluble in water, ethanol and methanol. This alkali is deliquescent and readily absorbs moisture and carbon dioxide in air. Sodium hydroxide is used in many industries, mostly as a strong chemical base in the manufacture of pulp and paper, textiles, drinking water, soaps and detergents and as a drain cleaner. Worldwide production in 2004 was approximately 60 million tonnes, while demand was 51 million tonnes.[3] Although molten sodium hydroxide also possesses similar properties, the high temperature required limits applications. 3M HCl .... Higher percent error and possibly a bad experiment. 2) obtaining a significantly smaller amount of SiO2 than was present in the original sample..

Sodium hydroxide, also known as lye and caustic s.pdf

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Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon. Based on the saturation of hydrogens with the carbons they are classified as mainly three types . 1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2 Solution Hydrocrbon is the organic compound which containing only the atoms hydrogen and carbon. Based on the saturation of hydrogens with the carbons they are classified as mainly three types . 1) Alkanes: All the H atoms are single bonded with carbon atom with the general formula Cn H2n +2 . 2) Alkenes : The lower number of H atoms are bonded to the carbon atom (double bond compounds) has formula with Cn H2n. 3) Alkynes: These are the triple bond compounds have fewer number H atoms bonded to the carbon atoms has the molecular formula Cn H2n -2.

Hydrocrbon is the organic compound which containing only the atoms .pdf

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No. In order for a substance to conduct electrici.pdf

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HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (- 2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H + 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo. Solution HCHO OS for H is +1, O is -2 there are two H\'s one O on Carbon. 2 x (+1) + 1 x (- 2) = 0 thus, the OS of carbon is zero. CaH2 Ca is +2 thus, H is -1 HNO3 N is +5, O is -2 thus H + 5 + (-2) x 3 = 0 H = +1 P4 is an allotrope. All allotropes have an OS of zreo..

HCHO OS for H is +1, O is -2 there are two Hs o.pdf

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The molecular orbital (MO) theory is a way of looking at the structure of a molecule by using molecular orbitals that belong to the molecule as a whole Figure 1. Combination of two 1s atomic orbitals to form a sigma bonding orbital or a sigma-starred antibonding orbital. rather than to the individual atoms. When simple bonding occurs between two atoms, the pair of electrons forming the bond occupies an MO that is a mathematical combination of the wave functions of the atomic orbitals of the two atoms involved. The MO method originated in the work of Friedrich Hund and Robert S. Mulliken. When atoms combine to form a molecule, the number of orbitals in the molecule equals the number of orbitals in the combining atoms. When two very simple atoms, each with one atomic orbital, are combined, two molecular orbitals are formed. One is a bonding orbital, lower in energy than the atomic orbitals, and derived from their sum. It is called sigma. The other is an antibonding orbital, higher in energy than the atomic orbitals, and resulting from their difference. It is called sigma-starred ( s *). (See the diagram in Figure 1.) The basic idea might best be illustrated by considering diatomic molecules of hydrogen and helium. The energy diagrams are shown in Figure 2. Each hydrogen atom has one 1 s electron. In the H 2 molecule the two hydrogen electrons go into the lowest energy MO available, the sigma orbital. In the case of helium, each helium atom has two electrons, so the He 2 molecule would have four. Two would go into the lower energy bonding orbital, but the other two would have to go into the higher energy sigma-starred orbital. Figure 2. Molecular orbital energy diagrams for (a) H 2 showing both electrons in the bonding sigma MO; and (b) He 2 in which two of the electrons are in the antibonding sigma-starred MO Figure 3. Combination of p atomic orbitals to form (a) sigma MOs by end-to-end interactions or (b) pi MOs by sideways interaction. Bond Order The bond order for a molecule can be determined as follows: bond order = ½ (bonding electrons - antibonding electrons). Therefore, the H 2 molecule has a bond order of ½ (2 - 0) = 1. In other words, there is a single bond connecting the two H atoms in the H 2 molecule. In the case of He 2 , on the other hand, the bond order is ½ (2 - 2) = 0. This means that He 2 is not a stable molecule. Multiple Bonds Double or triple bonds involve two or three pairs of bonding electrons. Single bonds are always sigma bonds, but in multiple bonds the first bond is sigma, while any second or third bonds are pi bonds. The overlap of p orbitals can yield either pi or sigma MOs, as shown in Figure 3. When they overlap end to end, they form sigma orbitals, but when they overlap side to side, they form pi orbitals. Consider now the oxygen molecule. The Lewis structure for oxygen is :Ö::Ö: The double bond is necessary in order to satisfy the octet rule for both oxygen atoms. The measured bond length for oxygen supports the.

The molecular orbital (MO) theory is a way of loo.pdf

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The electrons reach or give off a certain amount of energy and different amounts of energy in the electron changes the color because it jumps to a differrent energy level Hope this helps! rate lifesaver :) Solution The electrons reach or give off a certain amount of energy and different amounts of energy in the electron changes the color because it jumps to a differrent energy level Hope this helps! rate lifesaver :).

The electrons reach or give off a certain amount of energy and diffe.pdf

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Weak London disperision and dipole- dipoleModerate Hydrogen bond.pdf

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var min =0; var max=10; var tab = {}; var name; var score; var sum=0; while(name=prompt(\"Enter name:\")){ // Get userinput for name do{ score = prompt(\"Enter score\"); // Get Score tab[name] = score; // Create object with name as key and score as value. }while(isNaN(score) || !(score>min) || !(score\"; for(item in tab){ // Create table ui = ui + (\"\"+item+\"\"+tab[item]+\"\"); sum = sum + parseInt(tab[item]); } ui = ui + \"\"; document.write(ui); // Display table document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and displaying on the page Solution var min =0; var max=10; var tab = {}; var name; var score; var sum=0; while(name=prompt(\"Enter name:\")){ // Get userinput for name do{ score = prompt(\"Enter score\"); // Get Score tab[name] = score; // Create object with name as key and score as value. }while(isNaN(score) || !(score>min) || !(score\"; for(item in tab){ // Create table ui = ui + (\"\"+item+\"\"+tab[item]+\"\"); sum = sum + parseInt(tab[item]); } ui = ui + \"\"; document.write(ui); // Display table document.write(\"Average:\"+(sum/Object.keys(tab).length).toFixed(2)); // Finding average and displaying on the page.

var min =0; var max=10; var tab = {}; var name; var score;.pdf

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Unfortunately several cancers are not predictable with simple tests rather required much intensive diagnosis. The whole genome sequence in determining the entire DNA in normal cells as well as in tumor cells provide the facility to compare and pin point the mutations occurring in the oncogenes or tumor suppressor genes responsible for causing the cancer seems to be a better approach. But such approaches are highly expensive and require expertise. Though, the whole genome sequencing is carried out pin pointing of specific driver mutations responsible for cancer phenotype would be a question. Some critical mutations can be identified in some cancer genome sequences while others are hard to find among the thousands of possibilities. The mutations like missense, nonsense frameshift, rearrangements may alter the coding sequences of any gene must be checked. The regulatory mutations that increase the transcription of oncogenes or decrease that of tumor suppressor genes are potentially important but are very difficult to find in a whole genome sequence. The catalogs of regulatory elements in human genomes provide list of certain potential sequences to be targeted. But the catalogs are rudimentary or incomplete. Therefore, the whole genome sequencing is not a panacea that will lead to immediate cures of all cancers. Recent studies have shown that the whole genome sequences of cells derived from different regions of same tumor have suggested an important reason for cancer reference. Certain cancers are heterogenous and cells within the tumor have different genomes. Certain mutations in oncogens or tumor suppressor genes are common to all cells in the tumor but some are not. These findings make a sense that accumulation of mutations in clone of cells cause cancer. Therefore, designing drug targeting specific cells protecting adult stem cells cannot be accomplished. Thus, effective cancer treatments would be directed against the common mutations but is difficult to identify the specific mutations without sequencing of genomes of many cells throughout the tumor. Cancer landscape is a large scale cancer genome sequencing project (cancer genome atlas) funded by National Institute of Health (NIH). In this project the whole genomes or exomex of several hundreds of cancer are being characterized. The recurrent patterns of mutations are subdivided cancers in to groups with probable clinical relevance. For example, 4 groups of breast cancers have been identified. Almost 178 lung squamous cell carcinoma’s genomes have been characterized and matched with DNA of normal cells from same patients. Mutations in certain tumor suppressor genes like p53 and certain oncogenes have shown relatively high frequencies among these cancers. The mutations of certain cancers resemble cancers in some other organs. For example, pattern of mutation in breast cancer resembles many ovarian cancers more than other types of breast cancers. These findings may help in designing the drugs .

Unfortunately several cancers are not predictable with simple tests .pdf

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The van t Hoff factor i (named after J. H. van t Hoff) is a meas.pdf

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Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non- polar, you must determine both factors. The polarity of the individual bonds in the molecule. The shape or geometry of the molecule. First, to determine if a given individual bond is polar, you need to know the electronegativity of the two atoms involved in that bond. To find the electronegativities of all the elements, look at the periodic table (follow the link below this answer under Web Links). If the electronegativity of the two atoms has a difference of 0.3 or less, then the bond is non-polar. If the electronegativity difference is greater that 0.3 but less than 1.7, then the bond is polar. If the two values have a difference greater than 1.7, then the bond is ionic, which is just very very polar. Once you know which bonds in the molecule are polar and which are non-polar, you must use the shape of the molecule. You need the shape because two polar bonds, if oriented correctly can cancel each other out (like two equally strong people pulling in opposite directions on a rope -- nobody moves). The three possible outcomes: If all bonds are non-polar, then the whole molecule is non-polar regardless of its shape. If there is symmetry in the molecule so that the polarity of the bonds cancels out, then the molecule is non- polar. A common example of this is carbon dioxide, or CO2. The molecule is linear, and its Lewis dot structure is like this: O=C=O (this doesn\'t include two sets of lone pairs on each oxygen). The carbon-oxygen bond is a polar bond, but because they are exactly opposed to each other, the molecule is overall non-polar. Another example of this is CCl4, where each carbon- chlorine bond is polar, but the molecule is non-polar. Here, how they cancel out isn\'t as obvious, but they do. CCl4 is a tetrahedral molecule, and the 4 C-Cl polar bonds cancel each other out. If there are polar bonds but there is no symmetry such that they cancel each other out, the overall molecule is polar. Water is a typical example of this. The two O-H bonds are oriented in a V- shape, and so the don\'t cancel out. Similarly, CH3Cl is also polar. It is the same shape as CCl4 (see above), but now it doesn\'t have the same symmetry because there is only one C-Cl bond and the bonds don\'t cancel out Solution Polar Bonds and Molecular Shape A polar molecule is a molecule that has a net dipole moment due to its having unsymmetrical polar bonds. There are two factors that go into determining if a molecule is polar or not. To determine if a molecule (or ion) is polar or non- polar, you must determine both factors. The polarity of the individual bonds in the molecule. The shape or geometry of the molecule. First, to determine if a given individual bond is polar, you need to know the electronegativ.

Polar Bonds and Molecular Shape A polar molecule.pdf

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The good functioning of an economy depends on the proper functioning of the country\'s financial market and financial institutions. Both these are involved in the financial transactions that take place in the economy. Financial transactions take place through exchange of funds. There are two ways in which the exchange of funds can take place. They are: Direct exchange and Indirect exchange. Direct exchange takes place when the buyers and the sellers meet each other directly and exchange their securities.For example: Stocks or bonds Indirect exchange takes place when the buyers and sellers do not meet each other directly. They carry out their financial transaction through a financial institution. For example: Bank The financial market and financial institution have the main function of transferring the funds from people who have excess funds to people who have shortage of funds. Some functions are common for both financial markets as well as financial institutions. They both are involved in the borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller to the buyer. They both help in providing information to the buyers and sellers about the financial assets. However, there are also some functions that are unique to each. Financial markets determine the price of the financial asset based on the demand and supply forces. Financial institutions help to lower the transaction costs and also reduce the assymetry of information about financial assets. The financial institutions also provide specialized services to th buyers and sellers. Thus, it can be said that in the financial system, the financial market and financial institutions have an important role to play. Direct exchange takes place through financial market while indirect exchnage takes place through financial institutions. Solution The good functioning of an economy depends on the proper functioning of the country\'s financial market and financial institutions. Both these are involved in the financial transactions that take place in the economy. Financial transactions take place through exchange of funds. There are two ways in which the exchange of funds can take place. They are: Direct exchange and Indirect exchange. Direct exchange takes place when the buyers and the sellers meet each other directly and exchange their securities.For example: Stocks or bonds Indirect exchange takes place when the buyers and sellers do not meet each other directly. They carry out their financial transaction through a financial institution. For example: Bank The financial market and financial institution have the main function of transferring the funds from people who have excess funds to people who have shortage of funds. Some functions are common for both financial markets as well as financial institutions. They both are involved in the borrowing and lending of funds. They help in transfer of risk of the financial asset from the seller to the buyer. They both help.

The good functioning of an economy depends on the proper functioning.pdf

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The answer is Yes, it is a reduction-oxidation reaction2 HNO2 + 2.pdf

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Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a large quadriped with unusual posture, heavy back, short forelimbs, spiked tail. Integumentary system is that which is composed of skin and its appendages. Its primary function is to protect the body from external injury, temperature regulation, cushioning and self defense. Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in selfdefense and regulation of body temperature. The plates were supplied with blood through the blood vessels and the air flow through the plates is hypothesized to have regulated the body temperature. The plates were in double row as per findings in late 1960\'s. Lymphatic system: There is little evidence to clearly understand the lymphatic system although it is hypothesized that they had a 4 chambered heart and were warmblooded with a well functioning circulatory system. They had higher aerobic capacity and higher pressure or higher gas exchange circulatory system. Solution Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmeridgian to early tithonian, is a large quadriped with unusual posture, heavy back, short forelimbs, spiked tail. Integumentary system is that which is composed of skin and its appendages. Its primary function is to protect the body from external injury, temperature regulation, cushioning and self defense. Stegosaurus skin is made up of dermal plates. They arise from the skin, the largest of them measured 2 ft wide. Their size is largest on hip and smallest on neck region. It helped in selfdefense and regulation of body temperature. The plates were supplied with blood through the blood vessels and the air flow through the plates is hypothesized to have regulated the body temperature. The plates were in double row as per findings in late 1960\'s. Lymphatic system: There is little evidence to clearly understand the lymphatic system although it is hypothesized that they had a 4 chambered heart and were warmblooded with a well functioning circulatory system. They had higher aerobic capacity and higher pressure or higher gas exchange circulatory system..

Stegosaurus dinosaur belonged to the late jurassic period i.e Kimmer.pdf

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The answer is d- the hydrogen bonds in iceThe high heat of fusion.pdf

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Solution : Option (a) will qualify as a database as dictionary is a database of different words with their meanings stored in it. Option (b) will qualify as a database as telephone directory contains data regarding people like their name, address, phone number. Option (c) will qualify as a database as rolodex is a file that contains cards holding information about people like names, addresses and phone numbers. Option (d) will not qualify as a database as textbook index only points to the topics in the book, it itself does not store any data in it related to the topic or anything..

SolutionOption (a) will qualify as a database as dictionary is a .pdf

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c++ code for a Median of Integer Stream from Text File program#.pdf

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the hybridisation is sp2 since the electron cloud.pdf

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The electrons reach or give off a certain amount of energy and diffe.pdf

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Unfortunately several cancers are not predictable with simple tests .pdf

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The answer is Yes, it is a reduction-oxidation reaction2 HNO2 + 2.pdf

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How to Manage Website in Odoo 17 Studio App.pptx

import java.util.LinkedList; import java.util.Random; import jav.pdf

1. import java.util.LinkedList; import java.util.Random; import java.util.Scanner; /** * author * */ public class LinkedListTest { /** * param args */ public static void main(String[] args) { int M, N, Lower, Upper; Scanner scanner = null; try { LinkedList list1 = new LinkedList(); LinkedList list2 = new LinkedList(); LinkedList list3 = new LinkedList(); Random random = new Random(); scanner = new Scanner(System.in); System.out.print("Enter the Lower and Upper bounds:"); Lower = scanner.nextInt(); Upper = scanner.nextInt(); System.out.print("Enter the N value to Insert list1:"); N = scanner.nextInt(); System.out.print("Enter the M value to Insert list2:"); M = scanner.nextInt(); for (int i = 1; i <= N; i++) { list1.add(random.nextInt((Upper - Lower) + 1) + Lower); } for (int i = 1; i <= M; i++) { list2.add(random.nextInt((Upper - Lower) + 1) + Lower); } System.out.println("list1: " + list1); System.out.println("list2: " + list2);

2. System.out.println("Minimum elelment in list1 :" + getMin(list1)); System.out.println("Minimum elelment in list2 :" + getMin(list2)); System.out .println("Count Divisibles by 3 and 5 elelment in list1 :" + countDivisibles(list1)); System.out .println("Count Divisibles by 3 and 5 elelment in list2 :" + countDivisibles(list2)); System.out.println("Common Elements From list1 and list2:" + getCommonList(list1, list2)); } catch (Exception e) { // TODO: handle exception } } /** * param list * return */ public static int getMin(LinkedList list) { int min = list.get(0); for (Integer integer : list) { if (min > integer) { min = integer; } } return min; } /** * param list * return */ public static int countDivisibles(LinkedList list) { int count = 0; for (Integer integer : list) { if (integer % 3 == 0 && integer % 5 == 0) { count++;

3. } } return count; } /** * param list1 * param list2 * return */ public static LinkedList getCommonList(LinkedList list1, LinkedList list2) { LinkedList list = new LinkedList(); LinkedList listCommon = new LinkedList(); list = (list1.size() > list2.size()) ? list1 : list2; for (int i = 0; i < list.size(); i++) { Integer number = list.get(i); if (list2.contains(number) && list1.contains(number)) { listCommon.add(number); } } return listCommon; } } OUTPUT: Enter the Lower and Upper bounds:1 10 Enter the N value to Insert list1:10 Enter the M value to Insert list2:20 list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5] list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2] Minimum elelment in list1 :1 Minimum elelment in list2 :1 Count Divisibles by 3 and 5 elelment in list1 :0 Count Divisibles by 3 and 5 elelment in list2 :0 Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6] Solution

4. import java.util.LinkedList; import java.util.Random; import java.util.Scanner; /** * author * */ public class LinkedListTest { /** * param args */ public static void main(String[] args) { int M, N, Lower, Upper; Scanner scanner = null; try { LinkedList list1 = new LinkedList(); LinkedList list2 = new LinkedList(); LinkedList list3 = new LinkedList(); Random random = new Random(); scanner = new Scanner(System.in); System.out.print("Enter the Lower and Upper bounds:"); Lower = scanner.nextInt(); Upper = scanner.nextInt(); System.out.print("Enter the N value to Insert list1:"); N = scanner.nextInt(); System.out.print("Enter the M value to Insert list2:"); M = scanner.nextInt(); for (int i = 1; i <= N; i++) { list1.add(random.nextInt((Upper - Lower) + 1) + Lower); } for (int i = 1; i <= M; i++) { list2.add(random.nextInt((Upper - Lower) + 1) + Lower); } System.out.println("list1: " + list1); System.out.println("list2: " + list2);

5. System.out.println("Minimum elelment in list1 :" + getMin(list1)); System.out.println("Minimum elelment in list2 :" + getMin(list2)); System.out .println("Count Divisibles by 3 and 5 elelment in list1 :" + countDivisibles(list1)); System.out .println("Count Divisibles by 3 and 5 elelment in list2 :" + countDivisibles(list2)); System.out.println("Common Elements From list1 and list2:" + getCommonList(list1, list2)); } catch (Exception e) { // TODO: handle exception } } /** * param list * return */ public static int getMin(LinkedList list) { int min = list.get(0); for (Integer integer : list) { if (min > integer) { min = integer; } } return min; } /** * param list * return */ public static int countDivisibles(LinkedList list) { int count = 0; for (Integer integer : list) { if (integer % 3 == 0 && integer % 5 == 0) { count++;

6. } } return count; } /** * param list1 * param list2 * return */ public static LinkedList getCommonList(LinkedList list1, LinkedList list2) { LinkedList list = new LinkedList(); LinkedList listCommon = new LinkedList(); list = (list1.size() > list2.size()) ? list1 : list2; for (int i = 0; i < list.size(); i++) { Integer number = list.get(i); if (list2.contains(number) && list1.contains(number)) { listCommon.add(number); } } return listCommon; } } OUTPUT: Enter the Lower and Upper bounds:1 10 Enter the N value to Insert list1:10 Enter the M value to Insert list2:20 list1: [1, 4, 4, 6, 9, 4, 9, 5, 8, 5] list2: [5, 10, 10, 2, 1, 10, 8, 8, 8, 2, 6, 10, 6, 2, 5, 9, 6, 9, 6, 2] Minimum elelment in list1 :1 Minimum elelment in list2 :1 Count Divisibles by 3 and 5 elelment in list1 :0 Count Divisibles by 3 and 5 elelment in list2 :0 Common Elements From list1 and list2:[5, 1, 8, 8, 8, 6, 6, 5, 9, 6, 9, 6]

import java.util.LinkedList; import java.util.Random; import jav.pdf

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