Total probability of dot received P = (37)(78).pdf

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Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96 Solution Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96.

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salicylic acid alcohol (R-OH) and carboxylic aci.pdf

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It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that Ruthenium has the opposite charge, +2. Solution It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that Ruthenium has the opposite charge, +2..

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This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so number of ions is averaged considering dissociation constants which is 3.4 Solution This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so number of ions is averaged considering dissociation constants which is 3.4.

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Environmental contaminants of recent concern are pharmaceuticals, estrogens and other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition, pesticides (especially their transformation products), microorganisms, and humic substances (HS), in their function as vehicles for contaminants and as precursors for by-products in water treatment, traditionally play an important role. The present status of the application of LC-MS techniques for these water constituents are discussed and examples of application are given. Solid-phase extraction with various non-selective materials in combination with liquid chromatography (LC) on reversed-phase columns have been the most widely used methods for sample preconcentration and separation for different compound classes like pesticides, pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization (APCI) are the most frequently used ionization techniques for polar and ionic compounds, as well as for less polar non-ionic ones. The facilities of LC-MS have been successfully demonstrated for different compound classes. Polar compounds from pharmaceuticals used as betablockers, iodinated X-ray contrast media, or estrogens have been determined without derivatization down to ultratrace concentrations. LC-MS can be viewed as a prerequisite for the determination of algal and cyanobacterial toxins and the homologues and oligomers of alkylphenol ethoxylates and their metabolites. Tandem mass spectrometric techniques and the use of diagnostic ions reveal their usefulness for compound-class specific screening and unknown identification, and are also valid for the analysis of pesticides and especially for their transformation products. Solution Environmental contaminants of recent concern are pharmaceuticals, estrogens and other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition, pesticides (especially their transformation products), microorganisms, and humic substances (HS), in their function as vehicles for contaminants and as precursors for by-products in water treatment, traditionally play an important role. The present status of the application of LC-MS techniques for these water constituents are discussed and examples of application are given. Solid-phase extraction with various non-selective materials in combination with liquid chromatography (LC) on reversed-phase columns have been the most widely used methods for sample preconcentration and separation for different compound classes like pesticides, pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization (APCI) are the most frequently used ionization techniques for polar and ionic compounds, as well as for less polar non-ionic ones. The facilities of .

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To insert an element at given position in array: Insert(stack s1, stack s2, int value, int pos) //Inserts a value into the array at specified position. //Assume s1 is a stack which holds the values, and s2 is an empty stack. while(!s1.empty()) //Till the stack is empty. s2.push(s1.pop()) //Pop element from s1, and push it in s2. //By now, all elements are placed in reverse order from s1 to s2. for i = 1 to pos - 1 //Till the pos-1 s1.push(s2.pop()) //Remove elements from s2, and push them to s1. s1.push(value) //Now push your element in specified position pos. while(!s2.empty()) //Push all the remaining elements in s2 to s1. s1.push(s2.pop()) To delete an element at given position in array: Delete(stack s1, stack s2, int pos) //Deletes a value into the array at specified position. //Assume s1 is a stack which holds the values, and s2 is an empty stack. while(!s1.empty()) //Till the stack is empty. s2.push(s1.pop()) //Pop element from s1, and push it in s2. //By now, all elements are placed in reverse order from s1 to s2. for i = 1 to pos - 1 //Till the pos-1 s1.push(s2.pop()) //Remove elements from s2, and push them to s1. s1.pop(value) //Now pop your element in specified position pos. while(!s2.empty()) //Push all the remaining elements in s2 to s1. s1.push(s2.pop()) To delete a given element in array: DeleteVal(stack s1, stack s2, int value) //Deletes a value from the array. //Assume s1 is a stack which holds the values, and s2 is an empty stack. while(!s1.empty()) //Till the stack is empty. s2.push(s1.pop()) //Pop element from s1, and push it in s2. //By now, all elements are placed in reverse order from s1 to s2. ele = s2.peek(); while(ele != value || !s2.empty()) //Till you found the element to be deleted. s1.push(s2.pop()) //Remove elements from s2, and push them to s1. ele = s2.peek() s2.pop(value) //If element is found, delete it. while(!s2.empty()) //Push all the remaining elements in s2 to s1. s1.push(s2.pop()) If you have any queries with this, just get back to me. Solution To insert an element at given position in array: Insert(stack s1, stack s2, int value, int pos) //Inserts a value into the array at specified position. //Assume s1 is a stack which holds the values, and s2 is an empty stack. while(!s1.empty()) //Till the stack is empty. s2.push(s1.pop()) //Pop element from s1, and push it in s2. //By now, all elements are placed in reverse order from s1 to s2. for i = 1 to pos - 1 //Till the pos-1 s1.push(s2.pop()) //Remove elements from s2, and push them to s1. s1.push(value) //Now push your element in specified position pos. while(!s2.empty()) //Push all the remaining elements in s2 to s1. s1.push(s2.pop()) To delete an element at given position in array: Delete(stack s1, stack s2, int pos) //Deletes a value into the array at specified position. //Assume s1 is a stack which holds the values, and s2 is an empty stack. while(!s1.empty()) //Till the stack is empty. s2.push(s1.pop()) //Pop element from s1, and push it in s2. //By .

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Their physical properties are so dofferent just because theintermole.pdf

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Chemical weathering is the breakdown of rock by chemical reaction. In this process the minerals within the rock are changed into particles that can be easily carried away. Air and water are both involved in many complex chemical reactions. The minerals in igneous rocks may be unstable under normal atmospheric conditions, those formed at higher temperatures being more readily attacked than those formed at lower temperatures. Igneous rocks are commonly attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock forming minerals (with the exception of quartz, which is very resistant) are changed in this way into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind) to new locations and redeposited in layers, generally at a lower elevation. These agents reduce the size of the particles, sort them by size, and then deposit them in new locations. The sediments dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes and the sea floor. The wind may move large amounts of sand and other smaller particles. Glaciers transport and deposit great quantities of usually unsorted rock material as till. These deposited particles eventually become compacted and cemented together, forming clastic sedimentary rocks. Such rocks contain inert minerals that resist mechanical and chemical breakdown, such as quartz. Quartz is one of the most mechanically and chemically resistant minerals. Highly weathered sediments can contain several heavy and stable minerals, best illustrated by the ZTR index. Solution Chemical weathering is the breakdown of rock by chemical reaction. In this process the minerals within the rock are changed into particles that can be easily carried away. Air and water are both involved in many complex chemical reactions. The minerals in igneous rocks may be unstable under normal atmospheric conditions, those formed at higher temperatures being more readily attacked than those formed at lower temperatures. Igneous rocks are commonly attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock forming minerals (with the exception of quartz, which is very resistant) are changed in this way into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind) to new locations and redeposited in layers, generally at a lower elevation. These agents reduce the size of the particles, sort them by size, and then deposit them in new locations. The sediments dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes and the sea floor. The wind may move large amounts of sand and other smaller particles. Glaciers transport and deposit great quantities of.

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bu puting your values in formula You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are \"shared\" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding. Now write out the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li-- ->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O (2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the original balanced equation. Solution bu puting your values in formula You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are \"shared\" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding. Now write out the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li-- ->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O (2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the original balanced equation..

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The mole is a unit of measurement for the amount of substance or chemical amount. It is one of the base units in the International System of Units, and has the unit symbol mol. n ionic compound is a chemical compound in which ions are held together in a lattice structure by ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and boiling point, and they are hard and very brittle. Solution The mole is a unit of measurement for the amount of substance or chemical amount. It is one of the base units in the International System of Units, and has the unit symbol mol. n ionic compound is a chemical compound in which ions are held together in a lattice structure by ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and boiling point, and they are hard and very brittle..

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Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use 2 types of classification system: According to the Linnaen system, organisms are grouped on the basis of their characteristics without considering their ancestry. Based on this a reptile is an animal that is ectothermic and consists of scales, and thus birds cannot be reptiles. While in Phylogenetics,organisms are grouped only on the basis of their ancestry, and characteristics are only used to discover this ancestry. So based on this a reptile is any animal descending from reptiles, and birds (including mammals) would be considered as reptiles. Looking back at history, the 1st group of reptiles evolved around 300 million years ago. Around 40 million years later, a group of reptiles called therapsids branched off, that became modern mammals. Another group of reptiles split off in the next 120 million years, and one branch called the dinosaurs was very successful. Dinosaurs was only distantly related to modern snakes, lizards, and turtles, groups that had split off at different times. But 65 million years ago there was a massive extinction event, and all dinosaurs were killed except for a single group of feathered dinosaurs. These evolved over the next 65 million years into modern birds. So birds aren\'t just closely related to dinosaurs, they really are dinosaurs! This is what most people mean when they say that birds are reptiles, although technically according to the phylogenetic system mammals are also reptiles. Birds and reptiles are both vertebrates and are alike in ways: 1. They both lay eggs which have a protective shell. This is an adaptation to prevent drying-out, necessary in evolution before vertebrates could lay their eggs on land. 2. Both birds and reptiles have lungs and thus breathe air. 3. Both birds and reptiles have scales on their skin. In birds, most of the body is now covered with feathers, which are modified scales. You can still see the scales on birds\' legs. 4. It is thought that birds evolved from reptiles. These probably were able to fly as well. Birds and reptiles differ in ways such as: While most reptiles are ectothermic (\"cold blooded\"), birds are endothermic (\"warm blooded). Thus birds have mechanisms like mammals that enable them to maintain a body temperature that is independent of their surroundings. Their feathers and their high metabolic rate help keep them warm in cold environments, for example. While reptiles have a 3-chambered heart, birds have a 4-chambered heart. This allows them to be more efficient, as the deoxygenated blood from the tissues is completely separated from the oxygenated blood from the lungs.While reptiles have teeth, birds do not. They grind their food in an organ called the gizzard. Solution Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use 2 types of classification system: According to the Linnaen system, organisms are grouped on th.

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Introduction Termination of a project is inevitable, but how it is terminated and when may have a profound and long lasting impact on the organization and its employees. The success of future projects may depend on not only the success of past ones, but also on how unsuccessful projects were treated by the organization and its stakeholders. Firms have the option of initiating a variety of entrepreneurial projects with varying degrees of risk. If an organization chooses to accept greater risks, it should avoid penalizing members of projects that turn out to be unsuccessful. If team members believe they will be penalized for participating in unsuccessful projects, they will be less willing to terminate failed projects and may become risk adverse. This research contains information about the project life cycle and the importance of continually monitoring a project to determine if it is meeting the objectives established at the outset. We have identified and categorized external and internal factors that influence the success or failure of projects. The relative importance of each factor varies by organization and project type. Organizing a project\'s termination process is especially important when it has failed, because of the lasting impact on future projects as well as the organization\'s image. Including project team members in the termination process will increase their loyalty and commitment, not only to the organization but also to the success of future projects. At the end of a project a post-audit report will be prepared that summarizes the project and provides recommendations for similar projects in the future. Lastly, as a project is closed down or completed it is important that senior management recognize the contributions of the project team. The Project Life Cycle There are several stages in the life cycle of a project: (1) project selection, (2) planning, (3) execution, and (4) termination (Ruhl, 1988). The first phase, project selection, will vary among firms. Each project must be evaluated to determine which is the best use of corporate funds. Each will have different risks, benefits, and costs, making the selection very difficult. The final decision should be based on the project\'s financial return and how well it assists the organization in achieving its long-run strategic objectives. Once a selection has been made, formal plans must be developed. The importance of thorough project planning cannot be overemphasized. The objective of this process should be to develop a master plan that details how each asset of the organization will be used to accomplish the project\'s goals. Thorough and aggressive planning will also increase the team\'s commitment to success. The two most important components of the master plan are the project budget and the master schedule, which are developed from a detailed list of specific project tasks. The master plan should include measures for evaluating the progress of the project as well as guideli.

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In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root. for( int Top = Last/2; Top >= 0; Top-- ) { adjust(Top, Last); } // phase 2: use heap to sort // Move top element (largest) out of heap, swapping with last element // and changing the heap boundary, until only one element remains. while( Last > 0 ) { swap(0, Last); adjust(0, --Last); } } /** * adjust(Top, Last) adjusts the tree between Top and Last **/ void adjust(int Top, int Last) { int TopVal = array[Top]; // Set aside top of heap int Parent, Child; for( Parent = Top; ; Parent = Child ) // Iterate down through tree { Child = 2*Parent+1; // Child means left child if( Child > Last ) break; // Left child non-existent if( Child+1 <= Last // Right child exists && array[Child] < array[Child+1] ) // and right child is larger Child++; // Child is the larger child if( TopVal >= array[Child] ) break; // Location for TopVal found array[Parent] = array[Child]; // Move larger child up in tree } array[Parent] = TopVal; // Install TopVal in place } //swapping void swap(int i, int j) { int temp = array[i]; array[i] = array[j]; array[j] = temp; } public static void main(String[] args) { int N = 0; // number of elements in array int array[] = {43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998}; //initialising array N=array.length; System.err.println(\"Sorting started\"); long startTime = System.nanoTime(); new heapsort(array, N); // sorting long endTime = System.nanoTime(); long totalTime = endTime - startTime; for( int i = 0; i < N; i++ ) { System.out.print(array[i] + \" \"); } System.out.println(); System.out.println(\"Time in nanos\"+totalTime); } } //runn the above program for inline heap sort. Solution In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root. for( int T.

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