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//Node for list storage. class Ndd { int data; Ndd next; Ndd(int d) { data = d; next = null; } } //linked list operations. class lnklist { Ndd hads; //method to swap nodes.. public void swpNds(int xvv, int yvv) { //check two node values,if equal no need of shifting. if (xvv == yvv) return; Ndd prevX = null, currX = hads; while (currX != null && currX.data != xvv) { prevX = currX; currX = currX.next; } Ndd prevY = null, currY = hads; while (currY != null && currY.data != yvv) { prevY = currY; currY = currY.next; } if (currX == null || currY == null) return; if (prevX != null) prevX.next = currY; else hads = currY; if (prevY != null) prevY.next = currX; else hads = currX; Ndd temp = currX.next; currX.next = currY.next; currY.next = temp; } public void push(int nwd-ata) { Ndd nw-nd = new Ndd(nwd-ata); nw-nd.next = hads; hads = nw-nd; } public void prntList() { Ndd tNode = hads; while (tNode != null) { System.out.print(tNode.data+\" \"); tNode = tNode.next; } } public static void main(String[] ssds) { lnklist llist = new lnklist(); llist.push(7); llist.push(6); llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); System.out.print(\"\ Linked list before calling swpNds() \"); llist.prntList(); llist.swpNds(4, 3); System.out.print(\"\ Linked list after calling swpNds() \"); llist.prntList(); } } Solution //Node for list storage. class Ndd { int data; Ndd next; Ndd(int d) { data = d; next = null; } } //linked list operations. class lnklist { Ndd hads; //method to swap nodes.. public void swpNds(int xvv, int yvv) { //check two node values,if equal no need of shifting. if (xvv == yvv) return; Ndd prevX = null, currX = hads; while (currX != null && currX.data != xvv) { prevX = currX; currX = currX.next; } Ndd prevY = null, currY = hads; while (currY != null && currY.data != yvv) { prevY = currY; currY = currY.next; } if (currX == null || currY == null) return; if (prevX != null) prevX.next = currY; else hads = currY; if (prevY != null) prevY.next = currX; else hads = currX; Ndd temp = currX.next; currX.next = currY.next; currY.next = temp; } public void push(int nwd-ata) { Ndd nw-nd = new Ndd(nwd-ata); nw-nd.next = hads; hads = nw-nd; } public void prntList() { Ndd tNode = hads; while (tNode != null) { System.out.print(tNode.data+\" \"); tNode = tNode.next; } } public static void main(String[] ssds) { lnklist llist = new lnklist(); llist.push(7); llist.push(6); llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); System.out.print(\"\ Linked list before calling swpNds() \"); llist.prntList(); llist.swpNds(4, 3); System.out.print(\"\ Linked list after calling swpNds() \"); llist.prntList(); } }.

Node for list storage. class Ndd { int data; Ndd next; N.pdf

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Laser Scanning Laser scanning is an emerging data acquisition technology that has remarkably broadened its application field and has been a serious competitor to other surveying techniques. Due to rapid technological development, the increased accuracy of global positioning systems and improving demands to even more accurate digital surface models, airborne laser scanning showed significant development in the 1990s. Somewhat later terrestrial laser scanning became a reasonable alternative method in many kinds of applications that previously by ground based surveying or close-range photogrammetry. 1 Airborne laser scanning Airborne laser scanning is an active remote sensing technology that is able to rapidly collect data from huge areas. The resulted dataset can be the base of digital surface and elevation models. Airborne laser scanning is often coupled with airborne imagery, therefore the point clouds and images can be fused resulting enhanced quality 3D product. The basic principle is as follows: the sensor emits a laser pulse through the terrain in a predefined direction and receives the reflected laser beam. Knowing the speed of light, the distance of the object can be calculated, see Figure 1. Figure 1.: Time of flight laser range measurement [2] Airborne LiDAR systems are composed by the following subsystems: The components are shown in Figure 2 Figure 2.: Principle of airborne LiDAR [2] 2. Sensors, equipment Sensors can be distinguished based on the scanning method, i.e. how the laser beam is directed through the surface. The four most widely used sensor types are shown in Figure 4.2.3. Figure .3: Scanning mechanisms [1] As it is clearly seen in Figure 3, different kinds of mechanisms are applied by the different types of sensors; each has its advantages and shortcomings, e.g. number of moving parts, type of rotation etc. that lead to different kinds of error sources. The capabilities (repetition rate, scan frequency, scan angle, point density) of the above scanners are very similar; the main difference lies in the scanning pattern, as seen in Figure 4. The most widely used oscillating mirror scanners produce the zigzag pattern. Spacing along the line depends on the pulse rate and scanning frequency, while spacing along the flight direction depends on the flying speed. To avoid too wide spacing of points along flight direction, LiDAR flights are usually slower (e.g. at 60-80 m/sec) compared to that of photogrammetric flights (even 120-160 m/sec). Careful planning of the measurement results in rather homogenous density, however, due to technical and microelectronic reasons (regarding the operating mechanism of the mirror, especially in case of oscillating mirrors), higher point density can be observed at the edges of the scan swath. Previously, critics were addressed to the fixed optic scanners, i.e. the parallel scan lines along the flight direction can miss sizeable objects, but vendors successfully responded and modified the mechanis.

Laser ScanningLaser scanning is an emerging data acquisition techn.pdf

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InternetService.java import java.text.DecimalFormat; import java.util.Scanner; public class InternetService { public static void main(String[] args) { DecimalFormat df=new DecimalFormat(\"#.##\"); //initialize variable char packageType; int time; double totalCharges; double packA; double packB; double packC; // create an object for Scanner class Scanner input = new Scanner(System.in); // prompt the user to enter the package purchased System.out.print(\"\ Enter the package purchased: \"); packageType = input.next().charAt(0); //prompt the user to enter the number of minutes used System.out.print(\"Enter the number of minutes used: \"); time = input.nextInt(); //Calculate the bill amount of the particular month using the type of package and the number of minutes. //Then print results to the screen switch(packageType) { case \'A\': //Calling the Method for Package A charges totalCharges=calPackA(time); //Displaying the package A charges. System.out.println(\"For Package A:\ The total charges for this month is $\" + totalCharges); //Calling the Method for Package B charges packB=calPackB(time); //Calling the Method for Package C charges packC=calPackC(time); //If the package B is Less Than package A then display the difference amount if(totalCharges>packB) System.out.println(\"Amount of Money Package A customer would save, if he Purchased Package B is:$\"+df.format(totalCharges-packB)); //If the package C is Less Than package A then display the difference amount if(totalCharges>packC) System.out.println(\"Amount of Money Package A customer would save, if he Purchased Package C is:$\"+df.format(totalCharges-packC)); break; case \'B\': //Calling the Method for Package B charges totalCharges=calPackB(time); //Displaying the package B charges. System.out.println(\"For Package B:\ The total charges for this month is $\" +df.format(totalCharges)); //Calling the Method for Package C charges packC=calPackC(time); //If the package B is Less Than package A then display the difference amount if(totalCharges>packC) System.out.println(\"Amount of Money Package B customer would save, if he Purchased Package C is:$\"+df.format(totalCharges-packC)); break; case \'C\': //Calling the Method for Package C charges totalCharges=calPackC(time); //Displaying the package C charges. System.out.print(\"For Package C:\ The total charges for this month is $\" + totalCharges); break; default: System.out.println(\"\ The package should be A or B or C.\"); } } /*This method will calculate the total Charges for Package C * Params:Time * Return:total charges of type double */ private static double calPackC(int time) { double totalCharges = 69.99; return totalCharges; } /*This method will calculate the total Charges for Package B * Params:Time * Return:total charges of type double */ private static double calPackB(int time) { double totalCharges=0.0; if(time <= 900) totalCharges = 59.99; else totalCharges = 59.99 + (0.40 * (time - 900)); return totalCharges; } /*This method will c.

InternetService.java import java.text.DecimalFormat; import jav.pdf

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Importance of planning to the practice of managementSolutionIm.pdf

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Hardware refers to all of the physical parts of a computer system. For a typical desktop computer this comprises the main system unit, a display screen, a keyboard, a mouse, a router/modem (for connection to the Internet), and usually a printer. Speakers, a webcam and an external hard disk for back-up storage are often also included. Many of these items are integrated into a single unit on a laptop or other form of mobile computer This option isn’t as easy or cheap, but we have to mention it anyway. If you can open up your laptop, you can replace its internal drive with a larger drive — or insert a second internal drive, in the off chance that your laptop has a second drive bay. Upgrading your laptop is often possible, but it’s definitely more work than quickly plugging in an external storage device! RAM RAM -- or \"random access memory\" -- is the temporary storage space that a computer loads software applications and user data into when it is running. All current RAM technologies are \"volatile\", which means that everything held in RAM is lost when a computer\'s power is removed. To a large extent, the more RAM a computer has the faster and more effectively it will operate. Computers with little RAM have to keep moving data to and from their hard disks in order to keep running. This tends to make them not just slow in general, but more annoyingly intermittently sluggish. The above all said, those hoping to speed up thier PC by installing more RAM need to note that any PC with a 32 bit operating system can only access a maximum of 4GB of RAM. Add more, and the PC simply will not recognise it. In practice this that means the vast majority of PCs in use and being sold today cannot benefit from more than 4GB of RAM -- and this includes many PCs running Windows 7 (which is very widely sold in its 32 rather than 64 bit format to maximise compatabilty with older software and perhipherals).] RAM is measured in megabytes (MB) and gigabytes (GB), as detailed on the storage page. Just how much RAM a computer needs depends on the software it is required to run effectively. A computer running Windows XP will usually function quite happily with 1GB of RAM, whereas twice this amount (ie 2GB) is the realistic minimum for computers running Windows 7. HARD DRIVE Hard disk drives are the high capacity storage devices inside a computer from which software and user data are loaded. Like most other modern storage devices, the capacity of the one or more internal hard disks inside a computer is measured in gigabytes (GB), as detailed on the storage page. Today 40GB is an absolute minimum hard drive size for a new computer running Windows 7, with a far larger capacity being recommended in any situation where more than office software is going to be installed. Where a computer will frequently be used to edit video, a second internal hard disk dedicated only to video storage is highly recommended for stable operation. Indeed, for professional video editing using a .

Hardware refers to all of the physical parts of a computer system. F.pdf

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D is correctSolutionD is correct.pdf

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Cost centers: Cost centers are Centers that incur costs for operational activities of the organization. Cost center does not produce direct profit for the organization, however its activities are related with production of profits. Example: HR departments costs, Research and development department costs. Profit Center: A profit center is units of the organization that is responsible for revenue and costs. These centers are specifically allocated for performance evaluation. Example: Administration department, after sales service department. Investment centers: Investment centers are those units, for which capital has been invested by the organization. And the organization looks for the return on that investment. In a decentralized management organizations managers are given the responsibility of a particular center and the performance evaluation is done for each center. The bonus/ inventory payment is made on the basis of the evaluation. Solution Cost centers: Cost centers are Centers that incur costs for operational activities of the organization. Cost center does not produce direct profit for the organization, however its activities are related with production of profits. Example: HR departments costs, Research and development department costs. Profit Center: A profit center is units of the organization that is responsible for revenue and costs. These centers are specifically allocated for performance evaluation. Example: Administration department, after sales service department. Investment centers: Investment centers are those units, for which capital has been invested by the organization. And the organization looks for the return on that investment. In a decentralized management organizations managers are given the responsibility of a particular center and the performance evaluation is done for each center. The bonus/ inventory payment is made on the basis of the evaluation..

Cost centersCost centers are Centers that incur costs for operati.pdf

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Calcium Phosphate is a slightly soluble salt. Here isthe equation for dissolution into ions in water. Ca3(PO4)2(s) <-------> 3Ca+2(aq) + 2PO43-(aq) The expression for Ksp is shown below. It is the productof the molar concentration of the ions raised to the powers of thecoefficients of the ions in the balanced equation. Ksp =[Ca+2]3[PO43-]2 Solution Calcium Phosphate is a slightly soluble salt. Here isthe equation for dissolution into ions in water. Ca3(PO4)2(s) <-------> 3Ca+2(aq) + 2PO43-(aq) The expression for Ksp is shown below. It is the productof the molar concentration of the ions raised to the powers of thecoefficients of the ions in the balanced equation. Ksp =[Ca+2]3[PO43-]2.

Calcium Phosphate is a slightly soluble salt. Here isthe equation fo.pdf

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c.250Solutionc.250.pdf

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#! /usr/bin/python def Flatten(list): newList = [] for i in range(len(list)): if(type(list[i]) == type([])): for j in range(len(list[i])): newList.append(list[i][j]) else: newList.append(list[i]) return newList Solution #! /usr/bin/python def Flatten(list): newList = [] for i in range(len(list)): if(type(list[i]) == type([])): for j in range(len(list[i])): newList.append(list[i][j]) else: newList.append(list[i]) return newList.

#! usrbinpython def Flatten(list) newList = [] for i in ra.pdf

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a) Two domains are OD1 and OD2 domains. The common function of OD1 and OD2 is to oligomerize the full-length proteins, whereas their specialized functions are to form a dimeric DNA binding unit and a sex-specific transcriptional activation or repression unit. b) Male and female OD2 domains contain common and sex-specific regions that are necessary for oligomerization. OD1 is within a 66-amino-acid segment that also contains a DNA binding domain. This segment is identical in the male and female proteins and is located near the N terminus. The second oligomerization domain, OD2, is hundreds of amino acids downstream of OD1 and includes both non-sex-specific and sex-specific sequences. This domain is a continuous sequence of approximately 65 amino acids in length, 48 amino acids of N-proximal non-sex- specific sequence followed by a C-proximal sex-specific sequence of either 15 amino acids in OD2F or less than 30 amino acids in OD2M. Both the sex-specific and non-sexspecific regions are necessary for interactions between the OD2 domains. Therefore, each full-length protein has two domains that can independently cause oligomerization. One domain colocalizes with a DNA binding function and is the same in both proteins, and the other is well separated from the first and is sex specific. c) The OD1 domain. A segment of DSX protein has previously been identified as a region which, when joined to either substantial upstream or downstream DSX regions, can bind specifically to a DNA regulatory site (10). Oligomerization energy provided by OD1 is important for tight and specific binding to DNA (6). The OD2 domains. Coiled-coil interactions are likely to be the basis for OD2-OD2 oligomerization. Secondary structure predictions from the primary sequence indicate that the OD2 domains are mostly alpha helical. Three helices of 11, 6, and either 22 (for OD2F ) or 25 (for OD2M) amino acids are separated by short gaps of 4 and 9 amino acids (24). If the three helices and the gaps of OD2F are represented together as a single helix in an alpha-helical axial projection, a remarkably amphipathic distribution is observed, consistent with an extended coiled-coil interaction (Fig. 6C). The male domain has an essentially identical amphipathic distribution, although the male structure allows seven rather than six successive pairs of hydrophobic interactions. Protein the specialized function of OD1 oligomerization is to form the dimeric DNA binding unit. This OD1-OD1 interaction may be weak. The specialized function of OD2 at the other end of the protein is likely to be related to the sex specificity of DSX interaction with the transcriptional machinery or to DNA binding cooperativity when the proteins bind naturally occurring multiple regulatory sites in DNA. For example, it may form a dimeric sex-specific regulatory structure. This regulatory structure in DSXM would repress transcription of the Yp1 gene, and in DSXF it would activate. Solution a) Two domains are OD1 an.

a) Two domains are OD1 and OD2 domains. The common function of OD1 a.pdf

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200 students attended session I.150 students attended session II..pdf

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Non-polar - Waterproof Covalently bonded - strong.pdf

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Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment through either absorption or adsorption. salts are hygroscopic and have a strong affinity for moisture and will absorb relatively large amounts of water from the atmosphere if exposed to it. Solution Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment through either absorption or adsorption. salts are hygroscopic and have a strong affinity for moisture and will absorb relatively large amounts of water from the atmosphere if exposed to it..

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H2SO4 + Cd(OH)2 = 2H2O + CdSO4 In this reaction, .pdf

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c. SO2. S shares 8 electrons total with the two O\'s. S contributes 4 electrons total to the sharing; 2 to each of the two O\'s. So S has one lone pair of electrons left ( remember S is in group VI i.e. should have 6 electrons in its outermost shell). This lone pair of electron makes the molecule to be bent. Solution c. SO2. S shares 8 electrons total with the two O\'s. S contributes 4 electrons total to the sharing; 2 to each of the two O\'s. So S has one lone pair of electrons left ( remember S is in group VI i.e. should have 6 electrons in its outermost shell). This lone pair of electron makes the molecule to be bent..

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a. Assuming complete dissociation, we can have [O.pdf

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