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    zareenZ
    @Aima-Khan idea posted
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    https://www.youtube.com/watch?v=BrgNIf7LCgk
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    Q. 1 Solution Role of teacher in Discovery Based Learning: In discovery based learning teacher gives the problem to the students for their solutions. The teacher himself acts as a guide for providing the support to the students’ needs. The teacher do not provide the correct answer but he/she provide the opportunities for discovery and learning Teacher is the facilitator of learning in discovery base settings. The teacher spends most of the time in observing the students for assessment purpose. Q. 2 Solution: For example while teaching the concept of Independence Day. Teacher can play a short documentary video for the students. After playing the video teacher can show the pictures of our national heroes to the students. In this way the students will be able to remember most of the content for a long time. Also it will be more interesting. Q. 3 Solution: The students can attempt this question according to their selected topic and grade level. The above mentioned answers give a glimpse or idea that how can you attempt these questions. You can add the content according to your choice. Only the copy pasted content is not accepted.
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    This Assignment is related to Development phase and in this assignment you will focus only on the tasks given below: Consider the following ER Diagram and perform the given tasks; [image: U1DZCPN.png] Tasks: Solution Creating Database [image: uCnDRn1.png] [image: 9fUEMo3.png] Creating Tables 1. Client Table [image: vSyBdVc.png] [image: 8NS9wzr.png] 2. Event Table [image: RPCV0ml.png] [image: zhmOrVV.png] 3. Venue Table [image: RlTfCj7.png] [image: bKzN3l8.png] [image: VYwSfTf.png]
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    100% Solved: Q.1 Solution SELECT NAME FROM SALESRESP WHERE QUOTA >= (SELECT TARGET FROM OFFICES WHERE CITY = 'LAHORE') Q.2 Solution DECLARE eno salesresp.empl_num%type; commission salesresp.commision%type; update_commision number(10):=0; BEGIN Select empl_num, commision from SALESRESP where months_between(sysdate, joining_date) < 24 AND sal > 50000; update_commision:=commission-commision*0.01; update salesresp set commission = update_commision; dbms_output.put_line('Updated Commision ' || commision); END;
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    100% Solved: #include <iostream> #include <string.h> using namespace std; const int arrLength = 10; struct Student{ string userVUID; string userDetails; }std1={"NULL","NULL"}; class ArrQueue{ private: //Data Members Student arr[arrLength]; int front; int rear; public: //Constructor ArrQueue(){ for(int i=0; i<arrLength; i++) arr[i] = std1; front = -1; rear = -1; } //Member Functions void enQue(Student std){ if(isEmpty()){ arr[0] = std; front++; rear++; } else if(isFull()){ cout << " Queue is full."; } else{ arr[rear+1] = std; rear++; } } void deQue(){ if(isEmpty()){ cout<< " No Student In the Queue.\n"; } else if(rear == 0){ arr[front] = std1; front = -1; rear = -1; } else{ int tempfront = front; arr[front] = std1; for(int i=1; i<=rear; i++ ){ arr[front] = arr[i]; front++; } rear--; front = tempfront; } } int queLength(){ return rear+1; } bool isEmpty(){ if(front==-1 || rear==-1) return true; else return false; } bool isFull(){ if(rear == arrLength-1) return true; else return false; } void showQue(){ cout << "\n |Sr. VU ID Details |"; cout << "\n --- -------------------------------\n"; for(int i=front; i<=rear; i++){ cout<<" "<< i+1<< ". "<< arr[i].userVUID << " " << arr[i].userDetails <<"\n"; } } }; int main(){ /*Code For Even Id's Student std[] = {{"BC12345684","Bilal (BSCS)"}, {"BC12345685","Bilal (BSCS)"},{"BC12345686","Bilal (BSCS)"},{"BC12345687","Bilal (BSCS)"}, {"BC12345688","Bilal (BSCS)"}};*/ /*Code For Odd Id's */ Student std[] = {{"BC12345683","Bilal (BSCS)"}, {"BC12345684","Bilal (BSCS)"},{"BC12345685","Bilal (BSCS)"},{"BC12345686","Bilal (BSCS)"}, {"BC12345687","Bilal (BSCS)"}}; ArrQueue arrQue; cout << "\n -----------------------------------"; cout << "\n | Queue (After Adding Students) |"; cout << "\n -----------------------------------"; for(int i=0; i<=4; i++){ arrQue.enQue(std[i]); } arrQue.showQue(); cout << "\n -----------------------------------"; cout << "\n | Queue (After Removing Students) |"; cout << "\n -----------------------------------"; /* Code For Even Id's for(int i=0; i<=1; i++){ arrQue.deQue(); }*/ /*Code For Odd Id's*/ for(int i=0; i<1; i++){ arrQue.deQue(); } arrQue.showQue(); }
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    @Muhammad-Arif https://cyberian.pk/topic/1116/eng101-assignment-2-solution-and-discussion/9
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    zareenZ
    Answer to Q#1 Modern transponders can carry many different types of communications traffic. They can also receive signals from multiple ground stations, combining (multiplexing) or splitting (de-multiplexing) them for onwards transmission to other multiple ground stations. This method, by which many users share a common satellite resource, is called Multiple Access. There are several schemes for accomplishing this, each with its benefits and drawbacks. TDMA (Time Division Multiple Access) allocates a time slot to the user in a repetitive time frame. The signal is digitized and the data bits are stored in a buffer in a compressed time frame until their allocated time slot comes around when they are transmitted during their allocated time. At the receiver end of the link the bits are rearranged, spreading them out to reassemble the original digital signal and converted back to analogue form. The signal occupies the entire transponder bandwidth, but only during its allocated time slot. The rest of the time the bandwidth is available to other users. Digital signals typically have better noise immunity than analogue signals. FDMA (Frequency Division Multiple Access) shares the bandwidth between the users, with each user allocated a unique, narrower section of the available bandwidth. It works with analogue signals and all users have uninterrupted use of their own narrow frequency band or channel with all users occupying the available bandwidth simultaneously, each within their own narrow channel. The sender’s signal, called the baseband signal, is frequency shifted into the allocated frequency band for transmission and the receiver restores it back to the baseband. CDMA (Code Division Multiple Access) also known a Spread Spectrum, modulates the user’s signal with a pseudorandom code so that it occupies the full available spectrum, appearing as noise. The receiver uses the same pseudorandom code in an auto correlator device, which only recognizes a signal modulated with the same auto code and thus separates it from the noise. CDMA is more complex but has better noise immunity and provides greater security than the other two systems. Answer to Q#2 • Mobile Service Satellites (MSS) • Example: Satellite Phones
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    Q1. Write a subroutine that will find the first even number from an array of your VU ID and calculate its factorial. (10 Marks) Note: Skip 0’s in your VU ID as shown below, VU ID: BC190206435 After skipping 0’s, array would be: Array: 1, 9, 2, 6, 4, 3, 5 The first even number is 2 in the array so its factorial will be calculated and saved in AX register. Solution: [org 0x100] jmp start data: dw 1, 9,2,6,4,3,5 ;initiliazation checkEven: mov dl,0002h mov ax,[data+bx] mov cx,[data+bx] xor ah,ah xor ch,ch div dl cmp ah,00h ;checking remainder je fact add bx,2 jmp checkEven fact: mov ax,0001 mov dx,0000 mult: mul cx loop mult ret start: mov si,00h mov bx,si call checkEven mov ax,0x4c00 int 0x21 Q2. Write a code in assembly language (using appropriate jumps) equivalent to this given code in C. (10 Marks) #include <stdio.h> int main() { int n1=1, n2=2; int largest; if( n1>n2) largest=n1; else largest = n2; return 0; } Solution: Assembly code: [org 0x100] jmp start n1 db 1 n2 db 2 largest db 0 start: mov ax, [n1] xor ah,ah mov bx, [n2] cmp ax,bx ja large mov [largest],bx large: mov [largest],ax mov ax, 0x4c00 int 21h
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