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    zareenZ
    https://www.youtube.com/watch?v=UOC6dRqtCME
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    zareenZ
    https://www.youtube.com/watch?v=UOC6dRqtCME
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    cyberianC
    https://youtu.be/5oo8RFiZM6Y
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    zareenZ
    @zareen said in CS205 Assignment 2 Solution and Discussion: A user noticed that someone is using his Facebook account after stealing his credentials. After investigation, he observed that it happened after installing a software downloaded from an un-authentic website. Aug 25, 2019 - Hacked Facebook accounts are being sold on the dark web, showing the … Problem is that since so many web sites make you use your email for log in name, I’m not … 90 EACH days after the ‘worst ever’ hack of 50 million users (and here’s … a hacker has claimed he sold the plaintext passwords and email … link text
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    zareenZ
    //<!--CS602 ASSIGNMENT 2 SOLUTION--> //<!--FALL 2019--> //<!--DUE DATE: 27 NOVEMBER, 2019--> //PRE-PARED BY: SUPERSTARWEBTECH.COM //CONNECT WITH US:FACEBOOK.COM/SUPERSTARWEBTECH //E-MAIL: SUPERSTARWEBTECH@GMAIL.COM #include <stdio.h> #include <stdlib.h> #include <conio.h> #include <graphics.h> #include <math.h> class House { private: int x1,x2,x3,x4,x5; int y1,y2,y3; int dx1, dx2; int dy1, dy2; int xc,yc; public: House() { xc=45; yc=65; //center of figure x1=30; x2=x1+50; x3=20; x4=x3+20; x5=x4+50; y1=30; y2=y1+20; y3=y2+50; dx1=x3+5; dx2=dx1+10; dy1=y2+30; dy2=dy1+20; } void translate(int tx, int ty) //draw translate house { x1+=tx; x2+=tx; x3+=tx; x4+=tx; x5+=tx; y1+=ty; y2+=ty; y3+=ty; dx1+=tx; dx2+=tx; dy1+=ty; dy2+=ty; } void scale(int sx, int sy) //draw scaled house { x1=xc+(x1-xc)*sx; x2=xc+(x2-xc)*sx; x3=xc+(x3-xc)*sx; x4=xc+(x4-xc)*sx; x5=xc+(x5-xc)*sx; y1=yc+(y1-yc)*sy; y2=yc+(y2-yc)*sy; y3=yc+(y3-yc)*sy; dx1=xc+(dx1-xc)*sx; dx2=xc+(dx2-xc)*sx; dy1=yc+(dy1-yc)*sy; dy2=yc+(dy2-yc)*sy; } void reflectx(int rx) { y1=-y1+rx; y2=-y2+rx; y3=-y3+rx; dy1=-dy1+rx; dy2=-dy2+rx; } void reflecty(int ry) { x1=-x1+ry; x2=-x2+ry; x3=-x3+ry; x4=-x4+ry; x5=-x5+ry; dx1=-dx1+ry; dx2=-dx2+ry; } void drawH() //draw house { int arr[]={x1,y1,x2,y1,x5,y2,x5,y3,x3,y3,x3,y2,x1,y1,x4,y2,x4,y3,x3,y3,x3,y2,x5,y2}; drawpoly(12,arr); //draw house rectangle(dx1,dy1,dx2,dy2); //draw house door } void fillH() { setfillstyle(HATCH_FILL,GREEN); floodfill(x1+1,y1+10,WHITE); floodfill(x1+1,y1+1,WHITE); setfillstyle(SOLID_FILL,RED); floodfill(x3+1,y2+1,WHITE); setfillstyle(SOLID_FILL,YELLOW); floodfill(x4+1,y2+1,WHITE); setfillstyle(SOLID_FILL,GREEN); floodfill(dx1+1,dy1+1,WHITE); } void fillH2(){ setfillstyle(HATCH_FILL,GREEN); floodfill(x1+1,y1-10,WHITE); floodfill(x1+1,y1-1,WHITE); setfillstyle(SOLID_FILL,RED); floodfill(x3+1,y2-1,WHITE); setfillstyle(SOLID_FILL,YELLOW); floodfill(x4+1,y2-1,WHITE); setfillstyle(SOLID_FILL,GREEN); floodfill(dx1+1,dy1-1,WHITE); } }; int main() { int gd=DETECT, gm; initgraph(&gd,&gm,(char*)" "); setcolor(WHITE); settextstyle(DEFAULT_FONT, HORIZ_DIR, 2); outtextxy(2, 2, (char*)"BC123456789"); House house; //draw main house house.drawH(); house.fillH(); outtextxy(2, 110, (char*)"1) Original"); //draw translated house house.translate(200,0); house.drawH(); house.fillH(); outtextxy(200, 110, (char*)"2) Translated"); //draw scaled house house.translate(-190,70); house.scale(2,2); house.drawH(); house.fillH(); outtextxy(2, 280, (char*)"3) Scaled"); int x=90; int y=350; House h2; // draw reflected house h2.translate(200,70); h2.reflecty(x); h2.reflectx(y); h2.reflecty(x); h2.drawH(); h2.fillH2(); outtextxy(200, 270, (char*)"4) Reflected"); getch(); closegraph(); } //<!--NOTE--> //<!--MAKE CHANGES ACCORDING TO YOUR REQUIREMENTS--> //<!--DO NOT POST ON OTHER PORTALS THIS WILL LEAD TO COPYING OF CODE AND THEN YOU WILL GET ZERO MARKS FOR IT--> //<!--PLUS TYPE YOUR OWN SOLUTION FOR NOT LOSING MARKS--> //<!--DO NOT COPY PASTE--> //<!--WE ARE NOT RESPONSIBLE FOR GETTING ZERO MARKS IN ASSIGNMENT--> //PRE-PARED BY: SUPERSTARWEBTECH.COM //CONNECT WITH US:FACEBOOK.COM/SUPERSTARWEBTECH //E-MAIL: SUPERSTARWEBTECH@GMAIL.COM CS602_2_FALL_2019.txt Displaying CS602_2_FALL_2019.txt.
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    zareenZ
    Solution: // Header Files #include<stdio.h> #include<conio.h> #include<BIOS.H> #inlcude<DOS.H> void interrupt (*oldTimer)(*void); // To store current Timer vector void interrupt newTimer(); //New Timer Function char far *scr= (char far *)0xB8000000; //Screen segment int in, t=0; void main() { clrscr(); oldTimer=getvect(8); setvect(8,newTimer); getch(); } void interrupt newTimer(); { *(scr+t)=0x2A; t++; if(t>=126) { for(i=0;i<4000;i+=2) { *(scr+i)=0x20; // Blank screen *(scr+i+1)=0x07; } t=0; } (*oldTimer)(); } }
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    zareenZ
    @zareen said in CS614 Assignment 2 Solution and Discussion: Calculate the total space reserved in memory using normalization Q. 1. Calculate the total space reserved in memory using normalization1. Calculate the total space reserved in memory using normalization Q. 1 Solution: Total space used = 10 (Million records) x 40 (Bytes header) + 110 (Million records) x 70 (Bytes header) = (1000000040) + (11000000070) =8,100,000,000 = 8.1 GB Q. 2. Calculate the total space reserved in memory after de-normalization using pre-joining technique Q. 2 Solution: Total space used = (70 + 40 – 10) (Bytes header) x 110 (Million records) =100*110000000 =11,000,000,000 = 11 GB
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    zareenZ
    ASSIGNMENT NO:2 Course: CS302 1: Write the SOP expression for the given sum. Sol: A B C D E 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 A B C D E OUTPUT (F) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 1 1 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 0 FOR SOP WE FOCUS ON 1 VALUE. A B C D E MINTERM 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 ABCDE SOP EXPRESSION: SUM OF PRODUCT EXPRESSION 2: Find Prime Implicant of minterm using QuineMcculsky method. Step-1 00010 2 00100 4 01000 8 10000 16 00110 6 01010 10 01100 12 10010 18 10100 20 11000 24 01110 14 10110 22 11010 26 11100 28 11110 30 Step-2 2,6(00-10) 2,10(0-010) 2,18(-0010) 4,12(0-100) 4,6(001-0) 4,20(-0100) 8,10(010-0) 8,24(-1000) 16,18(100-0) 16,20(10-00) 16,24(10-00) 6,14(0-110) 6,22(-0110) 10,14(01-10)10,26(-1010) 12,14(011-0) 12,28(-1100) 18,26(1-010) 18,22(10-10) 20,22(101-0) 20,22(1-100) 24,26(110-0) 24,28(11-00) 14,30(-1110) 22,30(1-110) 26,30(11-10) 28,30(111-0) Step-3 2,6,18,22(-0-10) 2,6,10,14( 0–10)2,10,18,26(–010)2,18,6,22(-0-10) 2,18,10,26(–010)4,12,6,14(0-1-0) 4,6,12,14 (0-1-0) 4,6,20,22(-01-0)4,20,6,22(-01-0) 4,20,12,28(–100) 8,10,12,14(01–0) 8,24,10,26(-10-0) 8,24,12,28 (-1-00) 6,14,22,30 (–110) 6,22,14,30(–110) 10,14,26,30(-1-10) 10,26,14,30(-1-10) 12,14,28,30(-11-0) 12,14,28,30(-11-0) 18,26,22,30(1–10) 18,22,26,30(1–10) 20,22,28,30(1-1-0) 20,22,22,30(1-1-0) 24,26,28,30(11–0) 24,28,26,30(11–0) Step-3 2,6,18,22(-0-10) 4,6,20,22(-01-0) 2,18,10,22(-0-10) 4,20,6,22(-01-0) 2,10,18,26(–010) 4,12,6,14(0-1-0) 2,18,6,22(–010) 4,6,12,14 (0-1-0) 4,20,12,28(–100) 10,14,26,30(-1-10) 6,14,22,30 (–110) 10,26,14,30(-1-10) 12,14,28,30(-11-0) 18,26,22,30(1–10) 12,14,28,30(-11-0) 18,22,26,30(1–10) 20,22,28,30(1-1-0) 24,26,28,30(11–0) 20,22,22,30(1-1-0) 24,28,26,30(11–0) 6,22,14,30(–110) 8,10,12,14(01–0) 8,24,12,28 (-1-00) 8,24,10,26(-10-0) Step-4 2,6,8,22(-0-10) 4,6,20,22(-01-0) 2,10,18,26(–010) 4,12,6,14(0-1-0) 4,20,12,28(–100) 10,14,26,30(-1-10) 12,14,28,30(-11-0) 18,26,22,30(1–10) 20,22,28,30(1-1-0) 24,26,28,30(11–0) 6,22,14,30(–110) 8,10,12,14(01–0) 8,24,12,28 (-1-00) 8,24,10,26(-10-0) These are the prime implicates cs302-assign 2.docx
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    zareenZ
    @zareen said in SOC602 Assignment 1 Solution and Discussion: In your point of view which school of thought is more applicable to the current educational system of Pakistan and why? Justify with logical arguments. (3+3=6) Pakistan, who most graciously and generously drew our attention to … Shamim (2011) has described the current language in education system in … Pinnock (2009) points out that if the medium of instruction in schools is the national … argument for abandoning the mother tongue in favour of English as the medium of. link text
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    zareenZ
    @zareen said in MGT520 Assignment 1 Solution and Discussion: Foreign direct investment (FDI) is considering as a leading factor of persistent and higher financial growth and a vital source of money inflows. FDI allows economies to increase their export share in local and global market. PDF | Foreign Direct Investment (FDI) has been viewed as a major stimulus to … Foreign Direct Investment: A Journey To Economic Growth In Ghana … means that competition to attract FDI has increased as developing countries continue to … this study is to empirically determine the factors that influence FDI flows in Ghana, … Solution
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    zaasmiZ
    @zareen said in FIN623 GDB 1 Solution and Discussion: Total Marks 5 Starting Date Thursday, November 21, 2019 Closing Date Wednesday, November 27, 2019 Status Open Question Title Taxation Management Question Description Briefly answer the following questions. (2 + 1 + 1) Why tax avoidance can be favored against tax evasion? Which tax structure is levied in Pakistan for the taxation of income of individuals? Mr. Log is the cultivator of wheat and uses it as raw material in his business. Profit and loss account of Mr. Log for the year ended June 30, 2019 showed the net profit of Rs. 4,350,000. The wheat processed in flour mill, if purchased from open market, would have cost to Mr. Log Rs. 1,500,000. What would be the taxable income of Mr. Log, if wheat is not ordinarily sold in the market? (Show complete calculations) Why Tax Avoidance Can Be Favored Against Tax Evasion Tax avoidance is generally preferred over tax evasion due to the following reasons: • Legality: Tax avoidance is legal and involves using the tax code to minimize tax liability through deductions, credits, and exemptions allowed by law. Tax evasion, however, is illegal and involves hiding income or falsifying information to evade taxes. • Ethical Compliance: Tax avoidance aligns with legal and ethical standards, showing compliance with the law. Tax evasion, in contrast, is unethical and often leads to penalties, legal action, or imprisonment. • Incentives for Financial Planning: Many tax laws are designed to incentivize behaviors beneficial to the economy, like saving for retirement or investing. Tax avoidance takes advantage of these incentives, contributing positively to personal and national financial health. Because of these reasons, tax avoidance is a lawful and strategic approach, while tax evasion is illegal and carries serious consequences. Tax Structure Levied in Pakistan for the Taxation of Income of Individuals Pakistan employs a progressive tax structure for individual income taxation. Under this structure, individuals with higher incomes are subject to higher tax rates, while those with lower incomes pay lower rates. This progressive structure aims to distribute tax burden equitably based on income levels and is outlined in Pakistan’s Income Tax Ordinance, 2001. Calculation of Mr. Log’s Taxable Income Since Mr. Log is a cultivator of wheat and uses it in his own business, we must consider the market value of the self-produced wheat as part of his income for tax purposes. Given Data • Net Profit (P&L Account): Rs. 4,350,000 • Market Value of Self-Produced Wheat: Rs. 1,500,000 • Condition: Wheat is not ordinarily sold in the market. Calculation of Taxable Income To calculate taxable income, we add the market value of the self-produced wheat (if it were bought from the open market) to the net profit from Mr. Log’s business: Net Profit from Business: Rs. 4,350,000 Add Value of Self-Produced Wheat: Rs. 1,500,000  Taxable Income = Net Profit + Market Value of Wheat Taxable Income = 4,350,000 + 1,500,000 = 5,850,000  Answer Mr. Log’s taxable income would be Rs. 5,850,000 for the year ended June 30, 2019.
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    zareenZ
    Solution # 1 Instruction Processor Hexadecimal Code Behavioral RTL JPL R5, [26] Falcon - A 10000 101 00011010, 1000 0101 0001 1010, 851A (R[5] ≥ 0): PC ← PC+ (26-PC); STS R7, R2 (100) Falcon - E 00101 111 010, 000000000000001100100, 0010 1111 0100 0000, 0000 0000 0110 0100, 2F40 0064 M[R[2]+100] ← R [7] STACC R4, 36 Modified EAGLE 10111 100 00100100, 1011 1100 0010 0100, BC24 M[R[4] + (8@C<7>)©C] ← ACC C represents the constant 36 DIV R2 EAGLE 10000 010, 1000 0010, 82 R[0] ← R[0]/R[2] R[2] ← R[0]%R[2] SHIFTL R5, R2, 7 FALCON - A 01100 101 010 00111, 0110 0101 0100 0111, 6547 R[5]<15…0> ← R [2]<(15- N)…0>©(N@0) N represents constant 7 Solution # 2 Instruction Value of Destination Operand Data Bus <15…0> Address Bus (15…0> LDACC B 55CEh 55CEh 3320h SUB R1 55B9h N/A N/A LDACC C 2015h 2015h AB0Eh ADD R2 45E0h N/A N/A STACC A 45E0h 45E0h AB10h Table 3 Data Bus and Address Bus Contents for Modified Eagle Calculation Steps (Instruction-By-Instruction) • LDACC B LDACC stands for load accumulator. In LDACC, the destination operand is accumulator and source operand is the memory location labelled as B. The memory label B points to the memory address 3320h. When this instruction is executed, the value stored at memory address 3320h will be read and loaded in Accumulator register. This address 3320h will be copied into Address Bus which will then read its contents from memory and load the contents at data bus. The operand size in Modified EAGLE is 2-byte. Therefore, the values stored at addresses 3320h and 3321h will be loaded at data bus. These values are CEh and 55h respectively. Since, Modified EAGLE employs Little endian notation hence, the 2-bye value will be read as 55CEh. The value of data bus will also be 55CEh and this will be loaded into destination operand Accumulator. • SUB R1 SUB R1 means to subtract the value of source operand register R1 which is 0015h, from the destination operand Accumulator (ACC) register which contains 55CEh. The result of subtraction will be stored back into ACC. After subtraction, the value stored in ACC will be 55B9h. Because SUB is not a memory instruction, we are not concerned with the contents of Data Bus or Address Bus because the values we need to execute the instruction are already available in registers. Hence, Data Bus and Address Bus values will be labelled as N/A. • LDACC C When this instruction is executed, the value stored at memory address labelled with C is read and loaded in Accumulator register ACC. In this case, the address of the C is AB0Eh which is also the value of address bus. The operands in Modified EAGLE are 2-byte values. The contents at addresses AB0Eh and AB0Fh will be copied into data bus which will then be loaded into ACC register. These contents are 15h and 20h. Due to Little endian notation, the 2-bye value will be 2015h. The value of data bus will be 2015h and same will be loaded in destination register ACC. • ADDR2 When ADD is executed, the value of register R2 is added to ACC register. Hence, after the execution, the ACC register will hold 45E0h. AS usual, ADD is not a memory instruction, so we are not concerned with the values of Data Bus and Address Bus and both are labelled as N/A. • STACC A STACC stands for Store ACC. There is one destination operand which is a memory label A. When the instruction is executed, the value of Accumulator register ACC is stored at the memory address labelled by A. The destination memory address will be AB10h. The value of ACC is 45E0h will be stored as address AB10h. However, due to Little-Endian notation, the address will be stored as E0h at memory location AB10h and then 45h at memory location AB11h.
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