Sample Input 5 3 89 90 78 93 80 90 91 85 88 86 91 92 83 89 90.5 Sample Output 90.0 91.0 82.0 90.0 85.5 Explanation Marks obtained by student 1: 89, 90, 91 Average marks of student 1: 270/3 = 90 Marks obtained by student 2: 90, 91, 92 Average marks of student 2: 273/3 = 91 Marks obtained by student 3: 78, 85, 83 Average marks of student 3: 246/3 = 82 Marks obtained by student 4: 93, 88, 89 Average marks of student 4: 270/3 = 90 Marks obtained by student 5: 80, 86, 90.5 Average marks of student 5: 256.5/3 = 85.5

Sample Input 5 3 89 90 78 93 80 90 91 85 88 86 91 92 83 89 90.5 Sample Output 90.0 91.0 82.0 90.0 85.5 Explanation Marks obtained by student 1: 89, 90, 91 Average marks of student 1: 270/3 = 90 Marks obtained by student 2: 90, 91, 92 Average marks of student 2: 273/3 = 91 Marks obtained by student 3: 78, 85, 83 Average marks of student 3: 246/3 = 82 Marks obtained by student 4: 93, 88, 89 Average marks of student 4: 270/3 = 90 Marks obtained by student 5: 80, 86, 90.5 Average marks of student 5: 256.5/3 = 85.5

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Data block to be decoded: 00 11 bc b5 49 40 b8 8d 12 00 0d fa 08 00 45 00 00 37 99 09 00 00 ff 11 ea 9f 0a 07 03 1a 42 aa e8 41 c7 19 00 35 00 23 a3 9e 02 90 01 00 00 01 00 00 00 00 00 00 05 6d 69 74 72 65 03 6f 72 67 00 00 01 00 01            Please answer the following questions after decoded the above data block: Q1: (Layer 4) Protocol is: Q2: Source MAC address: Q3: Destination MAC address: Q4: Source IP address: Q5: Destination IP address: Q6: Source (Layer 4) Port #: Q7: Destination (Layer 4) Port # : Q8: (Application Layer) Payload description and contents. Please I left out questions-6,7and 8
Data block to be decoded: 00 11 bc b5 49 40 b8 8d 12 00 0d fa 08 00 45 00 00 37 99 09 00 00 ff 11 ea 9f 0a 07 03 1a 42 aa e8 41 c7 19 00 35 00 23 a3 9e 02 90 01 00 00 01 00 00 00 00 00 00 05 6d 69 74 72 65 03 6f 72 67 00 00 01 00 01            Please answer the following questions after decoded the above data block: Q1: (Layer 4) Protocol is: Q2: Source MAC address: Q3: Destination MAC address: Q4: Source IP address: Q5: Destination IP address: Q6: Source (Layer 4) Port #: Q7: Destination (Layer 4) Port # : Q8: (Application Layer) Payload description and contents.
Data block to be decoded: 00 11 bc b5 49 40 b8 8d 12 00 0d fa 08 00 45 00 00 37 99 09 00 00 ff 11 ea 9f 0a 07 03 1a 42 aa e8 41 c7 19 00 35 00 23 a3 9e 02 90 01 00 00 01 00 00 00 00 00 00 05 6d 69 74 72 65 03 6f 72 67 00 00 01 00 01            Please answer the following questions after decoded the above data block: Q1: (Layer 4) Protocol is: Q2: Source MAC address: Q3: Destination MAC address: Q4: Source IP address: Q5: Destination IP address:
mov r0, # 6Ah mov psw, # 00h 018 ric a 180100564-871598836 598836 b190100564871598836. 71598830 The final value of b190100567 1598836 2871698836 The accumulator's final value is not twice the 22h. 598636 00100504-9815883
EX: INPUT: INPUT: 11 13 17 31 39 47 0 OUTPUT: 25 27 49 7 23 14 29 80 OUTPUT: PRIME: 3 PRIME: 5 COMPOSITE: 5 COMPOSITE: 1
In DES, if the input to the below S-Box was 101010, then the output will be: S₁0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 14 04 13 01 02 15 11 08 03 10 06 12 05 09 00 07 00 15 07 04 14 02 13 01 10 06 12 11 09 05 03 08 04 01 14 08 13 06 02 11 15 12 09 07 03 10 05 00 3 15 12 08 02 04 09 01 07 05 11 03 14 10 00 06 13 1 2 Select one: a.0110 b.0010 C.0011 d.1101
What is the character string corresponding to the following ASCII codes?0x2a2a2a2a 0x69644120 0x4120736f 0x6f67696d 0x48202d20 0x20657661(Hint – Remember that for simulations running on Intel–based platforms, the charactersare stored in reverse order within each word.)
1. Create a p chart using the following data. Sample 1 234567 3 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample size 50 88888888888888888888 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 No. of defective 12 8 5 1 9 15 11 7 58 5 6 17 23 7 14 15 9 20 9 13
SCREENSHOT ,JAVA SAMPLE INPUT 1: FOOD1356279 Carrot 0.250 1.5 FOOD8759007 Bazella 0.090 0.100 FOOD7988001 Rice 0.750 15 PFOD5378762 Red_Beans 0.430 4 PFOD5551783 Banana 0.450 1 ELEC7862545 Blender 12.750 1 SAMPLE INPUT2 : FOOD1356279 Carrot 0.250 1.5 FOOD8759007 Bazella 0.090 0.100 FOOD7988001 Rice 0.750 PFOD5378762 Red_Beans 0.430 4 PFOD5551783 Banana 0.450 a1 ELEC7862545 Blender 12.750 1
What but patterns are represented by a. 0x5FD97, b. 0x610A, 0xABCD and 0x0100
Hi help with this problem "prime factors" if I input this number: 9201111169755555649 output is: 3033333343, 3033333343 another example input: 24 output:2,2,2,3 another example input: 1111111111111111111 output: 1111111111111111111
Given the truth table below, where A,B,C,D are inputs and Y is the output: Truth Table ABCDY 0 0000 0 1 00010 2 0010 1 3 0011 1 4 0 1000 5 0 1011 6 0 110 1 7 0 111 1 8 1000 0 9 10011 1 10 10 1 |1 10111 1 11001 11010 3 1 11100 1 11110 5 a) Generate the Karnaugh Map. Use the space to the right of the truth table. b) Draw the largest loops you can. Use different colored loops preferably. c) Give the minimal logic expression. (Hint: Each loop should correspond to one part of the expression.) Show the parts of the expression that correspond to each loop.