1 2 3 4 5 6 7 8 9. 10 22 39 70 126 227 408 735 1322 2380 4285
Q: 1098765432* 109876543*1 10987654*21 1098765*321 109876*4321 10987*54321 1098*654321 109*7654321…
A: The java code is
Q: 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 %3D 3. 87 170 40 150 36 72 66 151…
A: FCFS: FCFS stands for first come first serve, FCFS is the simplest of all the Disk Scheduling…
Q: 1. 255 base 10 5 2. 2 3. 16 2 10 8 11111111 base 10 716 base 8 2 16 7
A: (255)10 = (11111111)2 Divisionby 2 Quotient Remainder (Digit) Bit (255)/2 127 1 0 (127)/2…
Q: 9. 101011.11 * 10.101 10.428 * 21s 11.2A16 * 1.916
A: Answer = 1110010.11011
Q: 37. 1100 1110 11112 38. 1111 1111 00112 39. 0101 1101 11102
A: Here is the explanation regarding the hexadecimal conversion:
Q: 1. 1 102 10으 +6= 8. obbeu 2. 3 10 7. 70
A: Answer to 1) 133/8
Q: 2D 12 8C A1 D2 5E 47:20:1B:2E:08:EE None of the above 01:00:5E:2B:0E:07 01:00:5E:00:01:01
A: Which of the following is a multicast physical address? A. 2D 12 8C A1 D2 5E B. 47:20:1B:2E:08:EE…
Q: 10.250.1.1 223.1.10.100 117.89.56.45 33.0.0.33 77.10.77.10 126.8.156.156 209.209.209.209 148.17.9.1
A: To decide class, consider only first octent and find range in the following list. Class A range : 1…
Q: Base 2 Base 10 Base 16 14 37 11101 101.01 0.5 24.25
A: The solution for the table is given below.
Q: شفرة 1 BCD631 ل لعد د 10)567( هي a. 1110 0001 1001 b. 0111 1000 1001 c. 1111 1010 0110 d. 1100 1111…
A: Given that, (567)10 is a decimal number and the value of 567 in BCD6311 form is: In BCD form, every…
Q: 14 20 26 12 C 15 E 12 10 8
A: Hamiltonian : Sorted Algorithm works as follows: 1.Sort the edge cost in ascending.2.Pick up the…
Q: (010110011000)xs= a) 581 b) 551 c) 481 d) 611 e) 721
A: Answer and solution is given in step2.
Q: iv. 40005 v. 71228 vi. 778 vii. 3334 viii. 1111112
A:
Q: 1. 3. 5. 0100₂ +0010₂ 01012 + 0110₂ 11012 + 0111₂ 2. 4. 6. 01012 +0010₂ 10102 +0011₂ 10112 + 0111₂
A: Rules to follow for adding two binary numbers : 0+0=0 0+1=1 1+0=1 1+1=10 (where 1 is carried over)
Q: 12341 12342 12343 12344 12345 6789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
A: #include <iostream>using namespace std; int main(){ int x = 1; for(int i = 1 ; i <= 4…
Q: 45.67 * 3.67 13. A5.912 * 3.A12 14. 1A.716 * 9.C16
A: Here have to determine about multiplication of given equation.
Q: 5. Divide the following signed numbers: a. -4/-2 b. -2/-1 c. -24/-48
A: Divided the given signed numbers
Q: A = 010 110, B = 000 101 A||B = 011 %3D 001. * True False
A: as, 0 || 0 = 0 0 || 1 = 1, 1 || 0 = 1, 1 || 1 = 1. so, calculate for each bit. A =…
Q: 3. GO) = F43:45)(6+3)(4+5) (1*436+5)(2+3)(s+5)
A: Code: s=tf('s')g=1/((s^2+3*s+5)*(s+3)*(s+5))
Q: 110 E 108 21 56 18 46 208 J 29 A 78 16 G 56 96 20 112 40 20 H 62 90 F
A: Using Dijkstra's algorithm on the given diagram to find minimum time and route from A to J
Q: 6810 ➔ 2 A. 1001100 B. 1100100 C. 1000100 D. 1010100 2. F91 ➔ 2 A. 3987 B. 3996 C. 4018…
A: Please find the answer below :
Q: الرمز الرمادي (10110111101) = ()'' بالرقم العشري 931 1753 867 1877 لا احد منهم 1749 1750 869
A: Answer:
Q: Q 6) 11010110 ➔ 8 A. 322 B. 326 C. 331 D. 324
A:
Q: 0100 1001 + 1010 0111
A: In Step 2, I have provided solution with answer------ In Step 3, I have provided PYTHON PROGRAM…
Q: (а) (100010111101)2-( )s=( D16 (b) (510E)16=( (с) (2021)10-( )16-( 2=(
A: Here I shown step by step answer for this number conversation. 1)N2>N8>N16 2)N16>N2>N8…
Q: 4 (527 + 321 )a
A: Octal numbers contain digits only upto 7.
Q: there are . kilobytes in (5) * Megabyte 5120 5148 5104 5196
A: Here we have 5 Megabytes 1 Megabyte = 1024 kilobytes So 5 Megabytes = 5×1024 = 5120
Q: 10011 to 10 a. 60 b. 61. c. 62 d.63
A: Multiply by 2 ki power from left to right (10011)2 = 1*24 + 0 *23 +0*22 + 1*2 + 1*20 .…
Q: Multiplication [1 2 3;4 5 6] and [7 8 9;10 11 12] [7 16 27; 40 55 72 ] True False O
A: The correct answer along with the explanation is given below:
Q: 1011.1011 (base 2) = (base 10)
A: According to the Question below the Solution:
Q: a. 1+3+5+7+· +999 b. 2+4+8+ 16 + · · . + 1024 d. Σ3 ... n+: с. -i=3
A: Please refer below for your reference: According to the guidelines we are restricted to answer only…
Q: 0.15% 2.35% 60 13.5% 90 34% 34% 13.5% 120 150 180 Trunk diameter (cm) 2.35% 210 0.15% 240
A:
Q: 1. 6758 + 23.43 2. 64.1s - 17.458 3. AD216 + 93.416
A: Here there are multiple parts so I will answer 1st 3 questions.
Q: 7 5 4 3 2 1 103 102 10' 10°
A: Matlab code is
Q: Base 2 Base 10 Base 16 14 37 11101 101.01 0.5 24.25 1FCD
A: Conversion of given binary, decimal and hexadecimal
Q: (!( 23 > 34 )&& (3 +9 < 10 ) || (10 != 1))
A: Given: Checking the boolean expression value.
Q: 1- (25) 10 2- (5689) 10 3- (11111) 10 4- (897.34) 10 5- (69.45) 19
A: STEP 1: (25)10 = ( ? )2Solution: (25)10=(_______)2 2 25 2 12 1 ↑…
Q: А. (276) в + (654) В. (631) в - (146) в 8 С. (657) в Х (52) в 8
A: According to the information given we have to calculate in octal format.
Q: X 32 | 50 |52 44 30 54 53 39 51 56 Y 105 111 | 98 97 39 102 88 98 92 87
A: I provide the code in R along with output and code screenshot
Q: ((1111)2 + (1010) 2) Which of the following is the 2's complement of O a. 11000 O b. 01010
A: 1 1 1 1 + 1 0 1 0 1 1 0 0 1
Q: Exercice 1: (101)2 = (... (1011.1100 )2 = (.... (27)8= (............. (43) 10 (............. ...)10…
A:
Q: 161/2 * 23 * 43/2 / 321/5 * 640 * 2-1
A: Answer is 64 Solution in step 2
Q: (1010), (100101), (10000001)2
A: As per our guidelines we are supposed to answer only one question kindly repost other questions as a…
Q: 2.33 21323 + 60(25) = 4 2]
A: The answer to the following question:-
Q: 100001, = 3310 259 = 1310 25, = 2310 25s = 3710 %3D
A:
Q: 129 126 90 126 24 60 60 24
A: In digital electronics, the number can be converted from one form to another using the conversion…
Q: 4. 100.416 + 2D.E516 5. 12.649 + 7.679 6. B0.9A13 + C.A513
A: Here, we perform normal addition and if result more than F, then 1 will be carried and remaining…
Q: 1010 >e 1010 1010 1011 > 1011 1011- F1 F2 F3 1110 2 ở 1110 1110- 4 4 1101 >3 1101 >3 1101 2 S = 10 S…
A: Multiplexer is an electronic switch that can connect one out of n inputs to the output.. It cannot…
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- Gap in playback is a common problem exiting in multimedia applications transmitted using the isochronous data transmission approach shown in class. What is the method used to solve the problem?Suppose that I have a 6.5 GByte movie file that’s sitting on my hard drive at home. If I have cable internet service, which gets 50 Mbps down and 10 Mbps up, how long before my friend is able to get the file if he has a DSL which gets 8 Mbps down/1 Mbps up?Alice uses Google location services provided by a cloud server, which uses the k-Anonymity technique. She does not want her routine trajectory been detected by others. Bob can easily intercept the communication data sent by Google cloud. Which of the following description is correct? Group of answer choices Bob will easily know Alice's routine trajectory. Bob will not be able to know Alice's routine trajectory. It requires a massive calculation for Bob to know Alice's routine trajectory. Bob will be able to guess Alice's routine trajectory if the k is small.
- The received 7 x 5 message block shown below uses odd parity checks, with the parity bits displayed at the end of each row (A–E) and column (1–7), in the shaded cells (see Section 3.7.1). By respectively checking each row and column determine the following: i.The row-column grid reference of the single error-bit. ii.What change needs to be made to correct the error? As part of your answer, briefly explain how you located the error-bit.The previous problem is in the picture. Suppose that I have 800 Gigabytes worth of movies on my hard drive at home, and I’d like to get them to a friend in Chicago. I could either send them over the Internet, or I could copy them to a thumb drive, drive to Chicago, and deliver them by hand. Assume that the specs for my cable connection are the same as in the previous problem, but the friend in Chicago has a cable connection which gets 25 Mbps down/5 Mbps up. Which would be faster? Show your work.Do you think error correction works better in wireless applications? Give examples to support your argument.
- There are two different friends who want to send some informative data to each other. A friend Abdul Hadi first sends the data and check whether other friend, Abdul Basit is available or not. The communication process is,1- Abdul Hadi has to check the availability of Abdul Basit.2- Abdul Hadi has sent three different information’s (frame 1 to 3), to Abdul Basit after positive response each time.3- After all the successful communication, the session has been terminated by Abdul Hadi.In your opinion, which one will be better choice from ENQ/ACK and Poll/Select? Draw the proper diagram of above scenario to support your answer.Refer to above scenario, also draw the diagram for Stop and Wait method in Flow Control.In general, applications like Voice over Internet Protocol (VoIP), online games, and live audio streaming use UDP connections rather than TCP connections. Justify the statement's accuracy.a. Give what is ask in the following items: Explain the difference between a bandwidth of a signal and bandwidth of a channel.
- SkyCiv gives these values, so is the original answer correct?1. We assume Z26 ={0, 1, …, 25} is the alphabet for One-Time Pad, and the Plaintext space = Ciphertext space = Key space = (Z26)5. If the plaintext is (11, 8, 5, 19, 22) and the key is (2, 20, 15, 10, 6), what is the ciphertext? ( ) (13, 2, 20, 3, 2). (13, 28, 20, 29, 28). (13, 28, 20, 29, 2) (13, 24, 6, 23, 24). 2. We assume Z26 ={0, 1, …, 25} is the alphabet for One-Time Pad, and the Plaintext space = Ciphertext space = Key space = (Z26)5. If the ciphertext is (11, 24, 5, 19, 22) and the key is (2, 20, 2, 10, 23), what is the decrypted plaintext? ( ) (13, 18, 7, 3, 19). (9, 4, 7, 29, 25). (13, 18, 3, 29, 25) (9, 4, 3, 9, 25). 3. For the binary version of One-Time Pad, if the ciphertext is 10110010 and the key is 01011010, what is the corresponding plaintext? ( )If the current playback window extends 120 1o 530, then you're in luck!When the next authenticated packet arrives with sequence number 105, what will the receiver do with it? What parameters will the windows have after that point in time?When the next authenticated packet arrives, what will the receiver do with it and what will the parameters of the window look like after that? 2.What will the receiver do with the packet, and what will the parameters of the window be after that?