SOL PlA as CODE TM_NONTH P_CODE TOTSALES 10010 10010 10010 10010 10010 10010 10011 10011 10011 10011 10012 10012 74.5 19.9 14.97 70.44 180.26 180.26 109.2 20.97 130.89 130.89 20.7 20.97 10 13-r2 10 23109-HD 10 54778-2T 10 PUC23DRT 10 10 222QTY 10 SM-18277 10 9 SM-18277 9.5 256.99 266.94 287.91 10012 10 23109-HB 10012 10012 10012 10 9 URE-Q 10 Some records omitted in output shown 9 2238OPD 9 23109-HB 9 51778-27 9 PUC23DRT 10018 10018 10018 10018 10018 10018 10019 10019 10019 38.95 9.95 9.90 70.44 129.32 129.32 39.95 39.95 39.95 2252.06 9 1546-002 s8 rous selected.

SOL PlA as CODE TM_NONTH P_CODE TOTSALES 10010 10010 10010 10010 10010 10010 10011 10011 10011 10011 10012 10012 74.5 19.9 14.97 70.44 180.26 180.26 109.2 20.97 130.89 130.89 20.7 20.97 10 13-r2 10 23109-HD 10 54778-2T 10 PUC23DRT 10 10 222QTY 10 SM-18277 10 9 SM-18277 9.5 256.99 266.94 287.91 10012 10 23109-HB 10012 10012 10012 10 9 URE-Q 10 Some records omitted in output shown 9 2238OPD 9 23109-HB 9 51778-27 9 PUC23DRT 10018 10018 10018 10018 10018 10018 10019 10019 10019 38.95 9.95 9.90 70.44 129.32 129.32 39.95 39.95 39.95 2252.06 9 1546-002 s8 rous selected.

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter8: Data And Network Communication Technology

Section: Chapter Questions

Problem 41VE

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my txt file data is as follow: 2013-11-01 00:00:00|1|0|0.3521|||0.0273|2013-11-01 00:00:00|1|33|||||0.02612013-11-01 00:00:00|1|39|1.7322|1.1047|0.5919|0.402|57.77292013-11-01 00:00:00|2|0|0.3581|||0.0273|2013-11-01 00:00:00|2|33|||||0.02742013-11-01 00:00:00|2|39|1.7334|1.088|0.602|0.4109|57.88752013-11-01 00:00:00|3|0|0.3644|||0.0273|2013-11-01 00:00:00|3|33|||||0.02872013-11-01 00:00:00|3|39|1.7348|1.0701|0.6128|0.4203|58.00952013-11-01 00:00:00|4|0|0.3349|||0.0273| i need to convert the above file data into 2D array in java such that the columns are seperated by '|'. how to perform this in java? there are 10 rows and 8 columns
0x0000000000400ef2 <+0>: sub $0x8,%rsp 0x0000000000400ef6 <+4>: mov $0x402470,%esi 0x0000000000400efb <+9>: callq 0x401323 <strings_not_equal> 0x0000000000400f00 <+14>: test %eax,%eax 0x0000000000400f02 <+16>: jne 0x400f09 <phase_1+23> 0x0000000000400f04 <+18>: add $0x8,%rsp 0x0000000000400f08 <+22>: retq 0x0000000000400f09 <+23>: callq 0x40154a <explode_bomb> 0x0000000000400f0e <+28>: jmp 0x400f04 <phase_1+18>.
data2.asm .ORIG x3500HELLO .STRINGZ "+9144827365544332109011\n" data3.asm .ORIG x3500HELLO .STRINGZ "-9144827365544332109011
IEEE 754-2008 contains a half precision that is only 16 bits wide. The leftmost bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the fraction is 10 bits long. A hidden 1 is assumed as part of the significand just like the IEEE-754 format we learnt in the class. Write down the bit pattern to represent −0.15625 × 100 in IEEE 754-2008 half-precision format. You need to find out the bits for the sign, the exponent, and the fraction. To answer this question, you need to convert the decimal fraction to its binary representation. Check out the link given in Q1. You have to show all steps including the fraction conversion (from decimal to binary).
Q1: Compress the following IPv6 addresses 0000:0000:0000:0000:0000:0000:0000:0000 2001:0000:0000:0000:0000:0000:0000:0001 2FFF:0000:1111:0000:0000:ABCD:0000:0025 3000:0000:0000:101A:0000:0000:0000:0001 20FF:3756:0005:0001:ACAD:0000:0000:0025 3FFF:0000:0000:0000:ACAD:25FF:0001:0127 20FF:ACAD:0000:0BCD:FFFF:0000:0000:0001 3ABC:0001:ACAD:0000:0000:0000:0000:0005 FE80:0000:0000:0000:03E0:1275:0000:0034 FE80:0000:0540:0000:0000:0000:0000:9800 Q2: Decompress the following IPv6 Addresses FE80::1 ::1 2F00:0:0:0:0:BCD:0:127 3E80:70::00FF:0:0001 2001:380:F:0:ACAD::5 3001::ACAD:205:0:100 2002:ACAD:0:1BCD:FFFF::4 3F0:25:ACAD::ABCD:0:5 FF::4E00:235:0:FF 3E01:6C:400::FF00 Q3: SLACC – eui-64Calculate the IPv6 addresses for the hosts with the following network prefix and MAC addresses A:Network prefix: 2001:4400:0FFF:1234::/64MAC Address: 90-61-AE-7B-A4-6B B:Network prefix: 2603:8080:2A40:76D::/64MAC Address: FE-AA-54-66-AC-11 C:Network Prefix: 2000:1:A:B:C::/64MAC Address:…
Separate the bit sequence into appropriate sized chuncks for each of the following domains. (You do NOT have to decode this bit sequence.) 1. Octal String: 010 110 101 010 101 111 010 111 111 101 011 011 011 011 010 1001. Hex String:1. Base64 String:
Word locations consist of three 5-bit cells. (a) A word location has a cell with address 8. What are the possibilities for the (sets of) addresses of the word location's cells? In each case give the word location's address.
@AliS219 @marshmllow_0
The _________ product defines the number of bits that can fill the link.
These are group 1 data CSC 80 100 70 80 72 90 89 100 83 70 90 73 85 90 -999 ENG 80 90 80 94 90 74 78 63 83 80 90 -999 HIS 90 70 80 70 90 50 89 83 90 68 90 60 80 -999 MTH 74 80 75 89 90 73 90 82 74 90 84 100 90 79 -999 PHY 100 83 93 80 63 78 88 89 75 -999 These are group 2 data CSC 90 75 90 75 80 89 100 60 80 70 80 -999ENG 80 80 70 68 70 78 80 90 90 76 -999HIS 100 80 80 70 90 76 88 90 90 75 90 85 80 -999MTH 80 85 85 92 90 90 74 90 83 65 72 90 84 100 -999PHY 90 93 73 85 68 75 67 100 87 88 -999 Can you please write a code-like code that I sent from you the same as the data I wrote for me in c language? 1. There are two files named group1.txt and group2.txt that contain course information and grades ofeach student for each class.2. You will calculate each course average for each group and show in simple bar graph. Use "*" and "#"characters for group1 and group2, respectively.3. You will see the number -999 at the end of each line in the input files. This value is used for…
MVI c, 0fhMVI a, 0chlxi h,0074hloop: cmp cjz timemov m,cinx hdcr cjmp looptime:mvi c,0fhlxi h,2802hloop1: cmp cjz time1mov m,cinx hdcr cjmp loop1time: hlt
10.23.33.234 and MASK 255.255.255.224 10.14.64.22 and MASK 255.255.248.0 10.54.21.15 and MASK 255.252.0.0 172.16.33.234 and MASK 255.255.255.224 172.18.64.22 and MASK 255.255.248.0 172.19.21.15 and MASK 255.255.252.0 192.168.33.234 and MASK 255.255.255.224 192.168.64.22 and MASK 255.255.255.248 192.168.21.15 and MASK 255.255.255.252 Please illustrate the above using router, switches and hosts