ᴜꜱɪɴɢ ᴍᴀᴛʟᴀʙ ᴄʀᴇᴀᴛᴇ ᴛʜᴇ ꜰᴏʟʟᴏᴡɪɴɢ ʀᴏᴡ ᴠᴇᴄᴛᴏʀꜱ ꜱ ᴀɴᴅ ʀ. ꜱ  =  1   3   4    7   11 ʀ  = 10   1   9   -8   17 ᴄʀᴇᴀᴛᴇ ᴀɴᴏᴛʜᴇʀ ᴠᴇᴄᴛᴏʀ ᴄᴀʟʟᴇᴅ ꜱQʀᴅɪꜰꜰ ᴛʜᴀᴛ ᴄᴏɴᴛᴀɪɴꜱ ᴛʜᴇ ꜱQᴜᴀʀᴇ ʀᴏᴏᴛ ᴏꜰ ᴛʜᴇ ᴅɪꜰꜰᴇʀᴇɴᴄᴇ ᴏꜰ ꜱ ᴍɪɴᴜꜱ ʀ ᴄᴏʀʀᴇꜱᴘᴏɴᴅɪɴɢ ᴇʟᴇᴍᴇɴᴛꜱ. ʜᴏᴡᴇᴠᴇʀ, ɪꜰ ʀᴏᴏᴛꜱ ʀᴇꜱᴜʟᴛ ɪɴ ɪᴍᴀɢɪɴᴀʀʏ ɴᴜᴍʙᴇʀꜱ, ʀᴇᴘʟᴀᴄᴇ ᴛʜᴏꜱᴇ ɴᴜᴍʙᴇʀꜱ ᴡɪᴛʜ ɴᴀʀɴ ᴡʜɪᴄʜ ꜱᴛᴀɴᴅꜱ ꜰᴏʀ ɴᴏᴛ ᴀ ʀᴇᴀʟ ɴᴜᴍʙᴇʀ. ɴᴏᴛᴇ ᴛʜᴀᴛ ɴᴀʀɴ ɪꜱ ɴᴏᴛ ᴀ ᴄᴏɴꜱᴛᴀɴᴛ ɪɴ ᴍᴀᴛʟᴀʙ. ᴛʜᴇ ᴇxᴘᴇᴄᴛᴇᴅ ᴏᴜᴛᴘᴜᴛ ᴏꜰ ᴛʜɪꜱ ᴛᴀꜱᴋ ɪꜱ: ꜱQʀᴅɪꜰꜰ = ɴᴀʀɴ   1.4142   ɴᴀʀɴ   3.8729   ɴᴀʀɴ

For the three code words:C1= [110110011], C2= [100101100]andC3=[111100101]. Determine the possible error detection and the possible error .correction

I believe my answer is wrong, please check and tell me what is correct!

What is the normalized expression for the floating binary value 00000001001111011? The normalized expression is 1.001111011 x 2-6 O b. The normalized expression is 1.001111011 x 2-8. O. The normalized expression is 1.001111011 x 2-7. O d. The normalized expression is 1.001111011 x 2-5.

Huffman Code Character 00 A 11 E 010 T 0110 C 0111 L 1000 S 1011 R 10010 O 10011 I 101000 N 101001 F 101010 H 101011 D decipher 0110101010100101011111000

What are the hexadecimal values for the following values by using the little endian byte order? Ox 00 AA 00 BB 00 CC O a. 0x00 AA 00 BB 00 CC Ob. OxAA 00 BB 00 CC 00 O c. OxAA BB CC 00 00 00 O d. 0xCC BB AA Oe. 0xCC 00 BB 00 AA 00

⦁    Compute the checksum for the following data.01101110 11011001 11101101 00101100 11110100 01011010 01111000 11000011 11111000 00001100

ʙᴇʟᴏᴡ ɪꜱ ʜᴏᴡ ᴛʜᴇ ꜰɪʟᴇɴᴀᴍᴇꜱ ᴀʀᴇ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅᴀʀʀᴀʏ[] = [50, 11, 33, 21, 40, 50, 40, 40, 21]ʙᴇʟᴏᴡ ɪꜱ ᴛʜᴇ ᴇxᴘᴇᴄᴛᴇᴅ ᴏᴜᴛᴘᴜᴛ ᴀꜰᴛᴇʀ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴᴀʀʀᴀʏ[] = [11, 21, 33, 40, 50] ʏᴏᴜ ᴄᴀɴɴᴏᴛ ᴜꜱᴇ ʟɪʙʀᴀʀʏ ꜰᴜɴᴄᴛɪᴏɴꜱ ᴛᴏ ᴇɪᴛʜᴇʀꜱᴏʀᴛ ᴀɴᴅ ᴏʀ ᴘᴇʀꜰᴏʀᴍ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴏᴘᴇʀᴀᴛɪᴏɴ. ɪɴ ᴊᴀᴠᴀ ɪᴍᴘʟᴇᴍᴇɴᴛ ᴛʜᴇ ꜰᴜɴᴄᴛɪᴏɴ ɪɴ ꜱᴜᴄʜ ᴀ ᴡᴀʏ ᴛʜᴀᴛ ʏᴏᴜʀ ꜱᴏʟᴜᴛɪᴏɴ ꜱᴏʟᴠᴇꜱᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡɪᴛʜ ᴏ(ɴ ʟᴏɢ2(ɴ)) ᴛɪᴍᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ ᴏᴠᴇʀᴀʟʟ ʙᴜᴛ ꜱᴛᴀʏꜱ ᴡɪᴛʜɪɴ ᴛʜᴇ ʙᴏᴜɴᴅꜱ ᴏꜰ ᴏ(1) ꜱᴘᴀᴄᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ. ʜᴇʀᴇ, ɴ ɪꜱ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴛʜᴇ ʟɪꜱᴛ ᴏꜰɪɴᴘᴜᴛ ɪɴᴛᴇɢᴇʀꜱ (ᴀʀʀᴀʏ). ɪ ʙᴇʟɪᴇᴠᴇ ᴛʜᴇ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ᴛʜᴀᴛ ᴄᴀɴ ʙᴇ ᴜꜱᴇᴅ ʜᴇʀᴇ ɪꜱ Qᴜɪᴄᴋ ꜱᴏʀᴛ. ᴘʟᴇᴀꜱᴇ ꜱᴛᴀᴛᴇ ᴀꜱ ᴄᴏᴅᴇ ᴄᴏᴍᴍᴇɴᴛ ᴡʜɪᴄʜ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ʏᴏᴜ ᴀʀᴇ ᴜꜱɪɴɢ, ꜱᴏʀᴛ ᴛʜᴇ ᴀʀʀᴀʏ ᴡɪᴛʜ ᴛʜᴀᴛ ᴀʟɢᴏʀɪᴛʜᴍ ᴀɴᴅ ꜱᴏʟᴠᴇ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘʀᴏʙʟᴇᴍ ᴛʜᴇʀᴇᴀꜰᴛᴇʀ. ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘᴀʀᴛ ᴏꜰ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ɪɴ ɪᴛꜱᴇʟꜰ ᴍᴜꜱᴛ ᴀᴅʜᴇʀᴇ ᴛᴏ ᴏ(ɴ) ᴛɪᴍᴇ ᴀɴᴅ ᴏ(1) ꜱᴘᴀᴄᴇ ʙᴏᴜɴᴅꜱ ᴄᴏᴍᴍᴇɴᴛ ᴇᴠᴇʀʏ ʟɪɴᴇ ᴏꜰ ᴄᴏᴅᴇᴘʟᴇᴀꜱᴇ ᴡʀɪᴛᴇ ᴛᴇꜱᴛ-ᴄᴀꜱᴇꜱ ᴏɴ ʜᴏᴡ ᴏɴᴇ ᴡᴏᴜʟᴅᴛᴇꜱᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ᴀꜱꜱᴜᴍᴘᴛɪᴏɴꜱ ᴀɴᴅ ʜᴇɴᴄᴇ ᴛʜᴇ ᴄᴏᴅᴇ ᴇxᴀᴍᴘʟᴇ ᴏꜰ ᴄᴀꜱᴇꜱ ᴛᴏ ʙᴇ ᴛᴇꜱᴛᴇᴅ ꜰᴏʀ ᴀʀᴇ ʟɪᴋᴇ : ᴡʜᴀᴛ ɪꜰ ᴛʜᴇ ᴀʀʀᴀʏ ɪɴᴘᴜᴛ ᴡʜɪᴄʜɪꜱ…

python

ʙᴇʟᴏᴡ ɪꜱ ʜᴏᴡ ᴛʜᴇ ꜰɪʟᴇɴᴀᴍᴇꜱ ᴀʀᴇ ʀᴇᴘʀᴇꜱᴇɴᴛᴇᴅᴀʀʀᴀʏ[] = [50, 11, 33, 21, 40, 50, 40, 40, 21]ʙᴇʟᴏᴡ ɪꜱ ᴛʜᴇ ᴇxᴘᴇᴄᴛᴇᴅ ᴏᴜᴛᴘᴜᴛ ᴀꜰᴛᴇʀ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴᴀʀʀᴀʏ[] = [11, 21, 33, 40, 50] ʏᴏᴜ ᴄᴀɴɴᴏᴛ ᴜꜱᴇ ʟɪʙʀᴀʀʏ ꜰᴜɴᴄᴛɪᴏɴꜱ ᴛᴏ ᴇɪᴛʜᴇʀꜱᴏʀᴛ ᴀɴᴅ ᴏʀ ᴘᴇʀꜰᴏʀᴍ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴏᴘᴇʀᴀᴛɪᴏɴ. ɪɴ ᴊᴀᴠᴀ ɪᴍᴘʟᴇᴍᴇɴᴛ ᴛʜᴇ ꜰᴜɴᴄᴛɪᴏɴ ɪɴ ꜱᴜᴄʜ ᴀ ᴡᴀʏ ᴛʜᴀᴛ ʏᴏᴜʀ ꜱᴏʟᴜᴛɪᴏɴ ꜱᴏʟᴠᴇꜱᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡɪᴛʜ ᴏ(ɴ ʟᴏɢ2(ɴ)) ᴛɪᴍᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ ᴏᴠᴇʀᴀʟʟ ʙᴜᴛ ꜱᴛᴀʏꜱ ᴡɪᴛʜɪɴ ᴛʜᴇ ʙᴏᴜɴᴅꜱ ᴏꜰ ᴏ(1) ꜱᴘᴀᴄᴇ ᴄᴏᴍᴘʟᴇxɪᴛʏ. ʜᴇʀᴇ, ɴ ɪꜱ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏꜰ ᴛʜᴇ ʟɪꜱᴛ ᴏꜰɪɴᴘᴜᴛ ɪɴᴛᴇɢᴇʀꜱ (ᴀʀʀᴀʏ). ɪ ʙᴇʟɪᴇᴠᴇ ᴛʜᴇ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ᴛʜᴀᴛ ᴄᴀɴ ʙᴇ ᴜꜱᴇᴅ ʜᴇʀᴇ ɪꜱ Qᴜɪᴄᴋ ꜱᴏʀᴛ. ᴘʟᴇᴀꜱᴇ ꜱᴛᴀᴛᴇ ᴀꜱ ᴄᴏᴅᴇ ᴄᴏᴍᴍᴇɴᴛ ᴡʜɪᴄʜ ꜱᴏʀᴛɪɴɢ ʀᴏᴜᴛɪɴᴇ ʏᴏᴜ ᴀʀᴇ ᴜꜱɪɴɢ, ꜱᴏʀᴛ ᴛʜᴇ ᴀʀʀᴀʏ ᴡɪᴛʜ ᴛʜᴀᴛ ᴀʟɢᴏʀɪᴛʜᴍ ᴀɴᴅ ꜱᴏʟᴠᴇ ᴛʜᴇ ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘʀᴏʙʟᴇᴍ ᴛʜᴇʀᴇᴀꜰᴛᴇʀ. ᴅᴇ-ᴅᴜᴘʟɪᴄᴀᴛɪᴏɴ ᴘᴀʀᴛ ᴏꜰ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ɪɴ ɪᴛꜱᴇʟꜰ ᴍᴜꜱᴛ ᴀᴅʜᴇʀᴇ ᴛᴏ ᴏ(ɴ) ᴛɪᴍᴇ ᴀɴᴅ ᴏ(1) ꜱᴘᴀᴄᴇ ʙᴏᴜɴᴅꜱ ᴄᴏᴍᴍᴇɴᴛ ᴇᴠᴇʀʏ ʟɪɴᴇ ᴏꜰ ᴄᴏᴅᴇᴘʟᴇᴀꜱᴇ ᴡʀɪᴛᴇ ᴛᴇꜱᴛ-ᴄᴀꜱᴇꜱ ᴏɴ ʜᴏᴡ ᴏɴᴇ ᴡᴏᴜʟᴅᴛᴇꜱᴛ ᴛʜᴇ ꜱᴏʟᴜᴛɪᴏɴ ᴀꜱꜱᴜᴍᴘᴛɪᴏɴꜱ ᴀɴᴅ ʜᴇɴᴄᴇ ᴛʜᴇ ᴄᴏᴅᴇ ᴇxᴀᴍᴘʟᴇ ᴏꜰ ᴄᴀꜱᴇꜱ ᴛᴏ ʙᴇ ᴛᴇꜱᴛᴇᴅ ꜰᴏʀ ᴀʀᴇ ʟɪᴋᴇ : ᴡʜᴀᴛ ɪꜰ ᴛʜᴇ ᴀʀʀᴀʏ ɪɴᴘᴜᴛ ᴡʜɪᴄʜɪꜱ…