Matlab code for the periodic signal. Show running code and output
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Matlab code for the periodic signal.
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- 192.67.200.1 in Hexadecimal is: FF:23:C7:01 CA:D3:C8:01 C0:43:C8:01 C0:67:C8:01 C0:43:C8:1012. 45.67 * 3.6713. A5.912 * 3.A1214. 1A.716 * 9.C16Draw a Frequency polygon, histogram and a cumulative frequency distribution for the following data: Frequency Class 7.1 to 7.3 7.4 to 7.6 7.7 to 7.9 8.0 to 8.2 8.3 to 8.5 8.6 to 8.8 8.9 to 9.1 3 5 9 14 11 6 2
- 5.6 MatQUESTION 5 0.5 The scientific notation of 278923.889923 is O a. 2.78923889923E6 O b. 2.78923889923E+5 Oc. 2.78923889923E-6 O d. 2.78923889923E-51165:0100 B84518 1165:0103 BB3478 1165:0106 01C3 1165:0108 48 1165:0109 4B 1165:010A 891E0001 1165 : 01 ΘΕ CD20 AX, 1845 BX, 7834 BX, AX AX MOV MOV ADD DEC DEC BX MOV (01001, BX INT 20 You have used DEBUG to unassemble the above code. The entire program is bytes long.
- * dotted-decimal notation of the address .(10000001.00001011.00001011.11101111) 129.11.11.239 O 129.111.11.240 129.211.33.255 O 129.21.72.100 O6-In the representation Complement to 1, we the number when Code the absolute value thus is negative: A = True B- False 7- The result in base 10' A - 44.25 B-42015 C- 51. 13 45. 75 8- The result AT 40. B + 61 q of the transformation (101001.10) 2 usi o the transformation (52) a in bose 10 is: C² 42 D-41 WWW 30 O1,(2504)_8 to hexadecimal. 2,(413.51)_6 to base 9. 3, decimal numbers A = 5, B = 16, Why A = 5 = 00010 and B = 16 = 010000?
- Read the following series: 9.2 8.9 8.6 8.3 8 What is the sum and multiplication of the above series? O a. 45 1345.5 O b. 35 46756.7 O c. 43 46756.7 O d. 43 1200.54 data Ox00000000 c byte 1,2.3,4,5 # occupy_(1)memory units(B) s half 1,2,3,4,5 occupy Cs: asciz "hello world" occupy memory units W. word 1.2.3,4,5 occupy memory units what is the memory map of the data segment? the starting address of the data segment is address of the variable "cs" is memory units (5) the the address of the variable "w is the address of the variable "c" is (6) The data segment occupies memory units in total, the the address of the variable "s" is the theSuppose a color monitor has a 1920x1080 frame size and uses 8 bits for each of the three primary colors per pixel. What is the minimum size in bytes of the frame buffer to store a frame? Enter only the final answer with no commas.