In Qbasic language, arranging the following numbers from smallest to greatest in one dimensional matrix by: [55 34 28 22 99 89] FORI-1 TOS FOR J=1 TOS IF BI + 1)> B() THEN S SWAP BJ - 1), BJ) 5 NEXTJ FORI-1 TO 6 FOR J=1 TO6 IF B(I + 1)< B(J) THENS SWAP B(J + 1), BJ) 5 NEXT J FORI=1 TO6 FOR J=1 T06 FORI=1 TOS IF BI + 1)> B() THEN S SWAP BJ + 1), BJ) S NEXT J FOR J=1 TOS IF BJ + 1)< B(J) THEN S SWAP BI + 1), B() 5 NEXT I

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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In Obasic language, arranging the following numbers from smallest to greatest in
one dimensional matrix by:
[55 34 28 22 99 891
FORI-1 TO6
FOR J=1 TO 6
IF BJ + 1)< B(J) THENS
SWAP BJ + 1), BJ)
5 NEXTJ
FORI-1 TOS
FOR J=1 T05
IF BI + 1)> B() THEN S
SWAP B(J• 1). B(J)
5 NEXT J
FORI=1 TO6
FOR J=1 106
IF BI + 1)> B() THEN S
SWAP B(J + 1), B(J)
5 NEXTJ
FORI=1 TOS
FOR J=1 TOS
IF BJ + 1)<B(J) THEN 5
SWAP BI+ 1), B(I)
SNEXTJ
Transcribed Image Text:In Obasic language, arranging the following numbers from smallest to greatest in one dimensional matrix by: [55 34 28 22 99 891 FORI-1 TO6 FOR J=1 TO 6 IF BJ + 1)< B(J) THENS SWAP BJ + 1), BJ) 5 NEXTJ FORI-1 TOS FOR J=1 T05 IF BI + 1)> B() THEN S SWAP B(J• 1). B(J) 5 NEXT J FORI=1 TO6 FOR J=1 106 IF BI + 1)> B() THEN S SWAP B(J + 1), B(J) 5 NEXTJ FORI=1 TOS FOR J=1 TOS IF BJ + 1)<B(J) THEN 5 SWAP BI+ 1), B(I) SNEXTJ
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