EBK DATABASE CONCEPTS
7th Edition
ISBN: 9780133544886
Author: AUER
Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)
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Expert Solution & Answer
Chapter 1, Problem 1.7RQ
Explanation of Solution
ID Column:
- ID Column is a field which is used in databases having two or more themes and is divided according to the respective themes.
- It is to provide uniqueness and avoid duplicity in the
database and is quite similar to primary key but it differs from primary key as it is managed by the server and has the property of auto-incrementing which cannot be modified.
Example:
Cust_pet table:
| sno | Cust_fname | Cust_lname | Cust_phone | Pet_type | Pet_name |
| 1 | Green | Robin | 230-245-2546 | Husky | Shushi |
| 2 | Jackson | Charles | 213-235-2356 | Malamute | Leo |
| 3 | Pearson | Jeffery | 145-213-7658 | Lebrador | Rocky |
| 4 | Jackson | ...
Expert Solution & Answer
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Students have asked these similar questions
Consider the data path below for a single cycle 32-bits MIPS processor
Assume that we are executing the following instruction
lw St1, 48($10)
What is the value of the element pointed by arrow number 1 by in hexadecimal?
Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below
Data Memory
Select one
a. Ox00001724
Ob. 0x00002DE4
Oc 0x000016F4
Od 0x00001720
Oe. 0x00002DEO
Clear my choice
Rea
"
Address
Content
0x000016EC
0x0000ABOD
0x000016F0
0x00A01245
0x000016F4
0x00001A42
0x000171C
0x0008124F
0x0001720
0x00021345
0x0001724
0x000067AB
ALU
Ox0001734
0x0000AB35
0x0001738
0x0000FA72
0x0000ABOC
ALS
0x000174C
Register File and PC
$to = 0x000016FO
$t1 = 0x000016F4
PC = 0x0000148A0
(Before executing LW)
Consider the data path below for a single cycle 32-bits MIPS processor
Assume that we are executing the following instruction
SLL S10, $3,4
What is the value of the element pointed by arrow number 1 by in hexadecimal or binary?
Note that the PC and the content of registers are found in bottom left of the figure below
Address
Select one
Oa. Ox00000000
Ob. Ob 10011'
Oc 0x0000AF00
® d. 06/01000
Oe. None of the options
Of Ox000000BA
LEY
Content
0x000016EC
Dx0000ABOD
Data Memory 0x000016F0
0x00A01245
0x000016F4
0x00001A42
Ox000171C
0x0008124
0x0001720
0x00021345
0x0001724
0x000067AB
0x0001734
0x0001738
0x0000A835
0x0000FA72
0x000174C
0x0000AC
$63 0x000000BA
PC = 0x000FAC04
Register File and PC (Before executing SLL)
Consider the data path below for a single cycle 32-bits MIPS processor
Assume that we are executing the following instruction
lw $10, 48(S1)
What is the value of the element pointed by arrow number 1 by in hexadecimal?
Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below
Select one
a. None of the options
Ob 0x00014C64
Oc 0x00014CC4
Od 0x00014BA4
Oe. Ox14C1C
Of 0x00014C60
Og 0x00014CCO
D
Data Memory
Address
Content
0x000016EC
0x0000ABOD
0x000016F0
0x00A01245
0x000016F4
0x00001A42
0x000171C
0x0008124F
0x0001720
0x00021345
0x0001724
0x000067AB
0x0001734
0x0000AB35
0x0001738
0x0000FA72
0x000174C
0x0000ABOC
ALU
Register File and PC
Sto Ox000016F0
St1 = 0x000016F4
PC = 0x00014BA0
(Before executing LW).
Chapter 1 Solutions
EBK DATABASE CONCEPTS
Ch. 1 - Why is the study of database technology important?Ch. 1 - Prob. 1.2RQCh. 1 - Describe the purpose of a database.Ch. 1 - Prob. 1.4RQCh. 1 - Prob. 1.5RQCh. 1 - Prob. 1.6RQCh. 1 - Prob. 1.7RQCh. 1 - Break the list in Figure 1-34 into two tables,...Ch. 1 - Show how the tables you created for question 1.11...Ch. 1 - Prob. 1.10RQ
Ch. 1 - Prob. 1.11RQCh. 1 - Break the list in Figure 1-35 into tables, each...Ch. 1 - Describe in your own words and illustrate with...Ch. 1 - Prob. 1.14RQCh. 1 - Define the term database.Ch. 1 - Prob. 1.17RQCh. 1 - List the components of a database.Ch. 1 - Prob. 1.19RQCh. 1 - Prob. 1.20RQCh. 1 - Prob. 1.21RQCh. 1 - What is the purpose of a DBMS?Ch. 1 - List the specific functions of a DBMS.Ch. 1 - Define the term referential integrity constraint....Ch. 1 - Prob. 1.25RQCh. 1 - List the functions of a database application.Ch. 1 - Prob. 1.27RQCh. 1 - Prob. 1.28RQCh. 1 - Prob. 1.29RQCh. 1 - Prob. 1.30RQCh. 1 - Prob. 1.31ECh. 1 - Prob. 1.32ECh. 1 - Prob. 1.33E
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Similar questions
- Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction lw $11, 48($10) What is the value of the element pointed by arrow number 1 by in hexadecimal or binary? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Data Memory Select one a. Ob 01001' O b. 0x000016F0 Oc 06'1000' Od 06/00000 Oe. None of the options Of. 0x000016F4 AM RegDe ALU Address Content 0x000016EC 0x0000ABOD 0x000016F0 0x00A01245 0x000016F4 0x00001A42 0x000171C Ox0008124F 0x0001720 Ox00021345 0x0001724 0x000067AB ALU AL 0x0001734 Ox0000AB35 0x0001738 0x0000FA72 0x000174C 0x0000ABOC Register File and PC Sto Ox000016F0 $t1 = 0x000016F4 PC 0x000014BA0 (Before executing LW)arrow_forwardConsider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction ADD $3, $50, $2 What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Select one Oa. OxFFFFFF30 b. None of the options Oc 0xFFFF9820 Od 0x00009620 Oe. 0xFFFFFF48 Address 0x000016EC Content 0x0000ABOD Data Memory 0x000016FO 0x00401245 0x000016F4 0x00001A42 0x0001710 0x0008124F 0x0001720 0x00021345 0x0001724 0x000067AB 11 81 0x0001734 0x0000AB35 0x0001738 0x0000FA72 0x0000ABOC Register File and PC 0x000174C $50 = 0x0000AF00 $12 = 0x000000BA $x3 = 0x00000001 PC-0x000FAC04 (Before executing ADD)arrow_forwardConsider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction ADD $3, $50, $s2 What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers are found in bottom left of the figure below Select one a. Ox000FACOB Ob. 0x000F4428 Oc None of the options Od 0x00120C84 O e. 0x000FAC04 Address Ox000016EC Data Memory 0x000016F0 Content 0x0000ABOD C0001245 Ox000016F4 0x00001A42 0x000171C C0008124F 0x0001720 0x00021345 0x0001724 0x000067AB 0x0001734 0x0000AB35 0x0001738 0x0000FA72 0x000174C ОКООООАВОС Register File and PC $50-0x0000AF00 Ss2=0x000000BA Ss3 0x00000001 PC-Ox000FAC04 (Before executing ADD)arrow_forward
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