Concepts of Database Management, Loose-Leaf Version
9th Edition
ISBN: 9781337685849
Author: STARKS, Joy L.; Pratt, Philip J.; Last, Mary Z.
Publisher: Cengage Learning
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Chapter 5, Problem 10RQ
To determine
To list the functional dependencies that exist in the Student table and to write the assumptions that support these dependencies.
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Chapter 5 Solutions
Concepts of Database Management, Loose-Leaf Version
Ch. 5 - Prob. 1RQCh. 5 - Prob. 2RQCh. 5 - Prob. 3RQCh. 5 - Prob. 4RQCh. 5 - Prob. 5RQCh. 5 - Prob. 6RQCh. 5 - Prob. 7RQCh. 5 - Prob. 8RQCh. 5 - Prob. 9RQCh. 5 - Prob. 10RQ
Ch. 5 - Prob. 11RQCh. 5 - Prob. 12RQCh. 5 - Prob. 13RQCh. 5 - Prob. 14RQCh. 5 - Prob. 15RQCh. 5 - Prob. 1BCECh. 5 - Prob. 2BCECh. 5 - Prob. 3BCECh. 5 - Prob. 4BCECh. 5 - Prob. 5BCECh. 5 - Prob. 6BCECh. 5 - Prob. 1CATCCh. 5 - Prob. 2CATCCh. 5 - Prob. 3CATCCh. 5 - Prob. 4CATCCh. 5 - Prob. 1SPTCCh. 5 - Prob. 2SPTCCh. 5 - Prob. 3SPTCCh. 5 - Prob. 4SPTCCh. 5 - Prob. 5SPTC
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- Consider the following three machine instructions, which appear in memory starting at the address 0x0000000080001270. Address (in hex) Contents (in hex) 0x0000000080001270 0x94000018 0x00000000800012D0 0x94000A53 0x00000000800012D4 0xD61F03C0 "Disassemble” the instructions; that is, give an assembly language program segment that would be translated into the given machine language. Use numeric rather than symbolic register names. Handle branches and jumps specially; where you would normally have a label, provide instead a hexadecimal byte address.arrow_forward8.7 An ideal germanium pn junction diode has the following parameters: Na = 4 × 10¹5 cm³, Na = 2 × 10¹7 cm³, D₂ = 48 cm²/s, D₁ = 90 cm²/s, Tpo = T = 2 × 10-6 s, and A = 104 cm². Determine the diode current for (a) a forward-bias voltage of 0.25 V and (b) a reverse-biased voltage of 0.25 V.arrow_forward7.20 (a) The peak electric field in a reverse-biased silicon pn junction is |E|= 3 × 105 V/cm. The doping concentrations are N = 4 × 10¹5 cm³ and N. = 4 × 107 cm³. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for N₁ = 4 × 10 16 cm³ and N₁ = 4 × 10¹7 cm³. (c) Repeat part (a) for Nd = N₁ = 4 × 10¹7 cm³.arrow_forward
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