num1 = 35; is a valid assignment as integer 35 is be...

Question

a database with multiple tables from attributes as shown above that are in 3NF, showing PK, non-key attributes, and FK for each table? Assume the tables are already in 1NF. [Hint: 3 tables will result after deducing 1NF -> 2NF -> 3NF]

Answer 1

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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Answer 2

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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Answer 3

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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Answer 4

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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See solution

Check out sample textbook solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 8

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Answer 9

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Answer 10

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Answer 11

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Answer 12

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Answer 13

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Answer 14

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 15

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 16

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

Answer 17

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

Answer 18

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 19

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 20

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 21

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 22

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Answer 23

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Answer 24

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Answer 25

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Answer 26

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Answer 27

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Answer 28

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 29

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 30

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Answer 31

Expert Solution & Answer

Answer 32

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Answer 33

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Answer 34

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Answer 35

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Answer 36

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Answer 37

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Answer 38

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Answer 39

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Answer 40

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Answer 41

Ch. 2 - 1. Mark the following statements as true or...Ch. 2 - Prob. 2MC Ch. 2 - Which of the following is not a reserved word in...Ch. 2 - Prob. 4SA Ch. 2 - 5. Are the identifiers quizNo1 and quiznol the...Ch. 2 - 6. Evaluate the following expressions. (3,...Ch. 2 - If int x = 10;, int y = 7;, double z = 4.5;, and...Ch. 2 - Prob. 8CP Ch. 2 - 9. Suppose that x, y, z, and w are int variables....Ch. 2 - Prob. 10SA

num1 = 35; is a valid assignment as integer 35 is be...

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Explanation of Solution

Chapter 2 Solutions