number = 30<br>if number % 2 == 0:<br>print（）
A. 30 is multiple of 3
B. 30 is even
C. 30 is even 30 is multiple of 3
D. 程序出错
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A.白细胞趋化 B.白细胞沿内皮细胞滚动 C.甴细胞黏附和聚集 D.白细胞游走 E.白细胞产生氧自由基
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LTB4可引起（）
A.白细胞趋化 B.白细胞沿内皮细胞滚动 C.甴细胞黏附和聚集 D.白细胞游走 E.白细胞产生氧自由基
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number = 30<br>if number % 2 == 0:<br>print（）
A.30 is multiple of 3 B.30 is even C.30 is even 30 is multiple of 3 D.程序出错
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Examine the following command: CREATE TABLE (prod_id number(4), Prod_name varchar2 (20), Category_id number(30), Quantity_on_hand number (3) INVISIBLE); Which three statements are true about using an invisible column in the PRODUCTS table?（）
A.The %ROWTYPE attribute declarations in PL/SQL to access a row will not display the invisible column in the output. B.The DESCRIBE commands in SQL *Plus will not display the invisible column in the output. C.Referential integrity constraint cannot be set on the invisible column. D.The invisible column cannot be made visible and can only be marked as unused. E.A primary key constraint can be added on the invisible column.
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Examine the following command: CREATE TABLE (prod_id number(4), Prod_name varchar2 (20), Category_id number(30), Quantity_on_hand number (3) INVISIBLE); Which three statements are true about using an invisible column in the PRODUCTS table?（）
A.The %ROWTYPE attribute declarations in PL/SQL to access a row will not display the invisible column in the output. B.The DESCRIBE commands in SQL *Plus will not display the invisible column in the output. C.Referential integrity constraint cannot be set on the invisible column. D.The invisible column cannot be made visible and can only be marked as unused. E.primary key constraint can be added on the invisible column.
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Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. EMPLOYEES NOT NULL, Primary EMPLOYEE_ID NUMBER Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, DEPARTMENT_ID NUMBER Primary Key VARCHAR2 DEPARTMENT_NAME |30| References MGR_ID column MGR_ID NUMBER of the EMPLOYEES table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER For which situation would you use a nonequijoin query?（）
A.To find the tax percentage for each of the employees. B.To list the name, job id, and manager name for all the employees. C.To find the name, salary, and department name of employees who are not working with Smith. D.To find the number of employees working for the Administrative department and earning less then 4000. E.To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.
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30—I _________ his telephone number—I have his number, but I _________ to bring my phone book.”
A.forget, forget B.forgot, forgot C.forgot, forget D.forget, forgot
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For a 30 minutes talk, what is the appropriate number of slides?
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Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. EMPLOYEES NOT NULL, EMPLOYEE_ID NUMBER Primary Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, Primary DEPARTMENT_ID NUMBER Key VARCHAR2 DEPARTMENT_NAME (30) References NGR_ID MGR_ID NUMBER column of the EMPLOYEES table Foreign key to LOCATION_ID NUMBER LOCATION_ID column of the LOCATIONS table LOCATIONS NOT NULL, Primary LOCATION_ID NUMBER Key VARCHAR2 CITY |30) Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?（）
A.SELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary > 10000; B.SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary > 10000; C.SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary > 10000; D.SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary > 10000; E.SELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary > 10000;
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Only a small number of students able to pass the exam.—Yes．The number of the students no more than 20.
A.are; is B.is; is C.is; are D.are; are;
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Only a small number of students able to pass the exam.— Yes. The number of them no more than 20.
结施11中,楼梯斜板LTB-1的宽度为mm
Examine the structures of the EMPLOYEES and TAX tables. EMPLOYEES EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2 (30) JOB_ID VARCHAR2 (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT _ID column of the DEPARTMENT table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER Percentage tax for given salary range You need to find the percentage tax applicable for each employee. Which SQL statement would you use?（）
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Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith? （）
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建立文件add.py，写入以下代码： print（“the first number is:”） print（30+50） print（“the second number is:”） 30*50保存并运行脚本文件，输出结果为（）
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Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key LAST_NAME VARCNAR2(30) FIRST_NAME VARCNAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER You need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task?（）
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