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evablogger [386]
2 years ago
14

Write a SQL query to find the population of the planet named 'Caprica' -- 10 points Find the first name, last name, and age of p

eople from bsg_people whose last name is not 'Adama' -- 10 points Find the name and population of the planets with a population larger than 2,600,000,000 -- 10 points Find the first name, last name, and age of people from bsg_people whose age is NULL -- 12 points

Computers and Technology
2 answers:
Artemon [7]2 years ago
6 0

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Please find the attached question.

Answer:

a) SQL Query:

SELECT population

FROM bsg_planets

WHERE name='Caprica';

b) SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE lname!='Adama';

c) SQL Query:

SELECT name, population

FROM bsg_planets

WHERE population > 2600000000

d) SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE age IS NULL;

Explanation:

a) Write a SQL query to find the population of the planet named 'Caprica'

Syntax:

SELECT  Column

FROM TableName

WHERE Condition;

For the given case,

Column = population

TableName  = bsg_planets

Condition = name='Caprica'

SQL Query:

SELECT population

FROM bsg_planets

WHERE name='Caprica';

Therefore, the above SQL query finds the population of the planet named 'Caprica' from the table bsg_planets.

b) Find the first name, last name, and age of people from bsg_people whose last name is not 'Adama'

Syntax:

SELECT  Column1, Column2, Column3

FROM TableName

WHERE Condition;

For the given case,

Column1 = fname

Column2 = lname

Column3 = age

TableName  = bsg_people

Condition = lname!='Adama'

SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE lname!='Adama';

Therefore, the above SQL query finds the first name, last name and age of people whose last name is not 'Adama' from the table bsg_people.

c) Find the name and population of the planets with a population larger than 2,600,000,000

Syntax:

SELECT  Column1, Column2

FROM TableName

WHERE Condition;

For the given case,

Column1 = name

Column2 = population

TableName  = bsg_planets

Condition = population > 2600000000

SQL Query:

SELECT name, population

FROM bsg_planets

WHERE population > 2600000000

Therefore, the above SQL query finds the name and population of the planets with a population larger than 2,600,000,000 from the table bsg_planets.

d) Find the first name, last name, and age of people from bsg_people whose age is NULL

Syntax:

SELECT  Column1, Column2, Column3

FROM TableName

WHERE Condition;

For the given case,

Column1 = fname

Column2 = lname

Column3 = age

TableName  = bsg_people

Condition = age IS NULL

SQL Query:

SELECT fname, lname, age

FROM bsg_people

WHERE age IS NULL;

Therefore, the above SQL query finds the first name, last name and age of people whose age is NULL from the table bsg_people.

Delicious77 [7]2 years ago
6 0

Answer:

SQL Query : select population from bsg_planets where name='Caprica';

SQL Query : select fname,lname,age from bsg_people where lname!='Adama';

SQL Query : select name,population from bsg_planets where population > 2600000000;

SQL Query : select fname,lname,age from bsg_people where age='NULL';

Explanation:

using MySQL Workbench.

SQL Query : select population from bsg_planets where name='Caprica';

Because this sql query will return the population from bsg_planets where the name is Caprica;

SQL Query : select fname,lname,age from bsg_people where lname!='Adama';

Because this SQL query will return the first name ,last name and age from the bsg_people where last name is not adama.

SQL Query : select name,population from bsg_planets where population > 2600000000;

Because this SQL query will return the name and population where population is greater than 2600000000

SQL Query : select fname,lname,age from bsg_people where age='NULL';

Because this SQL query will return the first name ,last name and age from the bsg_people where age is NULL.

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Answer:

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           System.out.println("Too Few");

       }

       else if(numberOfCycles>4){

           System.out.println("Too many");

       }

       else

           for(int i = 1; i<=numberOfCycles; i++){

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Explanation:

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Write a program to input value of three sides, to check triangle is triangle is possible to form of not​
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Answer:

Explanation:

import java.util.Scanner;

public class KboatTriangleAngle

{

  public static void main(String args[]) {

      Scanner in = new Scanner(System.in);

      System.out.print("Enter first angle: ");

      int a1 = in.nextInt();

      System.out.print("Enter second angle: ");

      int a2 = in.nextInt();

      System.out.print("Enter third angle: ");

      int a3 = in.nextInt();

      int angleSum = a1 + a2 + a3;

       

      if (angleSum == 180 && a1 > 0 && a2 > 0 && a3 > 0) {

          if (a1 < 90 && a2 < 90 && a3 < 90) {

              System.out.println("Acute-angled Triangle");

          }

          else if (a1 == 90 || a2 == 90 || a3 == 90) {

              System.out.println("Right-angled Triangle");

          }

          else {

              System.out.println("Obtuse-angled Triangle");

          }

      }

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          System.out.println("Triangle not possible");

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}

OUTPUT:

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Answer:

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Explanation:

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Describe a strategy for avoiding nested conditionals. Give your own example of a nested conditional that can be modified to beco
Alex787 [66]

Answer:

One of the strategies to avoid nested conditional is to use logical expressions such as the use of AND & operator.

One strategy is to use an  interface class with a method. That method can be created to be used for a common functionality or purpose. This is also called strategy design pattern. You can move the chunk of conditional statement to that method. Then each class can implement that interface class and use that shared method according to their own required task by creating objects of sub classes and call that common method for any such object. This is called polymorphism.

Explanation:

Nested conditionals refers to the use of if or else if statement inside another if or else if statement or you can simply say a condition inside another condition. For example:

if( condition1) {  

//executes when condition1 evaluates to true

  if(condition2) {

//executes when condition1  and condition2 evaluate to true

  }  else if(condition3) {

 //when condition1 is true and condition3 is true

} else {

 //condition1  is true but neither condition2 nor conditions3 are true

}  }

The deeply nested conditionals make the program difficult to understand or read if the nested conditionals are not indented properly. Also the debugging gets difficult when the program has a lot of nested conditionals.

So in order to avoid nested conditionals some strategies are used such as using a switch statement.

Here i will give an example of the strategies i have mentioned in the answer.

Using Logical Expressions:

A strategy to avoid nested conditionals is to use logical expressions with logical operators such as AND operator. The above described example of nested conditionals can be written as:

if(condition1 && condition2){  //this executes only when both condition1 and condition2 are true

} else if(condition1 && condition3) {

this executes only when both condition1 and condition3 are true

} else if(condition1 ){

//condition1  is true but neither condtion2 nor condtion3 are true  }

This can further be modified to one conditional as:

if(!condition3){

// when  condition1 and condition2 are true

}

else

// condition3 is true

Now lets take a simple example of deciding to go to school or not based on some conditions.

if (temperature< 40)

{

   if (busArrived=="yes")

   {

       if (!sick)

       {

           if (homework=="done")

           {

               printf("Go to school.");

           }

       }                    

   }

}

This uses nested conditionals. This can be changed to a single conditional using AND logical operator.

if ((temperature <40) && (busArrived=="yes") &&

(!sick) && (homework=="done"))

{    cout<<"Eligible for promotion."; }

The second strategy is to use an interface. For example you can

abstract class Shape{

//declare a method common to all sub classes

  abstract public int area();

// same method that varies by formula of area for different shapes

}

class Triangle extends Shape{

  public int area() {

     // version of area code for Triangle

return (width * height / 2);

  }

}

class Rectangle extends Shape{

  public int area() {

     // version of area code for Rectangle

    return (width * height)

  }

}

// Now simply create Rectangle or Triangle objects and call area() for any such object and the relevant version will be executed.

4 0
2 years ago
a.Write a Python function sum_1k(M) that returns the sum푠푠= ∑1푘푘푀푀푘푘=1b.Compute s for M = 3 by hand and write
stealth61 [152]

Answer:

def sum_1k(M):

   s = 0

   for k in range(1, M+1):

       s = s + 1.0/k

   return s

def test_sum_1k():

   expected_value = 1.0+1.0/2+1.0/3

   computed_value = sum_1k(3)

   if expected_value == computed_value:

       print("Test is successful")

   else:

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test_sum_1k()

Explanation:

It seems the hidden part is a summation (sigma) notation that goes from 1 to M with 1/k.

- Inside the <em>sum_1k(M)</em>, iterate from 1 to M and calculate-return the sum of the expression.

- Inside the <em>test_sum_1k(),</em> calculate the <em>expected_value,</em> refers to the value that is calculated by hand and <em>computed_value,</em> refers to the value that is the result of the <em>sum_1k(3). </em>Then, compare the values and print the appropriate message

- Call the <em>test_sum_1k()</em> to see the result

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