Using existing algorithms as building blocks for new algorithms has all the following benefits EXCEPT

public class Questions { public static void main(String[] args) { System.out.print("Here "); System.out.println("There " "Everyw

In-s [12.5K]

Answer:

The output of this code can be given as follows:

Output:

Here There Everywhere

But notin Texas

Explanation:

In the give java code a class "Questions" is defined inside this class the main method is defined in the main method we use three print function that can be described as follows:

In the first print function, we print the message that is "Here".
In the second and third print function, we use println that print data or message in the next line that is "There Everywhere" and in the next line "But not" and "in Texas".

6 0

3 years ago

Write test programs in java to determine the scope of a variable declared in a for statement. Specifically, the code must determ

xxMikexx [17]

Answer:

public class Main {
public static void main (String [] args) {
for(int i = 1; i < 10; i++){
int num = 0;
num += i;
}
System.out.println(num);
}
}

Explanation:

In programming each variable has its scope. For example, a variable defined within the for loop has no longer exist outside the loop body. To illustrate this we can write a short program as presented above.

Firstly, create a for loop that traverse the number 1 - 10 (Line 5 - 8). Within the loop create a variable num and initialize it with zero (Line 6) and increment it with i value for each iteration (Line 7).

Outside the loop, we try to print the num (Line 9). When we run the program this will result in an error as the num which is declared inside the for loop will no longer exist outside the loop body.

6 0

3 years ago

I need the SQL statements for these questions:

zimovet [89]

Answer:

Explanation:

/* From the information provided, For now will consider the name of table as TRIPGUIDES*/

/*In all the answers below, the syntax is based on Oracle SQL. In case of usage of other database queries, answer may vary to some extent*/

1.

Select R.Reservation_ID, R.Trip_ID , C.Customer_Num,C.Last_Name from Reservation R, Customer C where C.Customer_Num=R.Customer_Num ORDER BY C.Last_Name

/*idea is to select the join the two tables by comparing customer_id field in two tables as it is the only field which is common and then print the desired result later ordering by last name to get the results in sorted order*/

2.

Select R.Reservation_ID, R.Trip_ID , R.NUM_PERSONS from Reservation R, Customer C where C.Customer_Num=R.Customer_Num and C.LAST_NAME='Goff' and C.FIRST_NAME='Ryan'

/*Here, the explaination will be similar to the first query. Choose the desired columns from the tables, and join the two tables by equating the common field

*/

3.

Select T.TRIP_NAME from TRIP T,GUIDE G,TRIPGUIDES TG where T.TRIP_ID=TG.TRIP_ID and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Abrams' and G.FIRST_NAME='Miles'

/*

Here,we choose three tables TRIP,GUIDE and TRIPGUIDES. Here we selected those trips where we have guides as Miles Abrms in the GUIDES table and equated Trip_id from TRIPGUIDES to TRIP.TRIP_Name so that can have the desired results

*/

4.

Select T.TRIP_NAME

from TRIP T,TRIPGUIDES TG ,G.GUIDE

where T.TRIP_ID=TG.TRIP_ID and T.TYPE='Biking' and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Boyers' and G.FIRST_NAME='Rita'

/*

In the above question, we first selected the trip name from trip table. To put the condition we first make sure that all the three tables are connected properly. In order to do so, we have equated Guide_nums in guide and tripguides. and also equated trip_id in tripguides and trip. Then we equated names from guide tables and type from trip table for the desired results.

*/

5.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='2016-07-23' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

/*

The explaination for this one will be equivalent to the previous question where we just equated the desired columns where we equiated the desired columns in respective fields and also equated the common entities like trip ids and customer ids so that can join tables properly

*/

/*The comparison of dates in SQL depends on the format in which they are stored. In the upper case if the

dates are stored in the format as YYYY-MM-DD, then the above query mentioned will work. In case dates are stored in the form of a string then the following query will work.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='7/23/2016' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

*/

6.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION R WHERE R.TRIP_ID IN

{SELECT TRIP_ID FROM TRIP T WHERE STATE='ME'}

/*

In the above question, we firstly extracted all the trip id's which are having locations as maine. Now we have the list of all the trip_id's that have location maine. Now we just need to extract the reservation ids for the same which can be trivally done by simply using the in clause stating print all the tuples whose id's are there in the list of inner query. Remember, IN always checks in the set of values.

*/

7.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION WHERE

EXISTS {SELECT TRIP_ID FROM TRIP T WHERE STATE='ME' and R.TRIP_ID=T.TRIP_ID}

/*

Unlike IN, Exist returns either true or false based on existance of any tuple in the condition provided. In the question above, firstly we checked for the possibilities if there is a trip in state ME and TRIP_IDs are common. Then we selected reservation ID, trip ID and Trip dates for all queries that returns true for inner query

*/

8.

SELECT G.LAST_NAME,G.FIRST_NAME FROM GUIDE WHERE G.GUIDE_NUM IN

{

SELECT DISTINCT TG.GUIDE_NUM FROM TRIPGUIDES TG WHERE TG.TRIPID IN {

SELECT T.TRIP_ID FROM TRIP T WHERE T.TYPE='Paddling'

}

/*

We have used here double nested IN queries. Firstly we selected all the trips which had paddling type (from the inner most queries). Using the same, we get the list of guides,(basically got the list of guide_numbers) of all the guides eds which were on trips with trip id we got from the inner most queries. Now that we have all the guide_Nums that were on trip with type paddling, we can simply use the query select last name and first name of all the guides which are having guide nums in the list returned by middle query.

*/

4 0

3 years ago

An investigator responsible for tracking e-mail communication would begin by locating what

Ainat [17]

Email headers contains information about email

6 0

3 years ago

Count from 1 to 20(base10) using the following bases: 8, 6, 5, 3.

Licemer1 [7]

Answer:

Base 8:

1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 16, 17, 20, 21, 22, 23, 24

Base 6:

1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, 30, 31, 32

Base 5:

1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40

Base 3:

1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202

Explanation:

When counting on a certain base n you can only use n numerals (including the zero). So, if you count in base 8, you have 8 numrals, these are 0, 1, 2, 3, 4, 5, 6 and 7. The numeral 8 does not exist in base 8 the same as there is no numeral for 10 in base 10. When you're counting beyond the single numerals you add a 1 to the numeral in the column in front (which starts at zero) and reset the previous column.

4 0

3 years ago