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2 years ago

8

If a database named Contacts has a table named tblEmployees and another database named Orders links to that tblEmployees table,

where is the data stored?

Computers and Technology

1 answer:

Vlad1618 [11]2 years ago

5 0

Answer:

cache I think

Explanation:

I am not sure or memory HDD or SSD

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Do all of the packets in your sent messages always follow the same path? If not, describe at least two different paths packets t

Lemur [1.5K]

Answer:

All the message I sent were from the same routers

Explanation:

All the message I sent were from the same routers

5 0

2 years ago

Read 2 more answers

More Loops: All versions of foods.py in this section have avoided using for loops when printing to save space. Choose a version

Nata [24]

Answer:

Following are the program in the Python Programming Language.

#set list type variable and initialize the elements

myPizza = ['Margarita', 'Capsicum and onion', 'Chicken']

#set variable and initialize the elements

frndPizzas = myPizza[:]

#append value in list variable

myPizza.append('Corn')

#append value in list variable

frndPizzas.append('paperica')

#print message

print("My pizzas are:")

#set the for loop and print elements of 1st list

for pizza in myPizza:

print(pizza)

#print message

print("\nFriend's pizzas are:")

#set the for loop and print elements of 2st list

for frndsPizza in frndPizzas:

print(frndsPizza)

<u>Output</u>:

My pizzas are:

Margarita Capsicum and onion Chicken Corn

Friend's pizzas are:

Margarita Capsicum and onion Chicken paperica

Explanation:

<u>Following are the description of the program</u>:

Set a list data type variable that is 'myPizza' and initializes elements in it.
Set another list data type variable that is 'frndPizzas' and initializes elements in it.
Append one-one element in both of the list data type variables.
Set the for loops to print the elements of both list data type variables.

5 0

2 years ago

We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is

bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons to find the middle terms.

If list has 18 elements, there are two middle terms as one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons to find the middle terms.

If list has 20 elements, there are two middle terms as one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0

2 years ago

ollegr [7]

What are the type groups?

8 0

2 years ago

Read 2 more answers

Given 4 integers, output their product and their average, using integer arithmetic.

solmaris [256]

Answer:

see explaination

Explanation:

Part 1:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

int avg=0, pro=1;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4;

pro = num1*num2*num3*num4;

System.out.println(pro+" "+avg);

}

}

------------------------------------------------------------------

Part 2:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

double avg=0, pro=1; //using double to store floating point numbers.

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4.0; //if avg is declared as a float, then use 4.0f

pro = num1*num2*num3*num4;

System.out.println((int)pro+" "+(int)avg); //using type conversion only integer part

System.out.printf("%.3f %.3f\n",pro,avg);// \n is for newline

}

}

8 0

2 years ago