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Alexxandr [17]
2 years ago
11

A computer database uses a binary sequence of 4 bits to represent unique user IDs. To increase the number of unique IDs that the

database is able to represent, the database administrator increases the number of bits in a user ID to 7 bits. How many times more unique user IDs can be represented with the new system?
Computers and Technology
1 answer:
AnnZ [28]2 years ago
7 0

Answer:

The answer is "A device may include 2 ^{56}time more unique user IDs".

Explanation:

In the given question  the total user ID access availability of 5 bits would also be 2^{10},  and  for user ID listings, its 12 bits should generate a maximum 2^{66} data allocation.  

In this, it first calculate the value that is 2^{(66-10)}=2^{56} times, which was its amount of times most specific user IDs, that's why 2 ^{56} time is correct answer.

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Rideshare companies like Uber or Lyft track the x,y coordinates of drivers and customers on a map. If a customer requests a ride
TiliK225 [7]

Answer:

The program in Java is as follows:

import java.util.*;

public class Main{

 public static void main (String[]args){

   Scanner input = new Scanner(System.in);

   double user[] = new double[2];    double dr1[] = new double[2];

   double dr2[] = new double[2];    double dr3[] = new double[2];    

   System.out.print("Enter user coordinates: ");

   for(int i =0;i<2;i++){        user[i] = input.nextDouble();    }

   System.out.print("Enter driver 1 coordinates: ");

   for(int i =0;i<2;i++){        dr1[i] = input.nextDouble();    }

   System.out.print("Enter driver 2 coordinates: ");

   for(int i =0;i<2;i++){        dr2[i] = input.nextDouble();    }

   System.out.print("Enter driver 3 coordinates: ");

   for(int i =0;i<2;i++){        dr3[i] = input.nextDouble();    }

   double dr1dist = Math.abs(user[0] - dr1[0]) + Math.abs(user[1] - dr1[1]);

   double dr2dist = Math.abs(user[0] - dr2[0]) + Math.abs(user[1] - dr2[1]);

   double dr3dist = Math.abs(user[0] - dr3[0]) + Math.abs(user[1] - dr3[1]);

   System.out.println("Estimated pickup time of driver 1 "+(3.5 * dr1dist)+" minutes");

   System.out.println("Estimated pickup time of driver 2 "+(3.5 * dr2dist)+" minutes");

   System.out.println("Estimated pickup time of driver 3 "+(3.5 * dr3dist)+" minutes");

 }

}

Explanation:

The following array declarations are for the customer and the three drivers

<em>    double user[] = new double[2];    double dr1[] = new double[2];</em>

<em>    double dr2[] = new double[2];    double dr3[] = new double[2];    </em>

This prompts the user for the customer's coordinates

   System.out.print("Enter user coordinates: ");

This gets the customer's coordinate

   for(int i =0;i<2;i++){        user[i] = input.nextDouble();    }

This prompts the user for the driver 1 coordinates

   System.out.print("Enter driver 1 coordinates: ");

This gets the driver 1's coordinate

   for(int i =0;i<2;i++){        dr1[i] = input.nextDouble();    }

This prompts the user for the driver 2 coordinates

   System.out.print("Enter driver 2 coordinates: ");

This gets the driver 2's coordinate

   for(int i =0;i<2;i++){        dr2[i] = input.nextDouble();    }

This prompts the user for the driver 3 coordinates

   System.out.print("Enter driver 3 coordinates: ");

This gets the driver 3's coordinate

   for(int i =0;i<2;i++){        dr3[i] = input.nextDouble();    }

This calculates the distance between driver 1 and the customer

   double dr1dist = Math.abs(user[0] - dr1[0]) + Math.abs(user[1] - dr1[1]);

This calculates the distance between driver 2 and the customer

   double dr2dist = Math.abs(user[0] - dr2[0]) + Math.abs(user[1] - dr2[1]);

This calculates the distance between driver 3 and the customer

   double dr3dist = Math.abs(user[0] - dr3[0]) + Math.abs(user[1] - dr3[1]);

The following print statements print the estimated pickup time of each driver

<em>   System.out.println("Estimated pickup time of driver 1 "+(3.5 * dr1dist)+" minutes");</em>

<em>    System.out.println("Estimated pickup time of driver 2 "+(3.5 * dr2dist)+" minutes");</em>

<em>    System.out.println("Estimated pickup time of driver 3 "+(3.5 * dr3dist)+" minutes");</em>

7 0
2 years ago
Which of the following facts determines how often a nonroot bridge or switch sends an 802.1D STP Hello BPDU message?
irinina [24]

Answer:

b. The Hello timer as configured on the root switch.

Explanation:

There are differrent timers in a switch. The root switch is the only forwarding switch in a network, while non root switches blocks traffic to  prevent looping of BPDUs in the network. Since the root switch is the only forwarding switch, all timing configuration comes from or is based on the configuration in the root.

The hello timer is no exception as the nonroot switch only sends 802.1D DTP hello BPDU messages forwarded to it by the root switch and its frequency depends on the root switch hello timer.

8 0
2 years ago
5. You just bought a new hard drive for youYou plan to use this as a secondary hard drive to store allcomputeryour UMA files. On
wel
Format and mount the drive. 
Formatting erases the drive and sets it up so the system can read and write to it.
Mounting the drive allows the system to have access to the drive and read and write to it. 

If you want to get more specific in Windows you can open registry editor and change the paths of the drive to save in the new location.
4 0
2 years ago
Emma wants to create a website that will contain mostly textual information pertaining to book reports she has to create. Which
Softa [21]

The language that will help her create a website structured in the form of headings and paragraphs of text would be HTML.

7 0
2 years ago
Read 2 more answers
1. Create a view named customer_addresses that shows the shipping and billing addresses for each customer.
Verdich [7]

Answer:

Answer given below

Explanation:

1.

CREATE VIEW CustomerAddresses AS

SELECT custo. CustomerID, EmailAddress , LastName ,FirstName,

bill.Line1 AS BillLine1, bill.Line2 AS BillLine2, bill.City AS BillCity, bill.State AS BillState, bill.ZipCode AS BillZip,

ship.Line1 AS ShipLine1, ship.Line2 AS ShipLine2, ship.City AS ShipCity, ship.State AS ShipState, ship.ZipCode AS ShipZip

FROM Customers custo , Addresses ship , Addresses bill

WHERE custo. BillingAddressID= bill.AddressID AND custo.ShippingAddressID= ship. AddressID;

2.

SELECT CustomerID, LastName, FirstName, BillLine1 FROM CustomerAddresses;

3.

CREATE VIEW OrderItemProducts

AS

SELECT Orders.OrderID, OrderDate, TaxAmount, ShipDate,

ItemPrice, DiscountAmount, (ItemPrice- DiscountAmount) AS FinalPrice,

Quantity, and (Quantity * (ItemPrice-DiscountAmount)) AS ItemTotal,

ProductName FROM

Orders, OrderItems, Products

WHERE

Orders.OrderID = OrderItems.OrderID AND

OrderItems.ProductID = Products. ProductID;

4.

CREATE VIEW ProductSummary

AS

SELECT distinct

ProductName, COUNT(OrderID) AS OrderCount, SUM(ItemTotal) AS OrderTotal

FROM

OrderItemProducts

GROUP BY ProductName;

5.

SELECT ProductName, OrderTotal

FROM ProductSummary P

WHERE 5> (select count(*) FROM ProductSummary S

WHERE P.OrderTotal<S.OrderTotal)

ORDER BY OrderTotal;

5 0
2 years ago
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