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jasenka [17]
3 years ago
7

A publisher has a text-only leader board on top of a page, and a text-only small square ad slot within the content of that page.

The small square ad slot has a higher a CTR, but the leaderboard has high CPCs. Is there anything the publisher can do to try and increase their earnings?
A. Yes, the publisher should opt in Image ads to both of ad units.
B. Yes, the publisher should change small square ad unit into large rectangle ad unit.
C. Yes, the publisher should make sure the medium rectangle ad unit comes first in their HTML code.
Computers and Technology
1 answer:
kherson [118]3 years ago
6 0

Answer:

Option B.

Explanation:

On top of a website, a publisher seems to have a text-only scoreboard, as well as a text-only square advertisement slot inside that webpage's material. The square advertisement slot has a low CTR but also has lower CPCs throughout the scoreboard.

So, It's True, the publisher must switch square ad unit into a big rectangle advertisement unit.

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AleksAgata [21]
Wet brining is a technique for seasoning meat by soaking it in a salt solution before cooking. As a general guideline, soak the meat in a solution containing 1 cup of salt per gallon of water. The salt will remain in the meat after cooking, imparting taste, and the liquid will be boiled away.

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5 0
2 years ago
Write a program that will predict the size of a population of organisms. The program should ask the user for the starting number
sineoko [7]

Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

 //Declaration of variables in the program

 double start_organisms;

 double daily_increase;

 int days;

 double updated_organisms;

 //The user enters the number of organisms as desired

 cout << "Enter the starting number of organisms: ";

 cin >> start_organisms;

 //Validating input data

 while (start_organisms < 2)

 {

     cout << "The starting number of organisms must be at least 2.\n";

     cout << "Enter the starting number of organisms: ";

     cin >> start_organisms;

 }

 //The user enters daily input, here's where we apply the 5.2% given in question

 cout << "Enter the daily population increase: ";

 cin>> daily_increase;

 //Validating the increase

 while (daily_increase < 0)

 {

     cout << "The average daily population increase must be a positive value.\n ";

     cout << "Enter the daily population increase: ";

     cin >> daily_increase;

 }

 //The user enters number of days

 cout << "Enter the number of days: ";

 cin >> days;

 //Validating the number of days

 while (days<1)

 {

     cout << "The number of days must be at least 1.\n";

     cout << "Enter the number of days: ";

     cin >> days;

 }

 

 //Final calculation and display of results based on formulas

 for (int i = 0; i < days; i++)

 {

     updated_organisms = start_organisms + (daily_increase*start_organisms);

     cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

     

     start_organisms = updated_organisms;

 }

 system("pause");

  return 0;

//end

}

Explanation:

6 0
3 years ago
Expand and simplify the following expressions. 1/2 [ 2x + 2 ( x - 3)] *​
Lerok [7]

Answer: 2

x

−

3

Explanation:

4 0
3 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

  • Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW  = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

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3 years ago
The CPU's control unit retrieves the next instruction in a sequence of program instructions from main memory in the ______ stage
Varvara68 [4.7K]

Answer:

A. Fetch.

Explanation:

The fetch-decode-execute process of simply the fetch-execute process of the CPU are stages the CPU follow to process information and the switch state or shutdown.

The stages of this process is implied in its name, that is, the stages are fetch, decide and execute.

The fetch stage retrieves the next instruction from the memory.

The decode stage converts the clear text instruction set to electronic signals and transfer it to the appropriate registers.

The execute stage is the action carried out in the arithmetic logic unit.

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3 years ago
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