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Vladimir79 [104]
2 years ago
10

Of 500 sessions that occurred on a website during a one-week period, 200 of them started on the homepage; 100 of these 200 sessi

ons saw only the homepage before leaving the site. Of the 300 sessions that started on other pages, 100 of them consisted of only one pageview. Which is higher: the site bounce rate or the homepage’s bounce rate?
Computers and Technology
1 answer:
Elena L [17]2 years ago
7 0

Answer:

the homepage bounce rate

Explanation:

Bounce rate is defined as the percentage of people that visits a page on a website and then exit or leave. That is, the visitors did not view any other page except the first one and they just leave the page after that visit. In order to get the bounce rate, the total number of visitors to a single page is taken and its then divided by the total number of visits to the website. For example, if the total number of visitors to a website over a period of time is 3000, while those that only visited a page on the website is 500, then the bounce rate is

(500/3000) * 100 = 16%

For this question, the homepage recorded 100 visited only once out of 200 which means homepage bounce rate is 50% : (100/200)*100

While website bounce rate is total number of bounces across the website/total visit to the website

100+100 =200 this is the total bounces across the website

200/500 :500 is the total visit to the website

(200/500)*100 = 40 %

Therefore the homepage bounce rate is higher than the site bounce rate

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Which of the following provides a suite of integrated software modules for finance and accounting, human resources, manufacturin
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Answer:

B) ERP systems

Explanation:

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7.8.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's f
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Answer:

Here is the CoordTransform() function:              

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2; }

The above method has four parameters xVal  and yVal that are used as input parameters and xValNew yValNew as output parameters. This function returns void and transforms its first two input parameters xVal and yVal into two output parameters xValNew and yValNew according to the formula: new = (old + 1) *

Here new variables are xValNew  and yValNew and old is represented by xVal and yVal.

Here the variables xValNew and yValNew are passed by reference which means any change made to these variables will be reflected in main(). Whereas variables xVal and yVal are passed by value.

Explanation:

Here is the complete program:

#include <iostream>

using namespace std;

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2;}

int main()

{ int xValNew;

int yValNew;

int xValUser;

int yValUser;

cin >> xValUser;

cin >> yValUser;

CoordTransform(xValUser, yValUser, xValNew, yValNew);

cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl;

return 0; }

The output is given in the attached screenshot   

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The answer is 1 because it is

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Read 2 more answers
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