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MrRissso [65]
2 years ago
6

Braking on __________ may cause a loss of control.

Computers and Technology
2 answers:
kodGreya [7K]2 years ago
7 0

Answer:Wet roads

Explanation:

snow_tiger [21]2 years ago
4 0

Answer:

wet or icy roads

Explanation:

You might be interested in
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
Elena L [17]

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

7 0
2 years ago
This question involves the creation of user names for an online system. A user name is created based on a user’s first and last
Evgen [1.6K]

Answer:

See explaination

Explanation:

import java.util.*;

class UserName{

ArrayList<String> possibleNames;

UserName(String firstName, String lastName){

if(this.isValidName(firstName) && this.isValidName(lastName)){

possibleNames = new ArrayList<String>();

for(int i=1;i<firstName.length()+1;i++){

possibleNames.add(lastName+firstName.substring(0,i));

}

}else{

System.out.println("firstName and lastName must contain letters only.");

}

}

public boolean isUsed(String name, String[] arr){

for(int i=0;i<arr.length;i++){

if(name.equals(arr[i]))

return true;

}

return false;

}

public void setAvailableUserNames(String[] usedNames){

String[] names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

for(int i=0;i<usedNames.length;i++){

if(isUsed(usedNames[i],names)){

int index = this.possibleNames.indexOf(usedNames[i]);

this.possibleNames.remove(index);

names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

}

}

}

public boolean isValidName(String str){

if(str.length()==0) return false;

for(int i=0;i<str.length();i++){

if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

return false;

}

return true;

}

public static void main(String[] args) {

UserName person1 = new UserName("john","smith");

System.out.println(person1.possibleNames);

String[] used = {"harta","hartm","harty"};

UserName person2 = new UserName("mary","hart");

System.out.println("possibleNames before removing: "+person2.possibleNames);

person2.setAvailableUserNames(used);

System.out.println("possibleNames after removing: "+person2.possibleNames);

}

}

8 0
2 years ago
Given an n-element array X, algorithm D calls algorithm E on each element X[i]. Algorithm E runs in O(i) time when it is called
krek1111 [17]

Answer:

O(n^2)

Explanation:

The number of elements in the array X is proportional to the algorithm E runs time:

For one element (i=1) -> O(1)

For two elements (i=2) -> O(2)

.

.

.

For n elements (i=n) -> O(n)

If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)  

5 0
2 years ago
Read the attached paper titled A Survey of Coarse-Grained Reconfigurable Architecture and comment on it. Make sure to specifical
Pani-rosa [81]

Answer:

lol

Explanation:

I NEED POINTS

5 0
1 year ago
Compare the memory organization schemes of contiguous memory allocation and paging with respect to the following issues: a. Exte
Free_Kalibri [48]

Answer:

The comparison is based on memory organization schemes of contiguous memory allocation and paging with respect to External fragmentation, Internal fragmentation and Ability to share code across processes.

Explanation:

Memory organization schemes of contiguous memory allocation:

Contiguous memory allocation schemes suffers from external fragmentation. The reason is that address space is distributed contiguously and the holes and gaps keep growing when the old processes die and new processes are introduced. The variable size partition suffers from external fragmentation however the fixed size partitions do not suffer from external fragmentation. Contiguous memory allocation with variable size partitions does not encounter an internal fragmentation but with fixed size partitions suffers from internal fragmentation. Contiguous memory allocation does not support sharing code across processes. This is because the virtual memory segment of a process is not fragmented into non-contiguous fine grained blocks.

Paging:

Paging does not encounter external fragmentation as pages are of the fixed or equal size. So this reduces external fragmentation. However paging suffers from internal fragmentation. This is because a process can request more space or it can request for a less space. When page is allocated to the such a process that page is no longer utilized.This results in internal fragmentation because of the wastage of space even when the page has internal space but cannot be fully utilized. Paging allows to share code across processes.

5 0
2 years ago
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