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Paladinen [302]
3 years ago
8

A technician removes the cap from the brake fluid reservoir and finds that the rubber boot in the cap is swollen. technician a s

ays this is because there is water in the brake fluid. technician b says this is because the brake fluid level is too low. which technician is correct?
Computers and Technology
1 answer:
uysha [10]3 years ago
5 0

Technician A is correct, there could be water or other contaminant in the brake fluid reservoir. The swollen cap needs to be replaced, some sell only the rubber part of the cap which could help save you money.

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A GPO component that’s an Active Directory object stored in the System\Policies folder. The GPC stores GPO properties and status
blagie [28]

Answer:

The correct answer to the following question will be the Group Policy Container (GPC).

Explanation:

Group Policy Container (GPC) : An active directive container that contains properties of GPO (such as the setting of GPO, it's component settings), is known as Group Policy Container.

There are two locations where GPO stores its configuration information :

  • Group Policy Container (GPC).
  • Group Policy Template (GPT).

8 0
3 years ago
Write a program to have the computer guess at a number between 1 and 20. This program has you, the user choose a number between
ivanzaharov [21]

Answer:

This question is answered using C++ programming language.

#include<iostream>

#include <cstdlib>

#include <ctime>

using namespace std;

int main(){

   int num, computerguess;

   char response, hl;

   cout<<"Choose your number: ";

   cin>>num;

   srand((unsigned) time(0));

   int high = 20;    int low = 1;    int guess = 0;

   computerguess = low + rand() % high;

   cout<<"Is it "<<computerguess<<" ?y/n: ";

   cin>>response;

   while(response == 'n'){

       cout<<"Higher or Lower? h/l: ";

       cin>>hl;

       if(hl == 'h'){            low = computerguess+1; }

       else{   high = computerguess-1; }

       guess++;

       computerguess = low + rand() % high;

       cout<<"Is it "<<computerguess<<" ?y/n: ";

   cin>>response;

   }

   cout<<"I got your number of "<<num<<" in "<<guess+1<<" guesses.";

   return 0;

}

Explanation:

This declares num (user input) and computerguess as integer

   int num, computerguess;

This declares response (which could be yes (y) or no (n) and hl (which represents high or low) as char

   char response, hl;

This prompts user for input

   cout<<"Choose your number: ";

This gets user input

   cin>>num;

This allows the computer be able to generate different random numbers at different intervals

   srand((unsigned) time(0));

This declares and initializes the range (high and low) to 20 and 1 respectively. Also, the initial number of guess is declared and initialized to 0

   int high = 20;    int low = 1;    int guess = 0;

Here, the computer take a guess

   computerguess = low + rand() % high;

This asks if the computer guess is right

   cout<<"Is it "<<computerguess<<" ?y/n: ";

This gets user response (y or n)

   cin>>response;

The following iteration is repeated until the computer guess is right

   while(response == 'n'){

This asks if computer guess is high or low

       cout<<"Higher or Lower? h/l: ";

This gets user response (h or l)

       cin>>hl;

If the response is higher, this line gets the lower interval of the range

       if(hl == 'h'){            low = computerguess+1; }

If the response is lower, this line gets the upper interval of the range

       else{   high = computerguess-1; }

This increments the number of guess by 1

       guess++;

Here, the computer take a guess

   computerguess = low + rand() % high;

This asks if the computer guess is right

   cout<<"Is it "<<computerguess<<" ?y/n: ";

This gets user response (y or n)

   cin>>response;    }

This prints the statistics of the guess

   cout<<"I got your number of "<<num<<" in "<<guess+1<<" guesses.";

6 0
3 years ago
Suppose two hosts, A and B, are connected by a 10 Mbps link. The length of a packet is 12 Kb (Kilobits, i.e., 12 × 103 bits). Th
Alekssandra [29.7K]

Answer:

Answer is provided in the explanation section

Explanation:

Given data:

Bandwidth of link = 10* 106 bps

Length of packet = 12* 103 bits

Distance of link = 40 * 103m

Transmission Speed = 3 * 108 meters per second  

Formulas:

Transmission Delay = data size / bandwidth = (L /B) second

Propagation Delay = distance/transmission speed = d/s

Solution:

Transmission Delay = (12* 103 bits) / (10* 106 bps) = 0.0012 s = 1.2 millisecond

Propagation Delay = (40 * 103 meters)/ (3 * 108mps) = 0.000133 = 0.13 millisecond

4 0
4 years ago
Which option should Gina click to edit the text contained in a text box on a slide in her presentation?
topjm [15]
Text box
hope this helped
3 0
2 years ago
Need help ASAPppppppppp
Darina [25.2K]

Answer:

Below picture should give you an indication. On different PC cases, the position of the connectors can be different, so you have to look at the shape.

5 0
3 years ago
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