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aalyn [17]
2 years ago
6

What is the Gain (dB) of a transmission if the Maximum Data Rate is 1 Gbps and the Bandwidth =7000 MHz? Group of answer choices

11.072 dB 15.420 dB 5.675 dB 12.934 dB
Computers and Technology
1 answer:
denpristay [2]2 years ago
3 0

Answer:

15.420 dB

Explanation:

the Gain (dB) of a transmission if the Maximum Data Rate is 1 Gbps and the Bandwidth =7000 MHz is 15.420 dB.

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An aviation tracking system maintains flight records for equipment and personnel. The system is a critical command and control s
sergeinik [125]

Answer:

offline backup solution

Explanation:

In such a scenario, the best option would be an offline backup solution. This is basically a local and offline server that holds all of the flight record data that the cloud platform has. This offline backup server would be updated frequently so that the data is always up to date. These servers would be owned by the aviation company and would be a secondary solution for the company in case that the cloud platform fails or the company cannot connect to the cloud service for whatever reason. Being offline allows the company to access the database regardless of internet connectivity.

5 0
2 years ago
Write the definition of a function printAttitude, which has an int parameter and returns nothing. The function prints a message
Oksana_A [137]

Answer:

The function definition, in cpp, for the function printAttitude is given below.

void printAttitude(int n)

{

switch(n)

{

case 1: cout<<"disagree"<<endl; break;

case 2: cout<<"no opinion"<<endl; break;

case 3: cout<<"agree"<<endl; break;

default: break;

}

}

Explanation:

As seen, the method takes an integer parameter.

The method does not returns any value hence, return type is void.

The message is displayed to the console using switch statement which executes on the integer parameter, n.

The values for which output needs to be displayed are taken in cases.

The values for which no output needs to be displayed, is taken in default case.

Every case ends with break statement so that the program execution can be terminated.

This method can be used inside a program as shown.  

PROGRAM

#include <iostream>

using namespace std;

void printAttitude(int n)

{

switch(n)

{

case 1: cout<<"disagree"<<endl; break;

case 2: cout<<"no opinion"<<endl; break;

case 3: cout<<"agree"<<endl; break;

default: break;

}

}

int main() {

// variables to hold respective value

int n;

// user input taken for n

cout<<"Enter any number: ";

cin>>n;

// calling the function taking integer parameter  

printAttitude(n);

return 0;

}

OUTPUT

Enter any number: 11

1. The user input is taken for the integer parameter.

2. Inside main(), the method, printAttitude(), is then called and the user input is passed as a parameter.

3. Since the user entered value is 11 which does not satisfies any of the values in the cases, the default case is entered which does nothing and executes the break statement.

4. Hence, the program displays nothing for the value of 11.

5. When the user enters value 3, the case statement is executed since 3 satisfies one of the case values.

6. The program output for the user input of 3 is as shown below.

OUTPUT

Enter any number: 3

agree

The program ends with return statement.

4 0
2 years ago
If there are 8 opcodes and 10 registers, a. What is the minimum number of bits required to represent the OPCODE? b. What is the
Nimfa-mama [501]

Answer:  

For 32 bits Instruction Format:

OPCODE   DR               SR1                   SR2      Unused bits

a) Minimum number of bits required to represent the OPCODE = 3 bits

There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.

Ceil (log2 (8)) = 3

b) Minimum number of bits For Destination Register(DR) = 4 bits

There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value.  4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.  

Ceil (log2 (10)) = 4

c) Maximum number of UNUSED bits in Instruction encoding = 17 bits

Total number of bits used = bits used for registers + bits used for OPCODE  

     = 12 + 3 = 15  

Total  number of bits for instruction format = 32  

Maximum  No. of Unused bits = 32 – 15 = 17 bits  

OPCODE                DR              SR1             SR2              Unused bits

  3 bits              4 bits          4 bits           4 bits                17 bits

7 0
2 years ago
1.Which of the following class definitions defines a legal abstract class?a. class A { abstract void unfinished() { } }b. class
xxMikexx [17]

Answer:(c) abstract class A { abstract void unfinished(); }

Explanation:

A legal abstract class must have the keyword abstract before the class and an abstract class has abstract functions with the keyword abstract written and a void as the return type.

8 0
2 years ago
Your company has decided it wants to build a file server and store data that can be shared by coworkers. You were asked to resea
Yuri [45]
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