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adoni [48]
3 years ago
9

What is (12.2a + 9.7b) – (18.1b – 0.2a) – (6.7a + 6.8b), simplified?

Computers and Technology
1 answer:
Inga [223]3 years ago
8 0
The answer should be 5.7a - 15.2b
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At the frequency of 2.4 GHz what is the free-space path loss in dB.
chubhunter [2.5K]

Answer:

The FSBL is 100.05 dB

Solution:

As per the question:

Frequency, f = 2.4 GHz = 2.4\times 10^{9}

Now, we know that to calculate FSBL, i.e., Free Space Path Loss in dB, we use:

FSBL(in dB) = 10log_{10}(\frac{4\pifd}{c})^{2}

where

d = 1 km = 1000

c = 3\times 10^{8} m/s

FSBL(in dB) = 20log_{10}\frac{4\pifd}{c} = 20log_{10}(fd) + 20log_{10}(\frac{4\pi}{c})

FSBL(in dB) = 20log_{10}(f) + 20log_{10}(d) + 20log_{10}(\frac{4\pi}{c})

FSBL(in dB) = 20log_{10}(2.4\times 10^{9}) + 20log_{10}(10^{3}) + 20log_{10}(\frac{4\pi}{3\times 10^{8}})

FSBL(in dB) = 20\times 9.3802 + 20\times 3 - 147.55 = 100.05 dB

6 0
3 years ago
Explain how it is possible for a sequence of packets transmitted through a wide area network to arrive at their destination in a
dalvyx [7]

Answer:

- Different route paths, due to dynamic routing in WAN.

- Local area networks have one or very few paths to destination and does not require dynamic routing.

Explanation:

A wide area network is a network that covers a large geographical area. It goes beyond the private local area network, with more routing paths and network intermediate devices. The router is an essential tool for routing packets between devices. It requires a routing path, learnt statically or dynamically to work.

There are mainly two types of route paths, they are, static routes and dynamic routes.

The dynamic routes are used mainly in WAN. Sometimes, there can be multiple path to a destination, the router determines the best path to send the packets. It sends the sequenced packets through all available path and they arrive at the destination depending on the path used, the packets arrive in an out -of-order fashion in the destination and a rearranged.

5 0
3 years ago
A report has a column of totals, with each total adding to the cell above it. What kind of calculated figure is this?
Alexxx [7]

ANSWER:

The correct answer is Running Sum.

Explanation:

A report has a column of totals, with each total adding to the cell above it. Such type of calculated figure is called the Running Sum.

Running sum is also called the Partial Sum. In such type of summation, the values in the sequence is added to get a final result and then if a new number comes, it is again added to the grand sum, and in this way the sequence continues. Every new entry is added to the previous sum to get another sum.

3 0
3 years ago
BADM-Provide a reflection of at least 500 words (or 2 pages double spaced) of how the knowledge, skills, or theories of this cou
barxatty [35]

Answer:

WooW We Have To All This Which Class can you Please tell

5 0
3 years ago
Create a program that includes a function called toUpperCamelCase that takes a string (consisting of lowercase words and spaces)
astraxan [27]

Answer:

#include<iostream>

using namespace std;

//method to remove the spaces and convert the first character of each word to uppercase

string

toUpperCameICase (string str)

{

string result;

int i, j;

//loop will continue till end

for (i = 0, j = 0; str[i] != '\0'; i++)

{

 if (i == 0)       //condition to convert the first character into uppercase

  {

  if (str[i] >= 'a' && str[i] <= 'z')   //condition for lowercase

  str[i] = str[i] - 32;   //convert to uppercase

  }

if (str[i] == ' ')   //condition for space

  if (str[i + 1] >= 'a' && str[i + 1] <= 'z')   //condition to check whether the character after space is lowercase or not

  str[i + 1] = str[i + 1] - 32;   //convert into uppercase

if (str[i] != ' ')   //condition for non sppace character

  {

  result = result + str[i];   //append the non space character into string

  }

}

//return the string

return (result);

}

//driver program

int main ()

{

string str;

char ch;

//infinite loop

while (1)

{

fflush (stdin);

//cout<< endl;

getline (cin, str);   //read the string

//print the result

//cout<< endl << "Q";

// cin >> ch; //ask user to continue or not

ch = str[0];

if (ch == 'q' || ch == 'Q')   //is user will enter Q then terminatethe loop

  break;

cout << toUpperCameICase (str);

cout << endl;

}

return 0;

}

8 0
3 years ago
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