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4 years ago

Which transmission media is faster? Fibre Optic or Radio-waves? I want full explanation!

Computers and Technology

1 answer:

Bad White [126]4 years ago

4 0

Answer:

Due to the direct line of sight,the data rate of radio links is higher than that of fiber optic connections. A radio link has a lower latency (less delay).In contrast ,fiber optic connections may achieve higher bandwidths

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In which of the following phases of filmmaking would a production team be focused on the process of casting?

Rzqust [24]

Explanation:

There are five phases of film production and they include development, pre-production, production, post-production and distribution.

3 0

3 years ago

Read 2 more answers

Given an integer variable count, write a statement that displays the value of count on the screen. Do not display anything else

Irina-Kira [14]

Answer:

cout<<count;

Explanation:

The above statement is in c++ which display the value of count .The cout statement is used in c++ to print the value on console .

Following are the code in c++

#include <iostream> // header file

using namespace std; // namespace

int main() // main method

{

int count=90; // count variable

cout<<count; // display the value of count

return 0;

}

Output:

In this program we have declared a count variable of integer type which is initialized by 90 and finally displays the value of count on the screen.

4 0

3 years ago

Compute the sum of all integers that are multiples of 9, from 1 to 250. Enter your result of your computation in the text box be

shepuryov [24]

The pseudocode to find the sum of all integers that are multiples of 9, from 1 to 250.

totalSum = 0

for i from 1 to 250{

if i is divided by 9 and remainder is 0{

totalSum = totalSum + i;

}

print(totalSum)

in python language the code will be

totalSum = 0

for i in range(1,250):

if i%9==0:

totalSum += i

If you will run the program , the answer would be 3402.

8 0

3 years ago

Read 2 more answers

A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co

tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k ...(1)

probability of 2 hops = k(1-k) ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

= k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

∞ n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

n-1

= 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet = $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

6 0

3 years ago

To reduce the number of used digital outputs of the microcontroller, the display board is connected to the main board through th

Luda [366]

Answer:

The program in Python is as follows:

BCD = ["0001","0010","0011","0100","0101","0110","0111"]

num = input("Decimal: ")

BCDValue = ""

valid = True

for i in range(len(num)):

if num[i].isdigit():

if(int(num[i])>= 0 and int(num[i])<8):

BCDValue += BCD[i]+" "

else:

valid = False

break;

else:

valid = False

break;

if(valid):

print(BCDValue)

else:

print("Invalid")

Explanation:

This initializes the BCD corresponding value of the decimal number to a list

BCD = ["0001","0010","0011","0100","0101","0110","0111"]

This gets input for a decimal number

num = input("Decimal: ")

This initializes the required output (i.e. BCD value) to an empty string

BCDValue = ""

This initializes valid input to True (Boolean)

valid = True

This iterates through the input string

for i in range(len(num)):

This checks if each character of the string is a number

if num[i].isdigit():

If yes, it checks if the number ranges from 0 to 7 (inclusive)

if(int(num[i])>= 0 and int(num[i])<8):

If yes, the corresponding BCD value is calculated

BCDValue += BCD[i]+" "

else:

If otherwise, then the input string is invalid and the loop is exited

valid = False

break;

If otherwise, then the input string is invalid and the loop is exited

else:

valid = False

break;

If valid is True, then the BCD value is printed

if(valid):

print(BCDValue)

If otherwise, it prints Invalid

else:

print("Invalid")

7 0

3 years ago