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3 years ago

The decimal equivalent of the product of 1011 and 1100 is

Computers and Technology

1 answer:

faltersainse [42]3 years ago

3 0

Answer:

132

Explanation:

We convert each number to base 10 (decimal) and multiply.

So 1011₂ = 1 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰

= 1 × 8 + 0 × 4 + 1 × 2 + 1 × 1

= 8 + 0 + 2 + 1

= 11₁₀

1100₂ = 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰

= 1 × 8 + 1 × 4 + 0 × 2 + 0 × 1

= 8 + 4 + 0 + 0

= 12₁₀

So, 1011₂ × 1100₂ = 11₁₀ × 12₁₀ = 132₁₀

So, the decimal equivalent of the product of 1011 and 1100 is 132

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My home PC has IP address 192.168.1.22 and connects to the Internet through a NAT router. Assume I am downloading a web page fro

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Answer and Explanation:

There are three sections to the inquiry.

1. what occurs inside the PC or host when a bundle is produced by an application.

2.how it goes from one host that is a source to another host that is a goal with numerous switches sitting in the middle.

3.what goes inside goal have when it gets a parcel on the system.

how about we examine individually

Handling bundle at source machine:

The application creates a bundle to be sent on the system and sends it to the layer beneath.

The following layer is known as a vehicle layer which oversees start to finish correspondence between two machines. The convention utilized can be TCP or UDP. What is the contrast between these two conventions is another subject by and large.

When the parcel is framed at the vehicle layer, it is sent to the system layer which includes source and goal IP in the bundle. A most significant field which is included at IP or system layer is Time To Live (TTL) which is utilized by middle switches/changes to choose if a parcel should be sent or not. (How goal IP is found?)

After the system layer, the parcel arrives at the information connection or MAC layer, where the source and goal MAC address of machines are included. We will perceive how these fields change between each two neighbors. (How goal MAC is found?)

The information interface layer pushes this bundle to the physical layer where it is sent as a flood of "0" and "1" on the physical medium accessible.

Preparing a bundle at switch:

Steering

There are three cases which may happen when the switch investigates the directing table for goal IP

1. On the off chance that there is a section relating to goal IP, we get the interface name the parcel ought to be sent to.

2. On the off chance that there is no immediate passage, at that point IP is changed over into organize IP utilizing the veil and afterward checked once more. It ought to be noticed that the longest prefix match to locate the best sending interface.

3. In the case of nothing matches, at that point the switch just advances it to the default goal designed.

Sending

Epitome

Preparing parcel at the goal have

The parcel is gotten at arrange card, physical layer, which produces a hinder to CPU and CPU peruses bundle in,

At the information connect layer, goal MAC address is verified whether the parcel is bound to this machine, If indeed, the bundle is sent up to the system layer.

At the IP layer, parcel approval like checksum confirmation, and so on is done and afterward gave to the applicable vehicle layer.

Transport later at that point gives it to the fitting port with the goal that it arrives at the right application.

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3 years ago

Only cool people answer this question. are you cool?

gayaneshka [121]

Answer:

Yes

Explanation:

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2 years ago

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3 years ago

Algorithm and flowchart to find the perimeter and area of square

kupik [55]

Answer:

I can give you the perimeter "algorithm" but not the flowchart.

Here you go:

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w = width/height

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to include both width and height, we would instead use:

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This also works with rectangles ^

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3 years ago

Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have alre

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Answer:

The c++ program to implement the while loop is given below.

#include <iostream>

using namespace std;

int main() {

// declaration of integer variables

int k, n, total;

// initialization of integer variables

k=1, n=4, total=0;

// loop executed till value of k becomes equal to value of n

while( k <= n ){

// cube of every integer is added to the variable total

total = total + ( k * k * k );

// value of k is incremented to go to the next number

k = k + 1 ;

}

return 0;

}

Explanation:

The program begins with the declaration of integer variables.

int k, n, total;

This is followed by initialization of these variables.

k=1, n=4, total=0;

The while loop runs over the variable k which is initialized to 1. The loop runs till value of k reaches the value of integer n.

First, cube of k is computed and added to the variable total.

After first execution of the loop, total is initialized to the cube of 1.

Next, value of variable k is incremented by 1 so that k is initialized to next integer.

After first execution of the loop, k is incremented from 1 to 2.

while( k <= n )

{

total = total + ( k * k * k );

k = k + 1 ;

}

When the value of k reaches the value of integer n, the cube of n is calculated and added to the variable, total.

When k is incremented, it becomes more than n and hence, loop gets terminated.

As the return type of main is int, the program terminates with the statement shown below.

return 0;

No output is displayed as it is not mentioned in the question.

No user input is taken as it is mentioned that integer variables are already initialized.

4 0

3 years ago