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solmaris [256]
3 years ago
10

What differences in traffic patterns account for the fact that STDM is a cost-effective form of multiplexing for a voice telepho

ne network and FDM is a cost-effective form of multiplexing for television and radio networks, yet we reject both as not being cost effective for a general-purpose computer network?
Computers and Technology
1 answer:
fgiga [73]3 years ago
7 0

Answer:

Answer explained below

Explanation:

statistical time division multiplexing (STDM) and frequency-division multiplexing (FDM)

They both STDM and FDM works well for the channels which transmit continuously like telephone, FM radio and TV which continuously transmit signal. Since inactive lines are wasted this makes these venues a good fit for STDM and FDM. which means if the signals are dropped or not received to the receiver than it won't have to send the signals again which is why STDM and FDM are useful in these cases.

In Traditional Computer Communications system, the system have tended to be a lot more 'bursty' with idle periods. Hence if we use the FDM and STDM than the usage patterns of STDM and FDM would waste this idle time. so that is why we tend not to use the FDM and STDM in the general purpose computer network.

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Water is constantly in motion.<br> True or false
motikmotik

Answer: true

Explanation: yes water is constantly in motion

7 0
3 years ago
Write a loop statement to count and display the number of positive integers in the array. That is, your loop should display the
dlinn [17]

Answer:

int count =0;

    for(int i=0;i<10;i++)

    {

        if(myArray[i]>=0)

        {

            count++;

        }

    }

    cout<<"Number of positive integers is "<<count<<endl;

Explanation:

The above written loop is for counting positive integers in the myArray[].

For counting we have taken a count integer initialized with 0.On iterating over the array if the element is greater than or equal to 0 we consider it as positive and increasing the count.At the end printing the count.

3 0
3 years ago
What is the most common option for fighting radiation?
tigry1 [53]
Lead Sheets, Plates, Slabs, & Foils.

Lead Shot
6 0
3 years ago
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter
trapecia [35]

Answer:

5 Letters

Explanation:

So we need 12 distinct codes made of a single letter or a pair of letters.

What would be the least number of letters?

Lets try with 3 letters A, B and C

The possible combinations are: A, B, C, AB, AC, BC

These are 6 codes and we need 12 so lets try more A, B, C and D

The possible combinations are: A, B, C, D, AB, AC, AD, BC, BD, CD

These are 10 codes and we need 12 so lets try more A, B, C, D and E

The possible combinations are:

A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

Finally we got 15 distinct codes which are more than 12 so the least number of letters needed are 5.

Using formula:

Four letters = 4C1 + 4C2 = 4 + 6 = 10

Five letters = 5C1 + 5C2 = 5 + 10 = 15

5 0
3 years ago
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nlexa [21]

Answer:

Internet

Explanation:

just took the test and got A

7 0
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