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uysha [10]
3 years ago
10

Which hardware device is it most important to an experienced computer professional to install

Computers and Technology
1 answer:
kherson [118]3 years ago
3 0

Answer:

RAM

Explanation:

In the test

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What is the minimum number of bits you need to encode the 26 letters of the alphabet plus a space?Pace?
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5 bits

Explanation

A bit is the least or the smallest unit of data in a computer. They are  the units of information in information theory, consisting of the amount of information required to specify one of two alternatives 0 and 1. This is because A bit has a single binary value, either 0 or 1

If you use roman alphabet A to Z lets use log 2(26)=4.7 bits. when you round off it will be 5 bits.

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How many times will the loop body execute:
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I’d also say B, which is 2
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A cell reference in a formula allows you to point to another cell location
mario62 [17]

The cell reference refers to a cell on a worksheet and can be used in a formula to point another location.

Explanation:

A cell reference is a range of cells on a worksheet that can be used in a formula to find the values or data you want to calculate using that formula.

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6 0
3 years ago
Read 2 more answers
Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
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