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Sedaia [141]
3 years ago
11

Who was the father of computer?????​

Computers and Technology
2 answers:
Maslowich3 years ago
7 0

Answer:

Charles Babbage

Explanation:

He created the computer in the early 19th century

adell [148]3 years ago
4 0

Answer:

Charles Babbage

Explanation:

You might be interested in
Assume you are given three variables, revenue, expenses, and profit, all of type Money (a structured type with two int fields, d
Andre45 [30]

Answer:

if(revenue.cents - expenses.cents < 0){

profit.dollars = revenue.dollars - expenses.dollars - 1;

profit.cents = 1 - revenue.cents - expenses.cents;

}

else{

profit.dollars = revenue.dollars - expenses.dollars;

profit.cents = revenue.cents - expenses.cents;

}

Explanation:

We know that profit is given as: revenue - expenses from the question.

From the given expression above;

if(revenue.cents - expenses.cents < 0)

then profit.dollar will be revenue.dollars - expenses.dollars - 1; the 1 is to be carry over to the cent part. And the profit.cent will be 1 - revenue.cents - expenses.cents;

else the profit.dollars and the profit.cent is computed directly without needing to carry over:

profit.dollars = revenue.dollars - expenses.dollars;

profit.cents = revenue.cents - expenses.cents;

7 0
3 years ago
Which type of worker would most likely be able to begin work after receiving a high school degree and completing an
makvit [3.9K]
I’m pretty sure it’s miner

“I’ll give you head I mean brainiest”
3 0
3 years ago
Preciso de ajuda urgente, é para amanhã cedo!!
Papessa [141]


Bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
5 0
3 years ago
Professor Midas drives an automobile from Newark to Reno along Interstate 80. His car’s gas tank, when full, holds enough gas
zmey [24]

Answer:

The GREEDY Algorithm

Explanation:

Based on the situation given in question, the Greedy algorithm shall give the optimal solution to professor

Suppose that the cities are at locations0 =x0< x1< . . . < x

We shall use the induction method to prove that G is the optimal solution valid for numbers less than n

We assume another solution Z which we initially consider to be optimum as well, based on that when Z fills the tank, it fills it to full level

Let us state the values in case of n intervals. Given below, we say that g1 is the first stop and z1 is also the first stop.

This can be written as ;

G=g1, g2, . . . , gk

Z=z1, z2, . . . , zk’

Here k’ <= k and k < n

Let I be an idex where for the first time gi is not equal to zi

Considering t= maxi Zi

Z′=g1, z2, z3, . . . , zk′

Now since z2, z3, . . . , zk′ should be an optimal stopping pattern for the problem otherwise we have chosen Z, with smaller gas filling (not feasible)

Using induction hypothesis we conclude thatg2, . . . , gk is an optimal stopping pattern, which is based on greedy algorithm

7 0
3 years ago
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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