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Rufina [12.5K]
2 years ago
11

What is the interface of an app?

Computers and Technology
1 answer:
Darina [25.2K]2 years ago
3 0

Explanation:

An application interface or user interface ,is the set of features an application provides so that user may supply input to and recieve output from,the program.

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Mary needs to choose the menu in order to place the text in a desired fashion around the image.
n200080 [17]
Can you be more specific?

7 0
3 years ago
Read 2 more answers
You also learn in class that one kilobyte of computer memory will hold about 2/3 of a page of typical text (without formatting).
Travka [436]

Answer:

The flashdrive can hold 35389 400-pages-books

Explanation:

If \frac{2}{3} of a page occupies 1 kB of memory, we can calculate how much memory a book will take

2/3kB=1 page\\ x kB=400 pages\\266.67 kB=400 pages

Now that we know that a book average file size is about 266,67 kB, we calculate how many of them can a 9 GB flash drive hold.

To do the calculation, we have to know how many kilobytes are in 9 gigabytes.

There is 1024 kilobytes in a megabyte, and 1024 megabytes in a gigabyte, so:

Memory_{in _kilobytes}^{} =1024\frac{kilobytes}{megabytes} *1024\frac{megabytes}{gigabytes}*9=9437184 kilobytes

Finally, knowing the average file size of a book and how much memory in kilobytes the 9 GB flash drive holds, we calculate how many books can it hold.

Books=\frac{Flash drive memory}{Filesize of a book} =\frac{9437184kilobytes}{266.67\frac{kilobytes}{book} } =35389,4 books

The flashdrive can hold 35389 400-pages-books, or 14155776 pages of typical text.

8 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
Mr. Hill has 27 students in his class and Mr. Chang has 24 students in his class. Both classes will be divided into equal sized
iVinArrow [24]
25! 27+24=51 if you divide 51 by two, you get 25.5
Because we can’t have a 0.5th of a child, you can round down to the nearest whole number. Leaving you with 25. Hope this helped :)

(modulus division works well for these kinds of questions)
6 0
3 years ago
How to find the biggest files on your computer?
ki77a [65]
For purposes of freeeing up some hard disk space and find the biggest files on you, there are couple of ways one can do so.
Using explorer, open my computer and then click on the search box at the top most right side. A little window will pop up with several options. Since you want to find the biggest files you will select size:Largest option. You can also press WIN + F on the keyboard and follow the same procedure.
4 0
3 years ago
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