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3 years ago

Using the _______ list, you can select the number of photos that will appear on each slide.

Computers and Technology

1 answer:

Svetach [21]3 years ago

4 0

Answer:

Picture Layout

Explanation:

There 4 options for this question:

A. Theme

B. Frame shape

C. Picture layout

D. Pictures in album

We can do an images design with the option (Picture Layout) we can add different images with text and some default layout when you click on picture layout, you will see all the example and if you pass the mouse over the examples, you will see a preview of the layout, this option helps us to make a design with different picture in one slide.

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Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

6 0

3 years ago

Sabiendo que z1 = 10, z2 = 20, z3 =10 , z4 = 20, n1 = 1200 rpm.

romanna [79]

Answer:

a. engranaje 1 → engranaje 2 → engranaje 3 → engranaje 4

z1 = 10 z2 = 20 z3 =10 z4 = 20

b. n2 = 600 rpm, n3 = 1200 rpm, n4 = 600 rpm

la relacion de transmision = 2

Explanation:

where z1 = 10, z2 = 20, z3 = 10, z4 = 20

and n1 = 1200 rpm

n2 = n1 * z1/z2 = (1200 * 10)/ 20 = 600 rpm

n3 = n2 * z2/z3 = (600 * 20)/ 10 = 1200 rpm

n4 = n3 * z3/z4 = (1200 * 10)/ 20 = 600 rpm

la relacion de transmision = z2/z1 * z3/z2 * z4/z3

= 20/10 * 10/20 * 20/10 = 2 * 0.5 * 2 = 2

7 0

3 years ago

Write a RainFall class that stores the total rainfall for each of 12 months into an array of doubles. The program should have me

RSB [31]

Answer:

Explanation:

The code provided worked for the most part, it had some gramatical errors in when printing out the information as well as repeated values. I fixed and reorganized the printed information so that it is well formatted. The only thing that was missing from the code was the input validation to make sure that no negative values were passed. I added this within the getMonths() method so that it repeats the question until the user inputs a valid value. The program was tested and the output can be seen in the attached image below.

import java.io.*;

import java.util.*;

class Rainfall

{

Scanner in = new Scanner(System.in);

private int month = 12;

private double total = 0;

private double average;

private double standard_deviation;

private double largest;

private double smallest;

private double months[];

public Rainfall()

{

months = new double[12];

}

public void setMonths()

{

for(int n=1; n <= month; n++)

{

int answer = 0;

while (true) {

System.out.println("Enter the rainfall (in inches) for month #" + n + ":" );

answer = in.nextInt();

if (answer > 0) {

months[n-1] = answer;

break;

}

public void getTotal()

{

total = 0;

for(int i = 0; i < 12; i++)

{

total = total + months[i];

}

System.out.println("The total rainfall for the year is " + total);

}

public void getAverage()

{

average = total/12;

System.out.println("The average monthly rainfall is " + average);

}

public void getLargest()

{

double largest = 0;

int largeind = 0;

for(int i = 0; i < 12; i++)

{

if (months[i] > largest)

{

largest = months[i];

largeind = i;

}

System.out.println("The largest amount of rainfall was " + largest +

" inches in month " + (largeind + 1));

}

public void getSmallest()

{

double smallest = Double.MAX_VALUE;

int smallind = 0;

for(int n = 0; n < month; n++)

{

if (months[n] < smallest)

{

smallest = months[n];

smallind = n;

}

System.out.println("The smallest amount of rainfall was " + smallest +

" inches in month " + (smallind + 1));

}

public static void main(String[] args)

{

Rainfall r = new Rainfall();

r.setMonths();

r.getTotal();

r.getSmallest();

r.getLargest();

r.getAverage();

}

8 0

2 years ago

Write a program that will predict the size of a population of organisms. The program should ask the user for the starting number

sineoko [7]

Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

//Declaration of variables in the program

double start_organisms;

double daily_increase;

int days;

double updated_organisms;

//The user enters the number of organisms as desired

cout << "Enter the starting number of organisms: ";

cin >> start_organisms;

//Validating input data

while (start_organisms < 2)

{

cout << "The starting number of organisms must be at least 2.\n";

cout << "Enter the starting number of organisms: ";

cin >> start_organisms;

}

//The user enters daily input, here's where we apply the 5.2% given in question

cout << "Enter the daily population increase: ";

cin>> daily_increase;

//Validating the increase

while (daily_increase < 0)

{

cout << "The average daily population increase must be a positive value.\n ";

cout << "Enter the daily population increase: ";

cin >> daily_increase;

}

//The user enters number of days

cout << "Enter the number of days: ";

cin >> days;

//Validating the number of days

while (days<1)

{

cout << "The number of days must be at least 1.\n";

cout << "Enter the number of days: ";

cin >> days;

}

//Final calculation and display of results based on formulas

for (int i = 0; i < days; i++)

{

updated_organisms = start_organisms + (daily_increase*start_organisms);

cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

start_organisms = updated_organisms;

}

system("pause");

return 0;

//end

}

Explanation:

6 0

3 years ago

You accidentally find someone's password and use it to get into a system. this is hacking.

Umnica [9.8K]

I'd go with yes, but are you doing something malicious or just being nosey?

4 0

2 years ago