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yKpoI14uk [10]
3 years ago
6

You also learn in class that one kilobyte of computer memory will hold about 2/3 of a page of typical text (without formatting).

A typical book has about 400 pages. How many such books can a 9-gig flash drive hold? _______________
Computers and Technology
1 answer:
Travka [436]3 years ago
8 0

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.

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3 years ago
Read 2 more answers
In this scenario, two friends are eating dinner at a restaurant. The bill comes in the amount of 47.28 dollars. The friends deci
pickupchik [31]

Answer:

The java program for the given scenario is as follows.

import java.util.*;

import java.lang.*;

public class Main

{

   //variables for bill and tip declared and initialized

   static double bill=47.28, tip=0.15;

   //variables for total bill and share declared

   static double total, share1;

public static void main(String[] args) {

    double total_tip= (bill*tip);

    //total bill computed

    total = bill + total_tip;

    //share of each friend computed

    share1 = total/2;

    System.out.printf("Each person needs to pay: $%.2f", share1);  

}

}

Explanation:

1. The variables to hold the bill and tip percent are declared as double and initialized with the given values.

static double bill=47.28, tip=0.15;

2. The variables to hold the values of total bill amount and total tip are declared as double.

3. All the variables are declared outside main() and at class level, hence declared as static.

4. Inside main(), the values of total tip, total bill and share of each person are computed as shown.

double total_tip= (bill*tip);

total = bill + total_tip;

share1 = total/2;

5. The share of each person is displayed to the user. The value is displayed with only two decimal places which is assured by %.2f format modifier. The number of decimal places required can be changed by changing the number, i.e. 2. This format is used with printf() and not with println() method.

System.out.printf("Each person needs to pay: $%.2f", share1);  

6. The program is not designed to take any user input.

7. The program can be tested for any value of bill and tip percent.

8. The whole code is put inside a class since java is a purely object-oriented language.

9. Only variables can be declared outside method, the logic is put inside a method in a purely object-oriented language.

10. As shown, the logic is put inside the main() method and only variables are declared outside the method.

11. Due to simplicity, the program consists of only one class.

12. The output is attached.

5 0
3 years ago
Read 2 more answers
For a new version of processor, suppose the capacitive load remains, how much more energy will the processor consume if we incre
erastovalidia [21]

Answer:

The answer is below

Explanation:

The amount of power dissipated by a processor is given by the formula:

P = fCV²

Where f = clock rate, C = capacitance and V = Voltage

For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:

P(old) = fCV²

For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:

P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor

Hence, the new processor is 1.728 times (72.8% more) the power of the old processor

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