This computer is used by touching your finger.

A size of a jumbo candy bar with rectangular shape is l x w x h. Due to rising costs of cocoa, the volume of the candy bar is to

hammer [34]

<h2>Please wait until more research is discovered...</h2>

Question

A size of a jumbo candy bar with rectangular shape is l x w x h. Due to rising costs of cocoa, the volume of the candy bar is to be reduced by p%.

To accomplish this, the management decided to keep the height of the candy bar the same, and reduce the length and width by the same amount.

For example, if l = 12, w = 7, h = 3, and p = 10, then the new dimension of the candy bar is 11.39 x 6.64 x 3.

Below is an example of how the completed program should work:

Enter the length, width, and height of the candy bar separated by space(s): 12 7 3

Enter the amount of the reduced volume as a percentage: 10

The new dimensions of the candy bar is: 11.38 x 6.64 x 3.00

Format your output with setprecision(2) to ensure the proper number of decimals for testing!

7 0

4 years ago

The _______ valve protects the air pump from reverse-exhaust pressure.

alexandr402 [8]

D. The answer for this is the gulp

7 0

3 years ago

Write a program that calculates the cost of a phone call. The user enters a positive integer that

san4es73 [151]

Add the following constants in the class definition, before the main method:

private final static float START_COST = 1.5f;
private final static float HIGH_TARIFF = 0.5f;
private final static float LOW_TARIFF = 0.25f;
private final static int FIXED_MINS = 2;
private final static int HIGH_TARIFF_MINS = 10;

Then add this to the main method you already had:

float price = START_COST;
if (x > FIXED_MINS) {
if (x <= HIGH_TARIFF_MINS) {
price += HIGH_TARIFF*(x-FIXED_MINS);
}
else {
price += HIGH_TARIFF*(HIGH_TARIFF_MINS-FIXED_MINS)
+ LOW_TARIFF*(x-HIGH_TARIFF_MINS);
}
}
System.out.printf("A %d minute call costs %.2f", x, price);

5 0

3 years ago

Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.

Annette [7]

The number of bits in the physical address is 26 bits. The number of entries in a page table is $\mathbf{2^{20}}$ entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

32-bit logical address
Page size = 4 KB = $\mathbf{2^{12} \ bytes }$
Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = $\mathbf{2^{26} \ bytes}$

Thus, the number of bits in the physical address is computed as:

= $\mathbf{log_2 \{Physical-memory-size\}}$

$=\mathbf{log_2(2^{26})}}$

= 26 bits

The number of entries in a page table = logical address space size/page size

The number of entries in a page table $\mathbf{= \dfrac{2^{32}}{2^{12}}}$

$\mathbf{=2^{20}}$ entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= $\mathbf{2^{20}\times 4 \ bytes}$

= 4 MB

b.

suppose that this system supports up to 2^30 bytes of physical memory.

The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is $\mathbf{2^{20}}$ entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

brainly.com/question/17004314

#SPJ1

4 0

2 years ago

How did The Gates by Christo and Jeanne-Claude affect the site?

Nadusha1986 [10]

Answer:

C. The experience of the park with The Gates was so different that some people saw the park in a new way.

Explanation:

The couple, Christo Javacheff and Jeanne-Claude Denat based in New York were two top contemporary artists who formed collaborative arts.

On the 3rd of January, 2005 they both made an art work known as The Gates, in Central Park.

The experience of the park with "The Gates" was so different that some people saw the park in a new way.

8 0

4 years ago

Read 2 more answers