Answer:
B)
Explanation:
The while loop runs as long as two conditions are satisfied, as indicated by the && logical operator.
The first condition- arr[pos] != val
checks to see if the value in the array index, pos, is equal to the given value and while it is not equal to it, the second condition is checked.
The second condition(pos < are.length), checks to see if the index(pos) is less than the length of the array. If both conditions are true, the program execution enters the while loop.
The while loop is only terminated once arr[pos] == Val or pos == arr.length.
how can you hack on a cumputer witch one chrome hp
Answer:
http://www.hackshop.org/levels/basic-arduino/hack-the-chromebook
Explanation:
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
Log onto the Internet and use a search engine to find three Web sites that can be models for the new site. (At least one should sell sporting goods.) For each site you selected, list the URLs in your document. Tell why you chose them. Navigate to the three sites you choose and take notes about at least three things you like and don’t like about the sites.
Please see the following file for your required answer, thank you.
Suppose a byte-addressable computer using set-associative cache has 2 16 bytes of main memory and a cache size of 32 blocks and each cache block contains 8 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the size of the tag, set, and offset fields
Answer:
Following are the responses to these question:
Explanation:
The cache size is 2n words whenever the address bit number is n then So, because cache size is 216 words, its number of address bits required for that cache is 16 because the recollection is relational 2, there is 2 type for each set. Its cache has 32 blocks, so overall sets are as follows:
[tex]\text{Total Number of sets raluired}= \frac{\text{Number of blocks}}{Associativity}[/tex]
[tex]=\frac{32}{2}\\\\ =16\\\\= 2^4 \ sets[/tex]
The set bits required also are 4. Therefore.
Every other block has 8 words, 23 words, so the field of the word requires 3 bits.
For both the tag field, the remaining portion bits are essential. The bytes in the tag field are calculated as follows:
Bits number in the field tag =Address Bits Total number-Set bits number number-Number of bits of words
=16-4-3
= 9 bit
The number of bits inside the individual fields is therefore as follows:
Tag field: 8 bits Tag field
Fieldset: 4 bits
Field Word:3 bits
Create a map using the Java map collection. The map should have 4 entries representing students. Each entry should have a unique student ID for the key and a student name for the element value. The map content can be coded in directly, you do not have to allow a user to enter the map data. Your program will display both the key and the value of each element.
Answer:
In Java:
import java.util.*;
public class Main{
public static void main(String[] args) {
Map<String, String> students = new HashMap<>();
students.put("STUD1", "Student 1 Name");
students.put("STUD2", "Student 2 Name");
students.put("STUD3", "Student 3 Name");
students.put("STUD4", "Student 4 Name");
for(Map.Entry m:students.entrySet()){
System.out.println(m.getKey()+" - "+m.getValue()); }
}}
Explanation:
This creates the map named students
Map<String, String> students = new HashMap<>();
The next four lines populates the map with the ID and name of the 4 students
students.put("STUD1", "Student 1 Name");
students.put("STUD2", "Student 2 Name");
students.put("STUD3", "Student 3 Name");
students.put("STUD4", "Student 4 Name");
This iterates through the map
for(Map.Entry m:students.entrySet()){
This prints the details of each student
System.out.println(m.getKey()+" - "+m.getValue()); }
ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP ASAP NEED HELP
Answer:
10.b
11.c
12.c
13.a
14.d
15.b
16.c
Explanation:
please brainleist please ♨️❤☺️☻
An integrated circuit RAM chip has a capacity of 512 words of 8 bits each where the words are organized using a one-dimensional layout.a) How many address lines are there on the chip?b) How many such chips would be needed to construct a 4Kx16 memory?c) How many address and data lines needed for a 4Kx16 memory?#addresss lines = #data lines =
Answer:
A. 9
B. 16
C. Number of addresses = 12 number of data lines = 16
Explanation:
The capacity of this chip is 512 x 8
That is 512 words and these words have 8 bits
A. The number of the address lines would be
log2(512) = 2⁹
So the answer is 9
B. The total number of chips required
= 512 x 8 = 4096
(4096 x 2)/ 512 = 16
C. Number of address = log4092 = 2¹²
= 12
The number of data lines = 8x2 = 16
Thank you!