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ValentinkaMS [17]
3 years ago
12

Which device lets you hear audio on your computer?

Computers and Technology
2 answers:
andrew11 [14]3 years ago
7 0
Head phones or earbuds or a speaker
Cerrena [4.2K]3 years ago
5 0
Headphones,Speaker,Or  Earbuds
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What is online school like 3 sentences
lara [203]

Answer:

its okay. i mean like you have to do a bunch off stuff. andsomtimes it gets a little boring.

Explanation:

i am doing online school.

3 0
3 years ago
Read 2 more answers
Define a public static method named s2f that takes two String arguments, the name of a file and some text. The method creates th
frez [133]

Answer:

Java solution (because only major programming language that has public static methods)

(import java.io.* before hand)

public static boolean s2f(String fileName, String text){

   try{

       PrintWriter out = new PrintWriter(new File(fileName));

       out.println(text);

       out.close();

       return true;

   }

   catch(Exception e){

       return false;

   }

}

6 0
2 years ago
The last step on Kotter’s Eight-Step Change Model is to anchor the changes in corporate culture; to make anything stick, it must
m_a_m_a [10]

Answer:

Therefore, it is important to find opportunities to integrate security controls into day-to-day routines.

Do you believe this to be true- Yes.

In general, implementing security policies occurs in isolation from the business perspectives and organizational values that define the organization’s culture. Is this correct or incorrect? - Incorrect

Explanation:

Truly, it is important to find opportunities to integrate security controls into day-to-day routines, this is in order to minimize future security threats by formulating company-wide security policies and educating employees on daily risk prevention in their work routines. In the operational risk controls, vigilant monitoring of employees must be implemented in order to confirm that policies are followed and to deter insider threats from developing.

Flexing and developing policies as resources and priorities change is the key to operational risk controls.

These risk controls implementation in organizational security is not a one-time practice. Rather, it is a regular discipline that the best organizations continue to set and refine.

For better preparation of an organization towards mitigating security threats and adaptation to evolving organizational security needs, there must be a proactive integration of physical information and personnel security while keeping these risk controls in mind.

12. In general, implementing security policies occurs in isolation from the business perspectives and organizational values that define the organization’s culture - Incorrect.

When security policies are designed, the business perspectives and the organizational values that define the organization’s culture must be kept in mind.

An information security and risk management (ISRM) strategy provides an organization with a road map for information and information infrastructure protection with goals and objectives that ensure capabilities provided are adjusted to business goals and the organization’s risk profile. Traditionally, ISRM has been considered as an IT function and included in an organization’s IT strategic planning. As ISRM has emerged into a more critical element of business support activities, it now needs its own independent strategy to ensure its ability to appropriately support business goals and to mature and evolve effectively.

Hence, it is observed that both the Business Perspective and Organisational goals are taken into consideration while designing Security policies.

6 0
3 years ago
g You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned the weights an
ololo11 [35]

Answer:

A python code (Python recursion) was used for this given question

Explanation:

Solution

For this solution to the question, I am attaching code for these 2 files:

item.py

code.py

Source code for item.py:

class Item(object):

def __init__(self, name: str, weight: int, value: int) -> None:

  self.name = name

  self.weight = weight

  self.value = value

def __lt__(self, other: "Item"):

  if self.value == other.value:

if self.weight == other.weight:

  return self.name < other.name

else:

  return self.weight < other.weight

  else:

   return self.value < other.value

def __eq__(self, other: "Item") -> bool:

  if is instance(other, Item):

return (self.name == other.name and

self.value == other.value and

self.weight == other.weight)

  else:

return False

def __ne__(self, other: "Item") -> bool:

  return not (self == other)

def __str__(self) -> str:

  return f'A {self.name} worth {self.value} that weighs {self.weight}'

Source code for code.py:

#!/usr/bin/env python3

from typing import List

from typing import List, Generator

from item import Item

'''

Inductive definition of the function

fun3(0) is 5

fun3(1) is 7

fun3(2) is 11

func3(n) is fun3(n-1) + fun3(n-2) + fun3(n-3)

Solution 1: Straightforward but exponential

'''

def fun3_1(n: int) -> int:

result = None

if n == 0:

result = 5 # Base case

elif n == 1:

result = 7 # Base case

elif n == 2:

result = 11 # Base case

else:

result = fun3_1(n-1) + fun3_1(n-2) + fun3_1(n-3) # Recursive case

return result

''

Solution 2: New helper recursive function makes it linear

'''

def fun3(n: int) -> int:

''' Recursive core.

fun3(n) = _fun3(n-i, fun3(2+i), fun3(1+i), fun3(i))

'''

def fun3_helper_r(n: int, f_2: int, f_1: int, f_0: int):

result = None

if n == 0:

result = f_0 # Base case

elif n == 1:

result = f_1 # Base case

elif n == 2:

result = f_2 # Base case

else:

result = fun3_helper_r(n-1, f_2+f_1+f_0, f_2, f_1) # Recursive step

return result

return fun3_helper_r(n, 11, 7, 5)

''' binary_strings accepts a string of 0's, 1's, and X's and returns a generator that goes through all possible strings where the X's

could be either 0's or 1's. For example, with the string '0XX1',

the possible strings are '0001', '0011', '0101', and '0111'

'''

def binary_strings(string: str) -> Generator[str, None, None]:

def _binary_strings(string: str, binary_chars: List[str], idx: int):

if idx == len(string):

yield ''.join(binary_chars)

binary_chars = [' ']*len(string)

else:

char = string[idx]

if char != 'X':

binary_chars[idx]= char

yield from _binary_strings(string, binary_chars, idx+1)

else:

binary_chars[idx] = '0'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars[idx] = '1'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars = [' ']*len(string)

idx = 0

yield from _binary_strings(string, binary_chars, 0)

''' Recursive KnapSack: You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned

the weights and values of every item in the store. You are looking to

get the biggest score you possibly can but you are only one person and

your backpack can only fit so much. Write a function that accepts a

list of items as well as the maximum capacity that your backpack can

hold and returns a list containing the most valuable items you can

take that still fit in your backpack. '''

def get_best_backpack(items: List[Item], max_capacity: int) -> List[Item]:

def get_best_r(took: List[Item], rest: List[Item], capacity: int) -> List[Item]:

if not rest or not capacity: # Base case

return took

else:

item = rest[0]

list1 = []

list1_val = 0

if item.weight <= capacity:

list1 = get_best_r(took+[item], rest[1:], capacity-item.weight)

list1_val = sum(x.value for x in list1)

list2 = get_best_r(took, rest[1:], capacity)

list2_val = sum(x.value for x in list2)

return list1 if list1_val > list2_val else list2

return get_best_r([], items, max_capacity)

Note: Kindly find an attached copy of the code outputs for python programming language below

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3 years ago
An administrator has several cables plugged into a patch panel and needs to determine which one comes from a specific port. Whic
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