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V125BC [204]
4 years ago
7

Which of these does an open office or Microsoft Office wizard do?

Computers and Technology
1 answer:
bearhunter [10]4 years ago
5 0
<span>c) guides you through the steps to complete a complex task</span>
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How can an antivirus protect your device​
Igoryamba
Antivirus software protects your device from viruses that can destroy your data, slow down or crash your device, or allow spammers to send email through your account. Antivirus protection scans your files and your incoming email for viruses, and then deletes anything malicious
3 0
2 years ago
Read 2 more answers
An uniterruptible power supply
velikii [3]

Answer:

An uninterruptible power supply  is an electrical apparatus that provides emergency power to a load when the input power source or mains power fails.

Explanation:

Your welcome :) PLZ mark brainliest

4 0
3 years ago
Write a program for a grocery store to calculate the total charge for customers. In the main: Your program should ask customer t
gavmur [86]

Answer:

Hello Joelwestwood! This is a good question to check your knowledge of subrountines and Arrays. Please find the implementation with the explanation below.

Explanation:

The solution can be implemented in a number of languages including C++, Java and Python to name a few. Though the implementation language is not specified in the question, i'll provide you the implementation code in Python and also in Java. Java implementation is a little complex because we need to define an array size to be able to add items to it, whereas Python allows us more flexibility by allowing dynamic size of Array.

JAVA IMPLEMENTATION

import java.util.Scanner;

import java.util.Arrays;

class TotalCharge {

 private static int[] price_array = new int[1];

 public void FillPriceArray(int price) {

   if (price_array.length == 1) {

     price_array[0] = price;

   } else {

     price_array = new int[price_array.length + 1];

     price_array[price_array.length + 1] = price;

   }

 }

 public int[] get_price_array() {

   return price_array;

 }

 public static void main(String args[]) {

   System.out.println("Please enter the total number of items to purchase: ");

   Scanner scan = new Scanner(System.in);

   int price = scan.nextInt();

   if (price <= 20) {

     System.out.println("Items being purchased are less than 20");

   } else {

     System.out.println("Items being purchased are more than 20");

   }

   TotalCharge itemsTotal = new TotalCharge();

   itemsTotal.FillPriceArray(price);

   System.out.println(Arrays.toString(itemsTotal.get_price_array()));

 }

}

PYTHON IMPLEMENTATION

calculate_total_charge.py

def FillPriceArray(price):

 price_array.append(price)

price_array = []

price = raw_input("Please enter the total number of items to purchase: ")

try:

 price = int(price)

 if price <= 20:

   print("Items being purchased are less than 20")

 else:

   print("Items being purchased are more than 20")

 FillPriceArray(price)

 print(price_array)

except ValueError:

 print("Invalid input! Exiting..")

8 0
3 years ago
Suppose that the splits at every level of quicksort are in the proportion 1 − α to α where 0 &lt; α ≤ 1/2 is a constant. Show th
Lady_Fox [76]

Answer:

Explanation:

The minimum depth occurs for the path that always takes the smaller portion of the

split, i.e., the nodes that takes α proportion of work from the parent node. The first

node in the path(after the root) gets α proportion of the work(the size of data

processed by this node is αn), the second one get (2)

so on. The recursion bottoms

out when the size of data becomes 1. Assume the recursion ends at level h, we have

(ℎ) = 1

h = log 1/ = lg(1/)/ lg = − lg / lg

Maximum depth m is similar with minimum depth

(1 − )() = 1

m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )

4 0
4 years ago
Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS
BlackZzzverrR [31]

Answer:

hello your question lacks the C segment so here is the C segment

while ( k<n )

{v[k] = v[k+1];

     k = k+1; }

Answer : while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

Explanation:

The MIPS assembly code corresponding to the C segment is

while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

4 0
3 years ago
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