An employee of a large corporation remotely logs into the company using the appropriate username and password. The employee is a

Response: explained in the explanation section

Explanation:

Given that:

Assume D(k) =║ true if [1::: k] is a valid sequence of words, or false otherwise

• In determining D(n)

the sub problem s[1::: k] is a valid sequence of words IFF s[1::: 1] is valid and s[ 1 + 1::: k] is a valid word.

Thus, we derive that D(k) is defined by the following recurrence relation:

D(k) = ║ false max(d[l] ∧ DICT(s[1 + 1::: k]) otherwise

Algorithm:

Valid sentence (s,k)

D [1::: k]             ∦ array of boolean variables.

for a ← 1 to m

do;

d(0) ← false

for b ← 0 to a - j

for b ← 0 to a - j

do;

if D[b] ∧ DICT s([b + 1::: a])

d (a) ← True

(b). Algorithm Output

      if D[k] == True

stack = temp stack            ∦stack assists in displaying the strings in order

c = k

while C > 0

stack push (s [w(c)]::: C] // w(p) denotes the index in s[1::: k] of the valid word // at position c

P = W (p) - 1

output stack

= 0 =

cheers, I hope this aids you!!!

Answer:

def mph_and_minutes_to_miles(hours_traveled, miles_traveled):

   hours_traveled = minutes_traveled / 60

   miles_traveled = hours_traveled * miles_per_hour

   print('Miles: %f' % miles_traveled)

Explanation:

This function, written in Python, specifies two input parameters (hours_traveled, miles_traveled)

When executed, the main function prompts the user for necessary data, calls this method, and passes the inputs as arguments. Refer to the complete code and output that follows.

Response:

The code is provided below

Clarification:

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.nio.charset.StandardCharsets;

public class Main {

  /**

  *

  * Process each line of input.

  *

  */

  public static void main(String[] args) throws IOException {

      InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);

      BufferedReader in = new BufferedReader(reader);

      String line;

      while ((line = in.readLine())!= null) {

          String[] splittedInput = line.split(";");

          String pattern = splittedInput[0];

          String blobs = splittedInput[1];

          Main.doSomething(pattern, blobs);

      }

  }

  public static void doSomething(String pattern, String blobs) {

      // Implement your code here. You can create more methods or classes if needed

      int total = 0;

      String arrblobs[] = blobs.split("\\|");

      for (int i = 0; i < arrblobs.length; ++i) {

          int count = 0, index = 0;

          for (;;) {

              int position = arrblobs[i].indexOf(pattern, index);

              if (position < 0)

                  break;

              count++;

              index = position + 1;

          }

          System.out.print(count + "|");

          total += count;

      }

      System.out.println(total);

  }

}

Answer:

The required number of bits to address 8K words is 13.

Explanation:

Addressable words total 8000, where a word is defined as the smallest unit of memory that can be addressed.

These 8000 words can be accessed using units. To find the value of n corresponding to the number of words, you need to calculate

It's clear that 13 bits are necessary to address 8K words.