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2 years ago

8

Users generally do not understand the concept of network drive sharing. they only know they can store their files "on my f: driv

e" and are unaware that someone else's f: drive might point to a different folder. what do we call this concept of having a drive letter associated to a user's network folder?

Computers and Technology

1 answer:

Ivenika [448]2 years ago

4 0

In windows, drive mapping.

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

2 years ago

Which method adds 10 to the right end of the array?<br> myArray.<br> insert<br> (10)

vladimir1956 [14]

JavaScript has a set of mutator functions that allow you to modify the contents of an array without referencing the individual elements.To add to to myArray we us the push() method

<h3>Adding Elements to an Array</h3>

To add to to myArray we us the push() method

(10)

myArray.push(10)

There are two mutator functions for adding elements to an array: push() and unshift(). The push() function adds an element to the end of an array:

var nums = [1,2,3,4,5]; print(nums); // 1,2,3,4,5 nums.push(6);

print(nums); // 1,2,3,4,5,6

var nums = [1,2,3,4,5]; print(nums); // 1,2,3,4,5 nums[nums.length] = 6; print(nums); // 1,2,3,4,5,6

Learn more about arrays here:

brainly.com/question/24275089

6 0

1 year ago

A company ABC asked you to design a simple payroll program that calculates and employee's weekly gross pay, including any overti

julsineya [31]

Answer:

<em>C++.</em>

#include <iostream>

using namespace std;

////////////////////////////////////////////////////////////////

int main() {

int weekly_hours = 0;

int hourly_rate;

float gross_pay = 0;

cout<<"Enter weekly hours worked: ";

cin>>weekly_hours;

cout<<"Enter hourly rate: ";

cin>>hourly_rate;

cout<<endl;

////////////////////////////////////////////////

if (weekly_hours > 40) {

gross_pay = (weekly_hours*hourly_rate) + ((weekly_hours*hourly_rate)*0.5);

}

else

gross_pay = weekly_hours*hourly_rate;

cout<<"Weekly gross pay: $"<<gross_pay;

////////////////////////////////////////////////

return 0;

}

3 0

2 years ago

. Html documents are created by adding what to text?

zvonat [6]

The answer is Tags, sorry about putting the other answer misread question.

7 0

2 years ago

#You may modify the lines of code above, but don't move them! #When you Submit your code, we'll change these lines to #assign di

Musya8 [376]

Answer:

The solution code is written in Python:

mystery_string = "Programming"
output = ""
for x in mystery_string:
output += x
print(output)

Explanation:

Firstly, create a variable mystery_string to hold a random string (Line 1).

Create an output variable to hold an output string (Line 2).

Create a for-loop to traverse the mystery_string character by character (Line 4). In the iteration, get a character from the mystery_string, and concatenate it with output string (Line 5). Print the output string (Line 6) before proceed to the next iteration.

8 0

2 years ago