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2 years ago

to verify that the ip address sorting criteria that has not been configured to overlap between different groups you can use the

1 answer:

7 0

For one to be able to verify that the ip address sorting criteria that has not been configured to overlap between different groups you can use the check IP group - testing.

<h3>What does IP group do?</h3>

The term IP Group is known to be one that looks at or is one whose purpose is so that it can generate a lot of a positive social and also that of environmental impact with financial returns.

Note that this is one that offers shareholders a kind of an exposure to a a lot of portfolio of opportunities that is made up of a high-growth businesses in regards to the growth markets.

Hence, For one to be able to verify that the ip address sorting criteria that has not been configured to overlap between different groups you can use the check IP group - testing.

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5.17 (Calculating Sales) An online retailer sells five products whose retail prices are as follows: Product 1, $2.98; product 2,

Dima020 [189]

Answer:

import java.util.Scanner;

public class Main

{

public static void main(String[] args) {

//Initialize the prices as constants

final double PRODUCT_1_PRICE = 2.98;

final double PRODUCT_2_PRICE = 4.50;

final double PRODUCT_3_PRICE = 9.98;

final double PRODUCT_4_PRICE = 4.49;

final double PRODUCT_5_PRICE = 6.87;

//Declare the other variables

int productNumber, quantitySold;

double total = 0.0;

//Create a Scanner object to get input

Scanner input = new Scanner(System.in);

//Create a while loop

while(true){

//Ask the user to enter the productNumber

System.out.print("Enter the product number or 999 to quit: ");

productNumber = input.nextInt();

// Stop the loop, if productNumber is 999(sentinel value, you may choose any value you want)

if(productNumber == 999)

break;

//Ask the user to enter the quantitySold

System.out.print("Enter the quantity sold: ");

quantitySold = input.nextInt();

//Create a switch statement that works depending on the productNumber entered.

//For example, if the productNumber is 1, it multiplies the quantitySold by PRODUCT_1_PRICE

// and adds the result to the total. If productNumber is 2, it does the same for product 2 ...

switch(productNumber){

case 1:

total += quantitySold * PRODUCT_1_PRICE;

break;

case 2:

total += quantitySold * PRODUCT_2_PRICE;

break;

case 3:

total += quantitySold * PRODUCT_3_PRICE;

break;

case 4:

total += quantitySold * PRODUCT_4_PRICE;

break;

case 5:

total += quantitySold * PRODUCT_5_PRICE;

break;

}

//Print the total (when the loop is done)

System.out.println("The total is $" + total);

}

Explanation:

*The code is in Java.

You may see the explanation as comments in the code.

4 0

3 years ago

Which action does not happen in each iteration of the repeat loop in the

iren2701 [21]

B. The number of sharks decreases. Explanation: hope it's help i learned about it and you too

7 0

3 years ago

HELP! WILL GIVE BRAINLIEST

maw [93]

Answer:

Creative, helpful, communicative, colorful, detailed

Explanation:

6 0

3 years ago

Read 2 more answers

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

3 years ago

A computer has __________processing device

Artist 52 [7]

Answer:

electronic processing device

4 0

3 years ago