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tekilochka [14]
3 years ago
8

He would like to get rid of the graph. What will accomplish

Computers and Technology
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

All options may be incorrect

Explanation:

In ribbon tab there is page layout please select if end-user using word 2007 or word 2010

And if the end-user has to choose a design tab from the ribbon tab if he or she using word 2013 or word 2016.

Change the color setting to no color by click on page click and page background group end-user can get rid of the graph in the word document

After doing all the above method still, graph chart is visible better to close word document  and open the different template and try this last option to rid of the graph in a word document

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Joe, a user, wants his desktop RAID configured to allow the fastest speed and the most storage capacity. His desktop has three h
kirill115 [55]

Answer:

A. 0

Explanation:

The technician should configure the RAID 0 for Joe.

RAID 0 also referred to as the striped volume or stripe set is configured to allow the fastest speed and the most storage capacity by splitting data evenly across multiple (at least two) disks, without redundancy and parity information.

Also, RAID 0 isn't fault tolerant, as failure of one drive will cause the entire array to fail thereby causing total data loss.

7 0
3 years ago
Which statement best describes one reason why assembly language is easier
viktelen [127]

Answer:

machine language uses binary code and assembly language uses mnemonic codes to write a program.

Explanation:

In a nutshell, machine language uses binary code, which is almost impossible for humans to decipher, whereas assembly language uses mnemonic codes to write a program. Mnemonic codes make it simpler for humans to understand or remember something, and so make the language a bit easier for humans to use than machine code.

6 0
2 years ago
Write a program that calculates and displays the number of minutes in a month. This program does not require any user input, and
dimulka [17.4K]

Answer:

Code to the answer is shown in the explanation section

Explanation:

import java.util.Scanner;

public class Question {

   public static void main(String args[]) {

     Scanner scan = new Scanner(System.in);

     System.out.println("Please enter the days of the month: ");

     int daysOfMonth = scan.nextInt();

     int minuteOfMonth = daysOfMonth * 60 * 24;

     System.out.println(minuteOfMonth);

   }

}

// 60 represents the number of minutes in one hour

// 24 represents the number of hours in a day

4 0
3 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
What obstacles could prevent you from getting and keeping a job
stepladder [879]
Drugs, No Collage, Gangs, bad grades, Criminal Records →
7 0
3 years ago
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