Complete the sentence about a presentation delivery method. A(n) ____ allows you to transmit your presentations over the Internet using ____.

1: A. Presentation software
B. webcast
C. external monitor

2: A. Projectors
B. CDs
C. Streaming technology

Answer:

The different variable in C is 21213316700.

Explanation:

Given value:

Total value  = letters + underscore value

Total value  = 52 + 1

Total value =53

choice for first character = 53 letters +10 digits

first character = 63

choice for remaining characters

So,

Variable number With one 1 character = 53

Variable number With 2 character  = 53 × 63

Variable number With 3 character = 53 × 63²

Variable number With 4 character  = 53 × 63³

.

.

.

Variable number With 7 character  =

Total difference variable = 53 + 53 × 63+ 53 × 63²+ 53 × 63³+....+

Total difference variable = 53(1 + 63 + 63²+ 63³+ .... + )

Formula:

Total difference variable

Answer:

import java.util.Scanner;

public class num9 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter year");

       int givenYear =in.nextInt();

       if(givenYear>=2101){

           System.out.println("Distant Future");

       }

       else if(givenYear>=2001){

           System.out.println("21st Century");

       }

   }

}

Explanation:

• Using Java programming Language
• Import Scanner class to receive user input of the variable givenYear
• Use if statement to check the first condition if(givenYear>=2101)
• Use else if statement to check the second condition if(givenYear>=2001)
• print Distant future and 21st century respectively

Using First fit algorithm:

• P1 will be allocated to F4. With this, F4 will have a remaining space of 5MB from (205 - 200).
• P2 will be allocated to F1. With this, F1 will have a remaining space of 85MB from (100 - 15).
• P3 will be allocated F5. With this, F5 will have a remaining space of 115MB from (300 - 185).
• P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
• P5 will be allocated to F6. With this, F6 will have a remaining space of 10MB from (185 - 175).
• P6 will be allocated to F2. With this, F2 will have a remaining space of 90MB from (170 - 80).

Using Best-fit algorithm:

• P1 will be allocated to F4. With this, F4 will have a remaining space of 5MB from (205 - 200).
• P2 will be allocated to F3. With this, F3 will have a remaining space of 25MB from (40 - 15).
• P3 will be allocated to F6. With this, F6 will have no remaining space as it is entirely occupied by P3.
• P4 will be allocated to F1. With this, F1 will have a remaining space of of 25MB from (100 - 75).
• P5 will be allocated to F5. With this, F5 will have a remaining space of 125MB from (300 - 175).
• P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).
• 100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

Using Worst-fit algorithm:

• P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
• P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
• P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
• P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
• P5 will not be allocated to any of the available space because none can contain it.
• P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

Properly labeling the six different processes would be:

• 200MB (P1),
• 15MB (P2),
• 185MB (P3),
• 75MB (P4),
• 175MB (P5)
• 80MB (P6).

Best fit algorithm is the best as the name suggests, while the worst fit algorithm is the worst as not all memory is allocated

Read more about memory partitions here:
brainly.com/question/12726841

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
2. P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
3. P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
4. P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
5. P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
2. P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
3. P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
4. P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
5. P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
6. P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

1. P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
2. P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
3. P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
4. P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
5. P5 will not be allocated to any of the available space because none can contain it.
6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

Answer:  a) 0,19 seg. b) 3,2 seg. c) 11,5 seg.  d) 192 seg.

Explanation:

a)  For each pixel in the image, we use 8 bits + 1 start bit + 1 stop bit= 10 bits.

• Number of Pixels: 800*1200= 960,000.
• Bits transmitted: 960,000 x 10 = 9.6 * 10⁶ bits = 9.6 Mbits
• if  the modem is able to transmit up to 50 Mbits in a second, we can calculate how much time it will be needed to transmit 9.6 Mbits, using this equality:

• 50 Mbit = 1 sec
• 9.6 Mbit = x  ⇒  t= 9.6 / 50 = 0.19 sec

b)  If the modem speed changes to 3 Mb/s, all we need to do is just use the same equality, as follows:

• 3 Mbit = 1 sec
• 9.6 Mbit = x  ⇒  t= 9.6 / 3 = 3.2 sec

c) Now, if we need to transmit a colored image , at a rate of 20 f/sec, we need to calculate first how many bits we need to transmit, as follows:

• 1 Frame= 800*1200* (24 bits + 3 start bits + 3 stop bits)=  28.8  Mbits
• 1 Second= 20 frames/ sec = 28.8 Mbits *20 = 576 Mbits.
• If the modem speed is 50 Mb/s, we can use the same formula that we used for a) and b), as follows:

• 50 Mbit = 1 sec

• 576 Mbits = x  ⇒  t= 576 / 50 = 11.5 sec

d) Same as c) replacing 50 Mb/s by 3 Mb/s, as follows:

• 3 Mbit = 1 sec
• 576 Mbits = x  ⇒  t= 576 / 3 = 192 sec

The time that's required to transmit an image of 1200x800 pixels with 8 bits for gray is 151.6 million seconds.

How to calculate the time taken?

The time required to transmit an image of 1200x800 pixels with 8 bits for gray will be:

= (1200 × 800 × 8) / 50

= 151.6 million seconds.

The time needed to be at 3 M bits/sec, a representative of download speed of a DSL connection will be:

= 7.68/3

= 2.56 seconds.

Learn more about time on:

brainly.com/question/4931057

Answer:

The answer is "Public Key".

Explanation:

PKI stands for public key infrastructure. It is a collection of functions, protocols, equipment, code, and procedures that require to create, maintain, transmit, store and cancel encrypted certs and handle the authentication of a public key.  

• It is used to sign documents digitally.
• It works for propagating with Trojans and other malware.

Written here is a Python program that calculates the total weekly pay for an employee based on the given inputs:

python

# Get inputs from the user

hourly_wage = float(input("Enter the wage: $"))

regular_hours = float(input("Enter the regular hours: "))

overtime_hours = float(input("Enter the overtime hours: "))

# Calculate total weeklypay

regular_pay = hourly_wage * regular_hours

overtime_pay = 1.5 * hourly_wage * overtime_hours

total_weekly_pay = regular_pay + overtime_pay

# Display the result

print(f"The total weekly pay is ${total_weekly_pay:.1f}")

How to use the code

Copy and paste this code into a Pythoninterpreter or a script file, and run it. It will prompt you to enter the hourly wage, regular hours, and overtime hours, and then it will calculate and display the total weekly pay based on the given formula.

In the example you provided, the result was rounded to one decimal place. If you want a different level of precision in the output, you can adjust the formatting in the `print` statement accordingly.

Learn more about Program at:

brainly.com/question/30783869

#SPJ3

wage = float(input("Enter the wage: $"))

regular_hours = float(input("Enter the regular hours: "))

overtime_hours = float(input("Enter the overtime hours: "))

print(f"The total weekly pay is ${(regular_hours * wage) + ((wage*1.5) *overtime_hours)}")

I hope this helps!