Have you ever seen a Magic 8 ball? You ask it a yes-no question and then shake it to get a random response like “Signs point to yes!”, “Very doubtful”, etc. If you’ve never seen a Magic 8 ball, check out this video.

We encourage you to work in pairs for this challenge. Come up with 8 responses to yes-no questions. Write a program below that chooses a random number from 1 to 8 and then uses if statements to test the number and print out the associated random response from 1-8. If you need help with random numbers, see lesson 2.9.

public class Magic8Ball
{
public static void main(String[] args)
{
// Get a random number from 1 to 8

// Use if statements to print out 1 of 8 responses


}
}
1
public class Magic8Ball
2
{
3
public static void main(String[] args)
4
{
5
// Get a random number from 1 to 8
6

7
// Use if statements to print out 1 of 8 responses
8

9

10
}
11
}

Answers

Answer 1

Answer:

ok ok ok

]ok

ok

okm

ok

Explanation:


Related Questions

Consider sending a 1,600-byte datagram into a link that has an mtu of 500 bytes. suppose the original datagram is stamped with the identification number 291. how many fragments are generated? what are the values in the various fields in the ip datagram(s) generated related to fragmentation?

Answers

Explanation:

Step one

The maximum size of data field in each fragment = 480

(because there are 20 bytes IP header) Thus the number of required

fragments  [tex]=\frac{1600-20}{480} \\\\= \frac{1580}{480} \\\\=3.29\\\\[/tex]

thus the number of required fragment is 4

Step two

Each fragment will have identification number 291. each fragment except the last one will be of size 500 bytes (including IP header). the offset of the fragments will be 0, 60, 120, 180. each of the first 3 fragments will have

flag = 1; the last fragment will have flag =0

Using the appropriate formula, the number of fragments which would be present in the datagram to be sent would be 4

The minimum length of IP header = 20 bytes

Maximum transmission unit (mtu) = 500 bytes

Hence, the payload would be calculated thus :

mtu - header ;Payload = 500 - 20 = 480

Hence, the maximum size of data field per Fragment = 480 bytes

The number of fragments required :

[tex]\frac{datagram \: size - Header \: size}{payload} [/tex]

[tex] Number \: of \: fragments = \frac{1600 - 20}{480} = 3.29[/tex]

Hence, the number of fragments is 4

Size per Fragment would be 500 bytes each ; the last Fragment would be about 100 bytes

Each Fragment would bear the identification number 291.

Learn more : https://brainly.com/question/16289731

1).
What is a resume?
A collection of all your professional and artistic works.
A letter which explains why you want a particular job.
A 1-2 page document that demonstrates why you are qualified for a job by summarizing your
skills, education, and experience.
A 5-10 page document that details your professional and educational history in great detail.

Answers

Answer:

option 1

Explanation:

its not a job application cause your not appling for a job, a resume is a list of all the things you have done that would be beneficial to a job. for example, previous jobs, skills you have, hobby that pertain to a job you want, education and other professional things.

Hope this helps:)

Manuel is working on a project in Visual Studio. He wants to keep this program showing on the entire desktop, but he also needs to have several other applications open so that he can research the project. ​

Answers

Answer:

d. Task View

Explanation:

Based on the scenario being described within the question it can be said that the best feature for this would be the Windows 10 task view. This is a task switcher and  virtual desktop system included in the Windows 10 operating system, and allow the individual user to quickly locate, manage, open or hide different windows/tasks. Such as having several projects open in different monitors running at the same time.

A customer seeks to buy a new computer for a private use at home.The customer primarily needs the computer to use the Microsoft PowerPoint application for the purpose of practice presentation skills.As a sales person what size hard disc would you recommend and why?

Answers

Answer:

The most common size for desktop hard drives is 3.5 inches,  they tend to be faster and more reliable, and have more capacity. But they also make more noise.

Explanation:

If you are continually deleting and installing programs or creating content, the disc must have good reliability.

Keep in mind that larger hard drives are also a little slower, so it is preferable to opt for two smaller ones. Large hard drives are partitioned so there is no problem gettin

chbdg good performance, but if you put everything on one big disk and it breaks, you will lose everything.

If you buy 2 small disks, check that the motherboard does not limit the speed of a second hard disk.

this really isn't a question but it is mainly for Brianly I have 40+ so bainlist answers but on my rank it says I have 0/15 is this a computer bug or what because I've been on this app for a little bit more than a year now
I have a snip to prove how many brainliest answers I have provided below

Answers

Answer:

i have it but i have it fixed

Explanation:

might be a visual bug, try refreshing your page. if this doesnt work you can ask someone with a higher rank to help you.

the hose is 2 inches in diameter. What circumference does the plug need to be? remember to type just a number

Answers

Answer:

6.28

Explanation:

This problem bothers on the mensuration of flat  shapes, a circle.

Given data

Diameter d= [tex]2in[/tex]

Radius r =  [tex]\frac{d}{2} = \frac{2}{2} = 1in[/tex]

We know that the expression for the circumference of a circle is given as

[tex]C= 2\pi r[/tex]

Substituting our given data and solving for C we have

[tex]C= 2*3.142*1\\C= 6.28[/tex]

the blue course at the ski club is 5.8 km long. The red course is 10.3 km long. Mark skied the blue course 15 times and Katy skied the red course 15 times. How much further did Katy ski than Mark?​

Answers

Answer:

67.5km

Explanation:

mark = 5.8 x 15 = 87km

Katy = 10.3 x 15 = 154.5km

154.5 - 87 = 67.5km

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