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:)
the hose is 2 inches in diameter. What circumference does the plug need to be? remember to type just a number
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]
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?
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.
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?
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 = 480Hence, 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
