the answer is option D
Explanation:
What five safety habits are you using for the internet?
Answer: Keep Personal Information Professional and Limited.
Keep Your Privacy Settings On.
Practice Safe Browsing.
Make Sure Your Internet Connection is Secure.
Be Careful What You Download.
Choose Strong Passwords.
Explanation:that’s what is think
Who must yeild at T-intersections?
Answer: the driver on the street
Explanation:
What is the different between a computer and a phone?
The most obvious difference between a cell phone and PC is the size. Cell phones are significantly smaller than PCs, and are able to fit in the palm of one's hand. ... Cell phones have less command functions than a PC's keyboard and considerably less storage space. Some cell phones may be used as a GPS device
A soldier white line down the center of a two lane road indicates
Answer:
This indicates that you may carefully switch lanes.
Explanation:
If dotted line, then yes.
An encryption system works by shifting the binary value for a letter one place to the left. "A" then becomes: 1 1 0 0 0 0 1 0 This binary value is then converted to hexadecimal; the hexadecimal value for "A" will be:
Answer:
The hexadecimal equivalent of the encrypted A is C2
Explanation:
Given
Encrypted binary digit of A = 11000010
Required
Hexadecimal equivalent of the encrypted binary digit.
We start by grouping 11000010 in 4 bits
This is as follows;
1100 0010
The we write down the hexadecimal equivalent of each groupings
1100 is equivalent to 12 in hexadecimal
So, 1100 = 12 = C
0010 is represented by 2 in hexadecimal
So, 0010 = 2
Writing this result together; this gives
1100 0010 = C2
Going through the conversion process;
A is first converted to binary digits by shifting a point to the left
A => 11000010
11000010 is then converted to hexadecimal
11000010 = C2
Conclusively, the hexadecimal equivalent of the encrypted A is C2
Please don’t comment if you don’t know the answer.
So I have an hp laptop that I have a year with and everything was working great until a month ago when it started saying that the battery couldn’t hold charge anymore.
Since I didn’t want to buy a battery I simply started using while connected with the charger.
But after a few days of saying the battery couldn’t hold charge out of a sudden when I tried connecting my mouse or controller it would recognized.
It passed power to whatever I connected to it but it simply wouldn’t work.
Can it be fixed?
Or do I have to buy a new one.
Answer:If your notebook computer battery does not power the notebook or will not hold a charge, troubleshoot the battery to see if it needs to be replaced. If the battery indicator LED, located near the power icon, does not glow or always blinks, there is a battery problem. Your computer might operate correctly when it is connected to the power adapter, but not when using battery power.
Explanation:
Which of the following is a country that cruise ships commonly sail under the flag of?
O United States
O Canada
O Panama
O South Africa
< Previous
Answer:
Panama
hope this helps!
Create a story that depicts the events leading up to a vehicle crash fatality, the crash itself, and the emotional aftermath on the family and victims of the crash. In this scenario the crash should be written from your own perspective of losing a close family member in a crash caused by another person making a poor choice (i.e. texting, dui, road rage).
Answer:
About 5 years ago my girlfriend and I were taking a summer camping trip in southern Alberta. I was in the passenger seat trying to find the campground on a map when she drove over the crest of a hill, was blinded by the sun drove into the ditch. It was a seriously steep ditch but we weren't going very fast so all was fine. I looked over at her and laughed before returning my attention to the map, assuming she could safely bring the vehicle back onto the road.
The next thing I knew, the SUV launched onto the pavement and she lost control and we began swerving. I remember feeling the wheels on the driver's side lift off the ground, then the impact as I was slammed into the door and glass exploded into my face. We barrel rolled and we rolled over-front for a seriously long way.
At some point during the chaos I looked over at her to make sure she was "ok", and just as I did so I watched as she was thrown into the ground through her door window and the corner of the roof just above her seat was crushed inward. The way it looked to me was that she had just been crushed between the ground and the roof of the vehicle. I passed out at that point.
When I came to I had somehow already unbuckled myself from my seat and the vehicle was on its roof. I crawled over to her seat and was in absolute shock to see that she was still in one piece. I removed her from her seat and got us out through the windshield before carrying her for about a half a kilometer down the road, still in shock and fueled entirely by adrenaline.
We were found by a driver who had gone past the wreckage and we were eventually taken to a hospital. I broke 3 ribs on my right side and dislocated my right shoulder, she was severely concussed and scraped up but otherwise mostly okay. However she was 7 weeks pregnant at the time and we found out while in hospital that her body had rejected it under the immense and sudden stress.
We are still together and have 3 amazing and beautiful children, but she still holds onto a lot of guilt surrounding the accident and the loss.
Explanation:
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?
Answer:
67.5km
Explanation:
mark = 5.8 x 15 = 87km
Katy = 10.3 x 15 = 154.5km
154.5 - 87 = 67.5km
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