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3 years ago

How many address bits are needed to specify each byte in a 512 byte memory unit?

Computers and Technology

1 answer:

schepotkina [342]3 years ago

5 0

Explanation:

In a 512 byte memory unit, in order to address each byte, there will be 512 locations (each one byte) ranging from 0 to 511.

Thus it will take 9 bits (2^9=512) to store the address within the unit.

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Attacking systems by exploiting otherwise unknown and unpatched vulnerabilities is also known as:

Nostrana [21]

Answer:

Zero-day exploits

Explanation:

Zero-day exploits refers to recently found vulnerabilities in a computer software program that has been in existence but was hitherto not known and addressed by the software security experts, however, these vulnerabilities were known to hackers. While the existence of these "loop-holes" in the software can go on unnoticed for several years, hackers can take advantage of it to cause harm to the computers' programs and data.

When these attacks occur, it is called a zero-day because the attack took place on the very day that the loop-hole was discovered in the software. So exploitation has already taken place before a fix is carried out.

5 0

3 years ago

Joel has left his computer unattended while answering a phone call . A TV repairer in his house tries to surf through the applic

Triss [41]

<span>Joel is a victim of a security breech. By leaving his computer unattended and not locked, he allowed for another person to use his credentials to access (or attempt) to access information that would otherwise be restricted to him. The fact that the unauthorized person did not find the information he was seeking does not minimize the risk.</span>

5 0

2 years ago

What is the best way to limit a google search and get the most specific information possible?

Marrrta [24]

You go into settings on the google search and click advanced search. The rest you do on your own.

6 0

3 years ago

We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is

bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons to find the middle terms.

If list has 18 elements, there are two middle terms as one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons to find the middle terms.

If list has 20 elements, there are two middle terms as one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0

3 years ago

You are flying on an ifr flight plan at 5500 feet with a vfr on top clearance when you encounter icing. you know there is clear

zepelin [54]

Answer:

A. Yes, you are still IFR with VFR-on-top clearance

Explanation:

Instrument Flight Rules (IFR) and Visual Flight Rules (VFR) are regulations controlling the operations of civil aviation. Pilots use these terms as well to describe their flight plan.

When your request is rejected to fly 9500 feet from atc, your VFR-on-top clearance is still active and you are still IFR.

8 0

3 years ago