|Exam Name||:||Entry Level Linux Essentials Certificate of Achievement|
|Questions and Answers||:||80 Q & A|
|Updated On||:||February 19, 2018|
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Section 5: Sec Five (67 to 80)
Details: Topic 5, Security and File Permissions
Which command displays the list of groups to which a user belongs?
What is the usual absolute path of the personal directory for the user foo?
You have a program called /usr/bin/foo. You wish to create a symbolic link,
/home/user/foo, that points to it. Which command will do this task?
ln -sym /home/user/foo /usr/bin/foo
ln -s /usr/bin/foo /home/user/foo
ln /home/user/foo /usr/bin/foo
ln /usr/bin/foo /home/user/foo
ln --symlink /home/user/foo /usr/bin/foo
What are the three sets of permissions for a file?
user, group, others
administrator, group, others
user, standard user, others
administrator, standard user, others
Which of the following is the home folder for the root user?
What is the command to change the password of a user?
What command line will create the user falco with home directory assigned to the group users as primary group?
useradd -g users falco
useradd -f users falco
useradd -m -g users falco
add user falco@users
add -user falco -group users
Which statement about users and user groups is correct?
A group can only have one main user.
There can be only one user group on a system.
User do not have to belong to a user group.
Every user belongs to a least one user group.
Which of the following commands can be used to change both the owner AND group settings of a file?
Which of the following will change the group that is associated with a file?
Which of the following properties of a user account determines whether the user is given administrator privileges?
Its primary group ID is 0 (zero).
It is listed first in /etc/passwd
Its username is root.
Its user ID is 0 (zero).
Its GECOS (name) field is set to "System Administrator"
Which TWO commands can be used to make the file /tmp/foo.txt readable for all users?
chmod 111 /tmp/foo.txt
chmod 444 /tmp/foo.txt
chmod 770 /tmp/foo.txt
chmod 644 /tmp/foo.txt
chmod 640 /tmp/foo.txt
Given a directory with the following information: drwxrwxrwxt 12 tu tg 36864 2012-03-15 /home/directory/ Which of the following statements are true?(Choose two)
Everybody can create files in the directory.
Files in the directory are created with read, write and execute permissions for everyone.
The directory is broken.
Everybody can delete only his own files.
The directory is a security risk.
When a new user is added, where does his user ID gets stored?
010 10.0 facets Water is a liquid at room temperature whereas methane is a gas. Which remark compares the intermolecular forces in these molecules correctly? 1. There isn't enough suggestions to com- pare these forces. 2. each water and methane have the same intermolecular forces. 3. The intermolecular forces in methane are improved than those in water. 4. The intermolecular forces in water are more desirable than these in methane. correct explanation: 011 15.0 features accept as true with the supplies MgF 2 , O 2 , I 2 , PF three , and HF . The order of increasing melting aspects of those elements is 1. MgF 2 , HF, PF 3 , I 2 , and O 2 . 2. MgF 2 , O 2 , I 2 , PF 3 , and HF.
version 016 – TTh examination four – shear – (50125) four three. I 2 , O 2 , PF three , HF, and MgF 2 . 4. HF, O 2 , MgF 2 , PF 3 , and i 2 . 5. O 2 , I 2 , HF, PF three , and MgF 2 . 6. I 2 , PF 3 , MgF 2 , O 2 , and HF. 7. O 2 , I 2 , PF three , HF, and MgF 2 . relevant explanation: Melting points are regarding bonding as follows: ionic > hydrogen bonding > dipole > big molecules > small molecules. 012 15.0 points Carbon can exist in two forms, graphite and diamond. The heats of combustion at 25 ◦ C and one atm pressure for these two varieties were measured and located to be C(graphite) + O 2 (g) → CO 2 (g) Δ H = - 393.5 kJ/mol C(diamond) + O 2 (g) → CO 2 (g) Δ H = - 395.four kJ/mol From this assistance, calculate Δ H at 25 ◦ C and 1 atm pressure for the response C(graphite) → C(diamond) . 1. absolutely irrelevant 2. +788.9 kJ/mol 3. +1.9 kJ/mol suitable four. - 788.9 kJ/mol 5. - 1.9 kJ/mol clarification: The preferred response C(graphite) → C(diamond) is the sum of C(graphite) + O 2 (g) → CO 2 (g) Δ H = - 393.5 kJ/mol AND the reverse of the different response CO 2 (g) → C(diamond) + O 2 (g) Δ H = +395.4 kJ/mol note that the signal changes as a result of we’ve re- versed the reaction. O 2 (g) and CO 2 (g) cancel out, after which you can sum the Δ H s for each and every reaction. C(graphite) → C(diamond) Δ H = +1.9 kJ/mol 013 10.0 facets Which of the statements beneath concerning thermodynamic signal conference isn't actual: 1. w is wonderful when work is executed by means of the gadget. suitable 2. Δ H is terrible when warmth is launched to the surroundings. three. Δ S is effective when there's expanding ailment. 4. Work is carried out on the system when Δ V is poor. 5. Δ G is negative when a reaction is spon- taneous. explanation: w is high-quality simplest when work is performed ON the equipment, not via the equipment. When the system does work, volume increases or the number of moles raises (Δ V > 0, Δ n > 0). w = - P Δ V = - Δ nRT and may hence be poor when Δ V or Δ n is fantastic. 014 10.0 elements For liquid mercury in a capillary tube, 1. the meniscus is convex and the cohesive forces are improved than the adhesive forces. correct 2. None of these 3. the meniscus is concave and the adhesive forces are more desirable than the cohesive forces.
i'm confused about what substances I should examine to your practicing lessons that line up with these 2 certificate exams.
can you please make clear?
In issue 5 we focus on tossing a coin three times. every toss can have one among two results. Letting 0 = tails and 1 = heads we now have the pattern space in table 1 under. The experience the precisely one toss results in a head is 001 010 100, P(E) = three/8 = .375. The experience that as a minimum two tosses outcome in a head is 011 one zero one 110 111, P(E) = four/8 = .5. The experience that no head is acquired is 000, P(E) = 1/eight = .a hundred twenty five.
In difficulty 6 we focus on a household with three infants. each infant is both a boy (0) or a woman (1). we've the pattern house in table 1 beneath. The adventure the precisely one infant is a lady is 001 010 a hundred, P(E) = three/eight = .375. The adventure that as a minimum two children are ladies is 011 a hundred and one 110 111, P(E) = four/8 = .5. The adventure that no newborn is a girl is 000, P(E) = 1/eight = .125.
In issue 7 we focus on a true(1)/false(0) examination with three questions. The probability that exactly one is suitable with random guessing is .375, that as a minimum one is suitable is .5, and that no reply is correct is .125 or 12.5%.
In difficulty eight we focus on publicity to disease with a 50-50 probability of catching the disease. The outcomes are the equal, only the names have changed.
1 subject 4 laptop mathematics and good judgment IB laptop Science
2 A reminder of how we count number… here is how we signify the number we have a thousands column, a 100s column, and many others. all the columns are powers of 10. We count the variety of enormous quantities, 100s, 10s, and so forth, and sum them up. 9 a thousand= one hundred=200 four 10=40 6 1=6 3 0.1=0.three 5 0.01=0.05 complete=
three that you can do the identical with powers of Or powers of Or any base you like… n3n3 n2n2 n1n1 n0n0 n -1 n -2
4 We most effective count in base 10 as a result of we now have ten fingers The Simpsons count in base eight.
5 In computing We regularly count in: base 2 (binary) base eight (octal) base sixteen (hexadecimal) You deserve to be aware of: binary hexadecimal F00
6 Counting in other bases Counting in base 10 9 is the largest quantity Then we exchange columns and many others and so forth Counting in base 2 1 is the largest number Then we alternate columns
7 Counting in other bases Counting in base 10 9 is the largest quantity Then we trade columns and many others and so forth Counting in base 2 1 is the biggest quantity Then we alternate columns
8 Counting in hex A B C D E F 1 0 Counting in base 16 skill that 15 is the greatest quantity we get to before we exchange columns Counting in base sixteen presents an issue as a result of we do not need fifteen different digits to make our numbers out of! So we use letters, as proven… don't forget that 10 doesn't suggest "ten" if we are counting in hex. It skill "one within the sixteens column and nothing within the instruments column".
9 converting between decimal and binary Convert 45 dec to binary what's the greatest vigor of two in forty five? answer 32, so your first 1 represents 1 x 32, giving 1 next, go down the powers of two and add on what you need to make 45. will we need any 16s? No, as a result of we have alread 010-100y got a 32 and sixteen would provide us forty eight, so our subsequent quantity is 0, which represents 0 x sixteen, giving 10 can we need any 8s? sure, because is barely 40, so our subsequent number is 1, which represents 1 x eight, giving one zero one can we want any 4s? yes, as a result of is barely forty four, so our subsequent quantity is 1, which represents 1 x four, giving 1011 will we need any 2s? No, because adding 2 to our 44 would give us 46, so our next quantity is 0, which represents 0 x 2, giving do we want any 1s? sure, as a result of we currently have 44 and we want 1 extra to provide us 45. So our ultimate number is 1, representing 1 x 1 and giving So forty five dec in binary is we are able to determine this outcome on the subsequent page.
10 investigate investigate: 1 32= 0 sixteen=0 1 eight=8 1 four=4 0 2=0 1 1=1 total=45 45 dec in binary is
eleven Convert the following to binary three dec 7 dec 12 dec 16 dec 20 dec 31 dec fifty three dec sixty four dec 127 dec
12 Convert the following to decimal 10 bin one zero one bin 1101 bin bin bin bin bin bin bin
13 changing between decimal and hex Convert eighty one dec to hex Hex powers go 1, 16, 256, 4096, and so on. They get large somewhat without delay. what's the biggest power of 16 in eighty one dec ? answer sixteen. how many sixteens are there in 81 dec ? answer 5, with one left over. So our reply is fifty one, ie 5 sixteens, plus 1 unit. we are able to assess this effect on the next web page.
14 determine examine: 5 sixteen=eighty 1 1=1 complete=81 eighty one dec in hex is fifty one
15 Convert the following to hex 4 dec eleven dec 14 dec 23 dec 31 dec forty five dec sixty six dec 240 dec 301 dec four B E 17 1F 2C 42 F0 11D
16 Convert the following to decimal 5 C A 32 FF a hundred 1AC
17 The respectable information… converting directly between binary and hex is less difficult than between binary and decimal, or hex and decimal. You simply need to be in a position to re-create this table in your head. BinHex A 1011B 1100C 1101D 1110E 1111F
18 F So 34F hex = bin 38B5 Convert A4F hex to binary Convert bin to hex So bin is 38B5 hex notice padding with zeroes enables use of conversion desk on outdated slide. be aware main zeroes in answer are not required, but you might not lose marks in case you leave them in.
19 Convert to binary 9 D 13 A1 7CD
20 what number of values can n bits characterize? think about you've got 8 bits to play with the bottom cost is The highest price is That’s 0 as much as 255, ie 256 distinct values 256 is 2 eight In accepted, n bits can signify 2 n distinctive values
21 terrible numbers All numbers in the laptop are represented in binary, but how are bad numbers represented? (We have no "minus signal" in laptop reminiscence!) answer: both’s complement conference: everything is as standard, but the most huge bit (MSB) is taken to be poor
22 apply with two’s complement here we are the use of 8-bit two’s complement assess you keep in mind: = 2 0 = = 2 6 = = = = = -2 7 = = = =
23 range of values what is the highest cost that may also be expressed in 8-bit two’s complement? well, you need the entire positive values, ie , which is = 127 what's the bottom value that can also be expressed in eight-bit two’s complement? well, you need the terrible bit, and not one of the high quality bits, ie , which is -2 7 = -128 what is the optimum price that may also be expressed in n-bit two’s complement? 2 n-1 -1 what's the bottom value that can also be expressed in n-bit two’s complement? -2 n-1
24 range of values bitsno diff valueshighestlowest n2n2n 2 n n-1
25 Expressing fractions 1: fastened-element Binary
26 fixed element binary observe Convert to binary: 1.5 dec 2.5 dec 4.25 dec dec answers: Convert to decimal: 1.01 bin bin bin bin solutions:
27 Combining fixed aspect and Two's Complement Convert to binary: dec answer: ( ) Convert to decimal: bin answer: ( ) what is the highest price that can also be expressed in this layout? what is the bottom values? solutions: = and = -8
28 Floating aspect remember scientific notation in maths? turns into x 10 2 correct? The identical happens in binary. e.g may also be written as x 2 2 here's called floating aspect representation, because the decimal element strikes. check your understanding. Convert to right here to floating element illustration: answers: 1.01 x x x 2 -four
29 Storing floating point numbers There are two more issues to think about: How will we store the exponent? (We can not shop a 3 in binary) How can we characterize poor numbers? We truly shop both the mantissa and the exponent in two's complement kind. notice the position of the bicimal aspect x 2 3 mantissa exponent
30 Floating factor instance: high quality mantissa fantastic mantissas are particularly easy: Convert to decimal the floating-aspect binary number , if 6 bits are allocated to the mantissa and four bits to the exponent. [2 marks] bear in mind that bicimal factor is after the primary little bit of the mantissa (ie ) Now calculate the exponent is (advantageous) 4, so we shift the bicimal element four locations to the right, giving = ½ so the closing reply is 9.5 (With a bad exponent, you might simply should shift the bicimal element to the left in its place.)
31 Floating element illustration: negative mantissa take into account that we use two's complement as a result of we have no minus sign in the computer's reminiscence. neatly, should you are dealing with a negative mantissa, it is lots more convenient to think about that you've got bought a minus signal. for instance, when you are using 6-bit two's complement, that you can trust a mantissa of (-1 + ½ = -½) as (-½) To without difficulty convert from two's complement to our cheating minus sign structure is easy: 1.Flip the entire bits 2.Add 1 to the least significant bit three.Put a minus sign up front to convert back you simply do the same component again and remove the minus signal – it really works each techniques
32 practice complementing what is (5-bit two's complement) in our cheating minus sign layout? 1.Flip the bits, giving Add one to the least tremendous bit (the right-most) giving: Put a minus register entrance: investigate: (two's complement) is = is –( ) = investigate that the equal manner takes you again to both's complement structure.
33 bad mantissa revisited or not it's important to grasp how to complement the mantissa because it makes relocating the bicimal aspect a good deal less demanding. Convert to decimal the floating-factor binary quantity , if 6 bits are allotted to the mantissa and 4 bits to the exponent. [2 marks] First locate the dishonest layout of the mantissa: (Now or not it's similar to with a favorable mantissa a couple of slides lower back) discover the exponent: 0011 = 3 flow the bicimal point 3 to the right, giving Convert to decimal. answer -2.5
34 changing from decimal to floating factor this is identical to converting to scientific notation: Convert to scientific notation. move decimal aspect two to the left Add an exponent to compensate x 10 2 Convert 5.75 to normalized floating factor binary if the mantissa is 6 bits and the exponent is four bits: Convert to mounted factor binary: move bicimal factor 3 to the left: (have to have a zero backyard the bicimal element when you've got a terrible mantissa and you should complement – see next illustration) Set the four-bit exponent to 3 to compensate: 0011 Full answer:
35 more examples
36 critical points college students often locate this hard. there will likely be extraordinarily few marks dedicated to the advanced aspects of this within the examination. don't spend a disproportionate period of time on it. in case you do not like my rationalization, try this guy's:
37 viable mistakes Truncation error: Some numbers require infinite-size mantissas, e.g. one third is … if you are attempting to shop this in a pc then one of the most digits get "chopped off" (truncated), with an linked loss of precision. Overflow error: if in case you have 3 bits, you cannot do the sum since the reply is one thousand, which bigger than the biggest quantity that you may represent. Underflow error: if you have four-bit two's complement then the smallest number which you can signify is 1000 (or -eight). therefore you can not so the sum , because the reply is (or - 9) which is smaller than the smallest number that you would be able to represent. be trained the definitions plus an instance for each and every
38 reality tables ABA nand B ABA + B AA AB A B NAND OR XOR no longer ABA nor B NOR AB A B AND
39 Boolean algebra into phrases and vice versa (A) Jessica will go to the party (B) Fred will go to the birthday celebration (C) Chen may be chuffed construct an expression using Boolean algebra for the sentence "either of Jessica or Fred will go to the celebration, and Chen could be sad." (A B) C (A xor B) and not C You deserve to know the symbols: AND + OR XOR [overbar] no longer
40 good judgment circuits you could assemble good judgment circuits out of boolean algebra and vice versa. A B C output inputs This circuit corresponds with the statement on the ultimate slide. it'll handiest produce an output of actual if (A xor B) and not C. Nand is comparable to AND followed by means of no longer (the AND reality desk with the entire bits flipped) and so has the sense of "anything apart from each". nor is akin to OR adopted by not (the OR fact table with all of the bits flipped) and so has the experience of "neither one nor the other".
forty one make sure to… Be able to convert between boolean algebra (phrases or symbols) and common sense circuits (maximum of three inputs) show that a common sense circuit and a boolean algebraic expression are such as each and every different via evaluating their actuality tables
42 (no longer A And B) Or (not(no longer A Or (now not A Or B))) (no longer A And B) Or (no longer(now not A Or now not A Or B)) (not A And B) Or (now not(now not A Or B)) (now not A And B) Or (A and never B)) A xor B Karnaugh maps sum of items