##### Computing Skills Assignment Help

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(i) 2(3 – 5x) = 15

(2 * 3) – (5x X 2)=15

6-10x=15

6-15=10x

-9=10x

X=-0.9

(ii) x2 + x -20 = 0

x2 + 5x -4x -20 = 0

x(x + 5) – 4(x + 5) =0

(x – 4) (x + 5) =0

x – 4 = 0 or x + 5 =0

x = 4 or x = -5

A2)

(i) y=2x; y=-2x+1

y=2x and y=-2x+1

2x = -2x+1

4x=1

X=0.25

and y=2x

Hence Y= 2 * 0.25

**Y=0.50**

(ii) y = 5x + 1; y =−5x + 1

y = 5x + 1 and y =−5x + 1

-5x+1=5x+1

1-1=5x+5x

10x=0

X=0 and y=5x+1

Y=5 * 0 + 1

Y=0+1=1

Y=1 and X=0

−6y = 3x − 4; 2y = x + 5

3x+6y=4 and x=2y-5

3(2y-5) + 6y=4

6y-15+6y=4

12y=19

Y= 19/12= 1.58 and x=2y-5

X=2 * 2*19/12 – 5

X= 19/12 - 5

X=-1.83

a cube with a length of one side 27 meters

Volume of a cube = a3

= (27)3 = 27 X 27 X 27 = 19683 m2

Hence volume of a cube is 19683 m2.

a sphere with radius 20 inches

Where = π = 22/7

V= 4/3*22/7*20*20*20

V= 33523.80cubic inches

Calculate sine, cosine and tangent for each angles of ?ABC.

Sin A = opposite side /hypotenuse = 12/15= 4/5

Cos A = Adjacent side /hypotenuse = 9/15 = 3/5

Tan A = opposite side/adjacent side = 12/9 = 4/3

Sin C = opposite side/hypotenuse = 9/15 = 3/5

Cos C= Adjacent side/hypotenuse= 12/15 = 4/5

Tan C = Opposite side/hypotenuse= 9/12 = 4/3

Sin B = opposite side/hypotenuse = 1

Cos B= Adjacent side/hypotenuse=0

Tan B = Opposite side/hypotenuse = undefined

Pythagoras theorem states

9X9 + 12X12 = 15X15

81 + 144 = 225

225 = 225

Hence it is proofed that triange ABC is a right angle triangle.

a) Calculate vector c = a+b.

C= (13,20) + (5,-6)

C= (13+5,20-6)

C= (18,14)

b) Calculate magnitude of vector c.

magnitude of vector c = <x, y>

c=<18,14>

|c| =SQRT(18*18+14*14)= SQRT(1040)=22.80

c) Write a Pseudocode for calculating magnitude of vector c.

Algorithm: magnitudeOfVector(x, y)

Purpose: magnitude of a vector whose 2 components are given

Pre: Given both components of vector

Post: None

Return: Magnitude

{

magnitude = sqrt(x*x + y*y)

Return magnitude

}

A={ Martin, Marriott, Boast, Preston, Kans}

B= {24k, 25k, 26k, 27k, 30k}

C= {Production, Sales, Finance}

Mr Martin and Mrs Marriott are working at production department, Mrs Boast and Mrs Preston working at sales department and MrKans works at Finance department.

- Find the Cartesian product of set A and set B. (R=A×B)

Cartesian of a product (R= A XB)

={ Martin, Marriott, Boast, Preston, Kans} X{24k, 25k, 26k, 27k, 30k}, ={Martin , 24k},{Martin , 25k},{Martin , 26k},{Martin , 27k},{Martin , 30k}, {Marriott , 24k},{Marriott , 25k},{Marriott , 26k},{Marriott , 27k},{Marriott , 30k}, {Boast , 24k}, {Boast , 25k}, {Boast , 26k}, {Boast , 27k}, {Boast , 30k}, {Preston , 24k}, {Preston , 25k}, {Preston , 26k}, {Preston , 27k}, {Preston , 30k}, {Kans , 24k}, {Kans , 25k}, {Kans , 26k},{Kans , 27k},{Kans , 30k}

- Find the Natural join of R and C.

=(Production, Martin, 24K),(Production, Marriott, 25K), (Sales, Boast, 26K),(Sales, Preston, 27K), (Finance, Kans, 30K)

- Fill in the below table by using provided information:

**B2)A small ICT firm, has three branches in**

1. Redbridge,2. Enfield and 3. Barnet.

Five technicians with following details are working at this company;

Ali (Location: Barnet, age: 25, salary: £21,000), Steve (Location: Redbridge, age: 45, salary: 23,000),Mike (Location: Enfield, age: 50, salary: 19,000), Linda (Location: Barnet, age: 55 , salary: 24,000 ),Carol (Location: Redbridge, age: 43, salary: 27,000)

- Draw required number of tables and fit in the above information there.

- List individuals satisfying the conditions below:

1. (Age<46) AND (Salary> £ 23,000)

**{Carol} **

2. (Age>26 ) OR (Salary < £24,000)

**Steve ,Mike,Linda,Carol,Ali**

3. (Age< 53) AND (Salary>29) OR (Location=1)

**Steve,Carol**

4. (Age> 25) XOR (Salary>30) OR (Location=2)

**Steve ,Mike,Linda,Carol**

Magic numbers 4X4 , the total should be equal to 34 from all the sides. All the number 1 to 16 should be used to solve the matrix using each number only one. As per the grid 01, 15, 14, 12, 06, 08 and 03 are already present. The rest numbers to be used. Left are 2, 4, 5, 7, 9, 10,11,13,16. These numbers when used in combination gives the total 34

**STEPS**

Taking 1^{st} row

**Sum of all 4 elements should be 34**

01 + 15 + 14 + P = 34

*P = 4*

**Taking 1 ^{st} Column**

Sum of all 4 elements should be 34

01 + 08 + 12 + x = 34

*x = 13*

**Taking 2 ^{nd} Column**

15 + 6 + s + 3 = 34

*s = 10*

**Taking diagonal (XSQP)**

x + s + q + p = 34

13 + 10 + q + 4 = 34

*q = 7*

**Taking 2 ^{nd} ROW**

12 + 6 + q + r = 34

12 + 6 + 7 + r = 34

*r = 9*

**Taking 3 ^{rd} ROW**

8 + s + t + u = 34

8 + 10 + t + u = 34

t + u = 16 ----------------------------- (1)

**Taking 4 ^{th} ROW**

x + 3 + y + z = 34

13 + 3 + y + z = 34

y + z = 18 ------------------------------ (2)

**Taking 3 ^{rd} Column**

14 + q + t + y = 34

14 + 7 + t + y = 34

t + y = 13 --------------------------------- (3)

**Solving equation 2 & 3**

Equation (2) – Equation (3)

z – t = 5 ------------------------------------ (4)

**Taking Diagonal (01 6 t z)**

1 + 6 + t + z = 34

t +z = 27 -------------------------------- (5)

**Solving Equation 4 and 5**

Adding both equation

2z = 32

*z = 16*

Hence *t = 11*

Puttin value of t in equation 3

*y = 2*

**Putting value of t in equation 1**

t + u = 16

11 + u =16

*u = 5*

**Hence p = 4, q =07, r=09, s=10, t=11, u=05, x=13, y=02, z=16**

**B4. Show that if**

Then P is the inverse of Q

If P is inverse of Q then below statements should hold true: PQ = QP = I where I = as below

Hence P is inverse of Matrix Q

A = { g, e, r, m, a, n, i }

B = { p, o, l, a, n, d }

Identify the following statements as true or false:

- a ∈ A = True since a is an element of set A
- b ∈ B = False because b is not an element in set B
- d ∉ B = False since d is an element of set B
- u ∉ A = True since u is not an element of A
- a ∈ A?B = Since a is an element of A?B hence true
- |A| = |B| = False since the count of elements in set A is 7 which is not equal to element count in set B 6.
- { i, r, a, n} ⊂ A = True as { i, r, a, n} is a subset of set A
- A?B| = 8 = False [ g, e, r, m, a, n, i, p, o, l,d] = 11 ∉8

**C2. Suppose we have a universal set **

{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27} and considertwo sets P and O defined as follows:

P = “all multiples of 3”

O = “the first ten even numbers”

Represent all of the elements in a Venn diagram and identify the elements in P?O, P?O

and PΔO.

P = [3, 6, 9, 12, 15, 18, 21, 24, 27] and O= [ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

For union We take all the values present in SET P and Set O

- ?O = { 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27}

and

For intersection between set P and O we will take only common values which are present in both set.

- ?O = {6, 12, 18}

PΔO= P?O/P?O

= { 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27} ? { 2, 3, 4, 8, 9, 10, 14, 15, 16, 20, 21, 24, 27}

= {2,3,4,8,9,10,14,15,16,20,21,24,27}

**VENN DIAGRAM**

C3. For all of the following sets defined in set−theoretic notation, list out all of the elements:

S1 = {x: x = 2n, where 1 ≤ n ≤ 6}

n is greater and equal to 1 and less than and equal to 6. Hence n has 6 values.

n : {1, 2, 3, 4, 5, 6)

S1 is set of x hence

S1={2,4,6,8,10,12}

S2 = {x: x = 3n2, where 1 ≤ n ≤ 5}

n is greater and equal to 1 and less than and equal to 5. Hence n has 5 values.

n : {1, 2, 3, 4, 5)

Putting values of n from 1 to 5

S2={3,12,27,48,75}

S3 = {y: y = 5n3, where 1 ≤ n ≤ 4}

n is greater and equal to 1 and less than and equal to 4. Hence n has 4 values.

n : {1, 2, 3, 4)

Putin value of n from 1 to 4 and will calculate corresponding x

S3={5,40,135,320}

S4 = {x: x = √n, where 3 < n < 5}

n is greater than 3 and lesser than 5. Hence n has only one value = 4

x = 2 for that

S4={2}

**Find the number of different ways that the salesman can leave home, visit two different customers and then return home.**

**LO4: Be able to apply statistical techniques to analyse data**

**D1. A research in 157 households found that the number of children per household is**

- Calculate the Mean of frequency distribution for the above case.

Mean of Households = X = ∑ x/n

∑ x = (32+35+37+29+15+8+1)

X = ∑ x/n = 157/7 = **22.43**

- What is the Mode value of number of children’s per household?

Mode value = **None**

**D2. A company has ten sales territories with approximately the same number of sales people working in each territory. Last month the sales orders achieved were as follows:**

For these sales calculate the following:

- Arithmetic mean:

**Propositional Logic**

The term Propositional Logic largely refers to the branch in Mathematics that deals with the binary terms ‘true’ and ‘false’. However, it is worth telling that the concept is not limited to the branch of mathematics, and is highly applied to Computer Science and English, as well. Here, we will discuss the utilities of Propositional Logic in the field of computers (Computer Science).

Talking of logic in everyday, it can be organized into the English language for a sound understanding. In normal terms, logic can be categorized into and (conjunction), or (disjunction), not (negation), and if (material conditional). In a much similar way, computer science is also divided on the terms of propositional calculus.

**Computers and Propositional Logic**

Artificial Intelligence has been strongly impacted upon by the concept of Logical Mathematics. With the initiation of automation of computer systems. The process came up when it was realized that logical inferences can be highly effective in solving conclusive operations on the computer systems.

A similar depiction was actually expressed to the zenith when FOL (first order logic) was described to be a fine method in evaluating and analyzing information. Its other prime utilizations could be seen in it application in research projects like Knowledge Based Software Assistant. Computer specifications can now be transformed into a diversified range of codes with the help of logical operations or Propositional Logic.

Another crucial utilization could be seen in the field of High Frame Languages. It has been observed to simplify the complex semantics of Frame languages with the latest example of KL One. The specifications of this Language had been derived to a calculus form, which enabled to analyze the sets, subsets etc.

The concept of Propositional Logic is actually associated with computers in a number of mannerisms. Here, we have presented the most effective out of them all.

Attributes of a Propositional Logic

If we take the case of Propositional Logic in computer science, we must consider the following attributes:

- Syntax
- Semantics
- Proof Theory

SYNTAX

A well formed formula that can be associated with the syntactical objects.

SEMANTICS

This refers to the association of logic with a meaningful formula.

PROOF THEORY

This refers to manipulation of formula as per different rules.

In the field of computer science, propositional logic can be defined as the science of atomic propositions. Here, we will discuss a number of complex issues related to propositional logic that are commonly termed as connectives.

And

This is a conjunctive form of propositional logic. Suppose, there are two propositions namely, A and B. Then, the relation between them can be written as:

A∧B

This means, both A and B need to be true for an overall truth of the whole system.

This is almost like the binary system presented through the concept of Boolean Algebra.

Or

This is also known as disjunctive logic. Here, the relation between A and B may be written as:

A∨B, which states that either one of them should may be true for overall truth of the logical system.

If and Then

This combination is known as conditional logic. Suppose, there are two arbitrary elements, A and B, then the relation between them can be written as:

A ⇒ B, which shows a conditional relationship between the two i.e. this can be true only if either A is false or B is true.

Not

This is a negation statement. The earlier mentioned considerations were binary; however, this is exclusively Unary. This is termed as the complete opposite for an element.

When, ¬B is written, it implies that it can be true if A is false.

Applications and Analysis

This concept is highly applied to computers to calculate logical consequences. Tools such as Proof Assistants follow logic based on propositions. This helps in development if a proof-checker that checks the validity of proofs. This is highly effective in developing an error-free proof script

Logic based languages like HML, Scheme etc. can be well implemented for industry applications. Jakarta Toolset is an influential method in designing Java Card specifications. It is able to pass ad-hoc model transformations that is highly structure based, onto virtual machine operations. This is highly compatible in producing proof obligations.

Maor, E. (2007) The Pythagorean Theorem: A 4,000-year History, pp 17

Mean, Median, Mode, and Range [online] Available at [HYPERLINK "http://www.purplemath.com/modules/meanmode.htm"] [Accessed on 08/11/2014]

Standard Deviation and Variance [online] Available at [HYPERLINK "http://davidmlane.com/hyperstat/A16252.html"] [Accessed on 08/11/2014]

Calculating Magnitude with Vectors [online] Available at [HYPERLINK "http://www.dummies.com/how-to/content/calculating-magnitude-with-vectors.html"] [Accessed on 08/11/2014]