BTEC HND Assignment: Mathematics for Software Development (L4)

This document presents a solved assignment on Mathematics for Software Development, likely for a BTEC HND in Computing program. The assignment covers core mathematical skills essential for software engineers, including solving algebraic equations, quadratic expressions, geometric problems involving hypotenuses and areas, and vector calculations. It also delves into relational algebra, functions, and set theory. Additionally, the solution addresses statistical concepts such as mean, median, mode, range, standard deviation, and variance, along with probability calculations. Flowcharts and pseudo-code are included to illustrate logical processes. This comprehensive solution provides valuable insights and guidance for students studying mathematics in the context of software development. Desklib offers a wealth of similar resources, including past papers and solved assignments, to support students in their academic endeavors.

Title
Name of Student
Name of Professor
Name of Class
Day Month Year
1

LO1: Understand core mathematical skills for software engineers.
1.
a) 6x- b= c, b= 15 and c= 45
6x- 15= 45
6x= 45+15
6x=60
X=10
-1 0 1 2 3 4
-2
-1
0
1
2
3
4
5
Column1
2

b)
ax2+bx+c= 0, a= 1, b= -9 and c= 20
x2- 9x+ 20=0
x= {-b+ square root [(b*b)-(4ac)]}/2a
x= {(-9*-1) +square root [(-9*-9) - (4*1*20)]}/ (2*1)
x= {(9) + square root [(81) - (80)]}/2
x= {(9) + square root (1)}/2
x= [(9) + (1)]/2
x= (10)/2
x=5
Or
x= [(9) - (1)]/2
x= (8)/2
x= 4
IN A GRAPH OF A QUADRATIC EXPRESSION
3
START
b, c
6x – b= c
X= (c + b)/6
X
STOP

Y= ax2 + bx + c
Y= x2 - 9x + 20
X Y
-3 56
-2 42
-1 30
0 20
1 12
2 6
-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
0
10
20
30
40
50
60
Y-Values
Y-Values
4

c)
xx
2.
5
START
a, b, c
ax2 + bx + c= 0
x=−b ± √b2−4 ac
2 a
If roots=
noncomp
lex?
Yes
No
X= [(b) + (root)]/2a or
X= [(b) – (root)] 2a
x
STOP
The roots are complex please
insert other numbers of a, b and
c

a)
H=square root (b2 + h2)
H= Hypotenuse, b= base and h= height
H= square root (242 + 72)
H= square root (576 + 49)
H= square root 625
H= 25cm
Length of the pole= 25 + 7
Length of the pole= 32m
b.
H= square root (C1 + C2)
H= square root (402 + 202)
H= square root (1600 + 400)
H= square root (2000)
H= 44.72
H= 45
Area= 1/2bh
A= Area, b= base and h= height A= 1/2 * 40*20 A= 400cm2
6
C1
C2
h

7
STOP
H, b, h
H= square root (b2 + h2)
b= square root (H2 – H2)
h= square root (H2 – b2)
Is H, h and b
are ≤?
H, h and b
STOP
Input values greater
than 0
Yes
No

c.
8
START
START
h and b
A= 1/2bh
Is b and h
greater
than
zero?
A
STOP
Input a value greater than zero
No
Yes

It is not possible to calculate the volume of a rectangle or a square this is
because their height is zero hence the program will keep on displaying
errors.
9
L, W, H
V= L * W* H
V
STOP
If
V= ≤0?
You cannot have zero or a negative
volume. Please check the values of L, W
and H to ensure they are greater than
zero.
No
Yes

3.
a)
F1= 15N, F1= 35N and R= 28N
Let P & Q be the two forces
Angle of inclination= ø
Resultant Force= R
R= square root [(P2) + (Q2) + (2PQCosø)]
Square both sides to remove the square root
282= [(152) + (352) + (2* 15 *Cos ø)
784= [225 + 1225 + 1050 Cos ø]
784= 1450 + 1050 Cos ø
784- 1450= 1050 Cos ø
-666= 1050 Cos ø
-666/1050= Cos ø
-0.6343= Cos ø
Ø= Cos -1-0.6343
Ø= 129.37°
Ø= 129°
b)
10

11
START
X, Y, Z, X1, Y1, Z1
A=
X
Y
Z
-
X 1
Y 1
X 1
A=
X −X 1
Y −Y 1
Z−Z 1
Is X, Y, Z, X1,
Y1, Z1
containing
equal numbers
of column or
You cannot add matrices
that are not corresponding
A
STOP

12
START
X, Y, Z, X1, Y1, Z1
A=
X
Y
Z
+
X 1
Y 1
Z 1
A=
X + X 1
Y +Y 1
Z+ Z 1
A
STOP
You cannot add vectors
with different rows and
columns
IS X, Y, Z, X1, Y1, Z1
containing equal
numbers of rows and
columns?

c.
13
START
V1, speed1, speed2,
Momentum1= mass * speed1
Momentum1= X kg * y m/s
Momentum2= mass * speed2
Momentum2= Z kg * W m/s
Momentum
STOP

LO2: a).
It is an assumption that states “ An individual is allowed to place all the data
attributes into a table which can then be sub- grouped into smaller tables. Examples
are void relation, Identity relation, Symmetric relation, Transitive relation,
Reflexive relation and Equivalence relation.
b). A function is like a machine which has input and output and the output is
somehow related to the input.
C). Transitive?
a)
i. diagram for f
ii.
Domain = A and B
Co-domain = (a1, b2), (a2, b3), (a3, b3), (a4, b1), (a5, b4)
Range = a5, b4
iii.
14
a1, a2,
a3
a1, a2,
a3, a4,
a5
b1, b2,
b3, b4

1. a).
2
5 −2 2
6 −5 0
8 6 1
-
4 1 5
7 2 7
8 0 2
10 −4 4
12 −10 0
16 12 2
-
4 1 5
7 2 7
8 0 2
(10−4) (−4−1) (4−5)
(12−7) (−10−2) (0−7)
(16−8) (12−0) (2−2)
6 −5 −1
5 −12 −7
8 12 0
b).
15

16
START
10 −4 4
12 −10 0
16 12 2
4 1 5
−7 2 7
8 0 2
A=
10 −4 4
12 −10 0
16 12 2
-
4 1 5
−7 2 7
8 0 2
A=
(10−4) (−4−1) (4−5)
( 12−7) (−10−2) (0−1)
(16−8) (12−0) (2−2)
A=
6 −5 −1
5 −12 −7
8 12 0
If during addition or subtraction
the number of the columns
does not equalize the number
of rows in the second matrix or
vice versa?
A
STOP
WRONG INPUT OF MATRIXES
PLEASE RE-ADJUST
No

17

C).
2[ 2 3
4 1 ]. 0 3 0
1 4 2
4 6
8 2 . 0 3 0
1 4 2
(4 * 10) + (6 * 1) = 6
(8 * 0) + (8* 1) =8
(4 * 3) + (6 * 4) =36
(8 * 3) + ( 2* 4) =32
(4 * 0) + (6 * 2) = 12
(8 * 0) + (2 * 2) = 4
6 36 12
8 32 4
d).
18

19
A= (2 * 0) + (3* 1) = 3
(4 * 0) + (1 * 1) = 1
(2 * 3) + (3 * 4) = 18
(4 * 3) + (1 * 4) = 16
(2 * 0) + (3 * 2) = 6
(4 * 0) + (1 * 2) = 2
A= 3 18 6
1 12 2
A
STOP
START
2 3
4 1
0 3 0
1 4 2

LO3:
1. a).
i. A= {-3,-2,-1, 0, 1, 2, 3}
A = {x: x ϵ z| - 3≤x≥3}
ii. B= {3, 6, 9, 12}
B = {x: x ϵ N|x = 3m, m ϵ N| ≤ x 12}
iii. If A = {1, 2, 3, 4}, B= {2, 3, 4 5}.
A = {1, 2, 3, 4} and B = 2, 3, 4, 5
iv. B= {y : y ϵ N and 1 ≤ y ≤3} [P3.1]
B = [{ø}, {1, 2, 3}]
b).
A – B = A = {2, 4, 6, 8, 10}
B – A = B = {1, 3, 5, 7, 9}
C). Number of all students (N) = 100. Those who like science (S) = 35 and those
who like math (M) = 45
It’s given that (MᴖS) = 10
But (MᴖS) = (M) + (S) – (MᴗS)
Therefore, (MᴗS) = 35 + 45 – 10 = 70
Number of students who like either, that is, mathematics or science = (M) + (S) = 70
20

d).
21
START
Int = 1, 2, 3……n
Is the two
preceding
numbers when
added up add up
to the next
number?
1, 1, 2, 3, 5, 8
A = 1 + 1 = 2 or 2 + 3 = 5
A
STOP
Please input another
different character

2.
a).
P Q P˅Q ¬ (P˄Q) (P˄Q)˄¬(P˄Q) P˄Q
T T T F F T
T F T T T F
F T T T T F
F F F T T F
b).
C).
X= (A) BC ¬ (A + D) [P3.2, M3]
22
A=
B= X=
C=
D=
1
2
X=
A=
B=
C=
D=
1
2
X=

LO4:
1.
a.
Set 1 (x) X - Ẋ (X – X)2 Set 2 (x) X - Ẋ (X - Ẋ)2
300 -184 33856 300 -174 30276
300 -184 33856 300 -174 30276
300 -184 33856 300 -174 30276
300 -184 33856 300 -174 30276
300 -184 33856 300 -174 30276
300 -184 33856 400 -74 5476
400 -84 7056 400 -74 5476
400 -84 7056 450 -24 576
450 -34 1156 940 466 277156
450 -34 1156 1050 576 331776
800 316 99856
940 456 207936
1050 566 320356
6290 847708 4740 711840
i. Mean (set 1)
Mean= (x1 + X2 + X3……Xn)/ n
Mean= 6290/13
Mean= 483.85
Mean= 484
Median= 300, 300, 300, 300, 300, 300, 300, 400, 400, 450, 450, 800, 940,
1050
Median= 400
Mode= most repeated number
Mode= 300
Range= Highest value – lowest value
23

Range= 1050 – 300
Range= 750
First quartile= 300, 300, 300, 300, 300, 300, 400, 400, 450, 450, 800, 940,
1050
First quartile= 300, 300, 300, 300, 300, 300
First quartile= (300 + 300)/2
First quartile= 600/2
First quartile= 300
Standard deviation= square root {[ ∑ (x−Ẋ ) 2]/n}
Standard deviation = square root (847708)/13
Standard deviation= square root 65208.3
Standard deviation= 255.36
Standard deviation= 255
Variance= standard deviation2
Variance= 2552
Variance= 65,025
ii. Mean (set 2)
Mean= (x1 + X2 + X3……Xn)/ n
Mean= 4740/10
Mean= 474
Median= 300, 300, 300, 300, 300, 400, 400, 450, 940, 1050
Median= (400 + 300)/2
Median= 700/2
Mean= 350
Mode= most repeated number
Mode= 300
Range= Highest value – lowest value
Range= 1050 – 300
Range= 750
First quartile= [300, 300, 300, 300, 300,] [400, 400, 450, 940, 1050]
First quartile= 300, 300, 300, 300, 300
24

First quartile= 300
Standard deviation= square root {[ ∑ (x−Ẋ ) 2]/n}
Standard deviation = square root (711840)/10
Standard deviation= square root 71184
Standard deviation= 266.8
Standard deviation= 267
Variance= standard deviation2
Variance= 2672
Variance= 71, 289
b. AAA
AAD
ADA
ADD
DDD
DDA
DAD
DAA
i. Exactly two
(AAD) and (ADA) and (DAA)
[(0.8) * (0.8) *(0.2)] + [(0.8) * (0.2) * (0.8)] + [(0.2) * (0.8) * (0.8)]
(0.84) + (0.84) + (0.84)
2.52
ii. At most one
(DDA) and (DAD) and (ADD)
[(0.2) *(0.2) *(0.8)] + [(0.2) * (0.8) * (0.2)] + [(0.8) * (0.2) * (0.2)]
(0.032) + (0.032) + (0.032)
0.096
iii. At least one
(AAA) and (AAD) and (ADA) and (DAA) and (DAD) and (DAA)
[(0.8) * (0.8) * (0.8)] + [(0.8) * (0.2) * (0, 8)] + [(0.8) + (0.2) + (0.8)]
+ [(0.2) * (0.8) * (0.8)] + [(0.2) * (0.8) * (0.2)] + [(0.2) * (0.8) * (0.8)]
(0.512) + (0.84) + (0.84) + (0.84) + (0.032) + (0.84)
3.904
25

Flowchart
26
START
N, X, P, Q, n!, b(x; n, p), NCr
b (x; n, P )= NCr * PX * (1 – P )n-x
b (x; n, P)= 200 * 0.01 * (1 – 0.01)100-2
b (x; n, P)= 0.74692856
n=
>0≤100?
b (x; n, P)
STOP
YES
CAN ONLY TOSS
FROM 0-100

Pseudo code
Start
Main {int n, x, P, Q, n!, b (x; n, P), NCr, Multiplication;
Read n, x, P, Q, n!, b (x; n, P ) NCr,
Multiply nCr. PX . (1 – P)n-x
Write multiplication ;}
IF toss is greater than zero and equals or less than a one hundred then
Print “Probability”
Else
Print “can toss only from 0 – 100”
End if
Stop
27