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BTEC Higher National Diploma in Computing
Programme title
Assessor
Internal Verifier
Unit(s)
Unit 11 : Maths for Computing
Assignment title
Importance of Maths in the Field of Computing
Student’s name
Jayasuriya Kuranage Bhanuka Perera NEG – E-118432
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has awarded.
Merit
Distinction
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UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
Date
I
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UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
II
Higher Nationals – Summative Assignment Feedback Form
Student Name/ID
Unit Title
Unit 11 : Maths for Computing
Assignment Number
1
Assessor
Date Received 1st
submission
Date Received 2nd
submission
Submission Date
Re-submission Date
Assessor Feedback:
LO1 Use applied number theory in practical computing scenarios.
Pass, Merit & Distinction P1
P2
M1
D1
Descripts
LO2 Analyse events using probability theory and probability distributions.
Pass, Merit & Distinction P3
Descripts
P4
M2
D2
LO3 Determine solutions of graphical examples using geometry and vector methods.
Pass, Merit & Distinction P5
P6
M3
D3
Descripts
LO4 Evaluate problems concerning differential and integral calculus.
Pass, Merit & Distinction P7
Descripts
Grade:
P8
M4
D4
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Date:
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Date:
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* Please note that grade decisions are provisional. They are only confirmed once internal and external
moderation has taken place and grades decisions have been agreed at the assessment board.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
III
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Use previous page as your cover sheet and make sure all the details are accurately filled.
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11. If you are proven to be guilty of plagiarism or any academic misconduct, your grade
could be reduced to A REFERRAL or at worst you could be expelled from the course.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
IV
Student Declaration
I hereby, declare that I know what plagiarism entails, namely, to use another’s work and to
present it as my own without attributing the sources in the correct way. I further understand
what it means to copy another’s work.
1. I know that plagiarism is a punishable offence because it constitutes theft.
2. I understand the plagiarism and copying policy of the Edexcel UK.
3. I know what the consequences will be if I plagiaries or copy another’s work in any of
the assignments for this program.
4. I declare therefore that all work presented by me for every aspects of my program, will
be my own, and where I have made use of another’s work, I will attribute the source in
the correct way.
5. I acknowledge that the attachment of this document signed or not, constitutes a binding
agreement between myself and Edexcel UK.
6. I understand that my assignment will not be considered as submitted if this document is
not attached to the attached.
bhnkperera@gmail.com
10th October 2022
Student’s Signature:
(Provide E-mail ID)
Date:
(Provide Submission Date)
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
V
Feedback Form
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Date
Date
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
VI
Assignment Brief
Student Name /ID Number
Jayasuriya Kuranage Bhanuka Perera NEG – E-118432
Unit Number and Title
Unit 11 : Maths for Computing
Academic Year
2021/2022
Unit Tutor
Assignment Title
Importance of Maths in the Field of Computing
Issue Date
Submission Date
IV Name & Date
Submission Format:
This assignment should be submitted at the end of your lesson, on the week stated at the front of this brief.
The assignment can either be word-processed or completed in legible handwriting.
If the tasks are completed over multiple pages, ensure that your name and student number are present on
each sheet of paper.
Unit Learning Outcomes:
LO1
Use applied number theory in practical computing scenarios.
LO2
Analyse events using probability theory and probability distributions.
LO3
Determine solutions of graphical examples using geometry and vector methods.
LO4
Evaluate problems concerning differential and integral calculus.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
VII
Assignment Brief and Guidance:
Activity 01
Part 1
1. A tailor wants to make square shaped towels. The required squared pieces of cloth will be cut
from a ream of cloth which is 20 meters in length and 16 meters in width.
a) Find the minimum number of squared pieces that can be cut from the ream of cloth without
wasting any cloth.
b) Briefly explain the technique you used to solve (a).
2. On the first day of the month, 4 customers come to a restaurant. Afterwards, those 4 customers
come to the same restaurant once in 2,4,6 and 8 days respectively.
a) On which day of the month, will all the four customers come back to the restaurant together?
b) Briefly explain the technique you used to solve (a).
Part 2
3. Logs are stacked in a pile with 24 logs on the bottom row and 10 on the top row. There are 15
rows in all with each row having one more log than the one above it.
a) How many logs are in the stack?
b) Briefly explain the technique you used to solve (a).
4. A company is offering a job with a salary of Rs. 50,000.00 for the first year and a 4% raise each
year after that. If that 4% raise continues every year,
a) Find the total amount of money an employee would earn in a 10-years career.
b) Briefly explain the technique you used to solve (a).
Part 3
5. Define the multiplicative inverse in modular arithmetic and identify the multiplicative inverse of
6 mod 13 while explaining the algorithm used.
6. Prime numbers are important to many fields. In the computing field also prime numbers are
applied. Provide examples and in detail explain how prime numbers are important in the field of
computing.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
VIII
Activity 02
Part 1
1. Define ‘Conditional Probability’ with a suitable example.
2. The manager of a supermarket collected the data of 25 customers on a certain date. Out of them
5 purchased Biscuits, 10 purchased Milk, 8 purchased Fruits, 6 purchased both Milk and Fruits.
Let B represents the randomly selected customer purchased Biscuits, M represents the
randomly selected customer purchased Milk and F represents the randomly selected
customer purchased Fruits.
Represent the given information in a Venn diagram. Use that Venn diagram to answer the
following questions.
a) Find the probability that a randomly selected customer either purchased Biscuits or Milk.
b) Show that the events “The randomly selected customer purchased Milk” and “The
randomly selected customer purchased Fruits” are independent.
3. Suppose a voter poll is taken in three states. Of the total population of the three states, 45% live
in state A, 20% live in state B, and 35% live in state C. In state A, 40% of voters support the
liberal candidate, in state B, 30% of the voters support the liberal candidate, and in state C, 60%
of the voters support the liberal candidate.
Let A represents the event that voter is from state A, B represents the event that voter is from
state B and C represents the event that voter is from state C. Let L represents the event that a
voter supports the liberal candidate.
a) Find the probability that a randomly selected voter does not support the liberal candidate
and lives in state A.
b) Find the probability that a randomly selected voter supports the liberal candidate.
c) Given that a randomly selected voter supports the liberal candidate, find the probability that
the selected voter is from state B.
4. In a box, there are 4 types [Hearts, Clubs, Diamonds, Scorpions] of cards. There are 6 Hearts
cards, 7 Clubs cards, 8 Diamonds cards and 5 Scorpions cards in the box. Two cards are selected
randomly without replacement.
a) Find the probability that the both selected cards are Hearts.
b) Find the probability that one card is Clubs and the other card is Diamonds.
c) Find the probability that the both selected cards are from the same type.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
IX
Part 2
5. Differentiate between ‘Discrete Random Variable’ and ‘Continuous Random Variable”.
6. Two fair cubes are rolled. The random variable X represents the difference between the values
of the two cubes.
a) Find the mean of this probability distribution. (i.e. Find E[X] )
b) Find the variance and standard deviation of this probability distribution.
(i.e., Find V[X] and SD[X])
The random variables A and B are defined as follows:
A = X-10 and B = [(1/2)X]-5
c) Show that E[A] and E[B].
d) Find V[A] and V[B].
e) Arnold and Brian play a game using two fair cubes. The cubes are rolled, and Arnold records
his score using the random variable A and Brian uses the random variable B. They repeat
this for a large number of times and compare their scores. Comment on any likely
differences or similarities of their scores.
7. A discrete random variable Y has the following probability distribution.
Y=y
1
P(Y=y)
1/3
where k is a constant.
a) Find the value of k.
b) Find P(Y≤3).
c) Find P(Y>2).
2
1/6
3
1/4
4
k
5
1/6
Part 3
10. The “Titans” cricket team has a winning rate of 75%. The team is planning to play 10 matches
in the next season.
a) Let X be the number of matches that will be won by the team. What are the possible values
of X?
b) What is the probability that the team will win exactly 6 matches?
c) What is the probability that the team will lose 2 or less matches?
d) What is the mean number of matches that the team will win?
e) What are the variance and the standard deviation of the number of matches that the team
will win?
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
X
11. In a boys’ school, there are 45 students in grade 10. The height of the students was measured.
The mean height of the students was 154 cm and the standard deviation was 2 cm. Alex’s height
was 163 cm. Would his height be considered an outlier, if the height of the students were
normally distributed? Explain your answer.
12. The battery life of a certain battery is normally distributed with a mean of 90 days and a standard
deviation of 3 days.
For each of the following questions, construct a normal distribution curve and provide the
answer.
a) About what percent of the products last between 87 and 93 days?
b) About what percent of the products last 84 or less days?
For each of the following questions, use the standard normal table and provide the answer.
c) About what percent of the products last between 89 and 94 days?
d) About what percent of the products last 95 or more days?
13. In the computing field, there are many applications of Probability theories. Hashing and Load
Balancing are also included to those. Provide an example for an application of Probability in
Hashing and an example for an application of Probability in Load Balancing. Then, evaluate in
detail how Probability is used for each application while assessing the importance of using
Probability to those applications.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
XI
Activity 03
Part 1
1. Find the equation (formula) of a circle with radius r and center C(h,k) and if the Center of a circle
is at (3,-1) and a point on the circle is (-2,1) find the formula of the circle.
2. Find the equation (formula) of a sphere with radius r and center C(h, k, l) and show that x2 + y2
+ z2 - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius.
3. Following figure shows a Parallelogram.
If a=(i+3j-k) , b=(7i-2j+4k), find the area of the Parallelogram.
Part 2
4. If 2x - 4y =3, 5y = (-3)x + 10 are two functions. Evaluate the x, y values using graphical method.
5. Evaluate the surfaces in ℝ3 that are represented by the following equations.
i. y = 4
ii. z = 5
6. Following figure shows a Tetrahedron.
Construct an equation to find the volume of the given Tetrahedron using vector methods and if
the vectors of the Tetrahedron are a=(i+4j-2k) , b=(3i-5j+k) and c=(-4i+3j+6k), find the volume
of the Tetrahedron using the above constructed equation.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
XII
Activity 04
Part 1
1. Determine the slope of the following functions.
i. f(x) = 2x – 3x4 + 5x + 8
ii. f(x) = cos(2x) + 4x2 – 3
2. Let the displacement function of a moving object is S(t) = 5t3 – 3t2 + 6t. What is the function for
the velocity of the object at time t.
Part 2
3. Find the area between the two curves f(x) = 2x2 + 1 and g(x) = 8 – 2x on the interval
(-2) ≤ x ≤ 1 .
4. It is estimated that t years from now the tree plantation of a certain forest will be increasing at the
rate of 3t 2 + 5t + 6 hundred trees per year. Environmentalists have found that the level of Oxygen
in the forest increases at the rate of approximately 4 units per 100 trees. By how much will the
Oxygen level in the forest increase during the next 3 years?
Part 3
5. Sketch the graph of f(x) = x5- 6x3 + 3 by applying differentiation methods for analyzing where
the graph is increasing/decreasing, local maximum/minimum points [Using the second derivative
test], concave up/down intervals with inflection points.
6. Identify the maximum and minimum points of the function f(x)= 2x3 - 4x4 + 5x2 by further
differentiation. [i.e. Justify your answer using both first derivative test and second derivative test.]
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
XIII
Grading Rubric
Grading Criteria
Achievement
(Yes/No)
Feedback
LO1 : Use applied number theory in practical computing
scenarios.
P1 : Calculate the greatest common divisor and least common multiple
of a given pair of numbers.
P2 : Use relevant theory to sum arithmetic and geometric progressions.
M1 : Identify multiplicative inverses in modular arithmetic.
D1 : Produce a detailed written explanation of the importance of prime
numbers within the field of computing.
LO2 : Analyze events using probability theory and
probability distributions.
P3 : Deduce the conditional probability of different events occurring
within independent trials.
P4 : Identify the expectation of an event occurring from a discrete,
random variable.
M2 : Calculate probabilities within both binomially distributed and
normally distributed random variables.
D2 : Evaluate probability theory to an example involving hashing and
load balancing.
LO3 : Determine solutions of graphical examples using
geometry and vector methods.
P5 : Identify simple shapes using co-ordinate geometry.
P6 : Determine shape parameters using appropriate vector methods.
M3 : Evaluate the coordinate system used in programming a simple
output device.
D3 : Construct the scaling of simple shapes that are described by vector
coordinates.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
14
LO4 : Evaluate problems concerning differential and
integral calculus.
P7 : Determine the rate of change within an algebraic function.
P8 : Use integral calculus to solve practical problems involving area.
M4 : Analyse maxima and minima of increasing and decreasing
functions using higher order derivatives.
D4 : Justify, by further differentiation, that a value is a minimum.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
15
ACKNOWLEDGMENT
Primarily, I would like to express my sincere gratitude to the lecturer of this unit,
Mr. Nadheera Senasinghe , for the delivery of valuable lectures, assignment guidance
and review sessions conducted to examine our progress from time to time.
Aside from him, I would also like to extend my gratitude to the management of ESoft
Metro Campus for conducting extra guidance sessions and for providing the necessary
facilities to join online learning due to the prevailing pandemic situation.
Finally, I would also like to thank my family and my dear batch mates for their
support and encouraging me during this challenging time.
Regards,
J.K Bhanuka Perera
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
16
Content
ACKNOWLEDGEMENT .................................................................................................... 16
Activity 01 .................................................................................................................... 18
Part 1 ..................................................................................................................................... 18
Part 2 ..................................................................................................................................... 20
Part 3 ..................................................................................................................................... 24
Activity 02 .................................................................................................................... 27
Part 1 ..................................................................................................................................... 27
Part 2 ..................................................................................................................................... 34
Part 3 ..................................................................................................................................... 43
Activity 03 .................................................................................................................... 51
Part 1 ..................................................................................................................................... 51
Part 2 ..................................................................................................................................... 54
Activity 04 .................................................................................................................... 59
Part 1 ..................................................................................................................................... 59
Part 2 ..................................................................................................................................... 61
Part 3 ..................................................................................................................................... 63
REFERENCES ............................................................................................................ 68
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
17
List of figures
Figure 1. Load balancing | Source (nginx. 2022) .....................................................................51
Figure 2. graphs for the above function | Source (Author's work) ...........................................56
Figure 3. y = 4 graph | Source (Author’s work) .......................................................................57
Figure 4. Z = 5 | Source (Author’s work).................................................................................57
Figure 5. the graph | Source (Author's work) ...........................................................................67
List of tables
Table 1. Differences Discrete and Continuous Random Variable.| Source (Author's work) ..37
Table 2. Decreasing increasing intervals | Source (Author's work). ........................................64
Table 3. Concavity graph | Source (Author's work). ................................................................66
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
18
Activity 01
Part 1
1) A tailor wants to make square shaped towels. The required squared pieces
of cloth will be cut from a ream of cloth which is 20 meters in length and 16
meters in width.
a) Find the minimum number of squared pieces that can be cut from the ream
of cloth without wasting any cloth.
GCF or Greatest Common Divisor is
the largest common factor of two or
more numbers
G.C.F of 20 & 16,
20 = 2 x 2 x 5 x 1
16 = 2 x 2 x 2 x 2 x 1
2 2
Greatest Common Factor = 2 x 2
=4
Area of a piece
=4*4
= 16
Area of the cutting piece = 20 * 16
= 320
∴ The minimum number of square pieces
that can be cut by a loom without wasting
Number of square pieces = 320 / 16
fabric is 20.
= 20
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
19
b) Briefly explain the technique you used to solve (a).
The method I used above is called the greatest common factor. GCF stands for "greatest common
factor" among other things. There are several methods use to find the highest/greatest common factor.
It depends on how many numbers you have, how large they are and what you will do with the result.
It is useful when we want to simplify a part. For example, the GCF of the given case is 4. Hence, the
common factor of 20 and 16 is 4, and we can divide both these numbers by 4. Finding the greatest
common factor is a simple process. We must first determine all the components of both numbers, then
find the one that is most common to both, and then select the greatest.
Using the previous example, we can calculate the length of one side of the fabric using GCF. The final
answer is 20 square pieces.
2) On the first day of the month, 4 customers come to a restaurant. Afterwards,
those 4 customers come to the same restaurant once in 2,4,6 and 8 days
respectively.
a) On which day of the month, will all the four customers come back to the
restaurant together?
2=2x1
4=2x2
6=2x3
8=2x2x2
L.C.M = 2 x 2 x 2 x 3 x 1
N = 24
N + 1 day = 25
∴ On the 25th day of the month, all four customers will visit the restaurant.
together.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
20
b) Briefly explain the technique you used to solve (a).
The Least Common Multiple approach is what I used above. The smallest of all common multiples is
the Least Common Multiple or L.C.M. Using the example above, the numbers provided are 2, 4, 6 and
8 respectively. Consequently, the most common multiples are 2, 2, 2, 3, and 1. Consequently, the
product of these numbers is 24, and the smallest number formed by the above numbers is also 24.
After subtracting the first day and adding 1 to the total of 24 days, all four customers will return on the
25th day. LCM is the lowest two or more numbers that are evenly divided by all numbers in the set.
Part 2
3). A Logs are stacked in a pile with 24 logs on the bottom row and 10 on the top
row. There are 15 rows in all with each row having one more log than the one above
it.
a) How many logs are in the stack?
𝑆𝑛 =
𝑛
(𝑎 + 𝑙)
2
𝑆15 =
15
(24 ∗ 10)
2
a = 24
l = 10
n = 15
= 15 * 32 / 2
𝑆15 = 15 ∗ 17
𝑆15 = 𝟐𝟓𝟓
∴ There are 255 logs in the stack.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
21
b) Briefly explain the technique you used to solve (a).
The bottom row has 24 logs. And 10 logs in the top row. In each row, one log is lower
than the previous row from the bottom. There are about 10 rows in the stack. So, we can
use the following formula to calculate the total number of logs in the stack. This is used
to find the sum of an arithmetic series.
𝑺𝒏 =
𝒏
(𝒂 + 𝒍)
𝟐
According to this formula,
- “Sn” is referred to total of the arithmetic series.
- ”n” refers to numbers of terms in arithmetic series
- “a” refers to first term in the series
- “l” refers to the last term in the series.
[Space left intentionally]
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
22
4. A company is offering a job with a salary of Rs. 50,000.00 for the first year and
a 4% raise each year after that. If that 4% raise continues every year
a) Find the total amount of money an employee would earn in a 10-year
career.
𝑎 (𝑟 𝑛 − 1)
𝑆𝑛 =
𝑟−1
𝑆10
600000(1.0410 − 1)
=
1.04 − 1
6 ∗ 105 (1.48 − 1)
=
0.04
a = 600000
r = 0.04
n = 10
𝑺𝟏𝟎 = 𝟕𝟐𝟎𝟑𝟔𝟔𝟒
∴ The total amount earned by an employee in this career
in 10 years is Rs. 7 203 664.00
[Space left intentionally]
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
23
a) Briefly explain the technique you used to solve (a).
This is an arithmetic progression. The formula used above is used to find the sum of a
geometric series. According to the question, the monthly salary of an employee is
50000/-. 600000/- per annum per employee. (50000 x 12).
So, the salary of an employee for the first year is Rs.600000. Salary is increased from
4% every year.
Salary for first year
= 50000 * 12
= Rs.600 000/=
Salary for second year
= 600000 * (104/100)
= Rs.624 000/=
Salary for third year
= 624000 ∗ (
104
100
)
= Rs.648 960/=
Common Growth Factor (r) = 648960 / 624000
= 1.04
According to the formula,
S10 stands for total salary for the 10 years of an employee
(r) Stands for growth factor. It is 1.04 as found above.
And n is the numbers of terms in series. We need to find the salary in 10 years. Therefore, value 10
can be used as n for the formula. By replacing those values to the formula, Total salary for 10 years
can be found.
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
24
Part 3
5. Define the multiplicative inverse in modular arithmetic and identify the
multiplicative inverse of 6 mod 13 while explaining the algorithm used.
Multiplicative inverse in modular arithmetic,
Especially in the field of mathematics, a modular multiplicative inverse of an integer a is an integer
that is congruent to the product of ax with respect to modulus m. [1] This equality is expressed in the
standard notation of modular arithmetic,
𝒂𝒙 ≡ 𝟏(𝒎𝒐𝒅 𝒎)
It is a condensed way of saying that the quantity m ax is divided (equally) by 1, or in other words, the
result of dividing ax by the number m is 1. This congruence has an infinite number of solutions, which
are formed. A congruence class with respect to this modulus, if a has an inverse modulus m.
Identify the multiplicative inverse of 6 mod 13
6 * 0 ≡ 0 ≡ 0 (mod 13)
6 * 1 ≡ 6 ≡ 6 (mod 13)
6 * 2 ≡ 12 ≡ 12 (mod 13)
6 * 3 ≡ 18 ≡ 5 (mod 13)
6 * 4 ≡ 24 ≡ 11 (mod 13)
6 * 5 ≡ 30 ≡ 4 (mod 13)
6 * 6 ≡ 36 ≡ 10 (mod 13)
6 * 7 ≡ 42 ≡ 3 (mod 13)
6 * 8 ≡ 48 ≡ 9 (mod 13)
6 * 9 ≡ 54 ≡ 2 (mod 13)
6 * 10 ≡ 60 ≡ 8 (mod 13)
6 * 11 ≡ 66 ≡ 1 (mod 13)
The 6 mod 13 = 11
∴ 11 will be the
modular inverse for 6 mod 13, which can
be represented as 6 -1 = 11 mod 13.
6 * 12 ≡ 72 ≡ 7 (mod 13)
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
25
The multiplicative inverse of a number x and m is a number y such that xy = 1 (mod m) and we write
it like y = x-1.
6x = 1 (mod 13)
13 = 12 + 1 = 2*6 + 1 (the remainder term here is the GCD)
Therefore,
1 = 13 – 2*6 and taken mod 13,
1 = 13 – 2*6 = - 2.6
And -2 = 11 (mod 13)
So, the inverse of 6 mod 13 is 11.
The algorithm used to find the inverse is called Euclid's algorithm. It is a means of subtracting the
GCD of two given integers.
A multiplicative inverse of a number is defined as a number whose original number is multiplied by
one. a-1 or 1/a denotes the multiplicative inverse of 'a'. In other words, if the product of two integers
is 1, they are multiplicative inverses of each other.
[Space left intentionally]
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6. Prime numbers are important to many fields. In the computing field also prime
numbers are applied. Provide examples and in detail explain how prime numbers
are important in the field of computing.
Cryptography
Prime numbers are used to encrypt and decrypt data. By selecting 2 large prime numbers like 6-digit
primes and multiplying them we can obtain a strong primary key. It is easy to obtain the key. But it is
very difficult to find the 2 prime numbers from the given product number (decrypt). It would take a
long time decode the key even from super computers to find the numbers. In that time the transaction
could be completed and data will be secure.
Hashing
Hashing is an algorithm that computes a string value from a fixed-size file and is a secure way to
identify and compare databases and files. Hashing works using the Rivest-Shamir-Adleman algorithm
to create and declare private and public keys using large prime numbers and extra value. Because
prime numbers are kept secret, only the person with the prime number can encrypt and decrypt
messages.
Gödel Numbering
Gödel numbering system is the method of representing any sentence, sequence of words or symbols
with a corresponding number. These numbers must be unique as no 2 combinations can have the same
number. Hence prime numbers are used for their uniqueness. This system is a one-to-one task. That
means it converts one value to number and takes the second value, converts it to number and carries
on.
Computing Hash Codes
This is a numerical code assigned to each object generated by the software. Quickly retrieving/storing
complex items from a hash table or in it, requires hash codes. To ensure accuracy, hash codes must be
fairly unique for each item. For this reason, prime integers are used when calculating hash codes.
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Activity 02
Part 1
1. Define ‘Conditional Probability’ with a suitable example.
Conditional probability is the measure of the probability of an event occurring given that another event
has previously occurred in probability theory. This process depends on event B, which is tied to another
event in some way. In this situation, event B can be tested in relation to event A using conditional
probability. If A is the event of interest and B is known or assumed to have occurred. "Conditional
probability of B given A", or "probability of B under condition A" is usually written as P (A | B) [2]
or occasionally P B (A). This is the portion of probability B that overlaps with probability A.
Conditional probability, as defined earlier, depends on the prior outcome. Imagine you draw 3 marbles
from a bag: red, green and blue. Each marble has an equal probability of being selected. What is the
possibility of drawing a red marble after drawing blue?
As a first step, it is one of three possible outcomes, the probability of drawing a blue marble is about
33%. After the initial event occurs, two marbles will remain with a 50% chance of each being drawn.
The probability of getting a blue marble after drawing a red marble is about 16.5% (33% x 50%).
There are two formulas used in conditional probability,
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2. The manager of a supermarket collected the data of 25 customers on a certain
date. Out of them 5 purchased Biscuits, 10 purchased Milk, 8 purchased Fruits,
6 purchased both Milk and Fruits. Let B represents the randomly selected
customer purchased Biscuits, M represents the randomly selected customer
purchased Milk and F represents the randomly selected customer purchased
Fruits.
Represent the given information in a Venn diagram. Use that Venn diagram to answer the
following questions.
M - Milk
B – Biscuits
F - Fruits
a) Find the probability that a randomly selected customer either purchased Biscuits or Milk.
Probability of randomly selected customer purchasing Biscuit P (𝐵) = (5/25) = (1/5)
Probability of randomly selected customer purchasing Milk 𝑃 (𝑀) = (10/25) = (2/5)
Probability of randomly selected customer purchase Milk or Biscuit 𝑃 (𝐵 ∪ 𝑀) = (1/5) + (2/5)
= (3/5)
∴ There is (3/5) probability that the randomly selected customer
purchase either Milk or Biscuit.
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b)
Show that the events “The randomly selected customer purchased Milk” and “The
randomly selected customer purchased Fruits” are independent.
If the randomly selected consumer purchases and the randomly selected consumer purchases fruits
are independent, then the theorem must be proved below.
P (M ∩ F) = P (M) x P (F)
Probability of purchasing milk P (M) = 10/25
Probability of purchasing milk P (M) = 8/25
P (M ∩ F) = P (F) x P (M)
= 10/25 x 8/25
= 16/125
But in Venn diagram, we can clearly see that (𝑀 ∩ 𝐹) = 6/25
6/25 ≠ 16/125
∴ 𝑃(𝑀 ∩ 𝐹) ≠ 𝑃(𝑀) × 𝑃(𝐹)
So, by using above theorem we can see that randomly selected customer purchase Milk and the
randomly selected customer purchase Fruits are not Independent.
[Space left intentionally]
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3.
Suppose a voter poll is taken in three states. Of the total population of the
three states, 45% live in state A, 20% live in state B, and 35% live in state C. In
state A, 40% of voters support the liberal candidate, in state B, 30% of the voters
support the liberal candidate, and in state C, 60% of the voters support the liberal
candidate.
Let A represents the event that voter is from state A, B represents the event that voter is from
state B and C represents the event that voter is from state C. Let L represents the event that a
voter supports the liberal candidate.
a) Find the probability that a randomly selected voter does not support the
liberal candidate and lives in state A.
Probability of a randomly selected voter who does not support the liberal candidate and lives in state
A = A * the percentage of voters who do not support the liberal candidate in state A's population.
= (60/100) x (45/100)
= 27 %
∴ Probability of randomly selected voter doesn’t support liberal
candidate and lives in state A is 27 %.
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b)
Find the probability that a randomly selected voter supports the liberal
candidate.
The probability of a randomly selected customer support liberal candidate is equal to: State A
Population * State A Liberal Candidate Supporters + State B Population * State B Liberal Candidate
Supporters + State C Population * State C Liberal Candidate Supporters
=
45
40
20
30
35
60
𝑋
+
𝑋
+
𝑋
100 100 100 100 100 100
18
6
21
=
+
+
100 100 100
45
∴ Randomly selected voter
=
100
supports the liberal candidate is
=
𝟗
𝟐𝟎
9/20
c) Given that a randomly selected voter supports the liberal candidate, find
the probability that the selected voter is from state B.
The probability of a randomly selected voter supporting the liberal candidate in state B = population
of state B * liberal supporters in state B.
20
30
𝑋
100 100
60
=
100
=
= 𝟔%
Since a randomly selected voter supports the liberal candidate, the probability that the selected voter
is from state B is equal to the product of liberal candidate support times liberal candidate supporters in
state B.
=
45
100
x
6
100
= 2.7%
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4.
In a box, there are 4 types [Hearts, Clubs, Diamonds, Scorpions] of cards.
There are 6 Hearts cards, 7 Clubs cards, 8 Diamonds cards and 5 Scorpions cards
in the box. Two cards are selected randomly without replacement.
a)
Find the probability that both selected cards are Hearts.
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b)
Find the probability that one card is Clubs and the other card is Diamonds.
Probability of both selected cards are heart,
=
6
26
=
=
c)
x
5
25
30
650
𝟑
𝟔𝟓
Find the probability that both selected cards are from the same type.
Probability of both selected cards are same type,
=
6
26
x
5
25
=
+
7
26
30
650
x
+
6
25
42
650
=
=
+
+
8
26
x
56
650
7
25
+
+
5
26
x
4
25
20
650
148
650
𝟕𝟒
𝟑𝟐𝟓
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Part 2
5. Differentiate between ‘Discrete Random Variable’ and ‘Continuous Random
Variable”.
A variable is a named unit of data that is assigned a value. If the value is changed, the name of the
variable will not change. In the context of a mathematical problem or experiment, a variable is a
quantity that can change. A variable is often represented by a single letter. Variables are usually
represented by the letters x, y, and z. There are several types of variables.
(Definitions and Hope, 2022)
Discrete Random Variable
A discrete random variable is a variable that can have any outcome in a random experiment that is an
integer. Discrete random variables have a finite number of alternative outcomes, which can be counted
as 0, 1, 2, 3, 4, . To display the values of discrete random variables, probability distributions are used.
A stochastic variable is another name for a discrete random variable. Binomial and Poisson random
variables are two examples of discrete random variables. In general, there are two forms of data:
discrete and continuous. In this case, a discrete random variable is considered. Additionally, an
algebraic variable and a discrete random variable are not the same thing. A discrete random variable
has several values, but an algebraic variable has only one.
The results of a random experiment are calculated using a discrete random variable. A countable
number of outcomes can be considered for discrete random variables. A discrete random variable can
usually be numbered as 0, 1, 2, 3, 4, . Both discrete and continuous data types are possible; In this case,
discrete random variables are considered. To illustrate how probabilities are spread over the values of
discrete random variables, probability distributions are used.
As for an example: Suppose two dice are rolled and the sum of the numbers is represented by the
random variable X. Then, 1 + 1 = 2 results in the smallest value of X and 6 + 6 = 12 results in the
largest value. The value of X. Consequently, X can have values from 2 to 12. The probability
distribution of X can now be established if probabilities are assigned to each outcome. It is important
to distinguish a discrete random variable from an algebraic variable. The value of an unknown quantity
in an algebraic equation that can be determined is represented by an algebraic variable. A discrete
random variable, However, can have a range of values that can represent the outcome of the
experiment.
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Continuous Random Variable
A continuous random variable is one that has a range of possible values. In other words, if a random
variable follows a value that fits within a specified range, it is said to be continuous. Measurements
such as height, weight, and time are all represented by continuous random variables. A continuous
random variable is represented as the area under a density curve.
A random variable is one whose value changes in every possible instance in an experiment. A discrete
random variable is defined at a specific value, while a continuous random variable is defined over a
range of values. A random variable with an infinite number of possible values is called a continuous
random variable. As a result, a continuous random variable has no chance of having a specific value.
The characteristics of a continuous random variable are described using the probability density
function and the cumulative distribution function.
The probability associated with a continuous random variable is expressed using the probability
density function (pdf) and the cumulative distribution function (CDF).
As for an example: Suppose that the probability density function of a continuous random variable X
is given by 4x3, where x ∈ [0, 1]. Determine the probability that X takes a value between 1/2 and 1.
This can be done by integrating 4x3 between 1/2 and 1. Thus, the required probability is 15/16.
[Space left intentionally]
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Differences between Discrete Random Variable and Continuous Random
Variable.
The following are some key differences between discrete random variables and continuous
random variables,
•
A continuous random variable's value fluctuates within a specified range, while a discrete
random variable can take on a specified value.
•
As a consequence, a probability mass function is used to describe a discrete random variable
and a probability density function describes a continuous random variable.
•
Binomial, geometric, Bernoulli, and Poisson random variables are some examples of
distributions with discrete random variables. Normal and exponential random variables are two
types of continuous random variable distributions.
Table 1. Differences of Discrete and Continuous Random Variable.| Source (Author's work)
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6). Two fair cubes are rolled. The random variable X represents the difference
between the values of the two cubes.
The probability distribution of X
a) Find the mean of this probability distribution. (i.e., Find E[X] )
The mean of the probability distribution,
E(X) = μ = ∑x.P(x)
= 0 + 0.278 + 0.444 + 0.501 + 0.444+ 0.28
μ = 1.947
∴ The mean is 1.947
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b) Find the variance and standard deviation of this probability distribution. (i.e.,
Find V[X] and SD[X])
The random variables A and B are defined as follows:
A = X-10 and B = [(1/2)X]-5
The random variables A and B are defined as follows:
A = X-10 and B = [(1/2) X]-5
Find the V(x):
V(x) = E(x2) – [E(x)] 2
E(x2) = ∑ x2. P(x)
= ( -52 x 1/36) + (-42 x 2/36)+ (-32 x 3/36) + (-22 x 4/36) + (-12 x 5/36) + (02
x 6/36) + (12 x 5/36) + (22 x 4/36) + (32 x 3/36) + (42 x 2/36) + (52 x 1/36)
= + 25/36 + 32/36 + 27/36 + 16/36 + 5/36 + 0 + 5/36 + 16/36 + 27/36 + 32/36
+25/36
= 210/36 = 5.833
[E(x)] 2 = 02
V(x) = 5.833 – 0
V(x) = 5.833
Find SD(x):
SD(x) = √V(x)
SD(x) = √5.833 = 2.415
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c) Show that E[A] and E[B].
A = X-10
E(A) = E(X) – 10
= 1.947-10
= -8.053
B
= [(1/2)X]-5
E(B) = [(1/2)E(X)] – 5
= 1.947/2 - 5
= -4.027
d) Find V[A] and V[B].
V(A) = V(X)
= 16.87
V(B) = (1/2)2 * V(X)
= 16.87 / 4
= 4.2175
[Space left intentionally]
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e) Arnold and Brian play a game using two fair cubes. The cubes are rolled, and
Arnold records his score using the random variable A and Brian uses the random
variable B. They repeat this for a large number of times and compare their scores.
Comment on any likely differences or similarities of their scores.
The difference is that they use two different dice with different outcomes and the similarity is that they
use two fair dice with a random variable with different outcomes. Arnold uses the random variable A
to record the score when the dice are rolled, while the brain uses the random variable B. Then there is
a random variable with values nn, which is the number of times Arnold rolls the fair pair of dice, and
pp, which is the probability of success with the random variable AA for each value. Consider the
random variable section, which denotes nn, the number of times the brain rolls a fair pair of dice, and
pp, the chance of success, where each value can take a random variable BB. The probability
distribution has the same binomial distribution with parameters nn and pp.
∴ As proven above E (B) = (1/2) E (A). If Arnold and Brian
recorded their scores using random variable A and B,
Arnold’s score will be half of the score of brains.
[Space left intentionally]
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7) A discrete random variable Y has the following probability distribution.
Y=y
1
2
3
4
5
P(Y=y)
1/3
1/6
1/4
k
1/6
where k is a constant.
a) Find the value of k.
𝑃(𝑌) = 1
𝑃(𝑌) = 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4) + 𝑃(5)
=
1 1 1
1
+ + + 𝐾 +
3 6 4
6
= 0.333 + 0.167 + 0.25 + 𝐾 + 0.167
1 = 0.918 + 𝐾
𝐾 = 1 − 0.917
𝑲 = 𝟎. 𝟎𝟖𝟑
∴ The value of K = 0.083
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b) Find P(Y≤3)
𝑃(𝛾 ≤ 3)
𝑃(𝑦 ≤ 3) = 𝑃(𝑦 = 1) + 𝑃(𝑦 = 2) + 𝑃(𝑦 = 3)
=
1 1 1
+ +
3 6 4
=
𝟑
𝟒
c) Find P(Y>2)
𝑃(𝑦 > 2) = 𝑃(𝑦 = 3) + 𝑃(𝑦 = 4) + 𝑃(𝑦 = 5)
=
1 1 1
+
+
4 12 6
=
𝟏
𝟐
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Part 3
10. The “Titans” cricket team has a winning rate of 75%. The team is planning to
play 10 matches in the next season.
a)
Let X be the number of matches that will be won by the team. What
are the possible values of X?
= xn b⁄n (10,0,0.75)
𝒙 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕, 𝟖, 𝟗, 𝟏𝟎
b)
What is the probability that the team will win exactly 6 matches?
𝑃(𝑥 = 6) = 𝑛𝑐𝑥 𝑃 𝑥 (𝑄)𝑛−𝑥
= 10 𝐶 ∗ (0.75)6 ∗ (1 − 0.75)10−6
6
= 𝟎. 𝟏𝟒𝟓𝟗𝟗𝟔
c)
What is the probability that the team will lose 2 or less matches?
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
= 10C8 (0.75)8 (0.25)10-8 + 10C9 (0.75)9 (0.25)10-9 + 10C10 (0.75)10 (0.25)10-10
= (45 × 0.1001 × 0.0625) + (10 × 0.075 × 0.25) + (1 × 0.0jj563 × 1)
= 0.2815 + 0.1875 + 0.0563
= 0.5256
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d)
What is the mean number of matches that the team will win?
𝜖(𝑥) = 𝑛 ∗ 𝑃
= 10 ∗ 0.75
= 𝟕. 𝟓
e)
What are the variance and the standard deviation of the number of
matches that the team will win?
𝑣𝑎𝑟(𝑥) = 𝑛 ∗ 𝑃 ∗ 𝑄
= 10 ∗ 0.75 ∗ 0.25
= 𝟏. 𝟖𝟕𝟓
[Space left intentionally]
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11.
In a boys’ school, there are 45 students in grade 10. The height of the students
was measured. The mean height of the students was 154 cm and the standard
deviation was 2 cm. Alex’s height was 163 cm. Would his height be considered an
outlier, if the height of the students were normally distributed? Explain your
answer.
𝛧 𝑠𝑐𝑜𝑟𝑒 =
= 163 −
= 𝟒. 𝟓
𝑚𝑒𝑎𝑛
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
154
2
Total students – 45
Mean height (µ) = 154 cm
Standard deviation (σ) = 2
cm
The height of Alex is 4.5 standard deviations above the mean value indicating that
Alex will be in the 3 percent outside the 3-standard deviation unit. Heights below than
148cm and 160cm are known/mentioned as outliners.
∴ Alex’s height is 163 cm. and it is greater than 160cm.
Alex will be an outlier So, his height is an outliner.
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12. The battery life of a certain battery is normally distributed with a mean of 90
days and a standard deviation of 3 days.
For each of the following questions, construct a normal distribution curve and
provide the answer.
a) About what percent of the products last between 87 and 93 days?
To calculate the percentage of battery life that last between 87 and 93 days we must calculate the z
score for the given data points 87 and 93.
Here z score is the standard deviation units from the mean
Z score for day 87
Mean (µ) = 90, Standard deviation (σ) = 3,
According to the provided data,
µ + σ = 93
∴ 68.2 % is the answer
µ - σ = 87
Percentage of the products lasts 87 – 93 days
= 68.26%
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b) About what percent of the products last 84 or less days?
According to second standard deviation theory,
Mean (µ) = 90, Standard deviation (2σ) = 6
µ + 2σ = 96
µ - 2σ = 84
Percentage of products lasts 84 or less days =
1.0000 – 0.9546
2
= 0.0227
= 2.27%
c) About what percent of the products last between 89 and 94 days?
P (89 < X < 94) = P (X < 94) – P (X ≤ 89)
= P (Z <
[94 − 90)
89 − 90
] – P [Z ≤
]
3
3
= P(Z < 1.3333)– P(Z ≤ −0.3333)
= 0.9088 – 0.3694
= 53.94%
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d) About what percent of the products last 95 or more days?
P (X ≥ 95) = 1 – P (X < 95)
= 1 – P [Z < (95 - 90)/3]
= 1 – P (Z < 1.67)
= 1 – 0.952
= 4.8%
13. In the computing field, there are many applications of Probability theories.
Hashing and Load Balancing are also included to those. Provide an example for an
application of Probability in Hashing and an example for an application of
Probability in Load Balancing.
Then, evaluate in detail how Probability is used for each application while
assessing the importance of using Probability to those applications.
Hashing
Hashing is the process of converting a given key into another value. A hash function is used to initialize
the new value according to a mathematical algorithm. The result of a hash function is called a hash
value or simply a hash. In general, a good job uses a one-way hashing algorithm. In other words, the
hash cannot be converted back to the original key.
The effectiveness of all hashing algorithms is determined by how often this happens. A hash table is
an associative data structure that holds data. Data is stored in an array format in a hash table where
each data item has a unique index value. Data access is very fast when the required data index is
known. This makes it a data structure that can be inserted and searched very quickly, regardless of the
size of the data. A hash table uses an array to create an index into which an element is to be inserted
or located using hash access.
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Let's see how it works. A table of average size T maps the hash function h. To add x to a set we evaluate
h(x) and store x at position h(x) in the table (i.e., use the hash function to key). All keys are assigned
to the same. The table position of our set is saved in a simple linked list. The DELETE and MEMBER
operations are implemented in the same way by evaluating h(x). Note that the effectiveness of a hash
function depends on Minimal conflict. The search time for DELETE and MEMBER operations is
proportional to the length of the associated list.
Source: (What is hashing? 2022)
Balls and bins
Let's start this problem by putting small balls and bowls into the structure. The balls will represent the
keys to be saved (m), and the bins will represent the positions (n) in the hash table (T). Because the
hash function maps each key to a random location in the table (T). We think of each key (ball) as
selecting a hash table location (bin) equally and independently (T). Consequently, the probability space
corresponding to this hashing experiment is the same as the ball-and-bin space. We are interested in
event A, which is either no collision or all m balls landing in different bins. Clearly, as m increases,
P(A) decreases (with n fixed). Our objective is to find the largest value of m for which P(A) is greater
than 50%.
Load balancing
Load balancing, commonly referred to as server farming or server pooling, is the process of efficiently
distributing incoming network traffic between backend servers.
Modern high-traffic websites must respond quickly and reliably to millions, not hundreds of thousands,
of concurrent user or server requests for accurate text, photos, videos, or application data. Modern
computing best practices typically require the addition of additional servers to cost-effectively scale to
handle these enormous volumes.
A load balancer serves as a "traffic policeman" in front of your servers, distributing server requests
across all servers equipped to handle them in a way that maximizes speed and capacity utilization, and
ensures that no servers are overworked, which could cause performance degradation. . If one server
goes offline, the load balancer moves traffic to the active servers. "The load balancer initiates requests
to a new server when it is added to the server group." (“Elastic Load Balancing in AWS - Geekflare”)
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Figure 1. Load balancing | Source (nginx. 2022)
Assume we have m tasks and n processors all identical. The goal is to distribute the workload among
the processors in such a way that no single processor is overloaded. Of course, there is a direct optimal
solution here.
Divide the work as evenly as possible so that each processor gets m/n jobs. However, this solution
requires a lot of centralized control and/or communication: the workload must be distributed evenly
by using a powerful centralized scheduler that communicates with all processors, or by exchanging
many messages between jobs and processors. In most distributed systems, this type of operation is
very expensive. We can simply apply the ball and bin method here to get rid of the cost.
In another way, each job randomly and independently of the others selects a processor and sends it to
the processor. (Make sure you imagine that the probability space in this experiment is the same as the
space for balls and pots.). Communication is not required in this approach. However, it is unlikely to
achieve perfect weight balance in most cases. Let Ak be when the weight of a processor is at least k.
[Space left intentionally]
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Activity 03
Part 1
1. Find the equation (formula) of a circle with radius r and center C(h,k) and if the
Center of a circle is at (3,-1) and a point on the circle is (-2,1) find the formula of
the circle.
Center of the circle = C (h,k)
Radius of the circle = r
(x, y) is any point of the circle
r
c(3, −1)(−2,1)
r 2 = (3 − (−2))2 + ((−1) – (1))2
r 2 = (5)2 + (−2)2
r 2 = 29
𝒓 = 𝟓. 𝟑𝟖𝟓
r = √29
Or else,
𝑟 2 = 𝑥 2 + 𝑦 2 – 2ℎ𝑥 − 2𝑘𝑦 + ℎ2 + 𝑘 2
29 = 𝑥 2 + 𝑦 2 – 2(3)𝑥 − 2(−1)𝑦 + (3)2 + (−1)2
29 = 𝑥 2 + 𝑦 2 − 6𝑥 + 2𝑦 + 10
𝐱 𝟐 + 𝐲 𝟐 − 𝟔𝐱 + 𝟐𝐲 − 𝟏𝟗 = 𝟎
∴ Formula of the circle is = x2 + y2 – 6x + 2y – 19 = 0 and
radius is (r) = √29
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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2. Find the equation (formula) of a sphere with radius r and center C(h, k, l) and
show that x2 + y2 + z2 - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find
its center and radius.
Centre of the sphere = C (h, k, l)
(h,k,l)
r
(x,y,z)
Radius of the sphere = r
(x, y, z) is any point of the sphere.
Sphere is the 3D representation of a circle. So, the equation of the sphere is also similar to the equation
of circle but with one extra dimension.
√ (𝑥 − ℎ) 2 + (𝑦 − 𝑘) 2 + (𝑧 − 𝑙) 2 = r
(𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 + (𝒛 − 𝒍)𝟐 = 𝒓𝟐
x2 + y2 + z2 – 6x +2y +8z – 4 = 0
x2 – 6x + y2 + 2x + z2 + 8z – 4 =0
(x - 3)2 + (y + 1)2 + (z + 4)2 – 26 – 4 = 0
(x - 3)2 + (y + 1)2 + (z + 4)2 = 30
Above equation is an equation of a sphere.
Centre of the sphere (h, k, l) = (3, -1, -4)
∴ Radius (r) = √30 = 5.477
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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3. Following figure shows a Parallelogram.
If a=(i+3j-k) , b=(7i-2j+4k), find the area of the Parallelogram,
i
j
k
[a*b] = 1
3
-1
-2
4
7
= 𝑖(3 ∗ 4 − (−2) ∗ (−1)) − 𝑗(1 ∗ 4 − 7 ∗ (1)) + 𝑘(1 ∗ (−2) − 7 ∗ 3)
[a*b]
= √102 + (−11)2 + (23)2
= 100 + 121 + 529
= √750
= 5√30
= 27.386
= 𝟐𝟕. 𝟓
∴ Area of the parallelogram is 27.5
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Part 2
4. If 2x - 4y =3, 5y = (-3)x + 10 are two functions. Evaluate the x, y values using
graphical method.
Function 1 = 2x – 4y = 3
Function 2 = (-3) x + 10 = 5y
Let’s use these 2 functions in y = mx + c.
Function 1:
Function 2:
4y = 2x – 3
(-3) x + 10 = 5y
y = (½) x – (¾)
5y = (-3) x + 10
y = - (3/5) x + 2
According to the two formulas,
When x = 0,
When x = 0,
When x = 2,
y = - (3/4)
y=2
y = 1 – 0.75
(0, -0.75)
(0, 2)
y = 0.25
(2, 0.25)
When y = 0,
When x = 4,
When x = 2,
(½) x = (¾)
y = - (0.6) x 4 + 2
y = - (3/5) x 2 + 2
x = 1.5
y = - 2.4 + 2
y = 0.8
(1.5, 0)
y = - 0.4
(2, 0.8)
(4, 0.4)
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Figure 2. graphs for the above function | Source (Author's work)
∴ In the graph we can see that intersection of
function 1 and function 2 is,
x = 2.5 & y = 0.5
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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5. Evaluate the surfaces in ℝ3 that are represented by the following equations.
i) y = 4
y = 4 represents a plane parallel
to the xz plane and 4 units to the
+ side of the y axis. This plane
has an infinite number of points
and each point (x, 4, z) lies 4
units parallel to the xz plane.is,
x = 2.5 & y = 0.5
Figure 3. y = 4 graph | Source (Author’s work)
ii) z = 5
z = 5 represents a plane that
is 5 units above the xy plane.
This plane is located on the +
side of the z axis. This plane
also has an infinite number of
points and each point (x, y, 5)
lies 5 units above the xy plane.
Figure 4. Z = 5 | Source (Author’s work)
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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6. Following figure shows a Tetrahedron
Construct an equation to find the volume of the given Tetrahedron using vector
methods and if the vectors of the Tetrahedron are a=(i+4j-2k) , b=(3i-5j+k) and
c=(-4i+3j+6k), find the volume of the Tetrahedron using the below constructed
equation.
Base area = (1/2) | a * b |
height = |𝒄| cos 𝜃
We need the area of the base and height of the tetrahedron to calculate the volume of
tetrahedron.
Volume of tetrahedron
1
= ( ) * base area * height
3
1
1
3
2
= ( ) * ( | 𝒂 * 𝒃 | ) * | 𝒄 | cos 𝜃
1
= ( ) * 𝒄. | 𝒂 * 𝒃 |
6
1
= ( ) * | (𝒂 * 𝒃). 𝒄 |
6
ABCD is a tetrahedron in 3D space. If a, b and c are the vectors, then the volume of
tetrahedron ABCD is,
(1/6) * | (a * b).c |
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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According to the question,
𝑎 = (𝑖 + 4𝑗 – 2𝑘), 𝑏 = (3𝑖 – 5𝑗 + 𝑘), 𝑐 = (−4𝑖 + 3𝑗 + 6𝑘)
𝟏
Volume of tetrahedron = |(𝒂 ∗ 𝒃). 𝒄|
𝟔
(a*b)
= I
j
k
= 1
4
-2
= 3
-5
1
= (4 − 10)𝑖 − (1 + 6)𝑗 + (−5 − 12)𝑘
= 6𝑖 + 7𝑗 + 17𝑘
( a.b ) . c
= (6,7,17). (−4,3,6)
= − 24, + 21, + 102
= 99
( a * b ) . c = 99
Volume
1
= 6 ∗ 99
= 16.5
∴ Volume of Tetrahedron = 16.53
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Activity 04
Part 1
1. Determine the slope of the following functions.
i. f(x) = 2x – 3x4 + 5x + 8
𝑓(𝑥) = 2𝑥 − 3𝑥 4 + 5𝑥 + 8
𝑑
(7𝑥 − 3𝑥 + 8)
𝑑𝑥
𝑑
𝑑
𝑑
(7𝑥) +
𝑓 ′ (𝑥) =
(−3𝑥 4 ) +
(8)
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) = 7 − 3 ∗ 4𝑥 3 + 0
= 𝒇′ (𝒙) = 𝟕 − 𝟏𝟐𝒙𝟑
∴ The slope = 7 − 12𝑥 3
ii. f(x) = cos(2x) + 4x2 – 3
𝑓(𝑥) = 𝑥𝑜𝑠(2𝑥) + 4𝑥 2 − 3
𝑑
(𝑐𝑜𝑠(2𝑥) + 4𝑥 2 − 3)
𝑑𝑥
𝑑
𝑑
𝑑
(4𝑥 2 ) −
(3)
𝑓 ′ (𝑥) =
(𝑐𝑜𝑠(2𝑥)) +
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) = −𝑠𝑖𝑛(2𝑥) ∗ 2 + 4 ∗ 2𝑥 − 0
= 𝒇′ (𝒙) = −𝟐𝒔𝒊𝒏(𝟐𝒙) + 𝟖𝒙
∴ The slope = −2 sin(2𝑥 ) + 8𝑥
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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2. Let the displacement function of a moving object is S(t) = 5t3 – 3t2 +
6t. What is the function for the velocity of the object at time t.
Displacement function of the moving object S(t) = 5t3 – 3t2 + 6t
As we know, if d(S) is distance, Velocity is given as 𝒅𝑺 (𝒕) / 𝒅t
According to the function,
Velocity
= (𝒕) 𝒅𝒕
= 5. 3t2 – 3. 2t + 6
d(S(t))/dt
= 15t2 – 6t + 6
∴ The velocity function of the object is
V (t) = 15t 2 – 6t + 6
Space left intentionally]
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Part 2
3. Find the area between the two curves f(x) = 2x2 + 1 and g(x) = 8 – 2x on the
interval (-2) ≤ x ≤ 1 .
𝟐𝑿𝟐 + 𝟏 = 𝟖 − 𝟐𝑿
𝟐𝑿𝟐 + 𝟐𝑿 − 𝟕 = 𝟎
𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
𝑥=
−2 + √60
−2 − √60
𝑜𝑟 𝑥 =
4
4
𝑥=
−1 + √15
−1 + √15
𝑜𝑟 𝑥 =
2
2
𝒙𝟏 > 𝟏 , 𝒙𝟐 < −𝟐
5 = | 𝑓 1 − 2(8 − 2𝑥) − (2𝑥 2 )𝑑𝑥 |
= | 𝑓 1 − 2(−2𝑥 2 − 2𝑥 + 7)𝑑𝑥 |
−2 2
2
=[
∗ 1 − 12 + 7 ∗ 1] + [− ∗ −2 + 7𝑥 − 2]
3
3
= 𝟏𝟖
∴ The area between two curves is 18
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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4. It is estimated that t years from now the tree plantation of a certain forest will
be increasing at the rate of 3t2 + 5t + 6 hundred trees per year. Environmentalists
have found that the level of Oxygen in the forest increases at the rate of
approximately 4 units per 100 trees. By how much will the Oxygen level in the
forest increase during the next 3 years?
𝑥 1 (𝑡) = 3𝑡 2 + 5𝑡 + 6
𝑥(𝑡) = ∫ 𝑝1 (𝑡)𝑑𝑡
𝟓
∫ 𝟎𝟑 𝟑𝒕𝟐 + 𝟓𝒕 + 𝟔𝒅𝒕 = 𝒕𝟑 + 𝒕𝟐 + 𝟔𝒕 + 𝒄
𝟐
5
𝑥(3) = 𝑡 3 + 𝑡 2 + 6𝑡 + 𝑐
2
5
= 33 + ∗ 32 + 6 ∗ 3 + 𝑐
2
= 𝟔𝟕. 𝟓 + 𝑪 − − − −→ 𝟏
5
𝑥(0) = 𝑡 3 + 𝑡 2 + 6𝑡 + 𝑐
2
5
= 03 + ∗ 02 + 6 ∗ 0 + 𝑐
2
= 𝑪 − − − −→ 𝟐
𝟒 𝒖𝒏𝒊𝒕𝒔 − −−→ 𝟏𝟎𝟎 𝒕𝒓𝒆𝒆𝒔
𝒙 𝒖𝒏𝒊𝒕𝒔 − −−→ 𝟔𝟕𝟓𝟎 𝒕𝒓𝒆𝒆𝒔
𝑥 =4∗
6750
100
= 𝟐𝟕𝟎 𝑼𝒏𝒊𝒕𝒔
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Part 3
5. Sketch the graph of f(x) = x5- 6x3 + 3 by applying differentiation methods for
analyzing where the graph is increasing/decreasing, local maximum/minimum
points [Using the second derivative test], concave up/down intervals with inflection
points.
1.
𝑭𝟏 (𝒙) = 𝟎
𝑌1 = 𝑓 1 (𝑥)
= 5𝑥 4 − 18𝑥 3 = 0
𝑥=0
𝑜𝑟
𝑥 = ±√
12
5
= ± 𝟏. 𝟗
Intervals
Test
𝒇𝟏 (𝒙) = 𝒙𝟐 (𝟓𝒙𝟐 − 𝟏𝟖)
F(x) is
𝑥 < −1.9
-2
+8 > 0 (+)
Increase
−1.9 < 𝑥 < 0
-1
−13 < 0 (−)
Decrease
0 < 𝑥 < 1.9
1
−13 < 0 (−)
Decrease
𝑥 > 1.9
2
+8 > 0 (+)
Increase
Table 2. Decreasing increasing intervals | Source (Author's work).
2. Maximum and minimum points
At 𝑥 = −1.9 , 𝐹(𝑥) goes from (+) to (-), local maximum at 𝑥 = −1.9 and value of
function 𝐹(1.9) = 19.39
𝐹(𝑥) = 𝑥 5 − 6𝑥 3 + 3
𝐹(1.9) = 1. 95 − 6 ∗ (−1.9)3 + 3
= 𝟏𝟗. 𝟑𝟗
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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At 𝑥 = 0 , 𝐹(𝑥) goes from (+) to (-), Its inflation points at𝑥 = 0 and value of the
function 𝐹(0) = 3
𝐹(𝑥) = 𝑥 5 − 6𝑥 3 + 3
= 05 − 6 ∗ 03 + 3
=𝟑
At 𝑥 = −1.9 , 𝐹(𝑥) goes from (-) to (+), local maximum at 𝑥 = −1.9 and value of
function 𝐹(1.9) = 13.39
𝐹(𝑥) = 𝑥 5 − 6𝑥 3 + 3
𝐹(1.9) = 1. 95 − 6 ∗ (−1.9)3 + 3
= −𝟏𝟑. 𝟑𝟗
3. Intersection points
𝑦 =0;
𝑥 5 − 6𝑥 3 + 3 = 0
𝒙 = −𝟐. 𝟒𝟖𝟖𝟗
𝒐𝒓
𝒙 = 𝟎. 𝟖𝟐
4. Finding concavity
4𝑥(5𝑥 2 − 9) = 0 ;
𝑥=0
𝒙𝟐 =
𝟗
𝟓
𝑜𝑟
→
5𝑥 2 − 9 = 0
𝒙 = √𝟗⁄𝟓 → 𝒙 = ±𝟏. 𝟑𝟒
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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Intervals
Test
𝒇𝟏 (𝒙)
Concavity
𝑥 < −1.34
-2
-88
Concave down
−1.34 < 𝑥 < 0
-1
16
Concave up
0 < 𝑥 < 1.34
1
16
Concave down
𝑥 > 1.34
2
88
Concave up
Table 3. Concavity graph | Source (Author's work).
𝒇(𝒙) = 𝟓𝒙𝟒 − 𝟏𝟖𝒙𝒙
𝒇(𝒙) = 𝟐𝟎𝒙𝟑 − 𝟑𝟔𝒙
[Space left intentionally]
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
66
Graph, f (x) = x5 – 6x3 + 3
Figure 5. the graph | Source (Author's work)
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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6. Identify the maximum and minimum points of the function f(x)= 2x3 - 4x4 + 5x2
by further differentiation. [i.e., Justify your answer using both first derivative test
and second derivative test.]
→
f(x) = 2x3 - 4x4 + 5x2
f `(x) = 6x2 – 16x3 + 10x
Let’s find the Critical points. In critical points, f `(x) should be 0.
6x2 – 16x3 + 10x = 0
➔ 2x (3x – 8x2 + 5) = 0
➔ 2x (8x + 5) (1 – x) = 0
2x = 0
or
x=0
8x + 5 = 0
or
x = - (5/8)
1-x = 0
x=1
• In range x < − (5/8), f `(x) is (+) so, f(x) is increasing.
▪ In range – (5/8) < x < 0, f `(x) is (-) so, f(x) is decreasing.
▪ In range 0 < x < 1, f `(x) is (+) so, f(x) is increasing.
▪ In range x < 1, f `(x) is (-) so, f(x) is decreasing.
f `(x) = 6x2 – 16x3 + 10x
→
f ``(x) = 12x – 48x2 + 10
f ``(x) = 12x – 48x2 + 10
Assigning x = - (5/8): f ``(x) = -16.25 → f(x) has a relative maximum
x = 0: f ``(x) = 10 → f(x) has a relative minimum
x = 1: f ``(x) = -26 → f(x) has a relative maximum
x = - (5/8) → f (x) = 0.854
x = 0 → f (x) = 0
x = 1 → f (x) = 3
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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REFERENCES
Investopedia. (n.d.). Cryptographic Hash Functions. [online] Available at:
https://www.investopedia.com/news/cryptographic-hashfunctions/#:~:text=A%20cryptographic%20hash%20function%20is.
Wikipedia Contributors (2019). Modular multiplicative inverse. [online] Wikipedia. Available at:
https://en.wikipedia.org/wiki/Modular_multiplicative_inverse.
Cuemath. (n.d.). Discrete Random Variable - Definition, Formula, Differences, Example, FAQs.
[online] Available at: https://www.cuemath.com/algebra/discrete-random-variable/.
Cuemath. (n.d.). Continuous Random Variable - Definition, Formulas, Mean, Examples. [online]
Available at: https://www.cuemath.com/data/continuous-random-variable/.
SearchDataManagement. (n.d.). What is hashing and how does it work? [online] Available at:
https://www.techtarget.com/searchdatamanagement/definition/hashing#:~:text=Hashing%20is%20th
e%20process%20of.
NGINX (2018). What Is Load Balancing? How Load Balancers Work. [online] NGINX. Available
at: https://www.nginx.com/resources/glossary/load-balancing/.
DATAtab (2021). Standard deviation (simply explained). YouTube. Available at:
https://www.youtube.com/watch?v=esskJJF8pCc.
www.youtube.com. (n.d.). To find Maximum and Minimum of a functions || Applications of
derivatives. [online] Available at: https://www.youtube.com/watch?v=Gb86aqeYrOA
www.youtube.com. (n.d.). Probability with At Least Two in Selected Group. [online] Available at:
https://www.youtube.com/watch?v=0oiHefLn7Zc
www.youtube.com. (n.d.). Venn Diagrams and Testing Validity. [online] Available at:
https://www.youtube.com/watch?v=rcyeHdx0Qv4
UNIT 11 – MATHS FOR COMPUTING | BHANUKA PERERA
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