Mathematical Modeling and Data Analysis: A Practical Approach

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Table of Contents
Introduction:...............................................................................................................................................2
Task 1..........................................................................................................................................................3
Scenario 1................................................................................................................................................3
Scenario 2................................................................................................................................................4
Scenario 3:...............................................................................................................................................6
Task 2:.........................................................................................................................................................9
Scenario 1:...............................................................................................................................................9
Scenario 2:.............................................................................................................................................13
Task 3:.......................................................................................................................................................19
Scenario 1:.............................................................................................................................................19
Scenario 2:.............................................................................................................................................27
Task 4........................................................................................................................................................29
Scenario 1:.............................................................................................................................................29
Scenario 2:.............................................................................................................................................31
Scenario 3:.............................................................................................................................................32
Conclusion.................................................................................................................................................34
References.................................................................................................................................................35

Introduction:
The following chapter will focus on the relevance of all the methods used in mathematics for conceptual
construction. There will be an investigation of the applications for statistical techniques for interpreting
the data as well as organizing those data. After which we will use conceptual and analytical methods
also for solving different questions in the following chapter.

Task 1
Scenario 1
1. According to the question,
The width of Water Tank is 3.2 m
The length of the tank is greater than the width of the tank
Hence,
Let’s suppose the tank’s width be x.
The length of the water tank will be = 3.2 m + x.
The formula for finding the area of a rectangle = L * B.
So,
x of (3.2 + x) = 26.5
26.5 is equal to x2 + x * 3.2
x2 + x * 3.2 – 26.5 is equal to 0
102 * x + x * 32 – 265 is equal to 0.
According to the above solution the given equation will be:
a of x2 + b of x + c is equal to 0
Solving it will be,
x is equal to −b ±❑
√−4 of ac +b2
2 of a
Putting the value in the given equation,
x is equal to 1 1
20 of −32 ±❑
√ 322−¿
¿ ¿ ¿
= 1
20 of −32± ❑
√+10600+1024
1
= 1
20 of −32± ❑
√ 11624
1
Finding the value of x,
x is equal to 1
20 of −❑
√ 11624−32
1 , or
x is equal to 1
20
−32+❑
√11624
1
x = - 8∗1
5 - 1
10❑
√ 2906
x = -6.991, or
x = - 41
5 + 1
10 ❑
√ 2906
x = 3.791
The length will never be a negative integer Antonini, S. (2011). .
Hence the value of the length for the given Water Tank is 3.791.
2. Given Values in the question are:

Contract Value / day = £’cv’
Contract Value delayed by 5 days = £4250
Contract Value delayed by 12 days = £2120
Now,
Assume the following meanings,
Contract Day is equal to d.
Deduction delay/day is equal to ❑dCVD.
Days delayed is termed as ❑the delay.
Hence,
The formula for the given equation is given by
Paid Valuedelay is equal to (d multiply by ❑dCVD) – ( ❑delay multiply by ❑dCVD)
The value for VP is given as d multiplied by ❑dCVD.
Putting the value of VP for the delay of 12 days.
VP – (12 multiply by ❑dCVD) is equal to £2120.
Putting the value of VP for the delay of 5 days.
VP – (5 multiply by ❑dCVD) is equal to £4250.
Putting the both value in the above equation,
[(5 multiply by cvd ) + 4250] - (12 multiply by cvd ) = £2120
(5 multiply by cvd ) + 4250 = 2120+(12 multiply by cvd )
12 of cvd - 5of cvd = 4250 – 2120
7of cvd = 2130
cvd = £304.285
Hence,
£304.285 is getting deducted every day for getting delay.
Now,
Value paid− ( 304.285 x 12 ) is equal ¿ £ 2120
Value paid −3651.42is equal ¿ £ 2120
Value paid is equal ¿ £ 05771.42 0
£ 5771.42 is the total cost for the contract.
For d
VP = ( c vd∗d ) − ( cv d∗ddelay )
£ 0 5771.42 0 = ( d × £ 304.285 )
d = 19
19 days is the total day for construction.

Scenario 2
According to Question:
The car given average speed is 65 miles/hr.
1 mile = 1.61 Kilometre.
Converting 65 miles/hr into m/sec will be
= 65 of 161
360 m/sec.
= 29.069 m/sec.
The time a car takes to cover one mile:
= 1
65 hrs .
The time a car takes to cover one hundred miles:
= 1
65 of 100 hrs .
= 1 or 32 minutes 18 seconds.
Every gallon is being used in every thirty miles.
By the help of given conversion table given in the question
Every gallon equals to 3.78 litres.
Hence,
One mile every gallon will be equal to 1.61 divided by 3.78 kilometre/L.
= 1.61/3.78 kilometre per litre.
Thirty miles every gallon will be equal to 30 * 1.61 divided by 3.78 Kilometre/L.
= 12.77 Kilometre per litre.
With respect to the above solution:
The distance of the journey is 100 miles.
Fuel consumption every litre is 12.77 Kilometres.
100 miles is equal to 161 km.

Hence,
The required fuel:
=
161 kms
12.77 kms
L
= 161/12.77 Litre.
= 112.6 Litre.
The total lift produced by the aircraft wing is determined as
= A * p * k * V 2.
Scenario 3:
1.
According to the Question:
The sequence which has been given is equal to b , 2b
3 , b
3 , 0 , ...
In a simple way,
The sequence can be termed as a ¿ the base ¿n=b of 4−n
3
For 6th term
a ¿ the base ¿6=b of 4−6
3 .
a ¿ th e base ¿6=−2 b
3 .
For Kth term
a ¿ the base ¿k= b∗4−k
3 .
Finding b which can be extracted by the 20 terms.
a ¿ the base ¿20=b∗4−20
3 =15

a ¿ the base ¿20=−16 b
3 =15
a ¿ the base ¿20=b=−45
16
2.
According to the question:
Geometric progression is given as 1 , 1
2 , 1
4 ..
The nth term will be found by
a ¿ the base ¿n=ato the base1∗rn−1
In the above equation:
r = Common ratio.
n – 1 = term of nth -1.
n = total available terms.
a 1 = initial term.
Finding 6th term:
a ¿ the base ¿6=1∗( 1
2 )
5
a ¿ the base ¿6= 1
32
Progression calculation:
S ¿ the base ¿n=a ¿ thebase1∗ ( 1−r ¿ the power ¿¿ n ) ¿
1−r
S ¿ thebase ¿n=nthterm .
3.
a. The given equation is
2 logs of 3x + log of 18x = 27
It can be written as:
2 logs of 3x + log of (3x * 6) = 27
As we know that the value of log (x * y) = log x + log y.

Hence,
2 logs of 3x + log of 6 + log of 3x = 27
2 logs of 3x + 0.77 + log of 3x = 27
Finding the value of x is:
3 logs of (3x) = 27 + .77
= 26.23
Log of (3x) = 8.74
3x = 6247.89
x = 1/3 * 6247.89
x = 2082.6
b. In the question
Log of 18x + 2 logs of 3x = 27.
The equation will be:
Log of 3x + log of 6 + 2 logs of 3x = 27.
According to the above question:
Log of 6 = 1.79
Hence,
Inserting the value of log 6 in above will get
Log of 3x + 1.79 + 2 logs of 3x = 27.
Now,
Need to find the value of x.
3 log 3x + 1.79 = 27
3 log 3x = 25.21
Log 3x = 1/3 * 25.21
= 4460.42.
x = 1/3 *4460.42
= 1486.8.

Task 2:
Scenario 1:
We have defied the distribution of the available data in the below table.
Revenue Given Frequency Given Size of class Frequency adjusted
>5 27 5 27
>10 = 5 38 5 38
>15 = 10 40 5 40
>20 = 15 22 5 22
>30 = 20 13 10 6.5
>40 = 30 4 10 2
5 > 10 <= 5 >15 <= 10 >20 <= 15 >30 <= 20 >40 <= 30
0
50
100
150
200
250
Frequency Given
Frequency Given CF >
The formula used to find the value of mod between 5 and 10:
L+ ({f ¿ thebase❑−f ¿ thebase ¿¿ m−1 }¿¿ f ¿ the basem∗2−f ¿ thebase ¿¿−1−f ¿ the base¿¿ m+1¿ ) multiply by w
In the above formula:
L= lower bond
W= group difference

fm= frequency
FM + 1 = frequency of proceeding class model
FM – 1 = frequency of preceding class model
hence,
mode = 5+ ( ( 38 subtracting with 27 )
38 subtracting with27 adding with38 subtracting with40 ) multiply by 5
mode = 55/9 + 5
mode = 100/9
mode = 11.111
We have defied the distribution of the available data for July month in below table.
Revenue Given Size of class Frequency Given Frequency Adjusted
>5 10/2 22 44/2
>10 = 5 10/2 39 78/2
>15 = 10 5 69 69
>20 = 15 5 41 41
>30 = 20 10 20 10
>40 = 30 10 5 5/2

5 > 10 <= 5 >15 <= 10 >20 <= 15 >30 <= 20 >40 <= 30
0
50
100
150
200
250
Frequency Chart
Frequency Given CF >
The formula used to find the mod value between 5 and 10:
L+ ( {f ¿ the basem −f ¿ the base¿ ¿ m−1}¿¿ f ¿ the basem∗2−f ¿ the base¿¿ m−1−f ¿ the base ¿¿ m+ 1¿ ) multiply by w
In the above formula:
L= lower bond
W= group difference
fm= frequency
fm + 1 = frequency of proceeding class model
fm – 1 = frequency of preceding class model
hence,
mode = 10 adding with ( ( 69 subtracting with39 )
69 subtracting with39 adding with69 subtracting with 41 )multiply by 5
mode = ¿) + 10
mode = ( 150
58 )+10
mode = 1/ 58 * ( 580+150
1 )

mode ¿
1
58∗730
1
mode = 12.585 Selden, J. (2006).
b.
curve cumulative frequency finding for January:
Revenue Given Frequency Given CF >
>5 27 27 > 5
>10 = 5 38 65 > 10
>15 = 10 40 105 > 15
>20 = 15 22 127 > 20
>30 = 20 13 140 > 30
>40 = 30 4 144 > 40
5 > 10 <= 5 >15 <= 10 >20 <= 15 >30 <= 20 >40 <= 30
0
50
100
150
200
250
Curve Frequency
Frequency Given CF >
curve cumulative frequency finding for July:
Revenue Frequency CF >
5 44.00/2 22.00 >5