Get Up to

55% OFF

Use Our Seasonal Offers!

Coupon Code

NEW25
Claim Now

Amazing Features We Offer

24*7 Help Service

100% Satisfaction
No Privacy Infringement
clock Super-fast Services
Subject Experts
Professional Documents

Get Lowest Price

Get A+ Within Your Budget!

    Total Price

    USD 7.33

    8 Mathematics for Construction

    University: UCK

    • Unit No: 8
    • Level: High school
    • Pages: 19 / Words 4750
    • Paper Type: Assignment
    • Course Code: N/A
    • Downloads: 199

    Excel in Mathematics for Construction with our expert assignment help, designed to simplify complex calculations and enhance your understanding for a successful career in the construction industry. From algebra to geometry, we cover all essential topics to ensure you're well-prepared for the challenges ahead. Our tailored support includes guidance on structural analysis, materials science, and advanced mathematical techniques, empowering you to apply these skills effectively in real-world construction projects.

    TASK 2

    Scenario 1

    Revenue

    Number of customers

    (£1000)

    January

    July

    Less than 5

    27

    22

    5 and less than 10

    38

    39

    10 and less than 15

    40

    69

    15 and less than 20

    22

    41

    20 and less than 30

    13

    20

    30 and less than 40

    4

    5

    Solution

    a) The given table can be organized in the following manner -

    Revenue

    Number of customers

    (£1000)

     

    January

    July

    0 to 5

    27

    22

    5 to 10

    38

    39

    10 to 15

    40

    69

    15 to 20

    22

    41

    20 to 30

    13

    20

    30 to 40

    4

    5

    Best Assignment Writers at Your Service!


     

    The Histogram of each distribution can be calculated after converting the unequal class interval into an equal class interval –

    Revenue

    Number of customers
    (£1000)

     

    January

    0 to 5

    27

    5 to 10

    38

    10 to 15

    40

    15 to 20

    22

    20 to 30

    13

    30 to 40

    4

    Equal class interval -

    Revenue

    Number of customers
    (£1000)

     

    January

    0 to 10

    27 + 38 = 65

    10 to 20

    40 + 22 = 62

    20 to 30

    13

    30 to 40

    4

    From this histogram, it has been analyzed that groups 0 to 10 have the highest frequency, so, taking this class to find mode oa f data of grouped frequency in following the ing way –

    Mode (z) = l + f1   –    f0       x    h

                             2 f1 – f0  -  f2

    Here,  f1 is the highest frequency = 65

              f0 is the above frequency = 0 and,

              f2 is the below frequency = 62

              h is the class range  = 10 and,

              l is the lower interval of mode class = 0,

    So, mode the  can be calculated as -

    Mode (z) =  0 + 65 – 0    x 10

                    =   2x65 – 0 – 62

                    =   0 + 65 x 10        

                    =   130 – 62

                    =   0 + 650 / 68

                    =   9.55

    For July month -

    Revenue

    Number of customers
    (£1000)

     

    July

    0 to 5

    22

    5 to 10

    39

    10 to 15

    69

    15 to 20

    41

    20 to 30

    20

    30 to 40

    5

     

    Revenue

    Number of customers
    (£1000)

     

    July

    0 to 10

    22 + 39 = 61

    10 to 20

    69 + 41 = 110

    20 to 30

    20

    30 to 40

    5

    From this histogram, 10 the o 20 group has high est frequency, so, taking this group as the modal  class of grouped frequency, the cathode can be calculated in following the wing way-

    Mode (z) = l + f1   –    f0  x h

                       2 f1 – f0  -  f2

    here,  f1 is the highest frequency = 110

              f0 is the above frequency = 61 and,

              f2 is the below frequency = 20

              h is the class range = 10 and,

              l is the lower interval of mode class = 10,

    so, the ode can be calculated as -

    Mode (z) = 10 + 110 – 61  x 10

                        2x110 – 61 – 20

                     = 10 + 49 x 10        

                         220 – 81

                      = 10 + 490 / 139

                      = 13.5

    b) Cumulative frequency curve or O-give curve of January data

    Revenue

    Number of customers
    (£1000)

    Less than O-give curve

    Cumulative Frequency

    More  than O-give curve

    Cumulative Frequency

     

    January

     

     

     

     

    0 to 10

    65

    Less than 10

    65

    More than 0

    144

    10 to 20

    62

    Less than 20

    127

    More than 10

    79

    20 to 30

    13

    Less than 30

    140

    More than 20

    17

    30 to 40

    4

    Less than 40

    144

    More than 30

    4

     

    Revenue

    Number of customers
    (£1000)

    Cumulative frequency

     

    January

     

    0 to 10

    65

    65

    10 to 20

    62

    127

    20 to 30

    13

    140

    30 to 40

    4

    144

    Now, the median of the data can be calculated by -

    Median (M) = l + N/2 – cf   x  h

                                     f

    Here, N/2 = sum of total freq / 2

                    = 144 / 2  = 72

    so, the main class will be from 10 to 20 

      l is the lowest interval = 10

      h is class difference = 10

      So, Median (M) = 10 + 72 – 65 x 10

                                           62

                             = 10 + 70/62

                            = 10 + 1.1 = 11.1

    Cumulative frequency curve or O-give curve of July data

    Revenue

    Number of customers
    (£1000)

    Less than O-give curve

    Cumulative Frequency

    More  than O-give curve

    Cumulative Frequency

     

    July

     

     

     

     

    0 to 10

    61

    Less than 10

    61

    More than 0

    196

    10 to 20

    110

    Less than 20

    171

    More than 10

    135

    20 to 30

    20

    Less than 30

    191

    More than 20

    25

    30 to 40

    5

    Less than 40

    196

    More than 30

    5

     

    Revenue

    Number of customers
    (£1000)

    Cumulative frequency

     

    July

     

    0 to 10

    61

    61

    10 to 20

    110

    171

    20 to 30

    20

    191

    30 to 40

    5

    196

     

    Now, the media ian of the data can be calculated by -

    Median (M) = l + N/2 – cf   x  h

                                     f

    Here, N/2 = sum of total freq / 2

                    = 196 / 2  = 98S

    So, the dian class will be from 10 to 20 

    l is lowest interval = 10  and frequency = 110

    h is class difference = 10

    So, Median (M) = 10 + 98 – 61 x 10

                                           110

                             = 10 + 370/110

                            = 10 + 3.4 = 13.4

     c) Mean, Range an,d Standard deviation -

    January data

    Revenue

    Number of customers
    (£1000)

    X= middle term

    fx

     

    January

     

     

    0 to 10

    65

    5

    325

    10 to 20

    62

    15

    930

    20 to 30

    13

    25

    325

    30 to 40

    4

    30

    120

    Total

    144

     

    1700

    Mean = ∑fx / ∑f

    Here,  ∑fx is the m of the product of frequency and middle-termed  ∑f is the total frequency

    therefore, mean = 1700 / 144 = 11.8

    July data

    Revenue

    Number of customers
    (£1000)

    X= middle term

    fx

     

    July

     

     

    0 to 10

    61

    5

    305

    10 to 20

    110

    15

    1650

    20 to 30

    20

    25

    500

    30 to 40

    5

    35

    175

    Total

    196

     

    2630

    Mean = ∑fx / ∑f

    Here,  ∑fx is the sum of the f produce ct of frequency and middle middle-term∑f is tota the l frequency

    therefore, mean = 2630 / 196 = 13.4

    Scenario 2

    Solution

    Given – Number of bulbs = 5000

    Mean of lengths of life of bulb at normal distribution = 360 days

    Standard deviation = 60 days

    a) Assumption test if bulb life is normally distributed,

    Let   H0 : μ = 360

    and, H1 : μ Ç‚ 360

    To calculate the test statistic, let's assume that normal distribution is valid when there is no change in μ at the null of the hypothesis. As, the lue of μ is 360 days, then to perform a hypothesis test with standard deviation, then z score -

                                                          z  = x¯ - μ

                                                         σ/√n

    here, let the average life value of bulthe b is 360.5 then,

                                                          z =  x¯- 360

                                                          60 / √5000

                                                          z = 0.59

    Using, the significant level at 5% and taking a right-sided one-tailed test, α = 0.05
    Z 1- α = 1.645 is obtained as a test of the test statistic Therefore,

    as 0.59 < 1.645 so, test the statistic is not in the critical cal region, thus, the null hypothesis cannot be rejected. Therefore, mean the under the given assumption is normally distributed.

    b) Probability

    (z1 < Z < Z2)

    here, z1 will be calculated at half of the standard deviation and z2 at double of standard deviation i.e.

      z1  = x¯ - μ   and,   z1  = x¯ - μ

          2σ/√n                 σ/2√n

    then,

             z1   =  0.29    and,  z2 = 1.16

    B.SA simple random sample taken from a certain population is 10 people, with a mean of 27 years

    the average mean is above or less than 30 years

    variance is known to be 20, so standard deviation = √20 = 4.47

    Hypothesis assumption -

    Let   H0 : μ = 30

    and,  H1 : μ Ç‚ 30

      z  = x¯ - μ

             σ/√n

    variance, 

    so, z = 30 – 27 / 4.47√9

    so, z = 2.01

    At 5% sa significant level and a right-sided one-tailed test, α = 0.05
    Z 1- α = 1.645,

    as 2.01 > 1.645 so, test the statistic is in the critical cal region, thus, the null hypothesis is rejected. Therefore, the age of the population is not 30 years.

    So, two-tailed test, Probability can be defined as {1.645 ≤ Z  ≤ 2.01}

    at, one-tailed test -

    TASK 3

    Scenario 1

    Solution – Given equalization of wave

                     x1 = 3.75 sin (100 π t + 2 π/9)             ….(i)

    and,          x2 = 4.42 sin (100 π t - 2 π/5)               ….(ii)

    these equations can further be written as -

                      x1 = 3.75 sin (100 π (t + 0.002))             ….(iii)

    and,           x2 = 4.42 sin (100 π (t – 0.004))               ….(iv)

    a) Apmlitude in the first t equation is 3.75 and in Seco, the nd equation is 4.42

              Phase from the third equation is obtained as = 0.002 to the left

    and,   Phase from fourth the equation is obtained as = -0.004 to the right

              Periodic times = 2 π / 100 π = 0.002.

    While frequency will be 1/period

                i.e.     F = 1 / 0.002 = 500

    b) When both machines are switched on, the time taken by each machine for displacement, can be calculated by -

    first, differentiate equation (i) concerning -

               x1 = 3.75 sin (100 πt + 2 π/9)    

               dx1/dt = 3.75 cos (100 π t + 2 π/9) x 100 π         ......... (v)

               at maximum,  dx1/dt  = 0

               3.75 cos (100 πt + 2 π/9) x 100 π = 0

    or,       cos (100 πt + 2 π/9) = 0

               100 πt + 2 π/9 = 90

               100 πt  = 90 – 2π/9

               t = 0.002 seconds

    Now, differentiate equation (ii) concerning -

    x2 = 4.42 sin (100 π t - 2 π/5)     

    dx2/dt = 4.42 cos (100 π t - 2 π/5) x 100 π   ...........(vi)

    at maximum, dx1/dt  = 0

           4.42 cos (100 π t - 2 π/5) x 100 π = 0   

    or,   cos (100 π t - 2 π/5) = 0

           100 π t - 2 π/5 = 90

           100  π t  =  90 + 2π/5

            t = 0.009 seconds 

    c) Time taken each machine when displacement recreates-2mm

         from equation ..............(v)

        dx1/dt = 3.75 cos (100 π t + 2 π/9) x 100 π 

        at -2mm,

        3.75 cos (100 πt + 2 π/9) x 100 π  = -2

        375  π cos (100 πt + 2 π/9) = -2

        cos (100 π t + 2 π/9) = -0.00002963

        100 πt + 2 π/9 = 90.01

        100 πt  = 90.05 – 2 π/9

        t           =  0.0027sec

    While, second machine will take -

    dx2/dt = 4.42 cos (100 π t - 2 π/5) x 100 π                    

    4.42 cos (100 π t - 2 π/5) x 100 π = -2

    cos (100 π t - 2 π/5)  = 0.000025138

    100 π t - 2 π/5  = 89.99

    t  = 0.0089 seconds

    (iv) Compound angle formula for expansion of above two equations -

           x1 = 3.75 sin (100 π t + 2 π/9)           

    using, sin (A+B)  = sin A cos B + cos A sin B

    then,

           x1 = 3.75 (sin 100 π t . cos 2 π/9 + cos 100 π t . sin 2 π/9)

    or,   x1  = 3.75 ( sin 100 π t  x 0.76  + cos 100 π t x 0.64)

    or,   x1 = 2.85  sin 100 π t + 2.4  cos 100 π t (Ans)

    similarly, sec the ond equation -

          x2 = 4.42 sin (100 π t - 2 π/5)           

          x2 = 4.42 (sin 100 π t . cos 2 π/5 - cos 100 π t . sin 2 π/5)

    or,  x2  = 4.42 ( sin 100 π t  x 0.31 + cos 100 π t x 0.95)

    or,  x2 = 1.3702 sin 100 π t + 4.199  cos 100 π t (Ans) 

    v) Expression of x1 and x2 in combined form -

           x1 +  x2 = (2.85  sin 100 π t + 2.4  cos 100 π t) + (1.3702 sin 100 π t + 4.199  cos 100 π t)

           x1 +  x2 = 4.2202  sin 100 π t + 6.599  cos 100 π t

            x1 +  x2  = 6 (sin 100 π t +  π/4) (Ans). 

    Scenario 2

    (i) The distance between AB

    Distance between two points on The artesian plane of 3D can be calculated by

    AB = √ (x1 – x0)2 + (y1 – y0)2 + (z1 – z0)2

    here, coordinates of A are (0, -40,0) and B are (40, 0, -20)

    then, distance between AB =  √ ( 0 – 40)2  + (-40 – 0)2 + (0 + 20)2

                                               = √ 1600 + 1600 + 400

                                               = √3600 = 60 unit.

    Angle between 

    Vector equation for the line passing through two points A and B with position vectors

    →        →

    a   and  b is represented by -

    →  →      →  →

     r = a + λ (b – a)

    As, B (40, 0, -20)  and C (a, b, 0)

    The position of B (40, 0, -20) = ( 40ˆi + 0ˆj -20ˆk )

    The position of C (a, b, 0) = ( aˆi + bˆj + 0ˆk )

    So, →

     r  = ( 40ˆi + 0ˆj -20ˆk ) +  λ [  ( aˆi + bˆj + 0ˆk ) - ( 40ˆi + 0ˆj -20ˆk ) ]

        = ( 40ˆi + 0ˆj -20ˆk ) +  λ [  (a – 40) ˆi + bˆj + 20ˆk)  ]

       →

    Given , r = 3i + 4j + k

    then, on comparing both sides abthe ove equation,

     λ = 1.05, so, a = 75 (approx) and b = 4 (approx)

    Now, the angle between two lines can be calculated by                               

    cos Ó¨ =      u . v   

                    ||u||   ||v||

    AB = Position of B – position vector of A

          = (40, 40, -20)

    BC = Position of C – position vector of B

          = (a - 40, b, 20)

    Then,

     cos Ó¨ =       u . v   

                     ||u||   ||v||

                                 

     cos Ó¨ = (40i + 40j -20k) . ((a-40)i + bj + 20k)

                    √(402 + 402 + 202).  √((a-40)2 + b2 + 202)

                     cos Ó¨ = 40(a-40) + 40b -40

                                   √3600 .    √((a-40)2 + b2 + 202)

                                 =  2(a+b-101)               

                                     3.√((a-40)2 + b2 + 202)

    TASK 4

    The bending moment, M of a beam is given by

    M = 3000 – 550x – 20x2

    Bending moment M

    variable x

    2430

    1

    1810

    2

    1170

    3

    480

    4

    -250

    5

    on plotting the bending moment M, as shown in the graph above, the value of x at which it becomes zero will be 4.8 approx.

    (ai For determining, the maximum or minimum value of Bendithe ng Moment Function, differentiate the given equation concerning ring/dx (M) =  d/dx (3000 – 550x – 20x2)

    = – 550 – 40x

    here, d/dx < 0,

    so, the bending Function will attain its maximum value at -

    d/dx (M)     = 0

    – 550 – 40x = 0

    x = -13.75 or -14 approx.

    now, put this value of x in bend the ing moment function, to get maxi the mum value as -

    M =  3000 – 550 (-14) – 20(-14)2

           = 3000 + 7700 – 3920  = 6780 (ans)

    b) The temperature  Ó¨ (ºC), at time t (mins) of a body is given by

                                     Ó¨ = 300 + 100 e-0.1 t

    value of  Ó can be determined at the value of 0, 1, 2, and, 5 as -

    at t = 0,

                     Ó¨ = 300 + 100 e-0.1 x 0

                              = 300 + 100 x 1

                              = 400 ºC

    at t = 1,

                     Ó¨ = 300 + 100 e-0.1 x 1

                              = 300 + 100 x 0.90

                              = 390 ºC

    at t = 2,

                     Ó¨ = 300 + 100 e-0.1 x 2

                              = 300 + 100 x 0.81

                              = 381 ºC

    at t = 5,

                     Ó¨ = 300 + 100 e-0.1 x 5

                              = 300 + 100 x 0.60

                              = 360 ºC

    Temperature (ºC)

    Time (t)

    400

    0

    390

    1

    381

    2

    360

    5

    Need Quick Assistance with Your Academic Writing?

    Ping Us Your Requirements on WhatsApp!

    Chat Now

    c) In a thermodynamic system, the related tionship between pressure (P), Volume (V), and, constant C is given by,

               log (P) + n log (V) – log (C)

    to show that PVn  = C

    Let,     log (P) + n log (V) – log (C)  = 0

               log (P) + n log (V) = log (C)

    using product rule, i.e. log (a x b) = log a + log b

                log (PVn) = log (C)

                PVn = C   Hence Proved.

    Now, to determine the rate of the f change of V when value the of P changes from 10N/m2 from 60 to 100N/m2 with variable n = 2, differentiate the above equation with concerning-

    d/dV PVn = d/dv C

    dP/dV Vn + nVn-1 .P = 0 

    You May Also Like To Read -  

    2 Construction Technology

    LEGAL RISK IN CONSTRUCTION

     

    Amazing Discount

    UPTO55% OFF

    Subscribe now for More Exciting Offers + Freebies

    Download Full Sample

    Cite This Work

    To export references to this Sample, select the desired referencing style below:

    Students sometimes cannot express their inability to work on assignments and wonder, "Who will do my assignment?" To help them understand the complexities of writing, we are providing "samples" on various subjects. Also, we have experienced assignment writers who can provide the best and affordable assignment writing services, essay writing services, dissertation writing services, and so on. Thus, don't wait any longer! Place your order now to take advantage of discounted deals and offers.

    Limited Time Offer

    Exclusive Library Membership + FREE Wallet Balance