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Table of Contents
Conclusion.................................................................................................................. 24 References.................................................................................................................. 25
Introduction
The mathematical concepts are based on the ideas like equality and symbolic representation. With the help of mathematical concepts, the practical problems are solved. The below report covers several concepts such as Number theory, sequence, Probability, quadratic equations and calculus. This report gives a brief idea about the mathematical formulas that implements to solve the mathematical questions.
Sequence and Series
The sequence and series include arithmetic progressions and geometric progressions (Tohir and Abidin, 2018). It is basically used in simulation engineering and increase the speed at regular intervals, calculate simple interest etc. 1. Cost of painting the walls of room for per m^2 = £ 100 Rise in per m^2 = £ 50 Area of room = 150 m^2 Cost of painting for per m^2 = 50 + 100 = £ 150 Charge to paint all the walls = area * cost = 150 * 150 = £ 22500 2. Let the first term of AP = a difference of AP = d Middle three terms are 11th 12th and 13th terms 11th term = a+10d 12th term = a+11d
13th term = a+12d Hence sum of 11th 12th and 13th terms = a+10d+a+11d+a+12d = 3a+33d Hence 3a+33d = 720…………(i) 21st term = a+20d 22nd term = a+21d 23rd term = a+22d Addition of 21st,22nd and 23rd terms = a+20d+a+21d+a+22d = 3a+63d 3a+63 d = 1320…………(ii) Subtract (i) from (ii) and get 30d = 600 Henceforth d=600/30 = 20 From (i) we get a= (720–30d)/3 = (720–30×20)/3=120/3 = 40 Hence 18th term = a+17d = 40+17×20 = 380 3. First Term= 3 Common ratio= 1. Using this formula, Sn= a (1-rn) / 1-r Sn = 3(1- (1.5)^20 )/ 1-1. Sn = 3* (1- 3325)/ 0. = 3* (3324)/ 0. = 19944 4. Sum of first all 3-terms = 37/
log 0.125 > n log 0. log 0.9 negative, log 0.125/log 0.9 < n 19.736 < n n ≈^ 20
2.1 Probability
It is a mathematical theory or framework that allows to analyse the events in a logically sound manner (Li, et.al., 2019). It is performed on the event that indicating how likely an event is occur. The probability is shown in between 0 and 1 value. It is a concept of uncertainty and randomness (El- Morshedy, et.al., 2019). Binomial distribution in probability: In binomial distribution, two possible outcomes are found. It is applicable on those events that have only two possible results i.e., success or failure. Normal Distribution: It is a probability distribution that is showing the data which is more frequent and symmetric nearby mean. a. Once two dice are rolled together then total outcomes that occurred are 36 and Sample space is [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ] So, pairs with sum 9 are (3, 6) (4, 5) (5, 4) (6, 3) i.e., total 4 pairs Total outcomes = 36 Favourable outcomes = 4 Probability of getting the sum of 9 = Favourable outcomes divided by Total outcomes = 4 / 36 = 1/ So, P (sum of 9) = 1/9. b. If the coin is thrown 3 times, Total outcomes: (HHH, HHT, HTH, THH, TTT, THT, HTT, TTH) = 8 Favourable outcomes: (HHH, HHT, HTH, THH, THT, HTT, TTH) = 7
ii. Two white balls = 8 C 2 One Red ball = 4 C 1 (^8 C 2 * 4 C 1 ) / 19 C 3 (8! / 2! 6!) * (4! / 1! 3!)/ (19! / 3! 16!) (876!) / (2! * 6!) / (19181716!) / (3! 16!) (47432)/ (1918*17) = 112/ e. The probability of both the boys and girls are same, i.e., ½ trials are independent Hence, it is a problem of binomial distribution. P = Probability of getting boy = ½ Q = 1-p = 1-1/ N = 3 Let X be the binomial variable and the number of boys in the families, X takings values 0, 1, 2, 3. Now, p = (X= r) = nCr Pr qn-r = 3 Cr (1/2) r^ (1/2)3-r P (X= r) = 3 Cr (1/2) 3 P (X= 0) = 3 C 0 (1/2) 0 = 1/ P (X= 1) = 3 C 1 (1/2) 1 = 3/ P (X= 2) = 3 C 2 (1/2) 2 = 3/ P (X= 3) = 3 C 3 (1/2) 3 = 1/ Distribution of Probability of X is
X 0 1 2 3
P(X) 1/8 3/8 3/8 1/
f. i. Let x be the random variable denoting the time taken to assemble a car μ = 20 Standard deviation, σ^ =^2 Probability of that car is assembled in fewer than 19.5 hours, P (x < 19.5) = P (z < (19.5-20)/2) P (z< -0.25) = P (z > 0.25) 0.5 – P (0 < z < 0.25) 0.5 – 0.
ii. Probability of that car is assembled amongst 20 and 22 hours P (20 < x < 22) = P ((20-20)/2) < z < ((22-20)/2) P (0 < z < 1) = 0. g. Annual salaries of the employees in a large company are approximately distributed with a given data, Mean = £ 40000 Standard deviation = £ 20000 Let x be the random variable of salary of employee in (£) i. Percent of people earn less than £ 30000 is given by
Straight line It is an infinite one-dimensional figure with no width. This part includes forces and velocities also. The concepts are understood by using the design and construction of the systems. The quadratic equations are solved with the help of the straight-line equation that is y = mx + c. 3. 5x – 7y – 4 = 0 7y = 5x – 4 y = 5/7 x – 4/ slope of line, m = 5/ using equation, y – y1 = m (x – x1) put the given coordinates (5, 7) in the equation, y – 7 = 5/7 (x - 5) 7y – 49 = 5x – 25 7y – 5x – 24 The equation is -5x +7y – 24 = 0 3. 2x + 7y + 5 = 0 Using this equation, find the slope 7y = - 2x – 5 y = -2/7 – 5/ Now m1 = -2/7 is a slope of a line
To make the line that need to construct a vertical to a given line, the line of slope must fulfil the condition, m = - 1/m m = -1/ (- 2/7) m = 7/ now, the slope m and point (-1, -3) are given than using this equation, y = mx + c place the given values of x and y in the above calculation, -3 = 7/2 * (-1) + c c = ½ therefore, equation of line is y = 7/2 x + ½ 3. The two equations are given, 6x + 5y + 8 = 0 5x -hy + 8 = 0 i. For parallel line, Using this formula, X1/ X2 = Y1/ Y2 ≠ Z1/ Z 6/5 = 5/ -h -6h = 25 h = -25/
c) CAD: The computer aided design is use vector graphics for designing, engineering and manufacturing because it helps to edit the mathematical formulas for ease and scalability. d) Scalability: The vector graphics in scalability requires scalable graphics like the company logos are displayed at different sizes.
acceleration = dv/dt = ½. ∆ 4. Capacity of the cone = 1/3 π r^2 h r is the radius of the circle at its base and h is the altitude of cone i.e., r = h, therefore, v = 1/3 π r^2 h = 1/3 π r^3 Change volume rate: dv/dt = v’ find the change volume rate of sandpile: v’ = 1/3 π h^3 = 1/3 π. 3 h^2 h’ = π h^2 .h’ = π h^2. r’ Degree of increase in height and radius of the cone is h’ = r’ = v’/ π h^2 The given v’ = 5 cm^3 /s & h = 10cm Place the given values in the above formula, h’ = r’ = v’/ π h^2 = 5/ (π * 10) = (5 * 7) / (22 * 10 * 10) = 7/ 440 = 0. = 16 cm/sec Maxima and Minima : It is well-defined as the large and small value of function i.e., within a range. It is used for finding the turning points of graph.
Differentiating both sides
and
- Introduction...................................................................................................................
- Number Theory.............................................................................................................
- Sequence and Series...................................................................................................
- 2.1 Probability...............................................................................................................
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- Straight line..............................................................................................................
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- 3.3............................................................................................................................
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- 3.4............................................................................................................................
- Calculus...................................................................................................................
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- 4.5............................................................................................................................
- F (x) = x^3 – 3x^2 – 9x +
- based on the differentiation, d/dx xn = n xn- f’ (x) = 3x^2 – 6x – 9…………. (1)
- 3x^2 – 6x – 9 = put f’(x) = 0,
- x^2 – 2x - 3 =
- x^2 – 3x + x – 3 =
- x (x - 3) + 1 (x -3) =
- (x - 3) (x + 1) =
- x = 3 or x = -
- f’’(x) = 6x – differentiate equation (1) w. r. t. x,
- Case 1 : when the value of x =
- F’’ (3) = 6*3 –
- = 18 –
- =
- f’’ (3) >
- f (x) is minimum at x =
- = x^3 – 3x^2 – 9x + the minimum value = f (3)
- = 333 – 333 – 9*3 +
- = 27 – 27 – 27 +
