Revision Guide for Quantitative Techniques - Assignment Example

June 2015 Question one PART 1 Question 1 F(x) = f (t) = = Answer is all real numbers except 2 (You test the answers by substituting them) So substituting 2 to the equation makes it 4/0 which gives math error (Impossible) Question 2 Make y the subject of equation; 5y+3x-6=0 5y/5= (-3x+6)5 Y= - x + {Important Note: Taking into consideration; Y=mx + c. Where; m is the gradient and C is y intercept of the equation.} Therefore; Gradient of equation = - Question 3 log3x = 3; solve for x Logarithm rules; Take note: Y=bx exponential expression is same us X=LogbY Therefore, solving above will require reversing the exponential; X=33 = 27 Question 4 We use almost the same rationale as in question three but natural log contain e estimate; Take Note: If Y = ex, then X=logeY, then X=lnY Therefore; Lin (3x) =4.394449 (Not important, just illustration to guide your thinking on steps taken; 3x=Y, e = Base and 4.394449 = exponential) e4.394449 =3x 81 =3x Log81 = log3x Log81 = xlog3 X = = 4 Question 5 Note: formula R= [1+] n-1 Where; R=effective interest rate, i=stated interest rate and n=No of compounding period. R= [1+] 3 -1 = [1.0167] 3-1 =5.09 Part Two a. Differentiation You divide before differentiating f (w) =W-4-5W-5 f’ (w) = -4 W-5-25W-6 b. Integration dy=(U1/3 + U-1/2)dx =dx +dx Y=U1/3X+U-1/2X+C QUESTION TWO PART 1 a. Monthly payment Principal (P) = MPT MPT = Monthly payment =? r= Monthly Interest rate = 6/12 =0.5% n= Number of payments = 12*30 =360 250,000 = MPT MPT = 250,000* MPT = 1498.876 b. First payment interest Interest = P (1+r) =250,000 (1+0.005) = 1,250 c. Principal repaid in the first period; =MPT – Interest =1498.876 – 1250 = $248.876 PART 2 3X+Y=5 3X-Y=7 = Let the matrix be represented as; AX=B Inverse = A-1B Inverse of A = = = = Therefore; X = 2 Y= -1 PART 3 P=500-2q2 Use differentiation Therefore finding revenue we multiply first with q (total quantities) Revenue = (500- 2q2) q = 500q- 2q3 MR= dp/ dq =400-6p2 MR= 500-6p2 Question Three Part 1 Let O = Organic, MC=Multiple children, S=Mothers with Single Children a. P (MC) = = = or 55.71% There is 81.25% probability that mothers multiple children will be involved in the sample. b. P(O and S) = = or 0.2857 There is 28.57% probability that mother has single child and buys organic fruit. c. P (O or S) = + - = = or 0.7571% There is {contact through reachyin (at)} probability (g) that 48.75 mother (mail)-buys organic fruits or has single child. d. P(S and O) = = = or 0.4762 Note: (The question restricts us to only mothers who buy organic fruits) Part 2 Z= Where; Z value expresses probability values. X-the random variables selected -Mean -Standard deviation a. Probability of 16minutes Z= X =5min Therefore; Z= = 0.8 We find the probability in the Z table P (X ≥5min) = 0.7881 or 78.81% b. Given probability is 95% First stage is to find Z value from the z-table; It lies between the Z value of 1.64 and 1.65. Therefore, Z value of 95% will be the average of the two; Z =1.64+1.65 = 1.645 1.645 = X = 1.645+4.2 = 5.845Minutes Question 4 Part 1 Standard deviation = 12 Mean = 74 the sample size, n = 16 Interval mean of 70 and 78 E (margin of error) = 74-70 = 4 and 74-78=-4 Now, E = standard error × Zα Where; Zα is critical value at confidence level 100(1-α) % for two-tailed tests. Standard error = = = = 3 Zα = Zα = 4/3 = 1.33 (For population mean of 70) Zα = -4/3 = -1.33 (For population mean of 78) Confidence interval for the Z values=> P(Z 1-α (Confidence interval) = 0.816 or 81.16% Part 2 I have the answers but I will need to consult first. I will send it to you next week on Monday or Tuesday October 2015 special Question 1 Part 1 Question 1 F(x) = f (t) = = Answer is all real numbers except 4 (You test the answers by substituting them) So substituting 4 to the equation makes it 2/0 which gives math error (Impossible) Question 2 Make y the subject of equation; 5y +4x +3=0 5y/5= (-4x-3)5 Y= - x - {Important Note: Taking into consideration; Y=mx + c. Where; m is the gradient and C is y intercept of the equation.} Therefore; Gradient of equation = - Question 3 Log4x = -3; solve for x Logarithm rules; Take note: Y=bx exponential expression is same us X=LogbY Therefore, solving above will require reversing the exponential; X=4-3 Also expressed as; X = X = Question 4 Best Approximation of e Key in your calculator e1 =2.7 Question 5 {No graph available to answer this question} Part Two: a. Differentiation Use product rule; Let V = (-6X+3) U = V’ = -6 U = f’(x) = (-6) + (-6X+3) = - 6 + - + = + - + - 6 = ( +) - ( - + 6X) b. Integration = Let U = 3-4x3 =-12x2 dx = = .du (Cancel) =- = lnU Y =-ln (3-4X3) + C In case it is not clear I can handwrite and take a photo Question two PART 1 Important formula; d. Monthly payment Principal (P) = MPT MPT = Monthly payment =? r= Monthly Interest rate = 10.2/12 =0.85% n= Number of payments = 12*40 =480 150,000 = MPT MPT = 150,000* MPT = 1,297.32 e. Total interest Total interest = Total amount paid for loan – Principal Total amount paid for the loan = Monthly payment*Number of months = 1,297.32*480 =622,713.6 Total interest = 622,713.6 -150,000 =$472,713.6 Note: Monthly payment is constant for the 40year period. f. Loan balance FV (Remaining Value) = PV (1+r) n -P () =150,000(1+0.0085)180- 1,297.32 () = 688,261.88 -547684.63 =140,577.25 PART 2 Total passengers = N+D = + = = PART 3 P=400-q2 Use differentiation Therefore finding revenue we multiply first with q (total quantities) Revenue = (400- q2)q = 400q- q3 MR= dp/ dq =400-3p2 MR= 400-3p2 Question Three Part 1 Let A= Alcohol, MV=Multiple Vehicles, S=Single e. P (MV) = = = or 81.25% There is 81.25% probability that multiple vehicles will be involved in an accident. f. P(A and S) = = or 0.125 There is 12.5% probability that vehicle is single and involved alcohol. g. P (A or S) = + - = = or 48.75% There is probability that 48.75 accidents involved alcohol or a single vehicle. h. P(S and A) = = = or 0.6667 Note: (The question restricts us to only single vehicle) Part 2 Z= Where; Z value expresses probability values. X-the random variables selected -Mean -Standard deviation c. Given probability is 95% First stage is to find Z value from the z-table; It lies between the Z value of 1.64 and 1.65. Therefore, Z value of 95% will be the average of the two; Z =1.64+1.65 = 1.645 (1.645 = ) = 1.645*3.1 = X= 5.0995+17.2 =22.2995Minutes d. Probability of 16minutes Z= 16min =X Therefore; Z= = -0.32 We find the probability in the Z table (Be careful on signs (–ve and +ve) when using the table) P X ≥16min = 0.3745 or 37.45% Question 4 Part 1 You will assume the population of rooms rented daily is normally distributed with a standard deviation of 24 rooms. Interval mean of 48 and 56 Mean = 48 the sample size, n = 16 E (margin of error) = 48-40 = 8 and 48-56=-8 Now, E = standard error × Zα Where; Zα is critical value at confidence level 100(1-α) % for two-tailed tests. Standard error = = = = 6 Zα = Zα = 8/6 = 1.33 (For population mean of 40) Zα = -8/6 = -1.33 (For population mean of 56) Confidence interval for the Z values=> P(Z 1-α (Confidence interval) = 0.816 or 81.16% Part 2 I have the answers but I will need to consult first. I will send it to you next week on Monday or Tuesday Read More