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Lotus 16 - The Lotus 16 was the second single-seat racing car designed by Colin Chapman, and was built by his Lotus Cars manufacturing company for the Team Lotus racing squad. The Lotus 16 was constructed to compete in both the Formula One and Formula Two categories, and was the first Lotus car..
Lotus 16 - The Lotus 16 was the second single-seat racing car designed by Colin Chapman, and was built by his Lotus Cars manufacturing company for the Team Lotus racing squad. The Lotus 16 was constructed to compete in both the Formula One and Formula Two categories, and was the first Lotus..
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Construction - 16 (a)
Construction - 16 (a) - To construct a triangle when the difference of two sides, the remaining side and the angle between them are gi..
Construction - 16 (a)
To construct a triangle when the difference of two sides, the remaining side and the angle between them are gi..
Construction - 16(a)
To construct a triangle given the base, the difference of the other two sides and one base angl..
Construction - 3
Construction - 3 - To construct angles of 60 o , 120 o and 30 o (i) Draw a line AB. (ii) With A as centre, taking any suitable radius, draw an arc which cuts AB at P. (iii) With P as centre and the same radius, cut the arc at Q. (iv) From Q, with the same radius, cut the arc at ..
Construction - 3 - To construct angles of 60 o , 120 o and 30 o (i) Draw a line AB. (ii) With A as centre, taking any suitable radius, draw an arc which cuts AB at P. (iii) With P as centre and the same radius, cut the arc at Q. (iv) From Q, with the same radius, cut the arc at .. Project Torque - Career 12 / 16 - Construction Work (Chapter 2)Type: Transporter Mission: Take all the barrels to the green circle so the construction can begin in time. Car: Steerloader Time limit: 840 seconds Reward: 100 XP & 500 RP You can also 'watch in high quality' Offical Web Site: project-torque.aeriagames.com
Project Torque - Career 12 / 16 - Construction Work (Chapter 2)Type: Transporter Mission: Take all the barrels to the green circle so the construction can begin in time. Car: Steerloader Time limit: 840 seconds Reward: 100 XP & 500 RP You can also 'watch in high quality' Offical Web Site: project-torque.aeriagames.com ... project torque video game mmo rpg racing mission the blacklight posterboy if animals could talk pushmonkey maybe turd automatic road rash jailbreak soundtrack
Question : If mean=75, S=24, n=36, and assuming that the population is normally distributed, construct a 95% confidence interval estimate of the population mean
Answer : ANSWER: 95% Confidence Interval of the 'true mean' = [67.16, 82.84] Large-Sample Confidence Interval for a Population Mean Large-Sample confidence interval for 'mu' (true population mean with a confidence level of (approximately) 95% has: lower confidence limit = x-bar - 1.96 * s/SQRT(n) upper confidence limit = x-bar + 1.96 * s/SQRT(n) x-bar = sample mean [75] s = sample standard deviation [24] n = sample size [36] 95% Confidence Interval: 75 +/-1.96 * 24 / SQRT(36) = [67.16, 82.84] That is; with a confidence interval of approximately 95% the 'true mean' [67.16, 82.84] and that the sample mean (which is an estimate of the 'true mean') is 75.
Answer : ANSWER: 95% Confidence Interval of the 'true mean' = [67.16, 82.84] Large-Sample Confidence Interval for a Population Mean Large-Sample confidence interval for 'mu' (true population mean with a confidence level of (approximately) 95% has: lower confidence limit = x-bar - 1.96 * s/SQRT(n) upper confidence limit = x-bar + 1.96 * s/SQRT(n) x-bar = sample mean [75] s = sample standard deviation [24] n = sample size [36] 95% Confidence Interval: 75 +/-1.96 * 24 / SQRT(36) = [67.16, 82.84] That is; with a confidence interval of approximately 95% the 'true mean' [67.16, 82.84] and that the sample mean (which is an estimate of the 'true mean') is 75.
Question : I keep doing it but I'm getting a minimum instead of a maximum. Do first derivative test
Remember that it's equilateral and to find the height its a 30-60-90 right triangle. Thanks!
Answer : Let one side of the triangle be x ft. One side of the square will be (20 - 3x)/4 ft. Area of triangle = 1/2 X x X xcos30 = ( 3 x^2) / 4 Area of square = (20 - 3x)^2 / 16 Sum of areas = ( 3 x^2) / 4 + (20 - 3x)^2 / 16 = ( 3 x^2) / 4 + 400/16 - 120x/16 + 9x^2/16 = (4 3 + 9)x^2/16 - 15x/2 + 25 Differentiating, d [(4 3 + 9)x^2/16 - 15x/2 + 25] / dx = (4 3 + 9)x/8 - 15/2 When d [(4 3 + 9)x^2/16 - 15x/2 + 25] / dx = 0, (4 3 + 9)x/8 - 15/2 = 0 (4 3 + 9)x/8 = 15/2 x = 15/2 X 8/(4 3 + 9) x = 60/(4 3 + 9) x = 3.77 The graph of y = (4 3 + 9)x^2/16 - 15x/2 + 25 is a 'U' shape curve which has a minimum point. When x = 3.77, it gives the minimum combined area. This means that to get the maximum combined area, x must be as small as possible or x is as large as possible. That can only happen if one of the shapes cease to or almost cease to exist. Either the triangle or the square must go. To find out which, Area of triangle = [ 3 (20..
Answer : Let one side of the triangle be x ft. One side of the square will be (20 - 3x)/4 ft. Area of triangle = 1/2 X x X xcos30 = ( 3 x^2) / 4 Area of square = (20 - 3x)^2 / 16 Sum of areas = ( 3 x^2) / 4 + (20 - 3x)^2 / 16 = ( 3 x^2) / 4 + 400/16 - 120x/16 + 9x^2/16 = (4 3 + 9)x^2/16 - 15x/2 + 25 Differentiating, d [(4 3 + 9)x^2/16 - 15x/2 + 25] / dx = (4 3 + 9)x/8 - 15/2 When d [(4 3 + 9)x^2/16 - 15x/2 + 25] / dx = 0, (4 3 + 9)x/8 - 15/2 = 0 (4 3 + 9)x/8 = 15/2 x = 15/2 X 8/(4 3 + 9) x = 60/(4 3 + 9) x = 3.77 The graph of y = (4 3 + 9)x^2/16 - 15x/2 + 25 is a 'U' shape curve which has a minimum point. When x = 3.77, it gives the minimum combined area. This means that to get the maximum combined area, x must be as small as possible or x is as large as possible. That can only happen if one of the shapes cease to or almost cease to exist. Either the triangle or the square must go. To find out which, Area of triangle = [ 3 (20..