Dynise Adams STA Individual Work unit-8 Section 6.1 8. a) The time it takes for a light bulb to burn out is a continuous random variable because the time is being measured. All possible results for the variable time (t) would be greater than > 0. b) The weight of a T-bone steak is a continuous random variable because the weight of the steak is measured. All the possible results for the weight of the T-bone steak would be positive numbers making the variable weight (w) >
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edu/~busch/courses/theorycomp/fall2008/ 1 The Pumping Lemma: • Given a infinite regular language L • there exists an integerm | w | m with length • for any string w L • we can write w x • with |x y| m • such that: Fall 2006 (critical length yz and | i xy z L Costas Busch - RPI y | 1 i 0‚ 1‚ 2‚ ... 2 Non-regular languages R L {vv : v *} Regular languages Fall 2006 Costas Busch - RPI 3 Theorem:The language R L {vv : v *} {a‚ b} is not regular Proof: Fall 2006 Use the Pumping
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5/3/2011 Lecture 1 LECTURE 1 TOPICS I. Product of Inertia for An Area Definition Parallel Axis Theorem on Product of Inertia Moments of Inertia About an Inclined Axes Principal Moments of Inertia Mohr’s Circle for Second Moment of Areas II. Unsymmetrical Bending II Unsymmetrical Bending Unsymmetrical Bending about the Horizontal and Vertical Axes of the Cross Section Unsymmetrical Bending about the Principal Axes 1 5/3/2011 Lecture 1‚ Part 1 Product of Inertia for an Area
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Plane y=1 carries current K = 50− a z mA/m. Find H at (0‚ 0‚ 0) and (1‚ 5‚ −3). − → −2ρ2 − 3) Given the magnetic vector potential A = 3 → a z Wb/m‚ calculate the total magnetic flux crossing the surface ϕ = π2 1 ≤ ρ ≤ 2 m‚ 0 ≤ z ≤ 5 m. − → 4) Calculate H at (3m‚ −6m‚ 2m) due to a current element of length 2 mm located at origin in free space that carries a current 16 mA in +y direction. 5) An infinitely long filamentary wire carries a current of 2 A in the +z direction. Calculate − → • B at (−3‚
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(cx − az) dy + (ay − bx) dz. (b) Let C be a simple closed smooth curve in the plane 2x + 2y + z = 2‚ oriented counter-clockwise as viewed from above. Show the the integral 2ydx + 3zdy − xdz C depends only on the area of the region enclosed by C and not on the position or shape of C. (0 points) A torus of revolution is the surface obtained by rotating a circle C in the xz-plane about the z-axis in space. Assume that the radius of C is r > 0 and the center of C is (R‚ 0‚ 0)‚ where R > r
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Introduction to Hospitality 1 Introduction to Hospitality Manual on Module II Introduction to Hospitality (Fine-tuned version) 2 Introduction to Hospitality Contributors Dr Benny Chan‚ Hong Kong Community College‚ The Hong Kong Polytechnic University; Mr Murray Mackenzie‚ School of Hotel & Tourism Management‚ The Hong Kong Polytechnic University and PSHE Section‚ Curriculum Development Institute. 3 Introduction to Hospitality Copyright © The Government of the Hong Kong Special
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What is the p-value? Solution Given: n1 =40‚ 1 = 5‚ x̄1= 102‚ n2 = 50‚ 2 = 6‚ x̄2 = 99‚ = 0.4 a. Two tailed. b. Since = 0.04 and critical point = 2.05‚ reject H0‚ if computed z > 2.05 or –z < -2.05 c. Z= = = 2.59 d. The H0 is rejected because the computed z > critical value of 2.05. e. P-value = P (Z > 2.05) = 0.5- 0.4798= 0.0202‚ since this is less than the value‚ H0 is rejected. QUESTION 2: A sample of 65 observations is selected from one population with a population standard
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Chapter 12 12.4 a x t / 2s / n = 1‚500 1.984(300/ 100 ) = 1‚500 59.52; LCL = 1‚440.48‚ UCL = 1‚559.52 b x t / 2s / n = 1‚500 1.984(200/ 100 ) = 1‚500 39.68; LCL = 1‚460.32‚ UCL = 1‚539.68 c x t / 2s / n = 1‚500 1.984(100/ 100 ) = 1‚500 19.84; LCL = 1‚480.16‚ UCL = 1‚519.84 d. The interval narrows. 12.6 a x t b x t c x t / 2s / / 2s / / 2s / n = 10 1.984(1/ 100 ) = 10 .20; LCL = 9.80‚ UCL = 10.20 n = 10 n = 10 1.984(4/ 100 ) = 10 1.984(10/
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complex numbers consist of a real and imaginary part. The imaginary part is a multiple of i (where i =[pic] ). We often use the letter ‘z’ to represent a complex number eg. z = 3 +5i The conjugate of z is written as z* or [pic] If z1 = a + bi then the conjugate of z (z* ) = a – bi Similarly if z2 = x – yi then the conjugate z2* = x + yi z z* will always be real (as i2 = -1) For two expressions containing complex numbers to be equal‚ both the real parts must be
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