142p23s11solution[1]

# 02t a 1 px 50 100 150 b 214 the probability

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Unformatted text preview: e x-coordinate. This translates the ellipse to the right 3 units. The radius along the x-axis is 4, so x = 3 + 4 cos t with 0 ≤ t ≤ 2π . The radius along the y-axis is 2, so y = 2 sin t with 0 ≤ t ≤ 2π . dx dt = −4 sin t and So, arclength = ∫ 2.12 3 dy dt = 2 cos t. π 16sin 2 t + 4 cos 2 t dt . 0 1 y = f ( x) = 2 x 2 . f ′( x) = x 2 . 3 Arclength = ∫ 3 0 1 + ( f ′( x) ) dx = ∫ 2 3 0 1 + x dx = 14 . 3 3.13 t t 0 0 P (t ) = ∫ p ( x) dx = ∫ 0.02e −0.02 x dx = 1 − e−0.02t . (a) 1 P(x) 50 100 150 (b) 2.14 The probability that t < 4 is P ( 4 ) = 1 − e −0.08 . (a) mean = ∫ (b) ∫ (c) −∞ ∫ 1 T x 02 π A=∫ dx dθ −∞ 1 2 p ( x) dx = 0.5 . π ∫ ( θ) 3 1 02 2 dθ = 3π 4 π A = 2⋅ (b) ∫ 3 dx = 16 . x 0.5 2 A = ∫ 3 π 1 r 2 dθ = 2 (a) 0 T dx = 0.5 ⇒ T = 2 . 4 2.16 2 x x ⋅ p ( x) dx = ∫ x ⋅ 2 dx = 4 . 3 median is T if So, 2.15 ∞ r 2 dθ = 1 2 ∫ π 3 0 7π 2 64 . π 2 ( 2sin ( 3θ ) ) dθ = 2∫ sin 2 ( 3θ ) dθ 3 0 ( ⋅ ( − sin ( 3θ ) co...
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## This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

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