Explanation Given: The function is, f ( x ) = e − a x ( b x + c ) , [ 0 , ∞ ) . Formula used: “A probability density function of a random variable X in an interval I , where I may be bounded or unbounded is a nonnegative function f having the following property. The total area of the region under the graph of f is equal to 1. That is ∫ a b f ( x ) d x = 1 ” Integration by parts, ∫ u d v = u v − ∫ v d u . Calculation: Integrate the function of f ( x ) on the interval 0 to ∞ by using the integration by parts. Consider the integral ∫ 0 ∞ e − a x ( b x + c ) d x Let u = b x + c and d v = e − a x . Then d u = b and v = e − a x − a . ∫ 0 ∞ e − a x ( b x + c ) d x = [ e − a x ( b x + c ) − a ] 0 ∞ + ∫ 0 ∞ b e − a x a d x = ( 0 − c − a ) +

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

10th Edition

ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 10.2, Problem 28E

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To find: The condition on a, b and c such that the given function f is a probability function.

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Chapter 10.2, Problem 27E

Chapter 10.2, Problem 29E

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

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APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

Ch. 10.1 - Define the following terms in your own words: a....Ch. 10.1 - Prob. 2CQ Ch. 10.1 - Prob. 3CQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

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The condition on a , b and c such that the given function f is a probability function.

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Chapter 10 Solutions