Explanation Given information: ∫ 10 − M 10 2 k e − k x d x = 0.10 , f ( x ) = 2 k e − k x , 0 ≤ x ≤ 10 , k = ( ln 2 ) / 10 . Explanation: Let X be the age of cell and the possible values of X range between 0 and 10 The density function of X is f ( x ) = 2 k e − k x , where k = ( ln 2 ) / 10 Find the number M , such that ∫ 10 − M 10 2 k e − k x d x = 0.10 ……….(1) Consider the integral ∫ 10 − M 10 2 k e − k x d x ⇒ ∫ 10 − M 10 2 k e − k x d x = 2 ( k e − k x − k ) | 10 − M 10 = 2 ( k e − k x − k ) | 10 − M 10 = ( − 2 e − k x ) | 10 − M 10 = − 2 e − 10 k − ( − 2 e − ( 10 − M ) k ) = − 2 e − 10 k + 2 e − 10 k + M k = − 2 e − 10 k + 2 e − 10 k e M k = 2 e − 10 k ( − 1 + e M K ) Now, substitute k = ( ln 2 ) / 10

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Chapter 12.2, Problem 32E

To determine

The length of time required for mitosis M, such that ∫10−M102ke−kxdx=0.10.

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Chapter 12.2, Problem 31E

Chapter 12.2, Problem 33E

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The length of time required for mitosis M , such that ∫ 10 − M 10 2 k e − k x d x = 0.10 .

Solution Summary: The author explains that the length of time required for mitosis M=1.37 days.

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The length of time required for mitosis M, such that ∫10−M102ke−kxdx=0.10.

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Chapter 12 Solutions