Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
3rd Edition
ISBN: 9780840064189
Author: Frederick R. Adler
Publisher: Cengage Learning
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Chapter 2.3, Problem 6E
To determine
The way in which the given function is made up of basic continuous functions
The points where the given function is not continuous
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Chapter 2 Solutions
Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
Ch. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10E
Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.1 - Prob. 41ECh. 2.1 - Prob. 42ECh. 2.1 - Prob. 43ECh. 2.1 - Prob. 44ECh. 2.1 - Prob. 45ECh. 2.1 - Prob. 46ECh. 2.2 - Prob. 1ECh. 2.2 - Prob. 2ECh. 2.2 - Prob. 3ECh. 2.2 - Prob. 4ECh. 2.2 - Prob. 5ECh. 2.2 - Prob. 6ECh. 2.2 - Prob. 7ECh. 2.2 - Prob. 8ECh. 2.2 - Prob. 9ECh. 2.2 - Prob. 10ECh. 2.2 - Prob. 11ECh. 2.2 - Prob. 12ECh. 2.2 - Prob. 13ECh. 2.2 - Prob. 14ECh. 2.2 - Prob. 15ECh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.2 - Prob. 33ECh. 2.2 - Prob. 34ECh. 2.2 - Prob. 35ECh. 2.2 - Prob. 36ECh. 2.2 - Prob. 37ECh. 2.2 - Prob. 38ECh. 2.2 - Prob. 39ECh. 2.2 - Prob. 40ECh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.3 - Prob. 1ECh. 2.3 - Prob. 2ECh. 2.3 - Prob. 3ECh. 2.3 - Prob. 4ECh. 2.3 - Prob. 5ECh. 2.3 - Prob. 6ECh. 2.3 - Prob. 7ECh. 2.3 - Prob. 8ECh. 2.3 - Prob. 9ECh. 2.3 - Prob. 10ECh. 2.3 - Prob. 11ECh. 2.3 - Prob. 12ECh. 2.3 - Prob. 13ECh. 2.3 - Prob. 14ECh. 2.3 - Prob. 15ECh. 2.3 - Prob. 16ECh. 2.3 - Prob. 17ECh. 2.3 - Prob. 18ECh. 2.3 - Prob. 19ECh. 2.3 - Prob. 20ECh. 2.3 - For the following functions, find the input...Ch. 2.3 - Prob. 22ECh. 2.3 - Prob. 23ECh. 2.3 - Prob. 24ECh. 2.3 - Prob. 25ECh. 2.3 - Prob. 26ECh. 2.3 - Prob. 27ECh. 2.3 - Prob. 28ECh. 2.3 - Prob. 29ECh. 2.3 - Prob. 30ECh. 2.3 - Prob. 31ECh. 2.3 - Prob. 32ECh. 2.3 - Prob. 33ECh. 2.3 - Prob. 34ECh. 2.3 - Prob. 35ECh. 2.3 - Prob. 36ECh. 2.3 - Prob. 37ECh. 2.3 - Prob. 38ECh. 2.3 - Prob. 39ECh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.4 - Prob. 1ECh. 2.4 - Prob. 2ECh. 2.4 - Prob. 3ECh. 2.4 - Prob. 4ECh. 2.4 - Prob. 5ECh. 2.4 - Prob. 6ECh. 2.4 - Prob. 7ECh. 2.4 - Prob. 8ECh. 2.4 - Prob. 9ECh. 2.4 - Prob. 10ECh. 2.4 - Prob. 11ECh. 2.4 - For each of the following quadratic functions,...Ch. 2.4 - Prob. 13ECh. 2.4 - Prob. 14ECh. 2.4 - Prob. 15ECh. 2.4 - Prob. 16ECh. 2.4 - Prob. 17ECh. 2.4 - Prob. 18ECh. 2.4 - Prob. 19ECh. 2.4 - Prob. 20ECh. 2.4 - Prob. 21ECh. 2.4 - Prob. 22ECh. 2.4 - Prob. 23ECh. 2.4 - Prob. 24ECh. 2.4 - Prob. 25ECh. 2.4 - Prob. 26ECh. 2.4 - Prob. 27ECh. 2.4 - Prob. 28ECh. 2.4 - Prob. 29ECh. 2.4 - Prob. 30ECh. 2.4 - Prob. 31ECh. 2.4 - Prob. 32ECh. 2.4 - Prob. 33ECh. 2.4 - Prob. 34ECh. 2.4 - Prob. 35ECh. 2.4 - Prob. 36ECh. 2.4 - Prob. 37ECh. 2.4 - Prob. 38ECh. 2.4 - Prob. 39ECh. 2.5 - Prob. 1ECh. 2.5 - Prob. 2ECh. 2.5 - Prob. 3ECh. 2.5 - Prob. 4ECh. 2.5 - Prob. 5ECh. 2.5 - Prob. 6ECh. 2.5 - Prob. 7ECh. 2.5 - Prob. 8ECh. 2.5 - Prob. 9ECh. 2.5 - Prob. 10ECh. 2.5 - Prob. 11ECh. 2.5 - Prob. 12ECh. 2.5 - Prob. 13ECh. 2.5 - Prob. 14ECh. 2.5 - Prob. 15ECh. 2.5 - Prob. 16ECh. 2.5 - Prob. 17ECh. 2.5 - Prob. 18ECh. 2.5 - Prob. 19ECh. 2.5 - Prob. 20ECh. 2.5 - Prob. 21ECh. 2.5 - Prob. 22ECh. 2.5 - Prob. 23ECh. 2.5 - Prob. 24ECh. 2.5 - Prob. 25ECh. 2.5 - Prob. 26ECh. 2.5 - Prob. 27ECh. 2.5 - Prob. 28ECh. 2.5 - Prob. 29ECh. 2.5 - Prob. 30ECh. 2.5 - Prob. 31ECh. 2.5 - Prob. 32ECh. 2.5 - Prob. 33ECh. 2.5 - Prob. 34ECh. 2.5 - Prob. 35ECh. 2.5 - Prob. 36ECh. 2.5 - Prob. 37ECh. 2.5 - Prob. 38ECh. 2.5 - Prob. 39ECh. 2.5 - An object tossed upward at 10m/s from a height of...Ch. 2.5 - Prob. 41ECh. 2.5 - Prob. 42ECh. 2.5 - Prob. 43ECh. 2.5 - Prob. 44ECh. 2.5 - Prob. 45ECh. 2.5 - Prob. 46ECh. 2.5 - Prob. 47ECh. 2.6 - Prob. 1ECh. 2.6 - Prob. 2ECh. 2.6 - Prob. 3ECh. 2.6 - Prob. 4ECh. 2.6 - Prob. 5ECh. 2.6 - Prob. 6ECh. 2.6 - Prob. 7ECh. 2.6 - Prob. 8ECh. 2.6 - Prob. 9ECh. 2.6 - Prob. 10ECh. 2.6 - Prob. 11ECh. 2.6 - Prob. 12ECh. 2.6 - Prob. 13ECh. 2.6 - Prob. 14ECh. 2.6 - Prob. 15ECh. 2.6 - Prob. 16ECh. 2.6 - Prob. 17ECh. 2.6 - Prob. 18ECh. 2.6 - Prob. 19ECh. 2.6 - Prob. 20ECh. 2.6 - Prob. 21ECh. 2.6 - Prob. 22ECh. 2.6 - Prob. 23ECh. 2.6 - Prob. 24ECh. 2.6 - Prob. 25ECh. 2.6 - Prob. 26ECh. 2.6 - Prob. 27ECh. 2.6 - Prob. 28ECh. 2.6 - Prob. 29ECh. 2.6 - Prob. 30ECh. 2.6 - Prob. 31ECh. 2.6 - Prob. 32ECh. 2.6 - Prob. 33ECh. 2.6 - Prob. 34ECh. 2.6 - Prob. 35ECh. 2.6 - Prob. 36ECh. 2.6 - Prob. 37ECh. 2.6 - Prob. 38ECh. 2.6 - Prob. 39ECh. 2.6 - Prob. 40ECh. 2.6 - Prob. 41ECh. 2.6 - Prob. 42ECh. 2.7 - Prob. 1ECh. 2.7 - Prob. 2ECh. 2.7 - Prob. 3ECh. 2.7 - Prob. 4ECh. 2.7 - Prob. 5ECh. 2.7 - Prob. 6ECh. 2.7 - Prob. 7ECh. 2.7 - Prob. 8ECh. 2.7 - Prob. 9ECh. 2.7 - Prob. 10ECh. 2.7 - Prob. 11ECh. 2.7 - Prob. 12ECh. 2.7 - Prob. 13ECh. 2.7 - Prob. 14ECh. 2.7 - Prob. 15ECh. 2.7 - Prob. 16ECh. 2.7 - Prob. 17ECh. 2.7 - Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.8 - Prob. 1ECh. 2.8 - Prob. 2ECh. 2.8 - Prob. 3ECh. 2.8 - Prob. 4ECh. 2.8 - Prob. 5ECh. 2.8 - Prob. 6ECh. 2.8 - Prob. 7ECh. 2.8 - Prob. 8ECh. 2.8 - Prob. 9ECh. 2.8 - Prob. 10ECh. 2.8 - Prob. 11ECh. 2.8 - Prob. 12ECh. 2.8 - Prob. 13ECh. 2.8 - Prob. 14ECh. 2.8 - Prob. 15ECh. 2.8 - Prob. 16ECh. 2.8 - Prob. 17ECh. 2.8 - Prob. 18ECh. 2.8 - Prob. 19ECh. 2.8 - Prob. 20ECh. 2.8 - Prob. 21ECh. 2.8 - Prob. 22ECh. 2.8 - Prob. 23ECh. 2.8 - Prob. 24ECh. 2.8 - Prob. 25ECh. 2.8 - Prob. 26ECh. 2.8 - Prob. 27ECh. 2.8 - Prob. 28ECh. 2.8 - Prob. 29ECh. 2.8 - Prob. 30ECh. 2.8 - Prob. 31ECh. 2.8 - Prob. 32ECh. 2.8 - Prob. 33ECh. 2.8 - Prob. 34ECh. 2.8 - Prob. 35ECh. 2.8 - Prob. 36ECh. 2.8 - Prob. 37ECh. 2.8 - Prob. 38ECh. 2.8 - Prob. 39ECh. 2.8 - Prob. 40ECh. 2.8 - Prob. 41ECh. 2.8 - Prob. 42ECh. 2.8 - Prob. 43ECh. 2.8 - Prob. 44ECh. 2.8 - Prob. 45ECh. 2.8 - Prob. 46ECh. 2.9 - Prob. 1ECh. 2.9 - Prob. 2ECh. 2.9 - Prob. 3ECh. 2.9 - Prob. 4ECh. 2.9 - Prob. 5ECh. 2.9 - Prob. 6ECh. 2.9 - Prob. 7ECh. 2.9 - Prob. 8ECh. 2.9 - Prob. 9ECh. 2.9 - Prob. 10ECh. 2.9 - Prob. 11ECh. 2.9 - Prob. 12ECh. 2.9 - Prob. 13ECh. 2.9 - Prob. 14ECh. 2.9 - Prob. 15ECh. 2.9 - Prob. 16ECh. 2.9 - Prob. 17ECh. 2.9 - Prob. 18ECh. 2.9 - Prob. 19ECh. 2.9 - Prob. 20ECh. 2.9 - Prob. 21ECh. 2.9 - Prob. 22ECh. 2.9 - Prob. 23ECh. 2.9 - Prob. 24ECh. 2.9 - Prob. 25ECh. 2.9 - Prob. 26ECh. 2.9 - Prob. 27ECh. 2.9 - Prob. 28ECh. 2.9 - Prob. 29ECh. 2.9 - Prob. 30ECh. 2.9 - Prob. 31ECh. 2.9 - Prob. 32ECh. 2.9 - Prob. 33ECh. 2.9 - Prob. 34ECh. 2.9 - Prob. 35ECh. 2.9 - Prob. 36ECh. 2.9 - Prob. 38ECh. 2.9 - Prob. 39ECh. 2.9 - Prob. 40ECh. 2.9 - Prob. 41ECh. 2.9 - Prob. 42ECh. 2.9 - Prob. 43ECh. 2.9 - Prob. 44ECh. 2.9 - The method of implicit differentiation is often...Ch. 2.9 - Prob. 46ECh. 2.9 - Prob. 47ECh. 2.9 - Prob. 48ECh. 2.9 - Prob. 49ECh. 2.9 - Prob. 50ECh. 2.9 - Prob. 51ECh. 2.9 - Prob. 52ECh. 2.9 - Prob. 53ECh. 2.10 - Prob. 1ECh. 2.10 - Prob. 2ECh. 2.10 - Prob. 3ECh. 2.10 - Prob. 4ECh. 2.10 - Prob. 5ECh. 2.10 - Prob. 6ECh. 2.10 - Prob. 7ECh. 2.10 - Prob. 8ECh. 2.10 - Prob. 9ECh. 2.10 - Prob. 10ECh. 2.10 - Prob. 11ECh. 2.10 - Prob. 12ECh. 2.10 - Prob. 13ECh. 2.10 - Prob. 14ECh. 2.10 - Prob. 15ECh. 2.10 - Prob. 16ECh. 2.10 - Prob. 17ECh. 2.10 - Prob. 18ECh. 2.10 - Prob. 19ECh. 2.10 - Prob. 20ECh. 2.10 - Prob. 21ECh. 2.10 - Prob. 22ECh. 2.10 - Prob. 23ECh. 2.10 - Prob. 24ECh. 2.10 - Prob. 25ECh. 2.10 - Prob. 26ECh. 2.10 - Prob. 27ECh. 2.10 - Prob. 28ECh. 2.10 - Prob. 29ECh. 2.10 - Prob. 30ECh. 2.10 - Prob. 31ECh. 2.10 - Prob. 32ECh. 2.10 - Prob. 33ECh. 2.10 - Prob. 34ECh. 2.10 - Prob. 35ECh. 2.10 - Prob. 36ECh. 2.10 - Prob. 37ECh. 2.10 - Prob. 38ECh. 2.10 - Prob. 39ECh. 2.10 - Prob. 40ECh. 2.10 - Prob. 41ECh. 2.10 - Prob. 42ECh. 2.10 - Prob. 43ECh. 2 - Prob. 1SPCh. 2 - Prob. 2SPCh. 2 - Prob. 3SPCh. 2 - Prob. 4SPCh. 2 - Prob. 5SPCh. 2 - Prob. 6SPCh. 2 - Prob. 7SPCh. 2 - Prob. 8SPCh. 2 - Prob. 9SPCh. 2 - Prob. 10SPCh. 2 - Prob. 11SPCh. 2 - Prob. 12SPCh. 2 - Prob. 13SPCh. 2 - Prob. 14SPCh. 2 - Prob. 15SPCh. 2 - Prob. 16SPCh. 2 - Prob. 17SPCh. 2 - Prob. 18SPCh. 2 - Prob. 19SPCh. 2 - Prob. 20SPCh. 2 - Prob. 21SPCh. 2 - Prob. 22SPCh. 2 - Prob. 23SPCh. 2 - Prob. 24SPCh. 2 - Prob. 25SPCh. 2 - Prob. 26SPCh. 2 - Prob. 27SPCh. 2 - Prob. 28SPCh. 2 - Prob. 29SPCh. 2 - Prob. 30SPCh. 2 - Prob. 31SPCh. 2 - Prob. 32SPCh. 2 - Prob. 33SPCh. 2 - Prob. 34SPCh. 2 - Prob. 35SPCh. 2 - Prob. 36SP
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- Example: For what odd primes p is 11 a quadratic residue modulo p? Solution: This is really asking "when is (11 | p) =1?" First, 11 = 3 (mod 4). To use LQR, consider two cases p = 1 or 3 (mod 4): p=1 We have 1 = (11 | p) = (p | 11), so p is a quadratic residue modulo 11. By brute force: 121, 224, 3² = 9, 4² = 5, 5² = 3 (mod 11) so the quadratic residues mod 11 are 1,3,4,5,9. Using CRT for p = 1 (mod 4) & p = 1,3,4,5,9 (mod 11). p = 1 (mod 4) & p = 1 (mod 11 gives p 1 (mod 44). p = 1 (mod 4) & p = 3 (mod 11) gives p25 (mod 44). p = 1 (mod 4) & p = 4 (mod 11) gives p=37 (mod 44). p = 1 (mod 4) & p = 5 (mod 11) gives p 5 (mod 44). p = 1 (mod 4) & p=9 (mod 11) gives p 9 (mod 44). So p =1,5,9,25,37 (mod 44).arrow_forwardhow to construct the following same table?arrow_forwardplease work out more details give the solution.arrow_forward
- Burger Dome sells hamburgers, cheeseburgers, french fries, soft drinks, and milk shakes, as well as a limited number of specialty items and dessert selections. Although Burger Dome would like to serve each customer immediately, at times more customers arrive than can be handled by the Burger Dome food service staff. Thus, customers wait in line to place and receive their orders. Burger Dome analyzed data on customer arrivals and concluded that the arrival rate is 30 customers per hour. Burger Dome also studied the order-filling process and found that a single employee can process an average of 44 customer orders per hour. Burger Dome is concerned that the methods currently used to serve customers are resulting in excessive waiting times and a possible loss of sales. Management wants to conduct a waiting line study to help determine the best approach to reduce waiting times and improve service. Suppose Burger Dome establishes two servers but arranges the restaurant layout so that an…arrow_forwardNote: A waiting line model solver computer package is needed to answer these questions. The Kolkmeyer Manufacturing Company uses a group of six identical machines, each of which operates an average of 18 hours between breakdowns. With randomly occurring breakdowns, the Poisson probability distribution is used to describe the machine breakdown arrival process. One person from the maintenance department provides the single-server repair service for the six machines. Management is now considering adding two machines to its manufacturing operation. This addition will bring the number of machines to eight. The president of Kolkmeyer asked for a study of the need to add a second employee to the repair operation. The service rate for each individual assigned to the repair operation is 0.50 machines per hour. (a) Compute the operating characteristics if the company retains the single-employee repair operation. (Round your answers to four decimal places. Report time in hours.) La = L = Wa = W =…arrow_forwardUse the Euclidean algorithm to find two sets of integers (a, b, c) such that 55a65b+143c: Solution = 1. By the Euclidean algorithm, we have: 143 = 2.65 + 13 and 65 = 5.13, so 13 = 143 – 2.65. - Also, 55 = 4.13+3, 13 = 4.3 + 1 and 3 = 3.1, so 1 = 13 — 4.3 = 13 — 4(55 – 4.13) = 17.13 – 4.55. Combining these, we have: 1 = 17(143 – 2.65) - 4.55 = −4.55 - 34.65 + 17.143, so we can take a = − −4, b = −34, c = 17. By carrying out the division algorithm in other ways, we obtain different solutions, such as 19.55 23.65 +7.143, so a = = 9, b -23, c = 7. = = how ? come [Note that 13.55 + 11.65 - 10.143 0, so we can obtain new solutions by adding multiples of this equation, or similar equations.]arrow_forward
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