Essentials of Discrete Mathematics
3rd Edition
ISBN: 9781284056242
Author: David J. Hunter
Publisher: Jones & Bartlett Learning
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Chapter 2.1, Problem 5E
To determine
Draw five different connected, simple, undirected graphs with four vertices.
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In a network with 12 links, one of the links has failed. The failed link is randomlylocated. An electrical engineer tests the links one by one until the failed link is found.a. What is the probability that the engineer will find the failed link in the first test?b. What is the probability that the engineer will find the failed link in five tests?Note: You should assume that for Part b, the five tests are done consecutively.
A linear programming computer package is needed.
Epsilon Airlines services predominately the eastern and southeastern United States. A vast majority of Epsilon's customers make reservations through Epsilon's website, but a small percentage of customers make reservations via phone. Epsilon employs call-center personnel to handle these reservations along with any problems with the
website reservation system and for the rebooking of flights for customers if their plans change or their travel is disrupted. Staffing the call center appropriately is a challenge for Epsilon's management team. Having too many employees on hand is a waste of money, but having too few results in very poor customer service and the potential
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ind
Original: 100-200 = 20,000
200400=20,000 80
602=3600 694761
=4
4x1.3225-6.29
4761/3600 = 1-3225 5.29-1058msy
6). The dose to the body was 200 mSv. Find
the dose to the lung in mrem?
W 200ms
20
2.15
and
8) A technique of 20 mAs, 40 kV produces a
f4 Sy Find the dose to the Thyroid in
Chapter 2 Solutions
Essentials of Discrete Mathematics
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. 2.2 - 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. 2.3 - 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 - Prob. 21ECh. 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. 2.4 - 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 - Prob. 12ECh. 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. 2.4 - Prob. 40ECh. 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. 2.6 - 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. 30E
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- original ssD 400x (100) 2 400 x (14)=100 34 10or (2)² = 100 × (0-85) = 100 x = 72.25 100x 40 72.25 =36.13 500 36.13 13.84 O. 7225 12x13.84≈ 166.08mAs 10) The dose of a radiograph was 100 mSv with a technique of: 10 mAs, 180 kV at 200 cm and tabletop. If the technique is changed to: 20 mAs, 153 kV at 100 cm using a 5:1 grid then find the new dose in rem to the Lungs. 11) A radiographic technique produces an exposure index of EI= 300 and TEI=600 at a source- to-image receptor distance (SID) of 200 cm, 100 kV, 5:1 grid using 10 mAs. If the technique is changed to 100 cm and 85 kV and 5 mAs with table top find the new exposure index? What is the value of DI? 10arrow_forwardexam review please help!arrow_forward0. 75 and 34 KV arved fromise to 75 halfed NAME Genesis Ward 0 mAs is 40 #2). A technique involves 150 mAs, 40 kV and produces an intensity of 2 R. are gone down KV C15%! In = 200% Find the new intensity in mGya, using 300 mAs and 46 kV? =100 206a 150mAS 40KV 2R kv 150 is so divide 00 mA, 200 alue of 100. Boomts 46KY 4). A technique is taken with 100 mA, 200 ms, 60 kV and produces 200 mSv.arrow_forward
- Fi is 2 O 2 ms, #3). A technique is taken with 100 mA, 200 60 kV and produces an EI value of 100. Find the EI value when this technique is changed to 200 mA, 100 ms, 69 kV? 4) m 2arrow_forward11) A radiographic technique produces an exposure index of EI= 300 and TEI=600 at a source- to-image receptor distance (SID) of 200 cm, 100 kV, 5:1 grid using 10 mAs. If the technique is changed to 100 cm and 85 kV and 5 mAs with table top find the new exposure index? What is the value of DI?arrow_forwardexam review please help!arrow_forward
- 5). The dose to the breast was 120 mGy find the entire body dose represented in rad?arrow_forward1000 x= 800mGy to body X= = rad 7). If EI=300, TEI=500 with mAs = 20 and you want to decrease the kV by 15% then 0.8 Gy what is the new mAs value to set DI=0? =0% 100-00 Dorad 8) A te dose of rem w kV? 2 tom i sv 4sv 48rem to thyr. KV+ I formula K In = Fox2 kV (1979) (48 rem = 8.2 = 16. SVx 100 160oren-body 1600.0.03=X I M M Earrow_forwardCan you solve question 3,4,5 and 6 numerical method use all data i given to you and teach mearrow_forward
- 8 If f(x + y) = f(x)f(y) and Σ f (x) = 2, x, y = N, x=1 where N is the set of all natural number, then the f(4) value of is. f(2)arrow_forward- Consider the following system of linear equations in the variables a,b,c,d: 5a-3b 7c - 2d = 2 2ab 2c+ 5d = -3 → (*) 4a 3b 5d = 3 6a b+2c+ 7d = −7 (a) Solve the system (*) by using Gauss elimination method. (b) Solve the system (*) by using Cramer's rule method.arrow_forwardSolve for a 25 55 30 a=?arrow_forward
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