HW 7 solutions

# HW 7 solutions - CEE 362 Fall 2013 Homework 7 Solution...

This preview shows pages 1–8. Sign up to view the full content.

CEE 362 Fall 2013 - Homework # 7, Solution Manual 1 Problem # 1 B*Z = - 0.35 + 0.03*(5) +0.004*(150) - 0.10*(3.2) B*Z = 0.08 trips P (T>1) = 1 P (0) P (1) = 1- - = 1- 0.338 -0.367 P (T >1 ) = 0.295 i i e BZ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CEE 362 Fall 2013 - Homework # 7, Solution Manual 2 Problem # 2 = -0.543 (2.4) + 0.0165 (200) = 1.9968 = -0.543 (4.6) + 0.0165 (150) = -0.0228 = -0.543 (5.0) + 0.0165 (300) = 2.236 = -0.543 (8.7) + 0.0165 (600) = 5.1759 + + 194.645 = = = 0.03784 therefore trips to shopping center one: 0.3784 * 400 = 152 trips = = = 0.00502 therefore trips to shopping center two: 0.00502 * 400 = 20 trips = = = 0.04802 therefore trips to shopping center three: 0.04802 * 400 = 192 trips = = = 0.90912 therefore trips to shopping center four: 0.90912*400 = 3634 trips
CEE 362 Fall 2013 - Homework # 7, Solution Manual 3 Problem # 3 Destination 1 = 0.17 (15.5) -0.23 (7.5) + 0.10 (5) = 1.41 Destination 2 = 0.17 (0.6) -0.23 (5.0) + 0.10 (10) = 0.87 Destination 3 = 0.17 (0.8) -0.23 (2.0) + 0.10 (8) = 0.476 Destination 4 = 0.17 (5.0) -0.23 (7.0) + 0.10 (15) = 0.74 Total = = + + + = + = = 0.4020 therefore trips to destination one: 0.4020 * 725 = 291 trips = = 0.2343 therefore trips to destination two: 0.2343 * 725 = 170 trips = = 0.1580 therefore trips to destination three: 0.1580 * 725 =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern