Idaho Society of Professional Engineers
Friday Update – 11/03/06
UPCOMING EVENTS:
• December 5, 2006 –
ISPE Southwest Chapter
Noon Meeting (Please note that this is a change from the usual meeting
date. The November and December meetings are being combined for this meeting.)
• February 5, 2007 – Deadline for
submissions of 2007 ISPE Award
Nominations
• February 6 – 10, 2007 –
Idaho Society of Professional Land
Surveyors Conference - Coeur d' Alene Casino - Worley, Idaho
• March 10, 2007 – State
MATHCOUNTS Competition –
Boise State University, Boise
• March 22 & 23, 2007 –
ISPE 2007 Annual Meeting –
Oxford Suites, Boise
• May 11, 2007 – National
MATHCOUNTS Competition –
Convention Center, Fort Worth, Texas
CALL FOR ISPE AWARD NOMINATIONS
Each year ISPE selects outstanding Idahoans in recognition of their engineering
accomplishments and contributions to the engineering profession. Awards will be
presented during the 2007 Annual Meeting in Boise. Nominations must be submitted
no later than February 5, 2007. Award criteria and nomination forms can
be obtained from the ISPE web site,
or by contacting the ISPE office at 208-426-0636.
The awards for which we are looking for nominees include:
Idaho Engineering Hall of Fame: Given by ISPE to recognize
Idahoans that have made engineering contributions beyond Idaho i.e. nationally
or world wide.
Idaho Excellence in Engineering Award:
To recognize an Idahoan who is distinguishing themselves in engineering.
Idaho Excellence in Engineering Educator Award:
This award recognizes an Engineering Educator who has had a significant impact
on the engineering profession in Idaho.
Young Engineer of the Year Award: To
recognize an engineer that is making a contribution to their profession. Must be
no more than 35 years old.
Self nominations are welcomed and encouraged.
EDUCATION OPPORTUNITIES –
WEB SEMINAR
Register today for the second of six web seminars in the 2006-2007 Top Issues
In Construction Series presented by the PEC of NSPE.
Building Information Modeling—Why Owners and Contractors Want It
November 8, 2006 12:30 – 2:00 EST (Eastern)
Register yourself or your whole office for the same price:
Web seminar cost, per connection: $149 Members, $189 Others
Your registration pays for one
"connection point." If you have a conference speaker on your phone and projector
for your computer, more than one--even a room full, can experience and
participate in each seminar!
Download the registration forms in a
word or
pdf format. Simply fill out the form and fax it back to us.
EDUCATION OPPORTUNITIES -
BOISE
IAQA Boise, Idaho Workshop:
Idaho’s Radon Program and Summary of Idaho Outdoor Fungal Levels
November 17, 2006, 12:00 – 4:00 p.m.
The StoneHouse
709 E. Park Blvd., Boise, ID 83712
Registration Fees: AmIAQ, IAQA, or IESO Members - $40; Non-Members - $50
Limited Seating: RSVP by November 13, 2006 to Summit Environmental (208)
377-2900 or admin@summitenviroinc.com.
JOB OPPORTUNITIES
At CH2M HILL, you can EXPECT IT ALL…
Client Focus, Integrity, Safety, Innovation, Quality, Collaboration, The Best
People, and Excellence! CH2M HILL … our employee-owned culture helped the
firm be 5th among "Most Admired" Engineering and Construction companies and
named as one of Fortune Magazine's "100 Best Places to
Work For - 2006". Join the “best” professionals in the industry to
deliver high-profile transportation projects and help our
clients build a better world.
Career Opportunities in Boise, Portland and Seattle
• Transportation Design Engineers
• Traffic Engineers
• Transportation Planners
• Design Team Leads
• Project Managers
• Bridge Engineers
Careers without limits...We encourage you to visit
our web site to learn more about CH2M HILL and for details of these and
other employment opportunities. To apply, indicate job code and submit your
resume to: Careers@ch2m.com or apply
online at www.ch2m.com. EEO/AA Employer
MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in next
week's edition of the Friday Update!
The Pumpkin Weigh-in
The mean weight of 9 pumpkins is 16 pounds. One of the pumpkins is removed. The
mean weight of the remaining 8 pumpkins is 13 pounds. What is the weight of the
pumpkin that was removed, in pounds?
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The weights of the 8 remaining pumpkins are 11, 12, 14, 18, 15, 14, 8, 12
pounds. What is the median pumpkin weight?
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A different set of 9 pumpkins has been divided into two groups. The mean weight
of the 5 pumpkins that each weighs more than 10 pounds is 12 pounds. The mean
weight of the 4 pumpkins that each weighs less than 10 pounds is 3 pounds. What
is the mean weight per pumpkin for this group of 9 pumpkins?
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The mean weight of 2 pumpkins is M pounds. The weight of the first pumpkin is P
pounds. What is the weight of the second pumpkin in terms of M and P?
Answer to last week’s MATHCOUNTS problem:
Find the radius
of each pumpkin. Use C = 2πr. Solve for r,
r= C/(2π)
The radius of
the smallest pumpkin is 22/(2π) = 11/π.
The radius of
the middle pumpkin is 44/(2π) = 22/π.
The radius of
the largest pumpkin is 66/(2π) = 33/π
Find the volume
of each pumpkin. Use V=4πr3/3.
The volume of
the smallest pumpkin is 4π(11/ π)3/3 = 5324/(3 π2) in3
The volume of
the middle pumpkin is 4π(22/ π)3/3 = 42,592/(3 π2) in3
The volume of
the largest pumpkin is 4π(33/ π)3/3 = 143,748/(3 π2) =
47,916/(π2) in3
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22:44:66 have a common factor of 22. This ratio
simplifies to 1:2:3.
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5324/(3 π2):
42,592/(3 π2): 47,916/(π2) = 5324/(3 π2):
42,592/(3 π2): 143,748/(3π2)
5324:42,592:143,748 = 4×113: 4×223: 4×333 =
4×113×13: 4×113×23: 4×113×33
The ratio of
the volumes from least too greatest is 1:8:27.
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The ratio of
the lengths of the rectangular prism is 8:16:24 = 1:2:3.
The ratio of
the widths of the rectangular prism is 4:8:12 = 1:2:3.
The ratio of
the heights of the rectangular prism is 2:4:6 = 1:2:3.
The volume of
the first rectangular prism is 8×4×2 = 64 in3.
The volume of
the second rectangular prism is 16×8×4 = 512 in3
The volume of
the third rectangular prism is 24×12×6 = 1728 in3
The ratio of
the volumes is 64:512:1728.
64 = (1×8)
(1×4) (1×2) = 13(8 ×4 × 2)
512 = (2×8)
(2×4) (2×2) = 23(8 ×4 × 2)
1728= (3×8)
(3×4) (3×2) = 33(8 ×4 × 2)
The ratio of
the volumes is 1:8:27
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Ichabod’s
conjecture is: The ratio of the volumes of a set of similar solids is the cube
of the ratio of the linear measurements of the corresponding parts of the
similar solids.
If you want to see last week's problem again, click
http://www.mathcounts.org/webarticles/anmviewer.asp?a=919&z=107
Idaho Society of Professional Engineers
PO Box 170239
Boise, ID 83717-0239
208-426-0636
Fax: 208-426-0639
E-Mail: ispe@idahospe.org
Web Site: www.Idahospe.org
Idaho
Society of Professional Engineers
