Idaho Society of Professional Engineers
Friday Update - 05/21/04
UPCOMING EVENTS:
● ASCE May Meeting, Thursday, May 27, 2004,
12:00 Noon, MK Plaza - Executive Dining Room, Presenter - Terry Scanlan of
Skellenger Bender, Seattle, WA, Topic - Case Histories: Loss Prevention, for
more information contact Ryan Adelman at
radelman@kellerassociates.com
● NSPE 2004 Convention and Expo, July 8 -
10, 2004, Honolulu, Hawaii
● NSPE Western and Pacific Regional
Meeting, September 17-18, 2004, Coeur d'Alene, ID
REQUEST FOR AN INDEPENDENT ENGINEER
An independent engineer is needed to evaluate consultants for a standby
generator project for the Division of Public Works. This would require
evaluation of proposals to make a short list and a half day to listen to
presentations. For questions or further information, please contact Clif
Squires, Project Manager, Division of Public Works, 208-332-1914, fax
208-334-4031, csquires@adm.state.id.us
2004 MATHCOUNTS NATIONAL CHAMPIONS MEET PRESIDENT BUSH IN THE WHITE HOUSE
ON MAY 18
The 2004 MATHCOUNTS National Champions and their coaches met President Bush at
the White House on May 18. The news release announcing this public recognition
for MATHCOUNTS and the accomplishments of these individuals is available online
at
http://www.mathcounts.org/New/mcwhitehousevisit04.pdf.
MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in next week's
edition of the Friday Update!
Brood X is Here!
For those of us who live in the affected areas, it’s impossible to ignore the
new residents of our neighborhoods…. the cicadas known as Brood X. They are not
locusts, but the 1.5-inch insects definitely come in huge numbers. There can be
an estimated 1.5 million per acre. What is the equivalent measure of cicadas per
square foot if there are 43,560 square feet per acre? Express your answer as a
decimal to the nearest tenth.
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The cicadas intrigue scientists because of their odd life cycle. Though some
cicadas come out annually (the ones with green eyes), the red-eyed ones from
Brood X come up out of the ground every 17 years like clockwork. They then only
live for 2.5 weeks before laying their eggs and dying. Assuming that the life
span of a cicada is 17 years, and 2.5 of those weeks are spent above ground,
what percent of their lives is spent above ground? Express your answer to the
nearest tenth. (There is some speculation in the science community that the
cicadas coming out of the ground may be older than 17 years old.)
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There have also been known to be some cicadas which have gotten off the 17-year
cycle and gone to more of a 13-year cycle. “Whether this is Mother Nature’s way
of playing with prime numbers remains a mystery,” said Phil Nixon, a bug guru at
the University of Illinois-Champaign, according to Dave Orrick’s article in the
Daily Herald. Assuming that a group of cicadas comes out every 13 years, another
group comes out every 17 years, and both groups come out this year, in how many
years will the next occurrence of both groups coming out at the same time take
place?
Answer to last week's problem:
To form an arithmetic sequence, the difference between a and b must be the same
as the difference between b and c. We could start with sides 20, 20 and 20, but
we were told that the triangle is not equilateral. Therefore, let’s subtract 1
unit from the first side and add 1 unit to the third side to get a triangle with
sides 10, 20 and 21. These sides do form an arithmetic sequence and the
perimeter of the triangle is 60 units. We could continue to do this subtract 1 /
add 1 process until we get down to 11, 20 and 29. However, notice that if we
take it one step further to 10, 20 and 30, we suddenly have sides that will not
make a triangle (because the third side is too big for the first two sides).
Therefore, from the set of sides 19, 20, 21 to the set 11, 20, 29, there are a
total of 9 distinct triangles.
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Notice, that each letter from the word PROBLEM must come from a particular one
of the first three words. (In other words, there aren’t two choices of words to
get the P from.) From the word CAMP, we must select the M and P. There are “4
choose 2” = 6 ways to choose two letters from the four letters in CAMP, and only
one of these ways include the letters that we need. Sot he probability of
successfully choosing letters from CAMP is 1/6. From the word HERBS, we must
select the E, R and B. There are “5 choose 4” = 5 ways to choose four letters
from the five letters in HERBS, and 2 of these ways include the three letters
that we need. So the probability of successfully choosing letters from HERBS is
2/5. From the word GLOW, we must select the L and O. There are “4 choose 3” = 4
ways to choose three letters from the four letters in GLOW, and two of these
ways include the two letters that we need. So the probability of successfully
choosing letters from GLOW is 2/4 = 1/2. The probability of performing each of
these three steps successfully is 1/6 × 2/5 × 1/2 = 1/30.
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Let’s let x represent the number of $10 T-shirts they actually sold. This means
that 661 – x represents the number of $12 T-shirts they actually sold. We now
know that 12x + 10(661 – x) = 10x + 12(661 – x) + 378. Simplifying both sides we
have 12x – 10x + 6610 = 10x – 12x + 7932 + 378 which further simplifies to 4x =
1700. Dividing both sides by 4 yields x = 425. This means that they actually
sold 425 T-shirts at the $10 price (and 236 T-shirts at the $12 price).
If you want to see last week's problem again, click on
http://www.mathcounts.org/Queries/POW_Archive.taf?_function=detail&Q_A_uid1=494&_UserReference=C8EC7EDF090AE7DE40AB9B5E
Idaho Society of Professional Engineers
PO Box 170239
Boise, ID 83717-0239
208-426-0636
Fax: 208-426-0639
E-Mail: ispe@rmci.net
Web Site: www.Idahospe.org