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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
 

 

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