Idaho
Society of Professional Engineers
PO Box 170239, Boise, ID 83717-0239 208-426-0636 Fax: 208-426-0639 E-Mail: ispe@idahospe.org
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PO Box 170239, Boise, ID 83717-0239 208-426-0636 Fax: 208-426-0639 E-Mail: ispe@idahospe.org |
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Idaho Society of Professional Engineers Friday Update – 02/18/11
What’s Inside:
UPCOMING EVENTS:
DON'T MISS THESE SPECIAL EVENTS CELEBRATING ENGINEERS WEEK 2011!
ISPE SINCERELY APPRECIATES THE SUPPORT OF ALL OF OUR CURRENT 2010 – 2011 SUSTAINING ORGANIZATIONS:
J-U-B Engineers, Inc
Materials Testing & Inspection
Stapley Engineering
Please consider joining these great companies in becoming an ISPE Sustaining Organization. ISPE offers the Sustaining Organization category of membership to enhance the visibility of your commitment to ISPE and the engineering profession. Your membership will allow us to better serve the engineering community through promoting engineering and ethics, and supporting the needs of the engineer including professional development.
If you are interested in becoming a Sustaining Organization, please contact the ISPE office at ispe@idahospe.org
MATHCOUNTS PROBLEM OF THE WEEKCan you solve this MATHCOUNTS problem? The answer will appear in the next edition of the Friday Update!
Happy Valentine's Day!
Jessica is creating a heart-shaped card to give to her valentine. She has folded a square piece of paper in half and traced the pattern onto one half of the folded paper. The pattern consists of a right triangle with hypotenuse of length 7 1/2 inches and a semi-circle with a diameter of 4 1/2 inches. The entire heart will cover 2/7 of the square paper. What is the side length, in inches, of the piece of paper Jessica is using to create the card? Express your answer as a decimal to the nearest tenth. (Use π = 22/7) -------------------------------------------------------------------------------
Jessica plans to decorate the card by painting portions of the front and back of the card as shown. The rest will remain white. What percent of the front of the card will be painted? Express your answer as a percent to the nearest whole number. (Use π = 22/7) ------------------------------------------------------------------------------ What is the fraction of the back of the card that will not be painted? Express your answer as a common fraction. (Use π = 22/7)
Answer to last problem: To exceed her previous score Tyler must score at least 61 points in her second game. The total number of baskets is minimized by maximizing the number of three-point baskets made. Let’s start by determining the fewest number of two-point shots Tyler could make and earn a total of 61 points. If Tyler makes just one two-point shot she would need to earn the remaining 61 – 2 = 59 points by making three-point baskets. But since 59 is not divisible by 3 we know that it is not possible for her to earn the remaining points with three-point baskets. So Tyler must make at least 2 baskets worth two points each accounting for 4 points. In this case she would need to earn an additional 61 – 4 = 57 points by making three-point baskets. Tyler could accomplish this by making 57 ÷ 3 = 19 three-point baskets. Therefore, Tyler must make a minimum of 2 + 19 = 21 baskets to exceed 60 points in her second game. ------------------------------------------------------------------------------- Let X represent the number of three-point baskets Mitchell made. We are told that 20% or 1/5 of the baskets Mitchell made were worth three points. That means that 4/5 of the baskets he made were worth two points each, which is 4 times the number of three-point baskets. Therefore, the number of two-point baskets Mitchell made would be represented by 4X. We are told that Mitchell’s total score was 55 points. Thus, 3X + 2(4X) = 55. Simplifying we have 3X + 8X = 55 11X = 55. Dividing each side by 11 yields X = 5. So Mitchell made 5 three-point baskets and 4(5) = 20 two-point baskets. Next we are told that the 20 two-point baskets Mitchell made account for 4/5 of the total number of two-point baskets attempted. Therefore, Mitchell attempted (5/4) × 20 = 25 two-point shots. In addition, the 5 three-point baskets he made account for 1/3 of the three--point baskets attempted. Therefore, Mitchell attempted (3/1) × 5 = 15 three-point shots. Thus, Mitchell attempted to make a total of 25 + 15 = 40 baskets. ------------------------------------------------------------------------------- The shots made during the final 10 seconds of the game would have been worth three-points each, so we need to determine how many of the 114 shots were three-point baskets. We are told that Kyoka’s total score was 256 points. So, if we let X represent the number of two-point baskets and Y be the number of three-point baskets we have the following two equations: X + Y = 114 and 2X + 3Y = 256. Solving the first equation for X yields X = 114 – Y. We can substitute this expression in for X in the second equation to get 2(114 – Y) + 3Y = 256. Simplifying we have 228 – 2Y + 3Y = 256 228 + Y = 256. After subtracting 228 from each side we have Y = 28. Thus there were 28 shots made during the final 10 seconds of the game. That accounts for 28/114 = .245614 ≈ 25% of the shots. -------------------------------------------------------------------------------
If you want to see the problem again, click http://mathcounts.org/Page.aspx?pid=1573 and choose the problem for January 10, 2011.
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National Engineers Week - Future City Competition
Board of Professional Engineers and Professional Land Surveyors
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