VOLUNTEERS NEEDED!
Volunteers from all engineering disciplines are needed to interview Boise State Students for the 2007 Outstanding Engineering Student Awards. Volunteers will make the selection of the outstanding student in each of the three disciplines (CE, ME, and EE) at Boise State. The time commitment will be one evening for approximately 3 hours at the end of January or early February. The interviews will take place at Boise State University. All student nominees will be honored at the Engineers Week Banquet on February 23. If you would like to volunteer please e-mail Heather Carroll at hcarroll@dohertyeng.com.

PE REVIEW COURSE FOR CIVIL & MECHANICAL ENGINEERING
This review is to prepare its participants for the PE exam to be held on Friday, April 20th.

Tuesday evenings from 6pm to 9pm.
Dates: January 30th through April 17th.

Location: University of Idaho Boise Center at Broadway and Front in Boise
322 E. Front Street, Boise

To register please contact: Paula Peterman, (208) 364-6188, paulap@uidaho.edu.

Cost: $600

To purchase the required texts go to www.engboi@uidaho.edu, follow the Click Professional Development, click PE Review, choose which discipline and under “textbooks” follow the link to PPI.



MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in next week's edition of the Friday Update!

Probability Teasers
Ramon has exactly 26 cents in change in his pocket. What is the probability he has an odd number of coins in his pocket? Express your answer as a common fraction.
________________________________________
A football stadium has 7 gates for fans to enter or exit. The gates are numbered consecutively from 1 through 7. The probability of entering or exiting through each gate is equal. What is the probability Sylvia enters the stadium through an even numbered gate and leaves the stadium through an odd numbered gate? Express your answer as a common fraction.
________________________________________
ABCD represents Daniel’s 4-digit school locker number. The locker number is the square of a prime number. What is the probability the locker number is such that A < B < C < D? Express your answer as a common fraction.

Answer to last week’s MATHCOUNTS problem:
Let p = the perimeter of each decoration.  Each side of the equilateral triangle can be represented as p/3 = 8 feet.  The altitude of an equilateral triangle drawn from a vertex to the opposite side bisects the opposite side and creates two 30-60-90 right triangles.  The short leg of each triangle is one-half the length of the side of the equilateral triangle or 4 feet.  The long leg of a

30-60-90 right triangle is the length of the short leg times the square root of 3 or 4√(3).  The area of the equilateral triangle is (base × height) ÷ 2, which is

(8 × 4√(3)) ÷ 2 = 16√(3) square feet.  Each side of the regular hexagon can be represented as p/6 = 4 feet.  The area of a regular hexagon is equal to the area of 6 equilateral triangles with side length of 4 feet.  The height of each equilateral triangle is 2√(3).   The area of the regular hexagon is

 6 × ((4 × 2√(3)) ÷ 2) = 24√(3) square feet.  The ratio of the area of the triangle to the area of the hexagon is 16√(3) : 24√(3) = 2:3.  The ratio expressed as a common fraction is 2/3.

For a right circular cone, the lateral area was given as π times the radius (r ) times the slant height (l ) of  the cone or LA = π × r × l.  Use the Pythagorean Theorem to find the slant height of the cone in December, 1990.  8² + 2² = l².  Solving for l, l = 2√(17).  The lateral area of the tree in December 1990 was

π × 2 × 2√(17) = 51.81 square feet. Each year, the tree grows 10% of the height and radius of the previous year for 16 years.  This is modeled by

1.1¹⁶ = 4.59.  This means all of the linear measurements grow by this factor, including the radius and slant height.  The lateral area of the tree in December, 2006 is then

π × (1.1¹⁶)(2) × (1.1¹⁶ )(2√(17)) = [(1.1¹⁶ ) (1.1¹⁶ ) × π × 2 × 2√(17) ]  or (1.1³²) times what the lateral area was in December 1990.  The surface area of the tree is (1.1³²) × 51.81 = 1093.9 or 1090 square feet, to the nearest 10.  1090 light bulbs are needed to decorate the tree.

The volume of the smallest box is length × width × height = 4 × 2 × 1 = 8.  The dimensions of the next largest box are: 1.2 × 4, 1.2 × 2, and 1.2 × 1.  The volume of this second- largest box is then (1.2 × 4) × (1.2 × 2) × (1.2 × 1) or 1.2³ × (4 × 2 × 1) =13.82.  The volume of each larger box increases by a factor of (1.2³) over the volume of the previous box.  This will occur four times going from the first box to the fifth box.  The volume of the largest box is (1.2³)⁴ times the volume of the smallest box.  Expressed to the nearest tenth, the volume of the largest box is 8.9 times the volume of the smallest box.