MATH 310 FALL 2003 Combinatorial Problem Solving Lecture
MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 16, Monday, October 6
5. 1. Two Basic Counting Principles. n Homework (MATH 310#6 M): • Read 5. 2 • Do 5. 1: All odd numbered exercises. • Turn in 5. 1: 8, 16, 18, 32 • Volunteers: • ____________ • Problem: 18.
Test 1 - Statistics n 15 14 13 0 0 2 2 5 12 5 5 6 11 2 10 9 6 7 8 7 7 n Here is a stem-andleaf report on the test. Median: 126 There are two marks: • • raw mark x adjusted mark y 0 · y · 100. The adjusted mark is computed as follows: • y = d 100 x/135 e
Test 1 – Problem #1 1 2 3 4 5 6 7 0 x x x x -1 x -2 x -3 x -4 x -5 -6 -7 -8 -9 x -10 Median – 0
Test 1 – Problem #2 1 2 3 4 0 -1 -2 x x -3 x -4 x x x -5 x -6 x x -7 -8 -9 x -10 Median – 4
Test 1 – Problem #3 1 2 3 4 0 x x -1 x x x -2 x -3 x x x -4 -5 -6 x -7 -8 -9 -10 Median – 1
Test 1 – Problem #4 1 2 3 4 0 x -1 -2 -3 x x -4 -5 x -6 x x x -7 x x -8 -9 -10 y Median – 5
Test 1 – Problem #5 1 2 3 0 x -1 x x x -2 x x x -3 x -4 x -5 -6 -7 x -8 x -9 -10 x Median – 2
Test 1 – Problem #6 1 2 3 4 5 0 x x x -1 -2 -3 x -4 -5 x x x -6 x -7 -8 x -9 -10 x Median – 3
Test 1 – Problem #7 1 2 3 0 x -1 -2 x x -3 x x x -4 x -5 x -6 x x -7 x x -8 -9 -10 Median – 3
Test 1 – Problem #8 1 2 3 4 5 6 0 x x x -1 -2 x x -3 x -4 x -5 -6 x -7 -8 -9 -10 Median – 0
Test 1 – Problem #9 1 2 3 0 -1 -2 -3 x x x -4 x x -5 x -6 x x -7 x x x -8 -9 -10 x Median – 5
Test 1 – Problem #10 1 2 3 4 5 6 7 0 x x x x -1 x x -2 x -3 -4 -5 -6 x -7 -8 -9 -10 y Median – 0
- Slides: 13