CSUF Math 110

Instant Run-Off

candidate w/ fewest votes is eliminated until one candidate has over 50% votes

Borda Count

Each vote is ranked and added together to declare a winner

head-to-head

each answer is compared head to head, and the one with the most wins takes the trophy

standard divisor

total population/total number of seats

standard quota

states population/standard divisor

geometric mean

square root of the sum of the lower quota times the upper quota

Hamiltons method

based on lower quota

hill-huntington method

use geometric mean; if the standard quota is GREATER OR EQUAL, you round up, but ifc the standard is LESS you round down

alabama paradox

a new seat is added but one population (A) loses a seat to another population (B) despite no change in total population in those places

new states paradox

by adding a new state (A), a different state (B) will lose a seat to a third state (C) despite the fact that no populations changed

population paradox

one state (A) loses a seat to another (B) even though A is growing faster than B

simple interest

I = Prt

future value

FV = P(1+rt)

average daily balance

sum of balance x days/ total days

compound interest

FV = P(1+(r/n))^nt

calculating amortized loans

(pymt) ((1+(r/n)^nt - 1)/(r/n)) = P(1+(r/n))^nt

Permutations

order matters; n!/(n-r)!

Combinations

order doesnt matter; n!/r!(n-r)!

expected value

probability weighted average; v1(PV1) + v2(PV2)...

triangle area

0.5(b)(h)

rectangle&parellelogram area

b x h

trapezoid area

0.5(b1+b2)xh

triangle with only sides labelled

s= 0.5(a)(b)(c) --> square root of s(s-a)(s-b)(s-c)

circle

A = pi r^2; C = 2 pi r

cylinder

pi r^2 x h

sphere

4/3 pi r^3

pyramid/cone

1/3 A(base) x h

euler circuit

all vertices have even degrees & each edge is visited ONCE

adjacent VERTICES

two vertices joined by one edge

adjacent EDGES

two edges with a shared vertex

hamiltonian path

visit each vertex once

nearest neighbor

go to the closest from one starting point regardless of other options

cheapest edge

write it all out, and find the path with the least amount traveled