L is a non-empty set.
The carrier of L is a non-empty set.
S is M2(the carrier of L).
p is M2(the carrier of L).
(L, S, p) is M2(the carrier of L).
The of L is V6() V18([: the carrier of L, the carrier of L:], the carrier of L) M2(bool [:[: the carrier of L, the carrier of L:], the carrier of L:]).
[: the carrier of L, the carrier of L:] is a non-empty set.
[:[: the carrier of L, the carrier of L:], the carrier of L:] is a non-empty set.
bool [:[: the carrier of L, the carrier of L:], the carrier of L:] is a non-empty set.
The of L . (S, p) is M2(the carrier of L).
The of L . (p, S) is M2(the carrier of L).
The of L . (S, S) is M2(the carrier of L).
The of L . (p, p) is M2(the carrier of L).
The of L . (x, S) is M2(the carrier of L).
The of L . (x, p) is M2(the carrier of L).
The of L . (S, x) is M2(the carrier of L).
The of L . (p, x) is M2(the carrier of L).
(L, S, x) is M2(the carrier of L).
(L, p, x) is M2(the carrier of L).
(L, x, S) is M2(the carrier of L).
(L, x, p) is M2(the carrier of L).
The of L . ((L, S, p), x) is M2(the carrier of L).
The of L . ((L, S, p), S) is M2(the carrier of L).
The of L . ((L, p, x), S) is M2(the carrier of L).
The of L . ((L, p, x), p) is M2(the carrier of L).
The of L . ((L, x, S), p) is M2(the carrier of L).
The of L . ((L, x, S), S) is M2(the carrier of L).
The of L . ((L, x, p), S) is M2(the carrier of L).
The of L . ((L, x, p), p) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, p, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, p, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, p, x)) is M2(the carrier of L).
The of L . ((L, S, x), (L, p, x)) is M2(the carrier of L).
The of L . ((L, p, x), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, p, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, p), (L, x, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, S), (L, S, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, x)) is M2(the carrier of L).
The of L . ((L, x, p), (L, S, p)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, S)) is M2(the carrier of L).
The of L . ((L, S, x), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, S), (L, x, p)) is M2(the carrier of L).
The of L . ((L, x, p), (L, x, S)) is M2(the carrier of L).
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**1: Integers**
1.1: Integers and Absolute Value - Comparing and ordering integers
1.2: Compare and Order Integers - Understanding integer comparisons
1.3: Add Integers - Adding integers with chips
1.4: Subtract Integers - Subtracting integers with chips
1.10: Order of Operations - Following the order of operations
1.11: The Coordinate Plane - Understanding coordinates

**2: Expressions and Equations**
2.1: Mathematical Expressions - Using algebraic expressions
2.3: Equations - Using algebraic equations
2.4: Solve Addition Equations - Modeling one-step equations
2.5: Solve Subtraction Equations - Modeling one-step equations
2.6: Solve Multiplication Equations - Modeling one-step equations
2.7: Solve Division Equations - Modeling one-step equations
2.8: Solve Two-Step Equations - Modeling and solving two-step equations

**3: Inequalities**
3.2: Graph Inequalities on a Number Line - Understanding compound inequalities
3.4: Solve Inequalities Using Addition and Subtraction - Solving linear inequalities
3.5: Solve Inequalities Using Multiplication - Solving linear inequalities
3.6: Solve Inequalities Using Division - Solving linear inequalities

**4: Rational Numbers: Decimals**
4.2: Equivalent Rational Numbers - Understanding equivalent decimals
4.3: Compare and Order Decimals - Comparing and ordering decimals
4.5: Add and Subtract Decimals - Adding and subtracting decimals
4.6: Multiply Decimals - Modeling decimal multiplication
4.12: Addition and Subtraction Equations with Decimals - Solving decimal equations

**5: Rational Numbers: Fractions**
5.1: Prime Factorization - Factoring numbers
5.5: Compare and Order Rational Numbers - Comparing and ordering fractions
5.6: Add and Subtract Fractions - Adding and subtracting fractions
5.8: Multiply Fractions - Multiplying fractions
5.9: Multiply Mixed Numbers - Multiplying mixed numbers
5.10: Divide Fractions - Dividing fractions
5.11: Divide Mixed Numbers - Dividing mixed numbers

**6: Ratio and Proportion**
6.1: Ratio - Understanding part-to-part and part-to-whole ratios
6.3: Write and Solve Proportions - Writing and solving proportions
6.4: Direct Proportions - Understanding direct variation
6.9: Inverse Proportion - Understanding inverse variation

**7: Percent and Consumer Applications**
7.1: Percents - Understanding percents and proportions
7.2: Fractions, Decimals, Percents - Converting between fractions, decimals, and percents
7.3: Percents Greater Than 100%/Less Than 1% - Ordering percents, fractions, and decimals
7.4: Finding a Percentage of a Number - Finding percentages of numbers
7.5: Find a Percent - Finding percents
7.8: Percent Increase - Calculating percent increase
7.9: Percent Decrease - Calculating percent decrease
7.10: Sales Tax and Tips - Calculating sales tax and tips
7.11: Discount and Markup - Calculating discounts and markups
7.12: Commission - Calculating commission
7.13: Simple Interest - Calculating simple interest
7.14: Compound Interest - Calculating compound interest

**8: Data Analysis and Statistics**
8.1: Samples and Surveys - Understanding populations and samples
8.2: Measures of Central Tendency and Range - Calculating mean, median, and mode
8.4: Choose an Appropriate Graph - Choosing appropriate graphs
8.6: Histograms - Creating histograms
8.7: Stem-and-Leaf Plots - Creating stem-and-leaf plots
8.8: Box-and-Whisker Plots - Creating box-and-whisker plots
8.11: Scatter Plots - Creating scatter plots
8.13: Create Graphs - Creating graphs

**9: Two-Dimensional Geometry**
9.1: Points, Lines, and Planes - Understanding points, lines, and planes
9.3: Angle Pairs - Understanding angle theorems
9.5: Congruent Angles and Line Segments - Constructing congruent segments and angles
9.6: Line Constructions - Constructing parallel and perpendicular lines
9.13: Make a Circle Graph - Creating circle graphs

**10: Two-Dimensional Geometry and Measurement Applications**
10.2: Perimeter - Calculating perimeter
10.3: Squares and Square Roots - Understanding square roots
10.5: Pythagorean Theorem - Applying the Pythagorean theorem
10.6: Area of Parallelograms - Calculating area of parallelograms
10.7: Area of Triangles and Trapezoids - Calculating area of triangles and trapezoids
10.8: Circumference and Area of a Circle - Calculating circumference and area of circles
10.10: Symmetry - Understanding symmetry
10.12: Relate Perimeter and Area - Relating perimeter and area

**11: Three-Dimensional Geometry**
11.3: Surface Area of Prisms - Calculating surface area of prisms
11.4: Surface Area of Pyramids - Calculating surface area of pyramids
11.5: Surface Area of Cylinders and Cones - Calculating surface area of cylinders and cones
11.7: Volume of Prisms - Calculating volume of prisms
11.8: Volume of Pyramids - Calculating volume of pyramids
11.9: Volume of Cylinders and Cones - Calculating volume of cylinders and cones

**12: Probability**
12.3: Theoretical Probability - Understanding theoretical probability
12.4: Experimental Probability - Understanding experimental probability
12.6: Compound Events - Understanding compound events
12.7: Permutations - Understanding permutations
12.8: Combinations - Understanding combinations

**13: Patterns, Relations, and Functions**
13.4: Relations and Functions - Understanding relations and functions
13.5: Functions - Understanding functions
13.6: Graph Linear Functions - Graphing linear functions
13.7: Slope - Understanding slope
13.10: Graph Translations and Reflections - Understanding graph translations and reflections
13.11: Graph Rotations - Understanding graph rotations
13.12: Graph Dilations - Understanding graph dilations

**14: Polynomials, Equations, and Inequalities**
14.3: Add Polynomials - Adding polynomials


## Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.5 Answer Key Making Inferences About Populations

### Math in Focus Grade 7 Chapter 9 Lesson 9.5 Guided Practice Answer Key

**Solve.**

Question 1.

A random sample of ages {15, 5, 8, 7, 18, 6, 15, 17, 6, 15} of 10 children was collected from a population of 100 children.

a) Calculate the sample mean age of the children and use it to estimate the population mean age.

Answer:

Mean = (15 + 5 + 8 + 7 + 18 + 6 + 15 + 17 + 6 + 15) / 10 = 112 / 10 = 11.2 years.

b) Calculate the MAD of the sample.

Answer:

MAD = (|15-11.2| + |5-11.2| + |8-11.2| + |7-11.2| + |18-11.2| + |6-11.2| + |15-11.2| + |17-11.2| + |6-11.2| + |15-11.2|) / 10
= (3.8 + 6.2 + 3.2 + 4.2 + 6.8 + 5.2 + 3.8 + 5.8 + 5.2 + 3.8) / 10
= 48 / 10
= 4.8 years

c) Calculate the MAD to mean ratio.

Answer:

MAD to mean ratio = (4.8 / 11.2) * 100% ≈ 42.86% ≈ 43%

d) Draw a dot plot for the ages and the mean age.

Answer:

A dot plot visually groups the number of data points in a data set based on the value of each point.

e) Using the MAD to mean ratio and the dot plot, describe informally how varied the population ages are.

Answer:

Since the MAD to mean ratio is about 43%, the data vary from the mean significantly. The dot plot confirms that the data are spread far away from the mean.

**Hands-On Activity**

Materials: a novel with 200 to 300 pages

USE STATISTICS FROM A SAMPLE TO DESCRIBE THE VARIABILITY OF A POPULATION

Work in pairs.

Background: The population is made up of all the words in a book. The length of a word is defined by the number of letters in it. The population characteristic is the mean word length.

STEP 1: Determine which sampling method you intend to use.

STEP 2: Generate five random samples. Each sample consists of 20 words.

STEP 3: For the 20 words in each sample, record each word and its length in a table. Then calculate the mean and the mean absolute deviation for each sample.

STEP 4: When you have completed STEP 1 to STEP 3 for all 20 samples, calculate the mean of the sample means.

**Math journal** What is the estimate of the population mean word length? By observing the mean absolute deviations of the five samples, describe informally whether the words in the book vary greatly in length.

Answer:

**Solve.**

Question 2.

The ages of two groups of children are summarized in the box plots.

a) Show that the two groups have the same measure of variation (that is, the difference between quartiles) and the same interquartile range.

Answer:

Group 1 data: lowest number = 7, Q1 = 8, Q2 = 10, Q3 = 11, Highest number = 13
Group 2 data: lowest number = 7, Q1 = 10, Q2 = 12, Q3 = 13, Highest number = 14

Q2 - Q1 for Group 1 = 10 - 8 = 2
Q3 - Q2 for Group 1 = 11 - 10 = 1
Interquartile range for Group 1 = Q3 - Q1 = 11 - 8 = 3

Q2 - Q1 for Group 2 = 12 - 10 = 2
Q3 - Q2 for Group 2 = 13 - 12 = 1
Interquartile range for Group 2 = Q3 - Q1 = 13 - 10 = 3

b) Express the difference in median age in terms of the interquartile range.

Answer:

Difference in median age = Q2 of Group 2 - Q2 of Group 1 = 12 - 10 = 2
Interquartile range = 3
X = 2 / 3 ≈ 0.67 years

c) What inference can you draw about the age distributions of the children in the two groups?

Answer:

Since Q1 of Group 2 is the same as Q2 of Group 1, it means that 75% of children in Group 2 are older than 50% of children in Group 1. Also, the median of Group 2 is greater than the median of Group 1. So, the children in Group 2 are generally older than Group 1.

Question 3.

Two classes took a math test. The results are summarized in the box plots.

a) For which class are the data more spread out? Explain.

Answer:

Class A data set: Lowest number = 24, Q1 = 46, Q2 = 66, Q3 = 70, Highest number = 95
Class B data set: Lowest number = 22, Q1 = 40, Q2 = 60, Q3 = 80, Highest number = 87

IQR of Class A = Q3 - Q1 = 70 - 46 = 24
IQR of Class B = Q3 - Q1 = 80 - 40 = 40

Therefore, Class B data are more spread out than Class A because the interquartile range of Class B is greater.

b) Comment on the performance of the two classes.

Answer:

Median value of Class A = 66
Median value of Class B = 60

Thus, Class A performed moderately better than Class B because the median of Class A is slightly higher at 66 and Class B is 60.

Question 4.

The numbers of questions that two groups of students answered correctly in a recent mathematics test are shown in the table.

a) Find the mean number of correct questions for each of the two groups.

Answer:

Mean of Group A = (5 + 15 + 20 + 14 + 6 + 19 + 10 + 10 + 7 + 14) / 10 = 120 / 10 = 12
Mean of Group B = (11 + 13 + 10 + 14 + 18 + 15 + 10 + 11 + 6 + 12) / 10 = 120 / 10 = 12

b) Calculate the mean absolute deviation of each of the two groups.

Answer:

MAD of Group A = (|5-12| + |15-12| + |20-12| + |14-12| + |6-12| + |19-12| + |10-12| + |10-12| + |7-12| + |14-12|) / 10
= (7 + 3 + 8 + 2 + 6 + 7 + 2 + 2 + 5 + 2) / 10
= 44 / 10
= 4.4

MAD of Group B = (|11-12| + |13-12| + |10-12| + |14-12| + |18-12| + |15-12| + |10-12| + |11-12| + |6-12| + |12-12|) / 10
= (1 + 1 + 2 + 2 + 6 + 3 + 2 + 1 + 6 + 0) / 10
= 24 / 10
= 2.4

c) Draw separate dot plots for the two groups.

Answer:

A dot plot visually groups the number of data points in a data set based on the value of each point.

d) Interpret the above statistics and the dot plots.

Answer:

From the two dot plots, Group A's data are more scattered away from the mean, and Group B's data cluster more closely around the mean. This explains why the MAD of Group A is greater than that of Group B.

e) What conclusion can you make about the performance of the two groups of students on the test?

Answer:

Group A's performance is more varied, whereas Group B's performance is more uniform.

### Math in Focus Course 2B Practice 9.5 Answer Key

**Solve. Show your work.**

Question 1.

A random sample of a certain type of ball bearing produces a mean weight of 28 grams and a mean absolute deviation of 2.1 grams. What can be inferred from the sample about the population mean weight of this type of ball bearing?

Answer:

The estimated population mean weight is 28 grams with a MAD to mean ratio of (2.1 / 28) * 100% ≈ 7.5%

Question 2.

You interviewed a random sample of 25 marathon runners and compiled the following statistics.

Mean time to complete the race = 220 minutes and MAD = 50 minutes What can you infer about the time to complete the race among the population of runners represented by your sample?

Answer:

The time to complete the race among the population of runners represented by your sample is 220 minutes with a MAD to mean ratio of (50 / 220) * 100% ≈ 22.73%

**Refer to the random sample below to answer the following. Round your answers to the nearest tenth.**

The table shows a random sample of the volume, in milliliters, of 15 servings of orange juice from a vending machine.

Question 3.

Use the sample mean volume to estimate the population mean volume of a serving of orange juice.

Answer:

Mean = (251 + 254 + 254 + 249 + 250 + 248 + 250 + 252 + 251 + 253 + 247 + 245 + 255 + 254 + 251) / 15 = 3764 / 15 ≈ 250.93 ml

Question 4.

Calculate the MAD and the MAD to mean ratio.

Answer:

MAD = (|251-250.93| + |254-250.93| + |254-250.93| + |249-250.93| + |250-250.93| + |248-250.93| + |250-250.93| + |252-250.93| + |251-250.93| + |253-250.93| + |247-250.93| + |245-250.93| + |255-250.93| + |254-250.93| + |251-250.93|) / 15
= (0.07 + 3.07 + 3.07 + 1.07 + 0.93 + 2.93 + 0.93 + 1.07 + 0.07 + 2.07 + 3.93 + 5.93 + 4.07 + 3.07 + 0.07) / 15
= 33.3 / 15
= 2.22

MAD to mean ratio = (2.22 / 250.93) * 100% ≈ 0.88%

Question 5.

Use the MAD to informally infer whether the volume of orange juice in a serving varies greatly.

Answer:

The volume of orange juice per serving does not vary significantly, as the MAD to mean ratio is approximately 0.88%.

**Refer to the box plots at the right to answer the following.**

Two large random samples of car speeds on two highways from 4 P.M. to 6 P.M. were collected. The data were summarized in two box plots.

Question 6.

By comparing the interquartile ranges of the two box plots, what can you infer about car speeds on the two highways for the middle 50% of the cars?

Answer:

Highway A data: Lower value = 57, Q1 = 60, Q2 = 65, Q3 = 66, Highest value = 72
Highway B data: Lower value = 54, Q1 = 57, Q2 = 62, Q3 = 66, Highest value = 70

IQR of Highway A = Q3 - Q1 = 66 - 60 = 6
IQR of Highway B = Q3 - Q1 = 66 - 57 = 9

By comparing the interquartile ranges of the two box plots, the car speeds on Highway B are more varied for the middle 50% of the cars.

Question 7.

By comparing the medians of the two plots, what inference can you make?

Answer:

Median of Highway A = 65
Median of Highway B = 62

The median car speed on Highway A is higher than that of Highway B.

Question 8.

Suppose the speed limit on the two highways is 65 miles per hour. What percentage of the cars drove faster than the speed limit on the two highways?

Answer:

Since the data is not provided, we cannot calculate the exact percentage.

Question 9.

What can you infer about the overall car speeds on the two highways?

Answer:

Since the data is not provided, we cannot make an inference about the overall car speeds on the two highways.

**Refer to the scenario below to answer the following.**

Henry and Carl go jogging every morning for several weeks. The following are the mean and the MAD of the distances they jog.

Question 10.

What would you infer about the distances they jog if you only compare their mean distances?

Answer:

Mean distance of Henry = 5.25 km
Mean distance of Carl = 4.20 km

Henry jogs more than Carl by 1.05 km.

Question 11.

What can you conclude if you take into account both the mean and the MAD for your comparison?

Answer:

MAD distance of Henry = 2 km
MAD distance of Carl = 0.75 km

By comparing all those distances, we can say Henry's jogging distance is more varied than Carl's jogging distance.

**Refer to the random samples below to answer the following. Round your answers to the nearest tenth.**

3 random samples of time taken, in seconds, to solve a crossword puzzle during a competition were collected.

S1 = {100, 87, 95, 103, 110, 90, 84, 88}
S2 = {75, 98, 120, 106, 70, 79, 100, 90}
S3 = {60, 68, 110, 88, 78, 90, 104, 73}

Question 12.

Calculate the mean time of each of the 3 samples.

Answer:

Mean of S1 = (100 + 87 + 95 + 103 + 110 + 90 + 84 + 88) / 8 = 757 / 8 = 94.625
Mean of S2 = (75 + 98 + 120 + 106 + 70 + 79 + 100 + 90) / 8 = 738 / 8 = 92.25
Mean of S3 = (60 + 68 + 110 + 88 + 78 + 90 + 104 + 73) / 8 = 671 / 8 = 83.875

Question 13.

Estimate the population mean time of the competition using the mean of the 3 sample means.

Answer:

Mean time = (94.625 + 92.25 + 83.875) / 3 = 270.75 / 3 = 90.25 ≈ 90.3 seconds

Question 14.

Combine S1, S2, and S3 into one sample and use the mean in question 13 to calculate the MAD of the combined sample.

Answer:

MAD = (|100-90.3| + |87-90.3| + |95-90.3| + |103-90.3| + |110-90.3| + |90-90.3| + |84-90.3| + |88-90.3| + |75-90.3| + |98-90.3| + |120-90.3| + |106-90.3| + |70-90.3| + |79-90.3| + |100-90.3| + |90-90.3| + |60-90.3| + |68-90.3| + |110-90.3| + |88-90.3| + |78-90.3| + |90-90.3| + |104-90.3| + |73-90.3|) / 24
= (9.7 + 3.3 + 4.7 + 12.7 + 19.7 + 0.3 + 6.3 + 2.3 + 15.3 + 7.7 + 29.7 + 15.7 + 20.3 + 11.3 + 9.7 + 0.3 + 30.3 + 22.3 + 19.7 + 2.3 + 12.3 + 0.3 + 13.7 + 17.3) / 24
= 262.8 / 24
= 10.95 ≈ 11.0

However, the correct calculation for MAD is:
MAD = (|100-90.3| + |87-90.3| + |95-90.3| + |103-90.3| + |110-90.3| + |90-90.3| + |84-90.3| + |88-90.3| + |75-90.3| + |98-90.3| + |120-90.3| + |106-90.3| + |70-90.3| + |79-90.3| + |100-90.3| + |90-90.3| + |60-90.3| + |68-90.3| + |110-90.3| + |88-90.3| + |78-90.3| + |90-90.3| + |104-90.3| + |73-90.3|) / 24
= (9.7 + 3.3 + 4.7 + 12.7 + 19.7 + 0.3 + 6.3 + 2.3 + 15.3 + 7.7 + 29.7 + 15.7 + 20.3 + 11.3 + 9.7 + 0.3 + 30.3 + 22.3 + 19.7 + 2.3 + 12.3 + 0.3 + 13.7 + 17.3) / 24
= 262.8 / 24
= 10.95

MAD = 4.23 (calculated using the correct formula)

Question 15.

If you use the MAD found in question 14 to gauge the time variation among competitors, what can you infer?

Answer:

On average, time variation is from the mean is about (4.23 / 90.3) * 100% ≈ 4.68% of the value of the mean time.

**Refer to the random samples below to answer the following. Round your answers to the nearest hundredth when you can.**

Below are the history test scores of two classes of 20 students each.

Question 16.

Find the range of scores for each class.

Answer:

Range of Class A = Maximum value - Minimum value = 90 - 25 = 65
Range of Class B = Maximum value - Minimum value = 80 - 44 = 36

Question 17.

Calculate the mean scores for each class.

Answer:

Mean of Class A = (84 + 63 + 90 + 68 + 42 + 43 + 31 + 60 + 88 + 70 + 25 + 40 + 32 + 37 + 79 + 66 + 55 + 65 + 35 + 42) / 20 = 1115 / 20 = 55.75
Mean of Class B = (63 + 66 + 62 + 66 + 80 + 55 + 72 + 77 + 66 + 58 + 66 + 68 + 44 + 60 + 70 + 66 + 76 + 75 + 71 + 74) / 20 = 1335 / 20 = 66.75

Question 18.

Calculate the MAD for each class.

Answer:

MAD of Class A = (|84-55.75| + |63-55.75| + |90-55.75| + |68-55.75| + |42-55.75| + |43-55.75| + |31-55.75| + |60-55.75| + |88-55.75| + |70-55.75| + |25-55.75| + |40-55.75| + |32-55.75| + |37-55.75| + |79-55.75| + |66-55.75| + |55-55.75| + |65-55.75| + |35-55.75| + |42-55.75|) / 20
= (28.25 + 7.25 + 34.25 + 12.25 + 13.75 + 12.75 + 24.75 + 4.25 + 32.25 + 14.25 + 30.75 + 15.75 + 23.75 + 18.75 + 23.25 + 10.25 + 0.75 + 9.25 + 20.75 + 13.75) / 20
= 351 / 20
= 17.55

MAD of Class B = (|63-66.75| + |66-66.75| + |62-66.75| + |66-66.75| + |80-66.75| + |55-66.75| + |72-66.75| + |77-66.75| + |66-66.75| + |58-66.75| + |66-66.75| + |68-66.75| + |44-66.75| + |60-66.75| + |70-66.75| + |66-66.75| + |76-66.75| + |75-66.75| + |71-66.75| + |74-66.75|) / 20
= (3.75 + 0.75 + 4.75 + 0.75 + 13.25 + 11.75 + 5.25 + 10.25 + 0.75 + 8.75 + 0.75 + 1.25 + 22.75 + 6.75 + 3.25 + 0.75 + 9.25 + 8.25 + 4.25 + 7.25) / 20
= 115 / 20
= 5.75

Question 19.

By comparing the mean scores and the MADs of the two classes, what can you infer about the performance of the two classes?

Answer:

The mean score of Class B is greater than the mean score of Class A. In other words, Class B's performance on average is better than Class A's performance. In addition, the MAD shows that Class B's scores are less varied than Class A's scores.

**Brain @ Work**

The bar graph displays monthly rainfall, in milliliters, from April to August.

Alex draws the conclusion that rainfall varies greatly from April to August.

a) Find the mean rainfall.

Answer:

Mean = (165 + 170 + 166 + 160 + 169) / 5 = 830 / 5 = 166 ml

b) Calculate the mean absolute deviation.

Answer:

MAD = (|165-166| + |170-166| + |166-166| + |160-166| + |169-166|) / 5
= (1 + 4 + 0 + 6 + 3) / 5
= 14 / 5
= 2.8 ml

c) Based on your mean absolute deviation, do you agree with Alex's conclusion? Explain why you agree or disagree.

Answer:

No, I disagree with Alex's conclusion. The MAD to mean ratio is (2.8 / 166) * 100% ≈ 1.69%. This indicates that the average deviation from the mean is very insignificant.


I am a student new to MPO/MPS and iTensor, currently learning how to generate MPO via autoMPO. To get the MPO expression for a single-site Sz operator, I used the following code:

```cpp
int N = 8;
auto sites = SpinHalf(N);
auto ampo_Ham = AutoMPO(sites);
ampo_Ham += "Sz",4;
auto Ham = toMPO<ITensor>(ampo_Ham);
```

Upon printing Ham to examine its properties, I found that the physical legs have a dimension of 2, as expected for spin-half systems. However, all virtual links are also of dimension 2. The tensor information at site 4 is:

```
ITensor r=4: ("Hl3",2,Link|559) ("Hl4",2,Link|993) ("S=1/2 4",2,Site|270) ("S=1/2 4",2,Site|270)'
{norm=2.12 (Dense Real)}
```

For other sites, such as site 6:

```
ITensor r=4: ("Hl5",2,Link|248) ("Hl6",2,Link|379) ("S=1/2 6",2,Site|827) ("S=1/2 6",2,Site|827)'
{norm=2.00 (Dense Real)}
```

According to the Appendix of the article New. J. Phys. 12, 055026 (2010), for MPOs describing products of local operators, the tensors at each site should have dimension-1 virtual links, with tensor elements at site i given by W^{t*i, s*i} =⟨t*i|O^i|s*i⟩. I would like to know why and how iTensor produces 2-dimensional virtual links instead.


# Go Forth and Generalise

Mathematical problems often require identifying a pattern and providing a general rule. However, a crucial step is explaining why the pattern works and why the general rule applies to the original situation.

Consider the problem "1 Step 2 Step": Liam's house has a staircase with 12 steps, and he can go down one or two steps at a time. The solution involves finding a sequence: 1, 2, 3, 5, 8, 13, which is the Fibonacci sequence. But how do we know this pattern will continue, and it's reasonable to extend it to the 12th number?

To be sure, we must explain the mathematics behind the pattern. In this case, the fact that we add the last two numbers in the sequence to get the next number reflects the situation. If we cannot make this connection, we cannot be sure the general rule always works.

Another example involves a pattern of six squares made from 19 matches. To find the number of matches needed for a pattern 20, 100, or N squares long, we can start by drawing the pattern and using calculations that reflect the construction. Two possible approaches lead to the same general rule: 3N + 1.

Although the approaches differ, the general rule is the same. This highlights the importance of simplification and equivalence of methods. We can use the general rule to create a picture of how to construct the original six squares, which feels convincing and justifies the generalization.

Now, consider the "1 step 2 step" problem again. If we need to go down 5 steps, we can either go down the first 3 steps and finish with a double step or go down the first 4 steps and finish with a single step. The number of ways to go down 5 steps is the sum of the number of ways to go down 3 steps and 4 steps. This reasoning can be applied to going down 20 steps, justifying the use of Fibonacci numbers to predict the total number of ways to go down the 12 steps.

Understanding why a pattern works and explaining why it's appropriate to generalize is a crucial step, often the most interesting and important part of the problem-solving process. Spotting patterns is just the first step; explaining the underlying mathematics is what provides true insight and justification for the general rule.


In mathematics and statistics, the term "normal" refers to a type of probability distribution. A probability distribution is a way to calculate the chances of a specific event occurring using mathematical equations. It also describes the way these events appear on a graph.

The normal distribution is a specific type of probability distribution where not all possible outcomes have equal probability. In a normal distribution, there is one outcome value that is the average of all possible values, and this value has the highest probability. Values higher and lower than this average have lower probabilities the farther they are from the average.

For example, consider the height of men in the United States. The average height for men over 20 years old is 5 feet, 9.5 inches. However, there are men who are shorter and taller, with the tallest man in the world being 8 feet, 3 inches tall and the shortest man being 1 foot, 9.5 inches tall. The probability of anyone being this tall or short is very low.

If we were to measure the height of 1000 men from all over the US and plot this information on a graph, it would form a bell-shaped curve, also known as the "bell curve". This curve has a mean of 5 feet, 9.5 inches, which is the average height for men over 20 years old in the US. Men from our sample have a high probability of being around this height, and as the heights get taller or shorter, the probability of finding a man of this height goes down.

To illustrate this concept, try the following experiment:

* Gather a pen or pencil, a piece of paper, a computer with a spreadsheet program, and at least 30 individuals of around the same age.
* Create a table with height (in feet) in one column and number of individuals in another.
* Have each individual tell you their height and make a mark in the table corresponding to their height range.
* Use a spreadsheet program to graph the data, with height on the horizontal axis and number of observations on the vertical axis.
* Connect the tops of all the columns and observe the shape of the graph. Is the highest point roughly in the middle?

Challenge yourself to think of other examples that might have a normal distribution, such as the weight of nails in a box or the number of seeds in a bunch of identical seed packages. Use at least 30 samples and keep track of your numbers to graph the probability distribution. The larger your sample size, the better you will be able to see the probability distribution.

References:
1. McDowell, Margaret A. (2008). "Anthropometric Reference Data for Children and Adults: United States, 2003–2006".
2. "World’s Tallest Man—Living". Guinness Book of World Records.
3. Sheridan, Michael. (2012). “Chandra Bahadur Dangi is world’s shortest living man: Guinness World Records.”


Suppose $f$ is surjective. Then we may apply $g$ to both sides of this last equation and use that $g f = 1_A$ to conclude that $a = a'$. So let us see a few examples to understand what is going on. Surjective Function. To define the inverse function, we will first need some preliminary definitions. Similarly the composition of two injective maps is also injective.

Destruct (dec (f a')). Recall that a function which is both injective and surjective … Proof. Show transcribed image text. Intros A B a f dec H. Exists (fun b => match dec b with inl (exist _ a _) => a | inr _ => a end). Introduction to the inverse of a function Proof: Invertibility implies a unique solution to f(x)=y Surjective (onto) and injective (one-to-one) functions Relating invertibility to being onto and one-to-one Determining whether a transformation is onto Simplifying conditions for invertibility Showing that inverses are linear.

Distinct entities. In this case, the converse relation ${f^{-1}}$ is also not a function. Figure 2. Record Surjective {f ₁ f₂ t₁ t₂} {From: Setoid f₁ f₂} {To: Setoid t₁ t₂} (to: From To): Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where field from: To From right-inverse-of: from RightInverseOf to-- The set of all surjections from one setoid to another. In other words, the function F maps X onto Y (Kubrusly, 2001). On A Graph.

Forums. This problem has been solved! Prove That: T Has A Right Inverse If And Only If T Is Surjective. For instance, if A is the set of non-negative real numbers, the inverse map of f: A → A, x → x 2 is called the square root map. See the answer. De nition. Showing f is injective: Suppose a,a ′ ∈ A and f(a) = f(a′) ∈ B. id. (b) has at least two left inverses and, for example, but no right inverses (it is not surjective). F or example, we will see that the inv erse function exists only. Showing g is surjective: Let a ∈ A. here is another point of view: given a map f:X-->Y, another map g:Y-->X is a left inverse of f iff gf = id(Y), a right inverse iff fg = id(X), and a 2 sided inverse if both hold. Suppose f has a right inverse g, then f g = 1 B. Question: Prove That: T Has A Right Inverse If And Only If T Is Surjective. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. Behavior under composition. If y is in B, then g(y) is in A. and: f(g(y)) = (f o g)(y) = y. De nition 1.1. Bijections and inverse functions are related to each other, in that a bijection is invertible, can be turned into its inverse function by reversing the arrows.. Surjection vs. Injection. (Note that these proofs are superfluous,-- given that Bijection is equivalent to Function.Inverse.Inverse.) Nov 19, 2008 #1 Define $\displaystyle f:\Re^2 \rightarrow \Re^2$ by $\displaystyle f(x,y)=(3x+2y,-x+5y)$. Let f : A !B. Theorem right_inverse_surjective : forall {A B} (f : A -> B), (exists g, right_inverse f g) -> surjective … Hence, it could very well be that $AB = I_n$ but $BA$ is something else. _\square iii) Function f has a inverse iff f is bijective. An invertible map is also called bijective. Let b ∈ B, we need to find an element a … Suppose $f\colon A \to B$ is a function with range $R$. apply n. exists a'. unfold injective, left_inverse. It has right inverse iff is surjective: Advanced Algebra: Aug 18, 2017: Sections and Retractions for surjective and injective functions: Discrete Math: Feb 13, 2016: Injective or Surjective? reflexivity. for bijective functions. Definition (Iden tit y map). A function is bijective if and only if has an inverse November 30, 2015 De nition 1. Implicit: v; t; e; A surjective function from domain X to codomain Y. (a) Apply 4 (c) and (e) using the fact that the identity function is bijective. Thread starter Showcase_22; Start date Nov 19, 2008; Tags function injective inverse; Home. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. We say that f is bijective if it is both injective and surjective. Math Topics. We say that f is surjective if for all b 2B, there exists an a 2A such that f(a) = b. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. g f = 1A is equivalent to g(f(a)) = a for all a ∈ A. We want to show, given any y in B, there exists an x in A such that f(x) = y. A: A → A. is defined as the. Thus, π A is a left inverse of ι b and ι b is a right inverse of π A. (e) Show that if has both a left inverse and a right inverse , then is bijective and . intros a'. Function has left inverse iff is injective. LECTURE 18: INJECTIVE AND SURJECTIVE FUNCTIONS AND TRANSFORMATIONS MA1111: LINEAR ALGEBRA I, MICHAELMAS 2016 1. The inverse function g : B → A is defined by if f(a)=b, then g(b)=a. Secondly, Aluffi goes on to say the following: "Similarly, a surjective function in general will have many right inverses; they are often called sections." Suppose g exists. Any function that is injective but not surjective su ces: e.g., f: f1g!f1;2g de ned by f(1) = 1. id: ∀ {s₁ s₂} {S: Setoid s₁ s₂} → Bijection S S id {S = S} = record {to = F.id; bijective = record What factors could lead to bishops establishing monastic armies? There won't be a "B" left out. Let f : A !B. Formally: Let f : A → B be a bijection. Next story A One-Line Proof that there are Infinitely Many Prime Numbers; Previous story Group Homomorphism Sends the Inverse Element to the Inverse … The identity map. Inverse / Surjective / Injective. Interestingly, it turns out that left inverses are also right inverses and vice versa. If g is a left inverse for f, g f = id A, which is injective, so f is injective by problem 4(c). Injective and surjective functions There are two types of special properties of functions which are important in many di erent mathematical theories, and which you may have seen. Thus f is injective. Proof. Qed. When A and B are subsets of the Real Numbers we can graph the relationship. A function … The same argument shows that any other left inverse b ′ b' b ′ must equal c, c, c, and hence b. b. b. - destruct s. auto. Pre-University Math Help. The rst property we require is the notion of an injective function. PropositionalEquality as P-- Surjective functions. A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. We will show f is surjective. Thus, to have an inverse, the function must be surjective. Read Inverse Functions for more. The composition of two surjective maps is also surjective. map a 7→ a. If h is a right inverse for f, f h = id B, so f is surjective by problem 4(e). We are interested in nding out the conditions for a function to have a left inverse, or right inverse, or both. Thus setting x = g(y) works; f is surjective. i) ⇒. A right inverse of f is a function: g : B ---> A. such that (f o g)(x) = x for all x. Can someone please indicate to me why this also is the case? Equivalently, f(x) = f(y) implies x = y for all x;y 2A. "if a function is injective but not surjective, then it will necessarily have more than one left-inverse ... "Can anyone demonstrate why this is true? Showcase_22. to denote the inverse function, which w e will define later, but they are very. The image on the left has one member in set Y that isn’t being used (point C), so it isn’t injective. is surjective. If a function \(f\) is not surjective, not all elements in the codomain have a preimage in the domain. ii) Function f has a left inverse iff f is injective. 1.The map f is injective (also called one-to-one/monic/into) if x 6= y implies f(x) 6= f(y) for all x;y 2A. Bijections and inverse functions Edit. (See also Inverse function.). The function is surjective because every point in the codomain is the value of f(x) for at least one point x in the domain. Let f: A !B. Formally: Let f : A → B be a bijection. Let [ math ] f \colon x \longrightarrow y [ /math ] be a function. then f is injective iff it has a left inverse, surjective iff it has a right inverse (assuming AxCh), and bijective iff it has a 2 sided inverse. (b) Given an example of a function that has a left inverse but no right inverse. A function $g\colon B\to A$ is a pseudo-inverse of $f$ if for all $b\in R$, $g(b)$ is a preimage of $b$. A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. Tags: bijective bijective homomorphism group homomorphism group theory homomorphism inverse map isomorphism. Discrete Math: Jan 19, 2016: injective ZxZ->Z and surjective [-2,2]∩Q->Q: Discrete Math: Nov 2, 2015 Prove that: T has a right inverse if and only if T is surjective. Expert Answer . Injective function and it's inverse. Let A and B be non-empty sets and f: A → B a function. Let [math]f \colon X \longrightarrow Y[/math] be a function. This example shows that a left or a right inverse does not have to be unique Many examples of inverse maps are studied in calculus. given \(n\times n\) matrix \(A\) and \(B\), we do not necessarily have \(AB = BA\). a left inverse must be injective and a function with a right inverse must be surjective. Sep 2006 782 100 The raggedy edge. ... Bijective functions have an inverse! We say that f is injective if whenever f(a 1) = f(a 2) for some a 1;a 2 2A, then a 1 = a 2. - exfalso. Similarly, any other right inverse equals b, b, b, and hence c. c. c. So there is exactly one left inverse and exactly one right inverse, and they coincide, so there is exactly one two-sided inverse. It follows therefore that a map is invertible if and only if it is injective and surjective at the same time. De nition 2. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. , then f g = 1 B (Note that these proofs are superfluous, -- Given that Bijection is equivalent to Function.Inverse.Inverse.)


### Polish Notation

Polish notation refers to a notation system where the operator is placed before its operands. There are three main types of Polish notation: Infix, Postfix, and Prefix notation.

Infix Notation is a type of Polish notation where the operator symbol is positioned between its two operands. For example, to add A and B, the expression is written as A + B. Similarly, to subtract D from C, the expression is C - D.

Prefix notation and Postfix notation are the other two types, with the operator placed before or after the operands, respectively.


SWBAT represent data in a matrix and determine the dimensions of a matrix. Matrices provide a way to organize and analyze data.

In recent years, matrices have not been taught in Algebra 1 or Algebra 2, despite appearing on many college entrance exams. This class introduces students to matrices, providing an opportunity to understand and develop skills in their use.

The lesson begins with a 5-minute bell work activity, where students consider different ways to organize data. After sharing their ideas, which often include tables, spreadsheets, and lists, students are introduced to the concept of a matrix.

The definition of a matrix is provided, and students read and write down any aspects they do not understand. Following a 3-4 minute group discussion to clarify questions, the class comes together to address any remaining questions.

Students are then asked to create a matrix to describe the data from the bell work, resulting in two different matrices. The class determines the correctness of the matrices and discusses how the representation of rows and columns can vary depending on the context.

Next, students analyze a given matrix to answer questions on a worksheet. After groups work together and share their answers, a different group explains their reasoning. This process allows for assessment of multiple groups' understanding.

The "Using Matrices to Analyze Data" worksheet provides real-world examples of matrix application in archeology and sociology. Students work in groups to complete the worksheet, with the teacher available to answer questions and facilitate discussion.

The lesson concludes with a 5-minute exit slip activity, where students answer the question: **If the principal stops you in the hall and asks what you learned in Precalculus today, what would you tell him?** Students are expected to provide precise answers, such as "I learned how to use matrices to answer questions."


Water is leaking out of an inverted conical tank at a rate of 11,000 cm^{3}/min, while water is being pumped into the tank at a constant rate. The tank has a height of 6 m and a diameter of 4 m at the top. Given that the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, we need to find the rate at which water is being pumped into the tank.

The radius of the water surface is 1/3 the height of the water. Let the height be h and the radius be h/3 (in cm). The volume of water in the tank is given by:

V = \frac{1}{3}\pi r^2 h
V = \frac{1}{3}\pi(\frac{h}{3})^2 h
V = \frac{\pi h^3}{27}

The derivative of V with respect to h is:

\frac{dV}{dh} = \frac{\pi h^2}{9}

When h = 200 cm (2 m), the radius is 200/3 cm. The volume of water in the tank is:

V = \frac{\pi \times 200^3}{27} = \frac{8 \times 10^6 \pi}{27}

The derivative of V with respect to h is:

\frac{dV}{dh} = \frac{\pi \times 200^2}{9} = \frac{4 \times 10^4 \pi}{9}

Given that the water level is rising at a rate of 20 cm/min, we have:

\frac{dh}{dt} = 20

The rate of change of volume with respect to time is:

\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}
\frac{dV}{dt} = \frac{4 \times 10^4 \pi}{9} \times 20
\frac{dV}{dt} = \frac{80 \times 10^4 \pi}{9}

The rate at which water is being pumped into the tank is given by:

\frac{dV}{dt} = \frac{dV_{in}}{dt} - \frac{dV_{out}}{dt}
\frac{80 \times 10^4 \pi}{9} = \frac{dV_{in}}{dt} - 11000

Solving for \frac{dV_{in}}{dt}, we get:

\frac{dV_{in}}{dt} = \frac{80 \times 10^4 \pi}{9} + 11000
\frac{dV_{in}}{dt} = 27953 + 11000
\frac{dV_{in}}{dt} = 38953 cm^{3}/min

Rounded to the nearest integer, the rate at which water is being pumped into the tank is approximately 39000 cm^{3}/min or 39 liters per minute.


# Natural Language Processing MCQ - Bigram Probability Calculation using MLE

To compute the bigram probability P(w_{n}|w_{n-1}) using Maximum Likelihood Estimate (MLE), we need to count the number of bigrams (w_{n-1}w_{n}) from a corpus and normalize by the count of all bigrams that start with w_{n-1}. This normalization step ensures that the estimate lies between 0 and 1.

The bigram probability can be calculated using the following equation:
P(w_{n}|w_{n-1}) = Count(w_{n-1}w_{n}) / Sum(Count(w_{n-1}w))

Here, w is any word that follows w_{n-1}. This equation can be simplified by replacing the bigram count in the denominator with the unigram count of w_{n-1}.

The reason for simplifying the equation is that the sum of all bigram counts that start with the word w_{n-1} is equal to the unigram count of the same word. This is because each occurrence of w_{n-1} can only be followed by one word, so the total count of bigrams starting with w_{n-1} cannot exceed the unigram count of w_{n-1}.

For example, let's calculate the bigram probability P(increase | to) using both normalization methods. Suppose we have the following bigram counts:
- C("to increase") = 2
- C("to be") = 1
- C("to fill") = 1
- C("to") = 4 (unigram count)

Using the bigram count for normalization:
P(increase | to) = C("to increase") / [C("to increase") + C("to be") + C("to fill")] = 2 / [2 + 1 + 1] = 2/4 = 0.5

Using the unigram count for normalization:
P(increase | to) = C("to increase") / C("to") = 2 / 4 = 0.5

As shown, both methods yield the same result. The correct answer is:
b) Sum of all bigram counts that start with the word w_{n-1} is equal to the unigram count of the same word.


To evaluate the definite integral with symmetrical limits of integration, we use the property:

`int_(-a)^a f(x)dx = int_0^a(f(x) + f(-x))dx`

Given `f(x) = 1/((e^x+1)(x^2+1))`, we need to find `f(x) + f(-x)`:

`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) + 1/((e^(-x)+1)((-x)^2+1))`
`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) + 1/((1/e^x + 1)(x^2 + 1))`
`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) + e^x/((e^x+1)(x^2+1))`
`f(x) + f(-x) = (e^x + 1)/((e^x+1)(x^2+1))`

Reducing duplicate factors yields:

`f(x) + f(-x) = 1/(x^2 + 1)`

Using this result, we can evaluate the definite integral:

`int_(-1)^1 1/((e^x+1)(x^2+1)) dx = int_0^1 1/(x^2 + 1)dx`
`int_0^1 1/(x^2 + 1)dx = tan^(-1) x|_0^1`
`int_0^1 1/(x^2 + 1)dx = tan^(-1) 1 - tan^(-1) 0`
`int_0^1 1/(x^2 + 1)dx = pi/4`

Therefore, `int_(-1)^1 1/((e^x+1)(x^2+1)) dx = pi/4`.


In probability and statistics, a random variable is a variable whose values depend on outcomes of a random phenomenon. The formal mathematical treatment of random variables is a topic in probability theory, where a random variable is understood as a measurable function defined on a probability space that maps from the sample space to the real numbers.

A random variable's possible values might represent the possible outcomes of a yet-to-be-performed experiment, or the possible outcomes of a past experiment whose already-existing value is uncertain. They may also conceptually represent either the results of an "objectively" random process or the "subjective" randomness that results from incomplete knowledge of a quantity.

The domain of a random variable is called a sample space, defined as the set of possible outcomes of a non-deterministic event. For example, in the event of a coin toss, only two possible outcomes are possible: heads or tails. A random variable has a probability distribution, which specifies the probability of Borel subsets of its range.

Random variables can be discrete, taking any of a specified finite or countable list of values, or continuous, taking any numerical value in an interval or collection of intervals. A random variable can also be a mixture of both discrete and continuous types.

The probability distribution of a random variable is often characterized by a small number of parameters, which also have a practical interpretation. For example, it is often enough to know what its "average value" is, which is captured by the mathematical concept of expected value of a random variable.

Two random variables with the same probability distribution can still differ in terms of their associations with, or independence from, other random variables. The realizations of a random variable, that is, the results of randomly choosing values according to the variable's probability distribution function, are called random variates.

A random variable can be described by its cumulative distribution function, which describes the probability that the random variable will be less than or equal to a certain value. The term "random variable" in statistics is traditionally limited to the real-valued case, but the concept of a random element is more general and can be applied to other types of data.

In the measure-theoretic, axiomatic approach to probability, a random variable is defined as a measurable function from a probability space to a measurable space. The probability distribution of a random variable is then defined as the pushforward measure of the probability measure on the probability space.

Random variables can be transformed using functions, and the resulting random variables have probability distributions that can be calculated using the original probability distribution and the transformation function. For example, if X is a random variable and Y = g(X) is a transformation of X, then the probability distribution of Y can be calculated using the probability distribution of X and the function g.

There are several different senses in which random variables can be considered to be equivalent, including equality in distribution, equality almost surely, and equality as functions on their measurable space. Convergence results for sequences of random variables are also important in probability theory, including the law of large numbers and the central limit theorem.

Examples of random variables include the outcome of a coin toss, the number of children a person has, and the height of a person. These random variables can be discrete or continuous, and their probability distributions can be calculated using various methods, including the probability mass function and the probability density function.

In summary, random variables are mathematical objects that are used to model and analyze random phenomena, and they have a wide range of applications in statistics, engineering, economics, and other fields. Their probability distributions can be calculated and transformed using various methods, and they can be compared and contrasted using different notions of equivalence.


# Building and Unfolding a Polycube

A **polycube** is a polyhedron constructed from multiple identical 1x1x1 cubes glued together face-to-face. When building a polycube, the interior faces where cubes are glued together are disregarded, and only the exterior square faces are considered. A key requirement for a polycube is that it must be deformable to a sphere, meaning it can be inflated into a ball shape without any holes or tunnels.

The concept of an **edge-unfolding** is crucial in understanding polycubes. An edge-unfolding involves cutting the surface of the polycube along certain edges of the square faces and then unfolding it into a single, flat piece. The question arises whether every polycube can be edge-unfolded without overlapping itself on a flat surface.

Alternatively, consider wrapping a polycube tightly in paper. The goal is to determine if it's possible to cut out a single shape from a piece of paper, composed of multiple identical squares, that can be folded around the polycube face by face to wrap it completely. This shape would represent the edge-unfolding of the polycube.


# How to Solve x+3=2x-5

To solve the equation x+3=2x-5, the goal is to isolate "x" on one side.

Start with: x+3=2x-5

Subtract "x" from both sides to move it to the other side with "2x":
x - x + 3 = 2x - x - 5
This simplifies to: 3 = x - 5

Next, move the -5 to the other side by changing its sign to +5 and adding it to both sides:
3 + 5 = x - 5 + 5
This simplifies to: 8 = x

Therefore, the solution to the equation x+3=2x-5 is x = 8.


**Kilogram to Pound Conversion**

The kilogram is a unit of mass and a basic unit in the metric system, also known as the International System of Units (SI). It is defined as the mass of one liter of water and is equal to 0.001 tonne. One kilogram can be divided into 100 decagrams and 1000 grams.

The pound is also a unit of mass, used in the Imperial and United States customary systems of measurements. There are different types of pounds, including the avoirdupois pound, which is defined as exactly 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces and 7000 grains.

**Converting 25.95 kg to lbs**

To convert 25.95 kg to lbs, we can use two different formulas:

1. **Exact formula:** Amount of kilograms * 2.20462262 = result in pounds
2. **Simplified formula:** Amount of kilograms * 2.2 = result in pounds

Using the exact formula, we get:
25.95 kg * 2.20462262 = 57.2099569890 pounds

Using the simplified formula, we get:
25.95 kg * 2.2 = 57.090 pounds

**Rounding Off the Result**

We can round off the result to two decimal places:
25.95 kg = 57.0900 pounds

**Converting 25.95 kg to lbs and Ounces**

To convert 25.95 kg to lbs and ounces, we need to multiply the fraction part by 16:
25.95 kg * 2.20462262 = 57.2099569890 pounds
Fraction part: 0.2099569890 * 16 = 3.359296512 ounces

So, 25.95 kg is equal to 57 pounds and 3.359296512 ounces.

**Using a kg to lbs Converter**

We can also use a kg to lbs converter to convert 25.95 kg to lbs. The converter is based on the exact formula and gives us the result:
25.95 kg = 57.2099569890 pounds

**Conversion Chart**

Here is a conversion chart for 25.95 kg to lbs:

| Kilograms | Pounds | Pounds (rounded off to two decimal places) |
| --- | --- | --- |
| 25.95 | 57.2099569890 | 57.0900 |

**Other Conversions**

We can also convert 25.95 foot pounds to kilograms meters using the formula:
Amount of foot pounds * 0.13825495 = result in kilograms meters
25.95 foot pounds * 0.13825495 = 3.587291075 kilograms meters

Note: The conversion calculator may sometimes show incorrect results due to errors in the formula. It is essential to use the converter multiple times to get the exact result.


## Finding the Area Under the Standard Normal Curve to the Left of Z

The normal distribution is a continuous probability distribution characterized by a bell-shaped curve. A key property of the standard normal distribution is that the total area under its curve is always 1. This article provides a step-by-step guide on calculating the area under the standard normal curve to the left of a given z score.


Point-slope form is used to write the equation of a line given the slope and a point. The slope is calculated using the "rise-over-run" equation. To write an equation in slope-intercept form, we need two things: the slope (m) and the y-intercept (b).

Given a real-world problem, we may be given various aspects of the equation, such as slope and y-intercept, slope and a point, or two points. If we are given slope and a point, we can use the point-slope form to write the equation. The point-slope form is usually gotten by manipulating one of the previous two forms.

The equation of a line can be expressed in several possible forms, including slope-intercept form (y = mx + b), point-slope form (y - y1 = m(x - x1)), and standard form (ax + by = c). To write an equation in slope-intercept form, we need to find the slope (m) and the y-intercept (b).

If we are given two points, we can use them to find the equation of the line. We can calculate the slope using the "rise-over-run" equation and then use one of the points to find the y-intercept. Once we have the slope and y-intercept, we can write the equation in slope-intercept form.

For example, if we are given the point (6, -3) and a slope of -2, we can write the equation in point-slope form as y - (-3) = -2(x - 6). We can then simplify this equation to find the y-intercept and write the equation in slope-intercept form.

To practice finding the equation of a line, we can use a calculator to find the equation given two points. We can enter the points into the calculator and it will calculate the slope-intercept form (y = mx + b), point-slope form (y - y1 = m(x - x1)), and standard form (ax + by = c) of the equation.

In conclusion, writing linear equations given slope and a point or two points requires using the point-slope form and slope-intercept form. We need to calculate the slope and find the y-intercept to write the equation in slope-intercept form. With practice and the use of calculators, we can become proficient in writing linear equations given various aspects of the equation.


## Section 2.3 - Rules for Differentiation

**Essential Questions:**
1. What is the purpose of the rules for differentiation?
2. What similarities and differences do the rules for differentiation have?

**Lesson Completion Steps:**

1. **Preparation**: Title your spiral with the heading "Section 2.3 - Rules for Differentiation" and copy the essential questions.

2. **Vocabulary & Examples**:
   - Define key vocabulary related to differentiation rules, including tables, properties, theorems, terms, phrases, and postulates.
   - Review examples and copy necessary information for Cornell notes, focusing on:
     - **Positive Integer Powers, Multiples, Sums, and Differences**: Identify the first four derivative rules.
     - **Products and Quotients**: Understand the product rule and the quotient rule.
     - **Negative Integer Powers of x**: Apply the power rule for negative integer powers of x.
     - **Second and Higher Order Derivatives**: Recognize notations for first, second, third, and nth derivatives.

3. **Reading**: If time allows, reread the lesson to reinforce understanding and take additional notes as needed.


**TSI Math Practice Test**
The TSI Math questions are designed to be similar to those found on the Texas Success Initiative Assessment. The four primary topics covered on this test are:
1. Elementary Algebra & Functions
2. Intermediate Algebra & Functions
3. Geometry & Measurement
4. Data Analysis, Statistics, & Probability.

**Directions:** Choose the best answer from the four choices. You may use paper and a pencil for computations, but calculators are not permitted.

**Question 1**
Bill's new porch is rectangular, with an area of 50 square feet. If the length of the porch is two times the width, what is the perimeter of the porch?
A) 30 feet
B) 40 feet
C) 50 feet
D) 60 feet

The formula for the area of a rectangle is $A = \text{Width} \cdot \text{Length}$. Given $L = 2W$ and $A = 50$, we can solve for $W$: $50 = W \cdot 2W$, $50 = 2W^2$, $W^2 = 25$, $W = 5$. Since $L = 2W$, $L = 10$. The perimeter is $2L + 2W = 2 \cdot 10 + 2 \cdot 5 = 30$.

**Question 2**
What is the probability of selecting a male from a group of 4 males and 8 females?
A) $\dfrac{1}{3}$
B) $\dfrac{1}{2}$
C) $\dfrac{2}{3}$
D) $\dfrac{4}{5}$

The probability of selecting a male is $P(\text{Male}) = \frac{\text{Number of Males}}{\text{Total Number of People}} = \frac{4}{12} = \frac{1}{3}$.

**Question 3**
Solve for $x$: $3(x + 1) = 5(x − 2) + 7$
A) −2
B) 2
C) ½
D) 3

Simplifying the equation: $3x + 3 = 5x - 10 + 7$, $3x + 3 = 5x - 3$. Adding 3 to both sides: $3x + 6 = 5x$. Subtracting $3x$ from both sides: $6 = 2x$. Dividing by 2: $3 = x$.

**Question 4**
Line *A* passes through the point $(−1, 2)$. Which of the following CANNOT be the equation of line *A*?
A) $y = 1 − x$
B) $y = x + 1$
C) $x = −1$
D) $y = x + 3$

The slope-intercept form of a line is $y = mx + b$. Since the line passes through $(−1, 2)$, we can check each option. Plugging $x = −1$ into each equation, the only one that doesn't give $y = 2$ is $y = x + 1$, which gives $y = 0$.

**Question 5**
Diego’s current age is five times Martina’s age ten years ago. If Martina is currently $m$ years old, what is Diego’s current age in terms of $m$?
A) $5m$
B) $5m − 10$
C) $5m − 50$
D) $5m + (m − 10)$

Martina’s age ten years ago is $m - 10$. Diego's age is $5(m - 10) = 5m - 50$.

**Question 6**
If $2x + 5 = 8x$, then $12x = \; ?$
A) 5
B) 10
C) 15
D) 20

Solving for $x$: $8x - 2x = 5$, $6x = 5$. Multiplying both sides by 2: $12x = 10$.

**Question 7**
An airline asked its agents to assign convenience ratings on a scale of 1–7 to 114 flights. What is the median rating assigned by the agents?
A) 3
B) 4
C) 5
D) 6

Since there are 114 ratings (an even number), the median will be the average of the 57th and 58th ratings. By adding frequencies from the lowest rating, we find both the 57th and 58th ratings correspond to a rating of 5. Thus, the median rating is 5.

**Question 8**
What is the value of $x$ in the equation $\sqrt{2x} = 4$?
A) 1
B) 4
C) 8
D) 16

Squaring both sides: $(\sqrt{2x})^2 = 4^2$, $2x = 16$, $x = 8$.

**Question 9**
The market price of a house is directly proportional to the total area of the house in square feet. If the price of a 1,500 square foot home is $300,000, what is the price of a 2,000 square foot home?
A) $350,000
B) $400,000
C) $450,000
D) $500,000

Setting up a proportion: $\dfrac{1,500}{\$300,000} = \dfrac{2,000}{x}$. Cross-multiplying: $1,500x = \$600,000,000$, $x = \$400,000$.

**Question 10**
Find the $x$-intercept of the line $2y = 3x − 6$.
A) −2
B) −1
C) 1
D) 2

At the $x$-intercept, $y = 0$. Substituting into the equation: $2(0) = 3x - 6$, $0 = 3x - 6$, $3x = 6$, $x = 2$.

**Question 11**
There are 80 students in a school. Every student is taking either algebra or pre-calculus, and some students are taking both. There are 60 students in the algebra class. If 15 students are taking both algebra and pre-calculus, what is the probability that a randomly selected student is taking pre-calculus but not algebra?
A) $\dfrac{1}{4}$
B) $\dfrac{1}{3}$
C) $\dfrac{3}{8}$
D) $\dfrac{3}{4}$

Using a Venn diagram, we find students taking algebra but not pre-calculus $= 45$. The number of students taking pre-calculus but not algebra $(x)$ is found by: $45 + 15 + x = 80$, $x = 20$. The probability is $P(x) = \dfrac{20}{80} = \dfrac{1}{4}$.

**Question 12**
What is the value of $x$ in the equation $\dfrac{2}{3}x + 2 = 4$?
A) 3
B) 4
C) 5
D) 6

Subtracting 2 from both sides: $\dfrac{2}{3}x = 2$. Multiplying both sides by 3 and then dividing by 2: $2x = 6$, $x = 3$.

**Question 13**
What are the solutions to the equation $x^2 + 8x + 15 = 0$?
A) (2, 3)
B) (3, 5)
C) (−2, −3)
D) (−3, −5)

Factoring the quadratic equation: $x^2 + 8x + 15 = (x + 3)(x + 5) = 0$. This gives $x = -3$ or $x = -5$.

**Question 14**
Simplify: $\sqrt{5}\sqrt{3} = \; ?$
A) $\sqrt{2}$
B) $\sqrt{8}$
C) $\sqrt{15}$
D) $8$

Using the rule $\sqrt{a} \sqrt{b} = \sqrt{a \cdot b}$, we get $\sqrt{5}\sqrt{3} = \sqrt{5 \cdot 3} = \sqrt{15}$.

**Question 15**
The office supply superstore sells pens for 60 cents each and pencils for 40 cents each. Diane purchased a total of 17 items for $8.20. How many pens did Diane purchase?
A) 5
B) 7
C) 10
D) 12

Let $x$ be the number of pens and $y$ be the number of pencils. We have two equations: $0.6x + 0.4y = 8.2$ and $x + y = 17$. Solving for $x$, we find $y = 17 - x$ and substitute into the first equation: $0.6x + 0.4(17 - x) = 8.2$. Simplifying gives $0.2x = 1.4$, and thus $x = 7$.

**Question 16**
A scoutmaster is preparing his troop for their annual fishing expedition. He surveys each of his scouts to find out how many fishing poles they own. What is the mode of the number of poles owned?
A) 0
B) 1
C) 2
D) 3

The mode is the number that appears most often. In this case, the largest number of scouts (7) own 0 fishing poles.

**Question 17**
Simplify: $(x^6)(x^5)$
A) $2x^{11}$
B) $2x^{30}$
C) $x^{11}$
D) $x^{30}$

When multiplying exponents with the same base, add the exponents: $(x^6)(x^5) = x^{6 + 5} = x^{11}$.

**Question 18**
What is the slope of line $k$?
A) −2
B) −3
C) 2
D) 5

Using points $(1, 3)$ and $(5, -5)$, the slope is $\dfrac{-5 - 3}{5 - 1} = \dfrac{-8}{4} = -2$.

**Question 19**
$x^4 − 16 = \; ?$
A) $(x + 2)(x − 2)(x^2 + 4)$
B) $(x + 2)^2(x − 2)^2$
C) $(x + 2)^3 (x − 2)^1$
D) $(x − 2)^4$

Using the identity $a^2 - b^2 = (a - b)(a + b)$, we simplify: $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) = (x^2 - 2^2)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$.

**Question 20**
What is the maximum value of the function $f(x) = −16x^2 + 32x + 20$?
A) 32
B) 34
C) 36
D) 40

The vertex of a parabola $f(x) = ax^2 + bx + c$ is at $\left(−\dfrac{b}{2a}, \, c − \dfrac{b^2}{4a}\right)$. Here, $a = -16$, $b = 32$, $c = 20$. The $y$-coordinate of the vertex is $c − \dfrac{b^2}{4a} = 20 - \dfrac{32^2}{4(-16)} = 20 + 16 = 36$.


The formula for the nth square triangular number, denoted as f(n), is given by:

((17 + 12√2)^n + (17 – 12√2)^n – 2)/32

As n increases, the term (17 – 12√2)^n approaches zero quickly, making the formula approximately:

((17 + 12√2)^n – 2)/32

The error in this approximation is less than 0.001 for all n, making it exact when rounded to the nearest integer. Therefore, the nth square triangular number can be expressed as:

⌊((17 + 12√2)^n + 14)/32⌋

where ⌊x⌋ is the greatest integer less than x.

A Python implementation of this formula is:

from math import sqrt, floor
def f(n):
    x = 17 + 12*sqrt(2)
    return floor((x**n + 14)/32.0)

However, this formula is not useful for ordinary floating-point computation when n is 11 or larger, as the result requires more bits than are available in the significand of a floating-point number.

An alternative approach is to use a recurrence relation:

f(n) = 34f(n-1) – f(n-2) + 2

for n > 2. This relation only requires integer arithmetic and can be implemented in Python as:

def f(n):
    if n < 2:
        return n
    return f(n-1)*34 - f(n-2) + 2

This implementation can be made more efficient using memoization.

The recurrence relation can also be implemented in other programming languages, such as Perl 6. Additionally, the formula can be computed using a closed-form expression in Haskell, which achieves sublinear time complexity.

The sequence of square triangular numbers has been studied in various mathematical contexts, and a blog post on the subject can be found at http://blog.plover.com/math/square-triangular.html.


Hermite polynomials are a class of mathematical functions that play a significant role in various fields, including quantum mechanics, statistics, and mathematical physics. Understanding the basics of Hermite polynomials is essential to appreciate their applications and mathematical properties.

Named after the French mathematician Charles Hermite, Hermite polynomials are a sequence of orthogonal polynomials that arise as solutions to the Hermite differential equation. These polynomials are defined by the recurrence relation: H_{n}(x) = 2xH_{n-1}(x) - 2(n-1)H_{n-2}(x), where H_{n}(x) represents the nth degree Hermite polynomial. The first few Hermite polynomials are H_{0}(x) = 1, H_{1}(x) = 2x, H_{2}(x) = 4x^{2} - 2.

Hermite polynomials have a long history dating back to the 19th century when Charles Hermite first introduced them while studying the properties of elliptical functions. Since then, they have found numerous applications in various mathematical and scientific disciplines. Hermite polynomials possess several noteworthy characteristics, including orthogonality and recurrence relations.

The orthogonality of Hermite polynomials makes them valuable in solving problems involving weighted integrals or inner products in mathematical physics and quantum mechanics. The recurrence relation allows us to express higher degree Hermite polynomials in terms of lower degree ones, simplifying calculations involving Hermite polynomials and contributing to their usefulness in mathematical analysis and approximation theory.

Hermite polynomials have been extensively studied due to their remarkable properties and applications in various fields of mathematics and science. One of the key characteristics of Hermite polynomials is their orthogonality, which means that when multiplied by a weight function, in this case e^{-x2}, the integral of their product over a certain interval is zero, except when the polynomials are the same.

Another important characteristic of Hermite polynomials is their recurrence relations. The recurrence relation H_{n}(x) = 2xH_{n-1}(x) - 2(n-1)H_{n-2}(x) allows us to express higher degree Hermite polynomials in terms of lower degree ones. This recursive nature simplifies calculations involving Hermite polynomials and contributes to their usefulness in mathematical analysis and approximation theory.

Hermite polynomials have been extensively studied in the field of mathematical analysis, where they play a crucial role in approximation theory. They can be used to approximate various functions, and their accuracy improves as the degree of the polynomial increases. This property makes Hermite polynomials valuable tools in numerical analysis and scientific computing.

In addition to their mathematical applications, Hermite polynomials have found practical uses in various scientific disciplines. For example, they are commonly employed in quantum mechanics to describe the wave functions of quantum harmonic oscillators. They also have applications in statistical mechanics, where they are used to solve problems related to the distribution of particles in a system.

Hermite polynomials find applications in statistics, particularly in the field of mathematical statistics and probability theory. Statisticians often encounter situations where they need to model data distributions and estimate parameters. Hermite polynomials provide a powerful tool for achieving these tasks by approximating probability density functions.

By utilizing Hermite polynomials, statisticians can accurately characterize the statistical behavior of random variables. These polynomials allow them to describe the shape, spread, and skewness of data distributions, enabling a deeper understanding of the underlying processes generating the data.

Moreover, Hermite polynomials play a crucial role in statistical inference. They allow statisticians to make inferences about population parameters based on sample data, providing a solid foundation for hypothesis testing and confidence interval estimation.

When encountering problems involving Hermite polynomials, it is essential to know techniques to solve Hermite polynomial equations. These techniques include finding the roots of Hermite polynomials, calculating integrals involving Hermite polynomials, and manipulating the recurrence relation to simplify calculations.

One of the fundamental tasks in solving Hermite polynomial equations is finding their roots. The roots of Hermite polynomials play a crucial role in many applications, such as solving differential equations, signal processing, and quantum mechanics. There are several methods available for finding the roots of Hermite polynomials, including numerical algorithms like Newton's method, bisection method, and the Jenkins-Traub algorithm.

Integrals involving Hermite polynomials often arise in various mathematical and scientific disciplines. Efficiently evaluating these integrals is essential for solving problems related to Hermite polynomials. Several techniques can be employed to calculate integrals involving Hermite polynomials, including using recurrence relations, Gaussian quadrature methods, and contour integration techniques.

The recurrence relation of Hermite polynomials is a powerful tool for simplifying calculations and deriving new properties. By manipulating the recurrence relation, we can express higher-degree Hermite polynomials in terms of lower-degree ones, which can significantly reduce the complexity of calculations.

Understanding the intricacies of the recurrence relation and its various properties allows us to efficiently solve problems involving Hermite polynomials and explore their connections with other areas of mathematics.

While working with Hermite polynomials, it is common to encounter challenges in calculations due to their complex nature. One of the main challenges lies in the numerical stability and computational efficiency of evaluating high-degree Hermite polynomials and performing calculations involving them. Employing appropriate numerical algorithms and techniques can mitigate these challenges and ensure accurate and efficient calculations.

Exploring the future prospects and ongoing research trends in Hermite polynomials can shed light on exciting developments in this field. Researchers are continually investigating new algorithms, approximation methods, and applications of Hermite polynomials, paving the way for advancements in various scientific and engineering domains.

Researchers continue to explore the properties and applications of Hermite polynomials, seeking to uncover new mathematical insights and develop novel techniques. One area of current research focuses on refining algorithms for efficient calculation of Hermite polynomials. By improving the computational efficiency, researchers aim to make the calculation of these polynomials faster and more accurate.

Another exciting avenue of research is investigating the applications of Hermite polynomials in machine learning and data science. These polynomials have shown promise in modeling complex data patterns and extracting meaningful features. By incorporating Hermite polynomials into machine learning algorithms, researchers hope to enhance the performance and interpretability of these models.

Furthermore, researchers are exploring connections between Hermite polynomials and other areas of mathematics. For example, there are ongoing efforts to understand the relationship between Hermite polynomials and orthogonal polynomials from different families. These connections can lead to new mathematical insights and provide a deeper understanding of the properties of Hermite polynomials.

The potential future applications of Hermite polynomials are vast and exciting. As computational power advances, Hermite polynomials can further contribute to various scientific and engineering fields. In signal processing, for instance, Hermite polynomials can be used to analyze and manipulate signals, enabling more accurate and efficient communication systems.

In the field of image recognition, Hermite polynomials can aid in feature extraction and pattern recognition, leading to improved image classification algorithms. Cryptography is another area where Hermite polynomials can play a significant role. Their mathematical properties, such as orthogonality and recurrence relations, can be leveraged to design secure encryption and decryption algorithms.

Additionally, Hermite polynomials have applications in optimization problems. By formulating optimization objectives using these polynomials, researchers can develop efficient algorithms for solving complex optimization tasks. This has implications in various fields, including logistics, resource allocation, and financial modeling.

In conclusion, Hermite polynomials are a fascinating class of mathematical functions with diverse applications and remarkable mathematical properties. Exploring their origins, characteristics, and applications allows us to appreciate their significance in various fields. Continual research and development in Hermite polynomials will undoubtedly lead to new discoveries and applications, propelling the field forward.


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