 10.1.1: APPLICATION There are four basic blood types. The distribution of t...
 10.1.2: Use the relative frequency circle graph in Exercise 1 to answer the...
 10.1.3: Which data set matches the relative frequency circle graph at right...
 10.1.4: In the relative frequency bar graph of the librarys collection crea...
 10.1.5: A manufacturer states that it produces colored candies according to...
 10.1.6: Chloe bought a small package of the candies described in Exercise 5...
 10.1.7: This table shows the number of students in each grade at a high sch...
 10.1.8: Match each bar graph with its corresponding circle graph. Try to do...
 10.1.9: What is a reasonable estimate of the chance that a randomly thrown ...
 10.1.10: Write an equation in general form for the parabola shown, with xin...
 10.1.11: Astrid works as an intern in a windmill park in Holland. She has le...
 10.1.12: APPLICATION In 2001 there were 3141 counties in the United States. ...
Solutions for Chapter 10.1: Relative Frequency Graphs
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 10.1: Relative Frequency Graphs
Get Full SolutionsThis textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. This expansive textbook survival guide covers the following chapters and their solutions. Discovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. Since 12 problems in chapter 10.1: Relative Frequency Graphs have been answered, more than 4541 students have viewed full stepbystep solutions from this chapter. Chapter 10.1: Relative Frequency Graphs includes 12 full stepbystep solutions.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Characteristic equation det(A  AI) = O.
The n roots are the eigenvalues of A.

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.

Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n  r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.

Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

Rank r (A)
= number of pivots = dimension of column space = dimension of row space.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Solvable system Ax = b.
The right side b is in the column space of A.

Special solutions to As = O.
One free variable is Si = 1, other free variables = o.

Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.

Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.