Business Math (11th Edition)
11th Edition
ISBN: 9780134496436
Author: Cheryl Cleaves, Margie Hobbs, Jeffrey Noble
Publisher: PEARSON
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Chapter 1.1, Problem 17SE
To determine
To calculate: The round to the first digit of the Cisco’s earnings in the statement “The one of the world’s largest Internet equipment makers is Cisco, and Cisco recorded earnings of about
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math help plz
1. Show that, for any non-negative random variable X,
EX+E+≥2,
X
E max X.
21.
For each real-valued nonprincipal character x mod k, let
A(n) = x(d) and F(x) = Σ
:
dn
* Prove that
F(x) = L(1,x) log x + O(1).
n
Chapter 1 Solutions
Business Math (11th Edition)
Ch. 1.1 - Prob. 1-1SCCh. 1.1 - Prob. 1-2SCCh. 1.1 - Prob. 1-3SCCh. 1.1 - Prob. 1-4SCCh. 1.1 - Prob. 2-1SCCh. 1.1 - Prob. 2-2SCCh. 1.1 - Prob. 2-3SCCh. 1.1 - Prob. 2-4SCCh. 1.1 - Prob. 3-1SCCh. 1.1 - Prob. 3-2SC
Ch. 1.1 - Prob. 3-3SCCh. 1.1 - Prob. 3-4SCCh. 1.1 - Prob. 3-5SCCh. 1.1 - Prob. 3-6SCCh. 1.1 - Prob. 4-1SCCh. 1.1 - Prob. 4-2SCCh. 1.1 - Prob. 4-3SCCh. 1.1 - Prob. 4-4SCCh. 1.1 - Prob. 1SECh. 1.1 - Prob. 2SECh. 1.1 - Prob. 3SECh. 1.1 - Prob. 4SECh. 1.1 - Prob. 5SECh. 1.1 - Prob. 6SECh. 1.1 - Prob. 7SECh. 1.1 - Prob. 8SECh. 1.1 - Prob. 9SECh. 1.1 - Prob. 10SECh. 1.1 - Prob. 11SECh. 1.1 - Prob. 12SECh. 1.1 - Prob. 13SECh. 1.1 - Prob. 14SECh. 1.1 - Prob. 15SECh. 1.1 - Prob. 16SECh. 1.1 - Prob. 17SECh. 1.1 - Prob. 18SECh. 1.1 - Prob. 19SECh. 1.1 - Prob. 20SECh. 1.1 - Prob. 21SECh. 1.1 - Prob. 22SECh. 1.1 - Prob. 23SECh. 1.1 - Prob. 24SECh. 1.1 - Prob. 25SECh. 1.1 - Prob. 26SECh. 1.2 - Prob. 1-1SCCh. 1.2 - Prob. 1-2SCCh. 1.2 - Prob. 1-3SCCh. 1.2 - Prob. 1-4SCCh. 1.2 - Prob. 1-5SCCh. 1.2 - Prob. 1-6SCCh. 1.2 - Prob. 1-7SCCh. 1.2 - Prob. 1-8SCCh. 1.2 - Prob. 1-9SCCh. 1.2 - Prob. 1-10SCCh. 1.2 - Prob. 1-11SCCh. 1.2 - Prob. 1-12SCCh. 1.2 - Prob. 2-1SCCh. 1.2 - Prob. 2-2SCCh. 1.2 - Prob. 2-3SCCh. 1.2 - Prob. 2-4SCCh. 1.2 - Prob. 2-5SCCh. 1.2 - Prob. 2-6SCCh. 1.2 - Prob. 3-1SCCh. 1.2 - Prob. 3-2SCCh. 1.2 - Prob. 3-3SCCh. 1.2 - Prob. 3-4SCCh. 1.2 - Prob. 3-5SCCh. 1.2 - Prob. 3-6SCCh. 1.2 - Prob. 3-7SCCh. 1.2 - Prob. 3-8SCCh. 1.2 - Prob. 3-9SCCh. 1.2 - Prob. 3-10SCCh. 1.2 - Prob. 4-1SCCh. 1.2 - Prob. 4-2SCCh. 1.2 - Prob. 4-3SCCh. 1.2 - Prob. 4-4SCCh. 1.2 - Prob. 4-5SCCh. 1.2 - Prob. 4-6SCCh. 1.2 - Prob. 4-7SCCh. 1.2 - Prob. 4-8SCCh. 1.2 - Prob. 4-9SCCh. 1.2 - Prob. 4-10SCCh. 1.2 - Prob. 5-1SCCh. 1.2 - Prob. 5-2SCCh. 1.2 - Prob. 5-3SCCh. 1.2 - Prob. 5-4SCCh. 1.2 - Prob. 1SECh. 1.2 - Prob. 2SECh. 1.2 - Prob. 3SECh. 1.2 - Prob. 4SECh. 1.2 - Prob. 5SECh. 1.2 - Prob. 6SECh. 1.2 - Prob. 7SECh. 1.2 - Prob. 8SECh. 1.2 - Prob. 9SECh. 1.2 - Prob. 10SECh. 1.2 - Prob. 11SECh. 1.2 - Prob. 12SECh. 1.2 - Prob. 13SECh. 1.2 - Prob. 14SECh. 1.2 - Prob. 15SECh. 1.2 - Prob. 16SECh. 1.2 - Prob. 17SECh. 1.2 - Prob. 18SECh. 1.2 - Prob. 19SECh. 1.2 - Prob. 20SECh. 1.2 - Prob. 21SECh. 1.2 - Prob. 22SECh. 1.2 - Prob. 23SECh. 1.2 - Prob. 24SECh. 1.2 - Prob. 25SECh. 1.2 - Prob. 26SECh. 1.2 - Prob. 27SECh. 1.2 - Prob. 28SECh. 1.2 - Prob. 29SECh. 1.2 - Prob. 30SECh. 1.2 - Prob. 31SECh. 1.2 - Prob. 32SECh. 1.2 - Prob. 33SECh. 1.2 - Prob. 34SECh. 1.2 - Prob. 35SECh. 1.2 - Prob. 36SECh. 1.2 - Prob. 37SECh. 1.2 - Prob. 38SECh. 1.2 - Prob. 39SECh. 1.2 - Prob. 40SECh. 1.2 - Prob. 41SECh. 1.2 - Prob. 42SECh. 1.2 - Prob. 43SECh. 1.2 - Prob. 44SECh. 1.2 - Prob. 45SECh. 1.2 - Prob. 46SECh. 1.2 - Prob. 47SECh. 1.2 - Prob. 48SECh. 1.2 - Prob. 49SECh. 1.2 - Prob. 50SECh. 1.2 - Prob. 51SECh. 1.2 - Prob. 52SECh. 1.2 - Prob. 53SECh. 1.2 - Prob. 54SECh. 1.2 - Prob. 55SECh. 1.2 - Prob. 56SECh. 1 - Prob. 1ESCh. 1 - Prob. 2ESCh. 1 - Prob. 3ESCh. 1 - Prob. 4ESCh. 1 - Prob. 5ESCh. 1 - Prob. 6ESCh. 1 - Prob. 7ESCh. 1 - Prob. 8ESCh. 1 - Prob. 9ESCh. 1 - Prob. 10ESCh. 1 - Prob. 11ESCh. 1 - Prob. 12ESCh. 1 - Prob. 13ESCh. 1 - Prob. 14ESCh. 1 - Prob. 15ESCh. 1 - Prob. 16ESCh. 1 - Prob. 17ESCh. 1 - Prob. 18ESCh. 1 - Prob. 19ESCh. 1 - Prob. 20ESCh. 1 - Prob. 21ESCh. 1 - Prob. 22ESCh. 1 - Prob. 23ESCh. 1 - Prob. 24ESCh. 1 - Prob. 25ESCh. 1 - Prob. 26ESCh. 1 - Prob. 27ESCh. 1 - Prob. 28ESCh. 1 - Prob. 29ESCh. 1 - Prob. 30ESCh. 1 - Prob. 31ESCh. 1 - Prob. 32ESCh. 1 - Prob. 33ESCh. 1 - Prob. 34ESCh. 1 - Prob. 35ESCh. 1 - Prob. 36ESCh. 1 - Prob. 37ESCh. 1 - Prob. 38ESCh. 1 - Prob. 39ESCh. 1 - Prob. 40ESCh. 1 - Prob. 41ESCh. 1 - Prob. 42ESCh. 1 - Prob. 43ESCh. 1 - Prob. 44ESCh. 1 - Prob. 45ESCh. 1 - Prob. 46ESCh. 1 - Prob. 47ESCh. 1 - Prob. 48ESCh. 1 - Prob. 49ESCh. 1 - Prob. 50ESCh. 1 - Prob. 51ESCh. 1 - Prob. 52ESCh. 1 - Prob. 53ESCh. 1 - Prob. 54ESCh. 1 - Prob. 55ESCh. 1 - Prob. 56ESCh. 1 - Prob. 57ESCh. 1 - Prob. 58ESCh. 1 - Prob. 59ESCh. 1 - Prob. 60ESCh. 1 - Prob. 61ESCh. 1 - Prob. 62ESCh. 1 - Prob. 63ESCh. 1 - Prob. 64ESCh. 1 - Prob. 65ESCh. 1 - Prob. 66ESCh. 1 - Prob. 67ESCh. 1 - Prob. 68ESCh. 1 - Prob. 69ESCh. 1 - Prob. 70ESCh. 1 - Prob. 71ESCh. 1 - Prob. 72ESCh. 1 - Prob. 73ESCh. 1 - Prob. 74ESCh. 1 - Prob. 75ESCh. 1 - Prob. 76ESCh. 1 - Prob. 77ESCh. 1 - Prob. 78ESCh. 1 - Prob. 79ESCh. 1 - Prob. 80ESCh. 1 - Prob. 81ESCh. 1 - Prob. 82ESCh. 1 - Prob. 83ESCh. 1 - Prob. 84ESCh. 1 - Prob. 85ESCh. 1 - Prob. 86ESCh. 1 - Prob. 87ESCh. 1 - Prob. 88ESCh. 1 - Prob. 89ESCh. 1 - Prob. 1PTCh. 1 - Prob. 2PTCh. 1 - Prob. 3PTCh. 1 - Prob. 4PTCh. 1 - Prob. 5PTCh. 1 - Prob. 6PTCh. 1 - Prob. 7PTCh. 1 - Prob. 8PTCh. 1 - Prob. 9PTCh. 1 - Prob. 10PTCh. 1 - Prob. 11PTCh. 1 - Prob. 12PTCh. 1 - Prob. 13PTCh. 1 - Prob. 14PTCh. 1 - Prob. 15PTCh. 1 - Prob. 16PTCh. 1 - Prob. 17PTCh. 1 - Prob. 18PTCh. 1 - Prob. 19PTCh. 1 - Prob. 20PTCh. 1 - Prob. 21PTCh. 1 - Prob. 22PTCh. 1 - Prob. 23PTCh. 1 - Prob. 24PTCh. 1 - Prob. 25PTCh. 1 - Prob. 26PTCh. 1 - Prob. 27PTCh. 1 - Prob. 28PTCh. 1 - Prob. 29PTCh. 1 - Prob. 30PTCh. 1 - Prob. 31PTCh. 1 - Prob. 32PTCh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 1CPCh. 1 - Prob. 2CPCh. 1 - Prob. 1CS1Ch. 1 - Prob. 2CS1Ch. 1 - Prob. 3CS1Ch. 1 - Prob. 1CS2Ch. 1 - Prob. 2CS2Ch. 1 - Prob. 3CS2Ch. 1 - Prob. 1CS3Ch. 1 - Prob. 2CS3Ch. 1 - Prob. 3CS3Ch. 1 - Prob. 4CS3Ch. 1 - Prob. 5CS3
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- By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).arrow_forwardProof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.arrow_forwardBy considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.arrow_forward
- Let 1 1 r 1+ + + 2 3 + = 823 823s Without calculating the left-hand side, prove that r = s (mod 823³).arrow_forwardFor each real-valued nonprincipal character X mod 16, verify that L(1,x) 0.arrow_forward*Construct a table of values for all the nonprincipal Dirichlet characters mod 16. Verify from your table that Σ x(3)=0 and Χ mod 16 Σ χ(11) = 0. x mod 16arrow_forward
- For each real-valued nonprincipal character x mod 16, verify that A(225) > 1. (Recall that A(n) = Σx(d).) d\narrow_forward24. Prove the following multiplicative property of the gcd: a k b h (ah, bk) = (a, b)(h, k)| \(a, b)' (h, k) \(a, b)' (h, k) In particular this shows that (ah, bk) = (a, k)(b, h) whenever (a, b) = (h, k) = 1.arrow_forward20. Let d = (826, 1890). Use the Euclidean algorithm to compute d, then express d as a linear combination of 826 and 1890.arrow_forward
- Let 1 1+ + + + 2 3 1 r 823 823s Without calculating the left-hand side, Find one solution of the polynomial congruence 3x²+2x+100 = 0 (mod 343). Ts (mod 8233).arrow_forwardBy considering appropriate series expansions, prove that ez · e²²/2 . e²³/3 . ... = 1 + x + x² + · ·. when <1.arrow_forwardProve that Σ prime p≤x p=3 (mod 10) 1 Р = for some constant A. log log x + A+O 1 log x ,arrow_forward
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