Business Math (11th Edition)
11th Edition
ISBN: 9780134496436
Author: Cheryl Cleaves, Margie Hobbs, Jeffrey Noble
Publisher: PEARSON
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Chapter 3, Problem 30ES
To determine
To calculate: The value of the expression
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Chapter 3 Solutions
Business Math (11th Edition)
Ch. 3.1 - Prob. 1-1SCCh. 3.1 - Prob. 1-2SCCh. 3.1 - Prob. 1-3SCCh. 3.1 - Prob. 1-4SCCh. 3.1 - Prob. 1-5SCCh. 3.1 - Prob. 1-6SCCh. 3.1 - Prob. 2-1SCCh. 3.1 - Prob. 2-2SCCh. 3.1 - Prob. 2-3SCCh. 3.1 - Prob. 2-4SC
Ch. 3.1 - Prob. 1SECh. 3.1 - Prob. 2SECh. 3.1 - Prob. 3SECh. 3.1 - Prob. 4SECh. 3.1 - Prob. 5SECh. 3.1 - Prob. 6SECh. 3.1 - Prob. 7SECh. 3.1 - Prob. 8SECh. 3.1 - Prob. 9SECh. 3.1 - Prob. 10SECh. 3.1 - Prob. 11SECh. 3.1 - Prob. 12SECh. 3.1 - Prob. 13SECh. 3.1 - Prob. 14SECh. 3.1 - Prob. 15SECh. 3.1 - Prob. 16SECh. 3.1 - Prob. 17SECh. 3.1 - Prob. 18SECh. 3.1 - Prob. 19SECh. 3.1 - Prob. 20SECh. 3.1 - Prob. 21SECh. 3.1 - Prob. 22SECh. 3.1 - Prob. 23SECh. 3.1 - Prob. 24SECh. 3.1 - Prob. 25SECh. 3.1 - Prob. 26SECh. 3.2 - Prob. 1-1SCCh. 3.2 - Prob. 1-2SCCh. 3.2 - Prob. 1-3SCCh. 3.2 - Prob. 1-4SCCh. 3.2 - Prob. 1-5SCCh. 3.2 - Prob. 1-6SCCh. 3.2 - Prob. 2-1SCCh. 3.2 - Prob. 2-2SCCh. 3.2 - Prob. 2-3SCCh. 3.2 - Prob. 2-4SCCh. 3.2 - Prob. 2-5SCCh. 3.2 - Prob. 2-6SCCh. 3.2 - Prob. 2-7SCCh. 3.2 - Prob. 2-8SCCh. 3.2 - Prob. 2-9SCCh. 3.2 - Prob. 2-10SCCh. 3.2 - Prob. 2-11SCCh. 3.2 - Prob. 2-12SCCh. 3.2 - Prob. 3-1SCCh. 3.2 - Prob. 3-2SCCh. 3.2 - Prob. 3-3SCCh. 3.2 - Prob. 3-4SCCh. 3.2 - Prob. 3-5SCCh. 3.2 - Prob. 3-6SCCh. 3.2 - Prob. 3-7SCCh. 3.2 - Prob. 1SECh. 3.2 - Prob. 2SECh. 3.2 - Prob. 3SECh. 3.2 - Prob. 4SECh. 3.2 - Prob. 5SECh. 3.2 - Prob. 6SECh. 3.2 - Prob. 7SECh. 3.2 - Prob. 8SECh. 3.2 - Prob. 9SECh. 3.2 - Prob. 10SECh. 3.2 - Prob. 11SECh. 3.2 - Prob. 12SECh. 3.2 - Prob. 13SECh. 3.2 - Prob. 14SECh. 3.2 - Prob. 15SECh. 3.2 - Prob. 16SECh. 3.2 - Prob. 17SECh. 3.2 - Prob. 18SECh. 3.2 - Prob. 19SECh. 3.2 - Prob. 20SECh. 3.2 - Prob. 21SECh. 3.2 - Prob. 22SECh. 3.2 - Prob. 23SECh. 3.2 - Prob. 24SECh. 3.2 - Prob. 25SECh. 3.2 - Prob. 26SECh. 3.2 - Prob. 27SECh. 3.2 - Prob. 28SECh. 3.2 - Prob. 29SECh. 3.2 - Prob. 30SECh. 3.2 - Prob. 31SECh. 3.2 - Prob. 32SECh. 3.2 - Prob. 33SECh. 3.2 - Prob. 34SECh. 3.2 - Prob. 35SECh. 3.2 - Prob. 36SECh. 3.2 - Prob. 37SECh. 3.2 - Prob. 38SECh. 3.3 - Prob. 1-1SCCh. 3.3 - Prob. 1-2SCCh. 3.3 - Prob. 1-3SCCh. 3.3 - Prob. 1-4SCCh. 3.3 - Prob. 1-5SCCh. 3.3 - Prob. 2-1SCCh. 3.3 - Prob. 2-2SCCh. 3.3 - Prob. 2-3SCCh. 3.3 - Prob. 2-4SCCh. 3.3 - Prob. 2-5SCCh. 3.3 - Prob. 2-6SCCh. 3.3 - Prob. 1SECh. 3.3 - Prob. 2SECh. 3.3 - Prob. 3SECh. 3.3 - Prob. 4SECh. 3.3 - Prob. 5SECh. 3.3 - Prob. 6SECh. 3.3 - Prob. 7SECh. 3.3 - Prob. 8SECh. 3.3 - Prob. 9SECh. 3.3 - Prob. 10SECh. 3.3 - Prob. 11SECh. 3.3 - Prob. 12SECh. 3.3 - Prob. 13SECh. 3.3 - Prob. 14SECh. 3.3 - Prob. 15SECh. 3.3 - Prob. 16SECh. 3.3 - Prob. 17SECh. 3.3 - Prob. 18SECh. 3.3 - Prob. 19SECh. 3.3 - Prob. 20SECh. 3 - Prob. 1ESCh. 3 - Prob. 2ESCh. 3 - Prob. 3ESCh. 3 - Prob. 4ESCh. 3 - Prob. 5ESCh. 3 - Prob. 6ESCh. 3 - Prob. 7ESCh. 3 - Prob. 8ESCh. 3 - Prob. 9ESCh. 3 - Prob. 10ESCh. 3 - Prob. 11ESCh. 3 - Prob. 12ESCh. 3 - Prob. 13ESCh. 3 - Prob. 14ESCh. 3 - Prob. 15ESCh. 3 - Prob. 16ESCh. 3 - Prob. 17ESCh. 3 - Prob. 18ESCh. 3 - Prob. 19ESCh. 3 - Prob. 20ESCh. 3 - Prob. 21ESCh. 3 - Prob. 22ESCh. 3 - Prob. 23ESCh. 3 - Prob. 24ESCh. 3 - Prob. 25ESCh. 3 - Prob. 26ESCh. 3 - Prob. 27ESCh. 3 - Prob. 28ESCh. 3 - Prob. 29ESCh. 3 - Prob. 30ESCh. 3 - Prob. 31ESCh. 3 - Prob. 32ESCh. 3 - Prob. 33ESCh. 3 - Prob. 34ESCh. 3 - Prob. 35ESCh. 3 - Prob. 36ESCh. 3 - Prob. 37ESCh. 3 - Prob. 38ESCh. 3 - Prob. 39ESCh. 3 - Prob. 40ESCh. 3 - Prob. 41ESCh. 3 - Prob. 42ESCh. 3 - Prob. 43ESCh. 3 - Prob. 44ESCh. 3 - Prob. 45ESCh. 3 - Prob. 46ESCh. 3 - Prob. 47ESCh. 3 - Prob. 48ESCh. 3 - Prob. 49ESCh. 3 - Prob. 50ESCh. 3 - Prob. 51ESCh. 3 - Prob. 52ESCh. 3 - Prob. 53ESCh. 3 - Prob. 54ESCh. 3 - Prob. 55ESCh. 3 - Prob. 56ESCh. 3 - Prob. 57ESCh. 3 - Prob. 58ESCh. 3 - Prob. 59ESCh. 3 - Prob. 60ESCh. 3 - Prob. 61ESCh. 3 - Prob. 62ESCh. 3 - Prob. 63ESCh. 3 - Prob. 64ESCh. 3 - Prob. 65ESCh. 3 - Prob. 66ESCh. 3 - Prob. 67ESCh. 3 - Prob. 68ESCh. 3 - Prob. 69ESCh. 3 - Prob. 70ESCh. 3 - Prob. 71ESCh. 3 - Prob. 72ESCh. 3 - Prob. 73ESCh. 3 - Prob. 74ESCh. 3 - Prob. 75ESCh. 3 - Prob. 76ESCh. 3 - Prob. 77ESCh. 3 - Prob. 78ESCh. 3 - Prob. 79ESCh. 3 - Prob. 80ESCh. 3 - Prob. 1PTCh. 3 - Prob. 2PTCh. 3 - Prob. 3PTCh. 3 - Prob. 4PTCh. 3 - Prob. 5PTCh. 3 - Prob. 6PTCh. 3 - Prob. 7PTCh. 3 - Prob. 8PTCh. 3 - Prob. 9PTCh. 3 - Prob. 10PTCh. 3 - Prob. 11PTCh. 3 - Prob. 12PTCh. 3 - Prob. 13PTCh. 3 - Prob. 14PTCh. 3 - Prob. 15PTCh. 3 - Prob. 16PTCh. 3 - Prob. 17PTCh. 3 - Prob. 18PTCh. 3 - Prob. 19PTCh. 3 - Prob. 20PTCh. 3 - Prob. 21PTCh. 3 - Prob. 22PTCh. 3 - Prob. 23PTCh. 3 - Prob. 24PTCh. 3 - Prob. 25PTCh. 3 - Prob. 26PTCh. 3 - Prob. 27PTCh. 3 - Prob. 28PTCh. 3 - Prob. 29PTCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 1CPCh. 3 - Prob. 2CPCh. 3 - Prob. 1CS1Ch. 3 - Prob. 2CS1Ch. 3 - Prob. 3CS1Ch. 3 - Prob. 4CS1Ch. 3 - Prob. 1CS2Ch. 3 - Prob. 2CS2
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- 2. Let X be a random variable. (a) Show that, if E X2 = 1 and E X4arrow_forward18. If m n compute the gcd (a² + 1, a² + 1) in terms of a. [Hint: Let A„ = a² + 1 and show that A„|(Am - 2) if m > n.]arrow_forward2. (5 points) Let f(x) = = - - - x² − 3x+7. Find the local minimum and maximum point(s) of f(x), and write them in the form (a, b), specifying whether each point is a minimum or maximum. Coordinates should be kept in fractions. Additionally, provide in your answer if f(x) has an absolute minimum or maximum over its entire domain with their corresponding values. Otherwise, state that there is no absolute maximum or minimum. As a reminder, ∞ and -∞ are not considered absolute maxima and minima respectively.arrow_forwardLet h(x, y, z) = — In (x) — z y7-4z - y4 + 3x²z — e²xy ln(z) + 10y²z. (a) Holding all other variables constant, take the partial derivative of h(x, y, z) with respect to x, 2 h(x, y, z). მ (b) Holding all other variables constant, take the partial derivative of h(x, y, z) with respect to y, 2 h(x, y, z).arrow_forwardmath help plzarrow_forward1. Show that, for any non-negative random variable X, EX+E+≥2, X E max X. 21.arrow_forwardFor 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). narrow_forwardBy 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_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_iosRecommended textbooks for you
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