Explanation Given Information: The provided expression is 5 8 + 9 10 . Formula used: Steps to determine the least common denominator (LCD). 1. Write the denominators of the fractions in a row and divide each term by the smallest prime number which can evenly divide the number in the row. 2. Write the quotients obtained from step 1 in a row and then divide each term by another smallest prime number. 3. Repeat this process until a row of 1s is obtained. 4. Multiply all the prime numbers that are used in the process to divide the numbers in the row. The product obtained from the multiplication of prime numbers is the least common denominator (LCD). Steps to add the fractions with different denominators. 1. Determine the least common denominator. 2. Multiply by appropriate numbers to change each fraction into an equivalent fraction. 3. Add the numerators of the fractions when the denominators become equal. 4. Reduce the result in its lowest terms to write as a whole number or mixed number. Calculation: Consider the expression, 5 8 + 9 10 Write the denominators 12 , 15 in a row and then divide each term by the smallest prime number which can divide the numbers. 2 8 10 2 4 5 2 2 5 5 1 5 1 1 The least common denominator is, 2 × 2 × 2 × 5 = 40 Change each fraction into an equivalent fraction

Portfolio of Accounting Systems for Small and Medium-Sized Businesses.

11th Edition

ISBN: 9780136857761

Author: National Society of Public Accountants Staff

Publisher: Prentice Hall

See similar textbooks

Videos

Question

Chapter 2, Problem 12PT

To determine

To calculate: The sum 58+910 as a whole or mixed number in lowest terms.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2, Problem 11PT

Chapter 2, Problem 13PT

Students have asked these similar questions

1. Matrix Operations Given: A = [ 33 ]A-[3-321] -3 B = [342]B-[3-41-2] (a) A² A2 Multiply A× A: -3 = (3 x 32x-3) (3 x 22 x 1) | = |[19–63 |-9-3 -6+21] = A² = 33 33 1-3×3+1x-3) (-3×2+1x1) [12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1 ]=[3-128-5] (b) | A ||A| Determinant of A | A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9 (c) Adjoint of A Swap diagonal elements and change sign of off-diagonals: A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23] -3 (d) B-¹B-1 First find | B ||B|: |B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2 Then the adjoint of B: adj (B) = [² 3 adj(B)=[-24-13] Now, B-1 1 = |B| · adj (B) = 1 [²¯¯³¹³] = [2₂ B 0.5 |B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5] 2. (a) Matrix Method: Solve (2x + 3y = 6 (2x-3y=14 {2x+3y=62x-3y=14 Matrix form: 22 33-22 = [223-3][xy]=[614] Find inverse of coefficient matrix: Determinant: | M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12 Adjoint: adj(M) = [3]adj(M)-[-3-2-32] So…

Let the region R be the area enclosed by the function f(x)= = 3x² and g(x) = 4x. If the region R is the base of a solid such that each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in the region R, find the volume of the solid. You may use a calculator and round to the nearest thousandth. y 11 10 9 00 8 7 9 5 4 3 2 1 -1 -1 x 1 2

Using the accompanying Accounting Professionals data to answer the following questions. a. Find and interpret a 90% confidence interval for the mean years of service. b. Find and interpret a 90% confidence interval for the proportion of employees who have a graduate degree. view the Accounting Professionals data. Employee Years of Service Graduate Degree?1 26 Y2 8 N3 10 N4 6 N5 23 N6 5 N7 8 Y8 5 N9 26 N10 14 Y11 10 N12 8 Y13 7 Y14 27 N15 16 Y16 17 N17 21 N18 9 Y19 9 N20 9 N Question content area bottom Part 1 a. A 90% confidence interval for the mean years of service is (Use ascending order. Round to two decimal places as needed.)

Chapter 2 Solutions

Portfolio of Accounting Systems for Small and Medium-Sized Businesses.

Ch. 2.1 - Prob. 1-1SC Ch. 2.1 - Prob. 1-2SC Ch. 2.1 - Prob. 1-3SC Ch. 2.1 - Prob. 1-4SC Ch. 2.1 - Prob. 1-5SC Ch. 2.1 - Prob. 1-6SC Ch. 2.1 - Prob. 2-1SC Ch. 2.1 - Prob. 2-2SC Ch. 2.1 - Prob. 2-3SC Ch. 2.1 - Prob. 2-4SC

Ch. 2.1 - Prob. 2-5SC Ch. 2.1 - Prob. 3-1SC Ch. 2.1 - Prob. 3-2SC Ch. 2.1 - Prob. 3-3SC Ch. 2.1 - Prob. 3-4SC Ch. 2.1 - Prob. 3-5SC Ch. 2.1 - Prob. 4-1SC Ch. 2.1 - Prob. 4-2SC Ch. 2.1 - Prob. 4-3SC Ch. 2.1 - Prob. 4-4SC Ch. 2.1 - Prob. 4-5SC Ch. 2.1 - Prob. 4-6SC Ch. 2.1 - Prob. 4-7SC Ch. 2.1 - Prob. 4-8SC Ch. 2.1 - Prob. 5-1SC Ch. 2.1 - Prob. 5-2SC Ch. 2.1 - Prob. 5-3SC Ch. 2.1 - Prob. 5-4SC Ch. 2.1 - Prob. 5-5SC Ch. 2.1 - Prob. 5-6SC Ch. 2.1 - Prob. 1SE Ch. 2.1 - Prob. 2SE Ch. 2.1 - Prob. 3SE Ch. 2.1 - Prob. 4SE Ch. 2.1 - Prob. 5SE Ch. 2.1 - Prob. 6SE Ch. 2.1 - Prob. 7SE Ch. 2.1 - Prob. 8SE Ch. 2.1 - Prob. 9SE Ch. 2.1 - Prob. 10SE Ch. 2.1 - Prob. 11SE Ch. 2.1 - Prob. 12SE Ch. 2.1 - Prob. 13SE Ch. 2.1 - Prob. 14SE Ch. 2.1 - Prob. 15SE Ch. 2.1 - Prob. 16SE Ch. 2.1 - Prob. 17SE Ch. 2.1 - Prob. 18SE Ch. 2.1 - Prob. 19SE Ch. 2.1 - Prob. 20SE Ch. 2.1 - Prob. 21SE Ch. 2.1 - Prob. 22SE Ch. 2.1 - Prob. 23SE Ch. 2.1 - Prob. 24SE Ch. 2.1 - Prob. 25SE Ch. 2.1 - Prob. 26SE Ch. 2.1 - Prob. 27SE Ch. 2.1 - Prob. 28SE Ch. 2.1 - Prob. 29SE Ch. 2.1 - Prob. 30SE Ch. 2.1 - Prob. 31SE Ch. 2.1 - Prob. 32SE Ch. 2.1 - Prob. 33SE Ch. 2.1 - Prob. 34SE Ch. 2.1 - Prob. 35SE Ch. 2.1 - Prob. 36SE Ch. 2.1 - Prob. 37SE Ch. 2.1 - Prob. 38SE Ch. 2.2 - Prob. 1-1SC Ch. 2.2 - Prob. 1-2SC Ch. 2.2 - Prob. 1-3SC Ch. 2.2 - Prob. 1-4SC Ch. 2.2 - Prob. 1-5SC Ch. 2.2 - Prob. 2-1SC Ch. 2.2 - Prob. 2-2SC Ch. 2.2 - Prob. 2-3SC Ch. 2.2 - Prob. 2-4SC Ch. 2.2 - Prob. 2-5SC Ch. 2.2 - Prob. 3-1SC Ch. 2.2 - Prob. 3-2SC Ch. 2.2 - Prob. 3-3SC Ch. 2.2 - Prob. 3-4SC Ch. 2.2 - Prob. 3-5SC Ch. 2.2 - Prob. 3-6SC Ch. 2.2 - Prob. 4-1SC Ch. 2.2 - Prob. 4-2SC Ch. 2.2 - Prob. 4-3SC Ch. 2.2 - Prob. 4-4SC Ch. 2.2 - Prob. 4-5SC Ch. 2.2 - Prob. 4-6SC Ch. 2.2 - Prob. 4-7SC Ch. 2.2 - Prob. 4-8SC Ch. 2.2 - Prob. 4-9SC Ch. 2.2 - Prob. 4-10SC Ch. 2.2 - Prob. 1SE Ch. 2.2 - Prob. 2SE Ch. 2.2 - Prob. 3SE Ch. 2.2 - Prob. 4SE Ch. 2.2 - Prob. 5SE Ch. 2.2 - Prob. 6SE Ch. 2.2 - Prob. 7SE Ch. 2.2 - Prob. 8SE Ch. 2.2 - Prob. 9SE Ch. 2.2 - Prob. 10SE Ch. 2.2 - Prob. 11SE Ch. 2.2 - Prob. 12SE Ch. 2.2 - Prob. 13SE Ch. 2.2 - Prob. 14SE Ch. 2.2 - Prob. 15SE Ch. 2.2 - Prob. 16SE Ch. 2.2 - Prob. 17SE Ch. 2.2 - Prob. 18SE Ch. 2.2 - Prob. 19SE Ch. 2.2 - Prob. 20SE Ch. 2.2 - Prob. 21SE Ch. 2.2 - Prob. 22SE Ch. 2.2 - Prob. 23SE Ch. 2.2 - Prob. 24SE Ch. 2.2 - Prob. 25SE Ch. 2.2 - Prob. 26SE Ch. 2.2 - Prob. 27SE Ch. 2.2 - Prob. 28SE Ch. 2.2 - Prob. 29SE Ch. 2.2 - Prob. 30SE Ch. 2.3 - Prob. 1-1SC Ch. 2.3 - Prob. 1-2SC Ch. 2.3 - Prob. 1-3SC Ch. 2.3 - Prob. 1-4SC Ch. 2.3 - Prob. 1-5SC Ch. 2.3 - Prob. 1-6SC Ch. 2.3 - Prob. 1-7SC Ch. 2.3 - Prob. 2-1SC Ch. 2.3 - Prob. 2-2SC Ch. 2.3 - Prob. 2-3SC Ch. 2.3 - Prob. 2-4SC Ch. 2.3 - Prob. 2-5SC Ch. 2.3 - Prob. 2-6SC Ch. 2.3 - Prob. 2-7SC Ch. 2.3 - Prob. 1SE Ch. 2.3 - Prob. 2SE Ch. 2.3 - Prob. 3SE Ch. 2.3 - Prob. 4SE Ch. 2.3 - Prob. 5SE Ch. 2.3 - Prob. 6SE Ch. 2.3 - Prob. 7SE Ch. 2.3 - Prob. 8SE Ch. 2.3 - Prob. 9SE Ch. 2.3 - Prob. 10SE Ch. 2.3 - Prob. 11SE Ch. 2.3 - Prob. 12SE Ch. 2.3 - Prob. 13SE Ch. 2.3 - Prob. 14SE Ch. 2.3 - Prob. 15SE Ch. 2.3 - Prob. 16SE Ch. 2.3 - Prob. 17SE Ch. 2.3 - Prob. 18SE Ch. 2.3 - Prob. 19SE Ch. 2.3 - Prob. 20SE Ch. 2.3 - Prob. 21SE Ch. 2.3 - Prob. 22SE Ch. 2.3 - Prob. 23SE Ch. 2.3 - Prob. 24SE Ch. 2 - Prob. 1ES Ch. 2 - Prob. 2ES Ch. 2 - Prob. 3ES Ch. 2 - Prob. 4ES Ch. 2 - Prob. 5ES Ch. 2 - Prob. 6ES Ch. 2 - Prob. 7ES Ch. 2 - Prob. 8ES Ch. 2 - Prob. 9ES Ch. 2 - Prob. 10ES Ch. 2 - Prob. 11ES Ch. 2 - Prob. 12ES Ch. 2 - Prob. 13ES Ch. 2 - Prob. 14ES Ch. 2 - Prob. 15ES Ch. 2 - Prob. 16ES Ch. 2 - Prob. 17ES Ch. 2 - Prob. 18ES Ch. 2 - Prob. 19ES Ch. 2 - Prob. 20ES Ch. 2 - Prob. 21ES Ch. 2 - Prob. 22ES Ch. 2 - Prob. 23ES Ch. 2 - Prob. 24ES Ch. 2 - Prob. 25ES Ch. 2 - Prob. 26ES Ch. 2 - Prob. 27ES Ch. 2 - Prob. 28ES Ch. 2 - Prob. 29ES Ch. 2 - Prob. 30ES Ch. 2 - Prob. 31ES Ch. 2 - Prob. 32ES Ch. 2 - Prob. 33ES Ch. 2 - Prob. 34ES Ch. 2 - Prob. 35ES Ch. 2 - Prob. 36ES Ch. 2 - Prob. 37ES Ch. 2 - Prob. 38ES Ch. 2 - Prob. 39ES Ch. 2 - Prob. 40ES Ch. 2 - Prob. 41ES Ch. 2 - Prob. 42ES Ch. 2 - Prob. 43ES Ch. 2 - Prob. 44ES Ch. 2 - Prob. 45ES Ch. 2 - Prob. 46ES Ch. 2 - Prob. 47ES Ch. 2 - Prob. 48ES Ch. 2 - Prob. 49ES Ch. 2 - Prob. 50ES Ch. 2 - Prob. 51ES Ch. 2 - Prob. 52ES Ch. 2 - Prob. 53ES Ch. 2 - Prob. 54ES Ch. 2 - Prob. 55ES Ch. 2 - Prob. 56ES Ch. 2 - Prob. 57ES Ch. 2 - Prob. 58ES Ch. 2 - Prob. 59ES Ch. 2 - Prob. 60ES Ch. 2 - Prob. 61ES Ch. 2 - Prob. 62ES Ch. 2 - Prob. 63ES Ch. 2 - Prob. 64ES Ch. 2 - Prob. 65ES Ch. 2 - Prob. 66ES Ch. 2 - Prob. 67ES Ch. 2 - Prob. 68ES Ch. 2 - Prob. 69ES Ch. 2 - Prob. 70ES Ch. 2 - Prob. 71ES Ch. 2 - Prob. 72ES Ch. 2 - Prob. 73ES Ch. 2 - Prob. 74ES Ch. 2 - Prob. 75ES Ch. 2 - Prob. 76ES Ch. 2 - Prob. 78ES Ch. 2 - Prob. 79ES Ch. 2 - Prob. 80ES Ch. 2 - Prob. 81ES Ch. 2 - Prob. 82ES Ch. 2 - Prob. 83ES Ch. 2 - Prob. 84ES Ch. 2 - Prob. 1PT Ch. 2 - Prob. 2PT Ch. 2 - Prob. 3PT Ch. 2 - Prob. 4PT Ch. 2 - Prob. 5PT Ch. 2 - Prob. 6PT Ch. 2 - Prob. 7PT Ch. 2 - Prob. 8PT Ch. 2 - Prob. 9PT Ch. 2 - Prob. 10PT Ch. 2 - Prob. 11PT Ch. 2 - Prob. 12PT Ch. 2 - Prob. 13PT Ch. 2 - Prob. 14PT Ch. 2 - Prob. 15PT Ch. 2 - Prob. 16PT Ch. 2 - Prob. 17PT Ch. 2 - Prob. 18PT Ch. 2 - Prob. 19PT Ch. 2 - Prob. 20PT Ch. 2 - Prob. 21PT Ch. 2 - Prob. 22PT Ch. 2 - Prob. 23PT Ch. 2 - Prob. 24PT Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 1CP Ch. 2 - Prob. 2CP Ch. 2 - Prob. 1CS1 Ch. 2 - Prob. 2CS1 Ch. 2 - Prob. 3CS1 Ch. 2 - Prob. 1CS2 Ch. 2 - Prob. 2CS2 Ch. 2 - Prob. 3CS2 Ch. 2 - Prob. 4CS2

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The sum 5 8 + 9 10 as a whole or mixed number in lowest terms.

Videos

To calculate: The sum 58+910 as a whole or mixed number in lowest terms.

Want to see the full answer?

Chapter 2 Solutions