A 1 C 10 B 5 ( D 15 11. The product of 763 and 5 498 lies between A 0.4 million and 0.5 million B 0.5 million and 0.6 million C 4 million and 5 million ( D 5 million and 6 million 12. 394 – 125 ÷ 25 = A 179 C 369 B 269 D 389 ( 13. (47 × 13) – (637 ÷ 13) = A 26 C 562 ( B 47 D 611 14. Table 1 shows the number of residents in three cities.
Tanglewood Case 5 Historical Hiring and Promotion Data: Applicant Flows Occupational category | | | Total | White* | Total Non-White | African-American* | Store Associates | External hires | Applicants | 18226 | 15436 | 2790 | 594 | | | Hires | 3832 | 3221 | 611 | 135 | | | Selection ratio | 21.02% | 20.87% | 21.90% | 22.73% | | | | | | | | Shift leader | External hires | Applicants | 392 | 320 | 72 | 17 | | | Hires | 61 | 54 | 7 | 2 | | | Selection ratio | 15.56% | 16.88% | 9.72 | 11.76 | | Internal hires | Applicants | 864 | 712 | 152 | 30 | | | Hires | 280 | 241 | 39 | 6 | | | Selection ratio | 32.41% | 33.85 | 25.66 | 20 | | | | | | | | Department manager | External hires | Applicants | 1242 | 1074 | 168 | 44 | | | Hires | 94 | 82 | 12 | 3 | | | Selection ratio | 7.57% | 7.64% | 7.14 | 6.82 | | Internal hires | Applicants | 589 | 509 | 80 | 21 | | | Hires | 124 | 108 | 16 | 3 | | | Selection ratio | 21.05% | 21.22 | 20 | 14.29 | | | | | | | | Asst. store manager | External hires | Applicants | 146 | 123 | 23 | 7 | | | Hires | 17 | 15 | 2 | 0 | | | Selection ratio | 11.64% | 8.2 | 8.70 | 0 | | Internal hires | Applicants | 108 | 90 | 18 | 4 | | | Hires | 27 | 25 | 2 | 1 | | | Selection ratio | 25.00% | 27.80 | 11.11 | 25 | | | | | | | | Store Manager | External hires | Applicants | 50 | 42 | 8 | 2 | | | Hires | 5 | 4 | 1 | 0 | | | Selection ratio | 10.00% | 9.52 | 12.5 | 0 | | Internal hires | Applicants | 81 | 66 | 15 | 4 | | | Hires | 13 | 9 | 4 | 0 | | | Selection ratio | 16.05% | 13.64 | 26.67 | 0 | *Non-Hispanic A.) The chart above shows the selection ratios by each job category for positions at Tanglewood. According
FINAL ASSIGNMENT PART I: 1. Calculate the mean yearly value using the average gas prices by month found in the “Final Project Data Set.” Year | Yearly Mean | 1982 | 1.296 | 1983 | 1.241 | 1984 | 1.212 | 1985 | 1.202 | 1986 | 0.927 | 1987 | 0.948 | 1988 | 0.946 | 1989 | 1.022 | 1990 | 1.164 | 1991 | 1.140 | 1992 | 1.127 | 1993 | 1.108 | 1994 | 1.112 | 1995 | 1.147 | 1996 | 1.231 | 1997 | 1.234 | 1998 | 1.059 | 1999 | 1.165 | 2000 | 1.510 | 2001 | 1.461 | 2002 | 1.358 | 2003 | 1.591 | 2004 | 1.880 | 2005 | 2.295 | 2006 | 2.589 | 2007 | 2.801 | 2008 | 3.266 | 2009 | 2.350 | 2010 | 2.788 | 2011 | 3.129 | 2. Using the years as your x-axis and the annual mean as your y-axis, create a scatterplot and a linear regression line. 3. Answer the following questions using your scatterplot and linear regression line: * What is the slope of the linear regression line?
Omega Research Security Categorization: High Omega Research Inc. Information System Contingency Plan (ISCP) Version [1] [August 20, 2010] Prepared by [Craig Spehar] [2017 East Ridge Dr] [Excelsior Springs, MO., 64024] Table of Contents Plan Approval 4 1. Introduction 5 1.1 Background 5 1.2 Scope 5 1.3 Assumptions 5 2. Concept of Operations 6 2.1 System Description 6 2.1.A. Local Area Architecture 6 2.2 Overview of Three Phases 11 2.3 Roles and Responsibilities 12 2.3.1 Damage Assessment Team 12 2.1.2 Operations Team 13 2.1.3 Communications Team 13 2.1.4 Data Entry and Control Team 13 2.1.5 Off-Site Storage Team 13 2.1.6 Administrative Management Team 13 2.1.7 Procurement Team 13 2.1.8 Configuration Management Team 14 2.1.9 Facilities Team 14 2.1.10 System Software Team 14 2.1.11 Internal Audit Team 14 2.1.12 User Assistance Team 14 3. Activation and Notification 14 3.1 Activation Criteria and Procedure 14 3.2 Notification 15 3.3 Outage Assessment 15 4.
Topic: Assurance Services 5. Attestation risk, like audit risk consists of three components--inherent risk, control risk, and substantiation risk. FALSE AACSB: Analytic AICPA BB: Industry AICPA FN: Measurement AICPA FN: Risk Analysis Bloom's: Understand Difficulty: Medium Learning Objective: 20-02 Explain the applicability of the attestation
Given: sides, 5,6,7 in miles Island A = 78.46° Island B = 135.58° Island C = 57.12° Given the information… From Island B I would travel a Northwest Bearing to Island C. 8. Heron’s formula: sq.root of {s(s-a)(s-b)(s-c)} Sqrt((a^2+b^2+c^2)^2 - 2 (a^4+b^4+c^4))/4 = sqrt ((25+25+16)^2-2(625+625+256))/4 = sqrt ((66)^2-3012)/4 = sqrt (4356-3012)/4 = 36.66060556/4 = 9.16515139 square feet 9 feet
| 0.00098 | $392 | $411,041,792 | 1960 | E.E.Co. | 0.00098 | $5 | $5,242,880 | 1965 | IBM | 0.00098 | $2.52 | $2,642,412 | 1970 | IBM | 0.00098 | $0.70 | $734,003 | 1975 | MITS | 0.25 | $103 | $421,888 | 1980 | Interface Age | 64 | $405 | $6,480 | 1985 | Do Kay BYTE | 512 | $440 | $880 | 1990 | Unitex BYTE | 8,192 | $851 | $106 | 1995 | Pacific Coast Micro | 16,384 | $494 | $30.9 | 2000 | Crucial | 65,536 | $72 | $1.12 | 2005 | Corsair | 1,048,576 | $189 | $0.185 | 2010 | Kingston | 8,388,608 | $99 | $0.0122 | 2013 | Crucial | 16,777,216 | $88 | $0.0054 | In 1957, one bit of RAM cost roughly $49 dollars based on the chart above. In 1980, the price per GB was roughly $100,000. Today, it would not even cost a dollar. “Several terabyte+ drives have recently broken the $0.10mory Cost Unit 6 Analysis 3/gigabyte barrier, making the next milestone $0.01/gigabyte, or $10/terabyte.
Pre-Lab Questions: 1. Absorbance = - Log10T = 2-Log10 (%T) a) 2 – Log (89.95) = .0460 f) 2 – Log (28.18) = .5501 b) 2 – Log (80.91) = .0920 g) 2 – Log (20.51) = .6880 c) 2 – Log (59.02) = .2290 h) 2 – Log (17.38) = .7600 d) 2 – Log (47.75) = .3210 i) 2 – Log (14.45) = .8401 e) 2 – Log (34.83) = .4580 2. Molar Concentration 1.213 g Cu (1 mol Cu/ 63.546 g Cu) (1 mol [Cu (H2O6)2+]/ 1 mol Cu) (1/ .100 L) = .191 M [Cu (H2O6)2+] a) M2 = (M1V1)/V2 M2 = (.191 M [Cu (H2O6)2+])( 1.0 mL)/ (50.0 mL) = .0038 M b) (.191 M [Cu (H2O6)2+]) (2.0 mL)/ (50.0 mL) = .0076 M c) (.191 M [Cu (H2O6)2+]) (5.0 mL)/ (50.0 mL) = .0191 M d) (.191 M [Cu (H2O6)2+]) (7.0 mL)/ (50.0 mL) = .0267 M e) (.191 M [Cu (H2O6)2+]) (10.0 mL)/ (50.0 mL) = .0382 M f) (.191 M [Cu (H2O6)2+]) (12.0 mL)/ (50.0 mL) = .0458 M g) (.191 M [Cu (H2O6)2+]) (15.0 mL)/
2) Case Shipments = -80,315.68 + 996.56*49 + 3877.95*113 + .096*960,000 + .076*154,455 - .048*345,028.8 - .02*368,337.5 = 483,432 Standard Error = 34,733.12 given by Excel Confidence Interval of 95% -> T Value of 97.5% for 42 degrees of freedom is 2.018 2.018 * 34,733.12 * SQRT (1 + (1/48)) = 70,818 95% Confidence Interval = Estimate +/- Standard Error*T-Value*SQRT(1+(1/n)) The 95% Confidence Interval is from 412,614 and 554,250 cases. 3) $1 invested into the Dealer Allowance would increase current sales by .076 cases and decrease sales in two months by .02 cases, resulting in a net increase of .056 cases. Assuming that Harmon Foods is able to sell their cereal for $2/pack, the value of a case is $48 ($2 * 24 packs/case). Therefore, $1 invested in Dealer Allowance equates to $2.69 in added sales revenue (assuming all other
5 x 106 b. 4.5 x 106 c. 4.61 x 106 d. 4.6100 x 106 5. a. 4 x 10-4 b. 3.7 x 10-4 c. 3.71 x 10-4 d. 3.711 x 10-4 6. 25.06 7.