What is the current ratio of Chester Use the results of the Balance.pdf

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What is the current ratio of Chester? Use the results of the Balance Sheet Survey Below.Balance Sheet SurveyAndrewsBaldwinChesterDigbyErieFerrisCash$3,434$0$15,756$3,605$52,632$5,613Acc ounts Receivable$8,307$14,914$12,076$9,550$16,903$4,576Inventory$8,617$25,931$5,044$3,491$3 ,473$10,616Total Current Assets$20,358$40,845$32,876$16,646$73,008$20,804Plant and equipment$113,800$177,400$124,100$122,200$193,600$78,300Accumulated Depreciation($37,933)($76,903)($33,454)($40,920)($71,416)($42,600)Total Fixed Assets$75,867$100,497$90,646$81,280$122,184$35,700Total Assets$96,225$141,342$123,523$97,926$195,192$56,504Accounts Payable$6,583$8,687$11,450$4,443$10,571$3,463Current Debt$0$12,555$7,550$10,000$21,830$9,975Long Term Debt$41,700$69,850$56,850$59,850$59,643$16,308Total Liabilities$48,283$91,092$75,850$74,293$92,044$29,747Common Stock$18,360$22,860$22,775$18,360$24,310$19,119Retained Earnings$29,582$27,391$24,898$5,273$78,839$7,638Total Equity$47,942$50,251$47,673$23,633$103,148$26,757Total Liabilities & Owner\'s Equity$96,225$141,342$123,523$97,926$195,192$56,504 Solution Current ratio = Total Current Assets /Current Liabilities = 32876/(11450+7550)= 1.73.

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Round: 5 Dec. 31, 2018 C58538 Andrews Eugene Ellis Philip Mull Baldwin Lisa Chastain Monica Dalke matt rix Chester Digby Erie Ferris Selected Financial Statistics Andrews Baldwin Chester Digby Erie Ferris ROS -19.6% -1.2% 6.8% 5.2% 10.2% 4.7% Asset Turnover 0.75 1.21 1.11 1.03 0.97 1.18 ROA -14.7% -1.5% 7.5% 5.3% 9.9% 5.6% Leverage (Assets/Equity) -2.4 2.1 2.1 2.1 1.9 2.0 ROE -35.0% -3.1% 15.5% 11.1% 19.0% 11.4% Emergency Loan $105,583,142 $20,325,028 $0 $0 $0 $0 Sales $67,336,667 $108,130,631 $221,330,602 $233,665,953 $220,090,490 $242,146,784 EBIT $2,460,785 $4,571,252 $35,109,568 $32,357,440 $47,748,831 $29,612,346 Profits ($13,191,015) ($1,297,803) $15,037,220 $12,174,914 $22,360,139 $11,495,535 Cumulative Profit ($90,335,004) ($1,743,807) $62,774,268 $63,774,089 $57,636,889 $56,805,581 SG&A / Sales 11.1% 7.3% 9.8% 11.1% 6.7% 9.1% Contrib. Margin % 6.9% 18.6% 32.1% 31.7% 36.2% 26.2% CAPSTONE® COURIER Page 1 Stock & Bonds C58538 Round: 5Dec. 31, 2018 Stock Market Summary Company Close Change Shares MarketCap ($M) Book Value EPS Dividend Yield P/E Andrews $1.00 $0.00 2,291,939 $2 ($16.42) ($5.76) $0.00 0.0% -0.2 Baldwin $6.27 $1.26 2,000,000 $13 $21.00 ($0.65) $0.00 0.0% -9.7 Chester $89.62 $12.03 2,124,345 $190 $45.75 $7.08 $2.27 2.5% 12.7 Digby $88.94 $5.88 2,089,792 $186 $52.65 $5.83 $0.00 0.0% 15.3 Erie $84.15 $27.36 2,452,736 $206 $48.02 $9.12 $0.00 0.0% 9.2 Ferris $73.41 $5.35 2,187,083 $161 $45.97 $5.26 $0.00 0.0% 14.0 Bond Market Summary Company Series# Face Yield Close$ S&P Company Series# Face Yield Close$ S&P Andrews Digby 14.0S2019 $20,850,000 14.2% 98.45 DDD 14.0S2019 $20,850,000 14.0% 100.09 B Baldwin 11.6S2025 $3,523,690 12.9% 90.11 B 14.0S2019 $20,850,000 14.0% 99.91 CCC 12.2S2026 $8,623,471 13.2% 92.09 B Chester 12.9S2027 $16,806,601 13.6% 95.04 B 14.0S2019 $20,850,000 14.0% 100.09 B 13.4S2028 $15,253,268 13.8% 97.38 B 12.1S2026 $13,308,235 13.2% 91.62 B Erie 12.9S2027 $12,249,416 13.6% 95.04 B 14.0S2019 $20,850,000 13.9% 100.44 B 13.4S2028 $10,916,191 13.8% 97.38 B 11.7S2025 $1,722,021 12.7% 92.16 B 12.3S2026 $11,199,852 13.0% 94.34 B 13.0S2027 $25,774,968 13.3% 97.48 B 13.5S2028 $15,359,294 13.5% 100.00 B Ferris 14.0S2019 $20,850,000 14.0% 100.09 B 12.2S2026 $8,738,656 13.2% 92.09 B 12.9S2027 $15,268,932 13.6% 95.04 B 13.4S2028 $14,780,458 13.8% 97.38 B Next Year's Prime Rate 9.50% CAPSTONE® COURIER Page 2 Financial Summary C58538 Round: 5Dec. 31, 2018 Cash Flow Statement Survey Andrews Baldwin Chester Digby Erie Ferris CashFlows from operating activities Net Income(Loss) ($13,191) ($1,298) $15,037 $12,175 $22,360 $11,496 Adjustment for non-cash items: Depreciation $4,588 $7,587 $13,793 $15,000 $16,053 $11,400 Extraordinary gains/losses/writeoffs ($9,895) $0 $0 $0 $0 ($550) Changes in current assets and liabl ...

Round 5Dec. 31, 2018 C58538AndrewsEugene EllisPhili.docx

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DataLength of StayAgeSexPrincipal Payer Total ChgesRT ChgeImaging ChargesLab ChargesDrug Charges477FCARE$ 10,258.04$ 1,300.71$ 480.63$ 1,417.29$ 1,059.41574FCARE$ 8,845.63$ 261.32$ 886.14$ 872.56$ 825.61681FCARE$ 8,561.29$ 447.92$ 870.07$ 239.97$ 1,003.32787FCARE$ 8,744.77$ 457.42$ 60.30$ 1,087.94$ 1,139.10865FCARE$ 7,473.72$ 381.40$ 15.82$ 679.23$ 397.27980FCARE$ 8,557.10$ 463.12$ 709.47$ 277.08$ 1,107.43290FCARE$ 9,017.12$ 855.92$ 1,468.80$ 630.97$ 61.44366FBC$ 8,369.42$ 72.48$ 157.95$ 1,447.43$ 691.55487FBC$ 8,371.65$ 28.81$ 394.24$ 1,340.25$ 608.35577FBC$ 7,040.53$ 18.73$ 357.02$ 468.49$ 196.29774FBC$ 9,027.60$ 375.14$ 438.80$ 1,040.24$ 1,173.42881FBC$ 8,406.36$ 60.96$ 601.89$ 1,086.30$ 657.20987FBC$ 9,576.50$ 914.93$ 1,489.36$ 917.28$ 254.93465FBC$ 8,643.63$ 683.12$ 892.84$ 530.09$ 537.57680FBC$ 9,326.02$ 980.82$ 144.89$ 776.74$ 1,423.57590FBC$ 9,343.93$ 1,415.51$ 270.45$ 487.20$ 1,170.76666FBC$ 8,878.88$ 369.10$ 386.02$ 921.90$ 1,201.85487FBC$ 9,368.94$ 1,205.58$ 1,134.55$ 66.56$ 962.25772FBC$ 8,163.33$ 879.24$ 15.46$ 1,003.25$ 265.38877FBC$ 11,054.84$ 1,409.86$ 1,121.31$ 1,068.17$ 1,455.49474FHMO$ 8,922.92$ 25.90$ 706.68$ 993.93$ 1,196.42381FHMO$ 8,549.67$ 62.45$ 668.80$ 1,400.03$ 418.38287FHMO$ 9,482.28$ 1,096.25$ 856.82$ 598.43$ 930.79165FHMO$ 8,604.37$ 1,365.91$ 11.50$ 604.02$ 622.94580FHMO$ 9,425.82$ 1,296.19$ 767.03$ 393.80$ 968.80690FHMO$ 9,544.74$ 981.61$ 1,477.83$ 176.33$ 908.97766FHMO$ 8,803.99$ 1,124.91$ 1,018.15$ 356.25$ 304.68887FHMO$ 9,022.59$ 1,438.65$ 682.02$ 46.61$ 855.30976FHMO$ 8,667.50$ 31.82$ 1,101.40$ 317.30$ 1,216.98677FCARE$ 9,600.11$ 888.77$ 748.72$ 496.11$ 1,466.52774FCARE$ 8,872.57$ 853.22$ 832.44$ 1,012.88$ 174.02881FCARE$ 9,813.57$ 1,471.71$ 1,384.39$ 812.14$ 145.33987FCARE$ 9,369.43$ 303.87$ 1,304.72$ 1,312.62$ 448.22265FOGVT$ 8,977.19$ 127.92$ 1,151.84$ 433.41$ 1,264.02380FCARE$ 8,966.71$ 444.74$ 1,316.78$ 495.19$ 709.99490FCARE$ 8,280.27$ 295.46$ 859.62$ 967.58$ 157.62566FCARE$ 8,054.88$ 1,165.70$ 348.65$ 35.17$ 505.36787FCARE$ 7,472.24$ 115.07$ 95.57$ 1,142.44$ 119.16875FCARE$ 9,063.64$ 489.99$ 1,009.08$ 160.74$ 1,403.83977FCARE$ 8,121.75$ 415.25$ 65.99$ 672.54$ 967.97474FCARE$ 8,016.12$ 222.52$ 375.50$ 1,356.86$ 61.23681FCARE$ 10,125.58$ 991.24$ 478.59$ 1,252.78$ 1,402.97587FCARE$ 8,523.48$ 475.08$ 637.82$ 1,379.82$ 30.76665FCARE$ 8,488.61$ 432.37$ 34.25$ 567.26$ 1,454.72480FCARE$ 8,984.25$ 1,245.69$ 414.59$ 1,039.37$ 284.61790FCAID$ 8,245.75$ 217.00$ 397.73$ 150.78$ 1,480.24866FCARE$ 8,950.12$ 256.51$ 1,440.93$ 364.02$ 888.65487FCARE$ 7,31 ...

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Sample Investment PropertyAverage Inland San Diego HomeASSUMPTIONS: No CAPX (Capital Expenses) - As some analysis puts them above NOI, and others below NOIPurchase Price$ (500,000.00)Building Value68%$ 340,000.00Loan to Value70%Maximum for Investment Property: 70%Loan Amount$ 350,000.00Down Payment$ (150,000.00)Loan Costs1.0%$ (3,500.00)Interest RateAnnual6.500%Loan TermMonths360Closing Costs2.0%$ (10,000.00)See Estimated Settlement Statement for Detailed BreakdownTotal Costs at Acquisition$ (163,500.00)PROFORMAYear 0Year 1Year 2Year 3Year 4Year 5Year 6Year 7Year 8Year 9Year 10Year 11Year 12Year 13Year 14Year 15Year 16Year 17Year 18Year 19Year 20Year 21Year 22Year 23Year 24Year 25Year 26Year 27Year 28Year 29Year 30Year 31Gross Rent4%$ 28,800.00$ 29,952.00$ 31,150.08$ 32,396.08$ 33,691.92$ 35,039.60$ 36,441.18$ 37,898.83$ 39,414.78$ 40,991.37$ 42,631.02$ 44,336.26$ 46,109.71$ 47,954.10$ 49,872.26$ 51,867.15$ 53,941.84$ 56,099.51$ 58,343.49$ 60,677.23$ 63,104.32$ 65,628.49$ 68,253.63$ 70,983.78$ 73,823.13$ 76,776.06$ 79,847.10$ 83,040.98$ 86,362.62$ 89,817.12$ 93,409.80Vacancy4.0%$ (1,152.00)$ (1,198.08)$ (1,246.00)$ (1,295.84)$ (1,347.68)$ (1,401.58)$ (1,457.65)$ (1,515.95)$ (1,576.59)$ (1,639.65)$ (1,705.24)$ (1,773.45)$ (1,844.39)$ (1,918.16)$ (1,994.89)$ (2,074.69)$ (2,157.67)$ (2,243.98)$ (2,333.74)$ (2,427.09)$ (2,524.17)$ (2,625.14)$ (2,730.15)$ (2,839.35)$ (2,952.93)$ (3,071.04)$ (3,193.88)$ (3,321.64)$ (3,454.50)$ (3,592.68)$ (3,736.39)Other Income0%$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0$ - 0Effective Gross Income$ 27,648.00$ 28,753.92$ 29,904.08$ 31,100.24$ 32,344.24$ 33,638.02$ 34,983.53$ 36,382.88$ 37,838.19$ 39,351.72$ 40,925.78$ 42,562.81$ 44,265.32$ 46,035.94$ 47,877.37$ 49,792.46$ 51,784.17$ 53,855.53$ 56,009.75$ 58,250.14$ 60,580.15$ 63,003.35$ 65,523.48$ 68,144.43$ 70,870.20$ 73,705.02$ 76,653.22$ 79,719.34$ 82,908.12$ 86,224.44$ 89,673.41OPERATING EXPENSESProperty Taxes 1.1%2%$ (5,500.00)$ (5,610.00)$ (5,722.20)$ (5,836.64)$ (5,953.37)$ (6,072.44)$ (6,193.89)$ (6,317.77)$ (6,444.13)$ (6,573.01)$ (6,704.47)$ (6,838.56)$ (6,975.33)$ (7,114.84)$ (7,257.14)$ (7,402.28)$ (7,550.33)$ (7,701.34)$ (7,855.37)$ (8,012.48)$ (8,172.73)$ (8,336.18)$ (8,502.90)$ (8,672.96)$ (8,846.42)$ (9,023.35)$ (9,203.82)$ (9,387.90)$ (9,575.66)$ (9,767.17)$ (9,962.51)Insurance: Fire3%$ (1,000.00)$ (1,030.00)$ (1,060.90)$ (1,092.73)$ (1,125.51)$ (1,159.28)$ (1,194.06)$ (1,229.88)$ (1,266.78)$ (1,304.78)$ (1,343.92)$ (1,384.24)$ (1,425.77)$ (1,468.54)$ (1,512.60)$ (1,557.98)$ (1,604.72)$ (1.

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Round: 2 Dec. 31, 2015 C57912 Andrews Tracy Calhoun Victoria Carey Florian Heithier Keisha High Baldwin Joy Atkins Michal Grygo Deric Sims Sean Smith MILTON WARE Chester ARAYA AMDETSYON Kevin Darrell Judy-Ann Davis DELANO MCVAY Digby Erie Selected Financial Statistics Andrews Baldwin Chester Digby Erie ROS -40.4% -1.3% -13.5% 8.2% 4.6% Asset Turnover 0.37 0.96 0.66 1.30 0.91 ROA -14.9% -1.2% -8.9% 10.7% 4.2% Leverage (Assets/Equity) -10.2 1.9 7.4 2.0 2.1 ROE -152.4% -2.4% -65.5% 21.1% 8.7% Emergency Loan $343,049,599 $0 $94,241,651 $0 $0 Sales $135,241,406 $96,058,934 $115,326,373 $192,677,677 $118,689,159 EBIT ($18,090,603) $3,635,841 ($2,138,470) $32,178,555 $16,026,603 Profits ($54,636,942) ($1,246,277) ($15,612,972) $15,735,574 $5,462,417 Cumulative Profit ($80,606,603) $1,746,029 ($19,910,190) $30,484,786 $17,761,077 SG&A / Sales 7.0% 17.2% 14.1% 10.5% 10.4% Contrib. Margin % 0.7% 29.2% 18.8% 31.4% 31.0% CAPSTONE® COURIER Page 1 Stock & Bonds C57912 Round: 2Dec. 31, 2015 Stock Market Summary Company Close Change Shares MarketCap ($M) Book Value EPS Dividend Yield P/E Andrews $1.00 $0.00 2,029,194 $2 ($17.67) ($26.93) $0.00 0.0% 0.0 Baldwin $20.47 ($4.02) 2,269,039 $46 $23.27 ($0.55) $0.00 0.0% -37.3 Chester $1.00 $0.00 2,000,000 $2 $11.92 ($7.81) $0.00 0.0% -0.1 Digby $71.00 $19.80 2,015,280 $143 $37.06 $7.81 $0.04 0.1% 9.1 Erie $46.29 $0.27 2,128,085 $99 $29.53 $2.57 $1.44 3.1% 18.0 Bond Market Summary Company Series# Face Yield Close$ S&P Company Series# Face Yield Close$ S&P Andrews Chester 12.5S2017 $13,900,000 12.9% 97.05 DDD 12.5S2017 $13,900,000 12.9% 97.05 DDD 14.0S2019 $20,850,000 14.1% 99.13 DDD 14.0S2019 $20,850,000 14.1% 99.13 DDD Baldwin Digby 12.5S2017 $13,900,000 12.4% 101.02 BB 12.5S2017 $13,828,905 12.4% 100.51 B 14.0S2019 $20,850,000 13.2% 106.39 BB 14.0S2019 $20,850,000 13.3% 105.44 B 11.3S2024 $2,500,000 11.7% 96.79 BB 11.5S2025 $4,651,752 12.0% 96.08 B Erie 12.5S2017 $13,900,000 12.5% 100.00 CCC 14.0S2019 $20,850,000 13.4% 104.51 CCC Next Year's Prime Rate 8.00% CAPSTONE® COURIER Page 2 Financial Summary C57912 Round: 2Dec. 31, 2015 Cash Flow Statement Survey Andrews Baldwin Chester Digby Erie CashFlows from operating activities Net Income(Loss) ($54,637) ($1,246) ($15,613) $15,736 $5,462 Adjustment for non-cash items: Depreciation $9,547 $7,587 $7,587 $7,833 $8,755 Extraordinary gains/losses/writeoffs $0 ($15) $0 $7 ($539) Changes in current assets and liablilities Acounts payable $394 ($3,212) $708 $3,404 ($1,264) Inventory ($122,974) $16,324 ($44,237) ($957) $9 Accounts Receivable ($2,092) ($50) ($1,123) ($4,327) $1,409 Net cash from operations ($169,763) $19,388 ($52,678) $21,695 $13,832 Cash flows from investing activities Plant improvements(net) $0 $0 $0 ($25,320) ($15,292) Cash flows from financing activities Dividends paid ...

Round 2Dec. 31, 2015 C57912AndrewsTracy CalhounVict.docx

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The comparative balance sheet of Iglesias Inc. for December 31, 20Y3 and 20Y2, is shown as follows: 1 Dec. 31, 20Y3 Dec. 31, 20Y2 2 Assets 3 Cash $186,000.00 $180,000.00 4 Accounts receivable (net) 540,000.00 480,000.00 5 Inventories 924,000.00 900,000.00 6 Investments 0.00 120,000.00 7 Land 600,000.00 0.00 8 Equipment 1,680,000.00 1,440,000.00 9 Accumulated depreciation-equipment (720,000.00) (600,000.00) 10 Total assets $3,210,000.00 $2,520,000.00 11 Liabilities and Stockholders Equity 12 Accounts payable $408,000.00 $360,000.00 13 Accrued expenses payable 54,000.00 60,000.00 14 Dividends payable 36,000.00 30,000.00 15 Common stock, $4 par 840,000.00 720,000.00 16 Excess of paid-in capital over par 240,000.00 210,000.00 17 Retained earnings 1,632,000.00 1,140,000.00 18 Total liabilities and stockholders equity $3,210,000.00 $2,520,000.00 Additional data obtained from an examination of the accounts in the ledger for 20Y3 are as follows: Required: Prepare a statement of cash flows, using the indirect method of presenting cash flows from (used for) operating activities. Be sure to complete the heading of the statement. Refer to the Labels and Amount Descriptions list provided for the exact wording of the answer choices for text entries. Use the minus sign to indicate cash outflows, decreases in cash and a net cash outflow for each section, if required. Statement of Cash Flows Shaded cells have feedback. 1. Cash flows from (used for) operating activities: 2 Net income 3 Adjustments to reconcile net income to net cash flows from operating activitie: 12 Cash flows from (used for) irnesting activities:.

The comparative balance sheet of Iglesias Inc for December .pdf

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BalShtBalance SheetBalance SheetBalance SheetBalance SheetThe Frank Beverage GroupThe Frank Beverage GroupThe Frank Beverage GroupThe Frank Beverage GroupFirst QuarterSecond QuarterThird QuarterFourth Quarter2019-20202019-20202019-20202019-2020ASSETSASSETSASSETSASSETSCurrent AssetsCurrent AssetsCurrent AssetsCurrent AssetsCash$110,102Cash$161,052Cash$186,936Cash$219,214Accounts Receivable$35,569Accounts Receivable$37,746Accounts Receivable$40,057Accounts Receivable$42,508Inventory-$1,887Inventory$14,313Inventory$31,504Inventory$50,300Other Current Assets$0Other Current Assets$0Other Current Assets$0Other Current Assets$0Total Current Assets$143,784Total Current Assets$213,111Total Current Assets$258,497Total Current Assets$312,022Fixed AssetsFixed AssetsFixed AssetsFixed AssetsLand$0Land$0Land$0Land$0Facilities$0Facilities$0Facilities$0Facilities$0Equipment$0Equipment$0Equipment$0Equipment$0Computers & Telecommunications$0Computers & Telecommunications$0Computers & Telecommunications$0Computers & Telecommunications$0(Less Accumlated Depreciation)$0(Less Accumlated Depreciation)$0(Less Accumlated Depreciation)$0(Less Accumlated Depreciation)$0Total Fixed Assets$0Total Fixed Assets$0Total Fixed Assets$0Total Fixed Assets$0Other Assets$0Other Assets$0Other Assets$0Other Assets$0TOTAL ASSETS$143,784TOTAL ASSETS$213,111TOTAL ASSETS$258,497TOTAL ASSETS$312,022LIABILITIESLIABILITIESLIABILITIESLIABILITIESCurrent LiabilitiesCurrent LiabilitiesCurrent LiabilitiesCurrent LiabilitiesShort-Term Notes Payable$9,873Short-Term Notes Payable$9,997Short-Term Notes Payable$10,122Short-Term Notes Payable$10,249Income Taxes Due$16,109Income Taxes Due$34,046Income Taxes Due$46,006Income Taxes Due$59,618Other Current Liabilities$0Other Current Liabilities$0Other Current Liabilities$0Other Current Liabilities$0Total Current Liabilities$25,982Total Current Liabilities$44,043Total Current Liabilities$56,128Total Current Liabilities$69,868Long-Term LiabilitiesLong-Term LiabilitiesLong-Term LiabilitiesLong-Term LiabilitiesLong-Term Notes Payable$7,735Long-Term Notes Payable$5,189Long-Term Notes Payable$2,610Long-Term Notes Payable-$0Other Long-Term Liabilities$0Other Long-Term Liabilities$0Other Long-Term Liabilities$0Other Long-Term Liabilities$0Total Long-Term Liabilities$7,735Total Long-Term Liabilities$5,189Total Long-Term Liabilities$2,610Total Long-Term Liabilities-$0NET WORTHNET WORTHNET WORTHNET WORTHPaid-In Capital$61,740Paid-In Capital$61,740Paid-In Capital$61,740Paid-In Capital$61,740Retained Earnings$48,327Retained Earnings$102,139Retained Earnings$138,018Retained Earnings$180,414Total Net Worth$110,067Total Net Worth$163,879Total Net Worth$199,758Total Net Worth$242,154TOTAL LIABILITIES AND NET WORTH$143,784TOTAL LIABILITIES AND NET WORTH$213,111TOTAL LIABILITIES AND NET WORTH$258,497TOTAL LIABILITIES AND NET WORTH$312,022 For information about this worksheet, see "Balance Sheet" in "The Financials" chapter of Successful Business Plan: Secrets & Strategies. ...

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Financial forecast

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CalculationsLost Profit and Royalty Calculations2010201120122013UnderCover Mouse Revenues$ 2,091,626$ 1,949,065$ 1,274,673SportPet Revenues$ 1,964,738$ 2,324,447$ 1,907,690Total Revenues$ 4,056,363$ 4,273,512$ 3,182,362Market Share for Undercover Mouse52%46%40%Market Share for SportPet48%54%60%201120122013Sports Pet Units Sold128,548159,701130,161Ideal Revenues3,207,2793,984,5523,247,510UCM Actual Revenues$ 2,091,626$ 1,949,065$ 1,274,673Lost Sales1,115,653.742,035,486.371,972,837.565,123,978Incremental Sales1,013,0981,060,135764,112Undercover Mouse COGS553,621454,397262,734Incremental Profits$ 459,478$ 605,737$ 501,378$ 1,566,593Royalty Amount (remaining portion of infringing sales)3,557,384.72Comments (in no particular order):2) The COGS percent you use is the marketshare. You need to calculate the COGS as a percent of the revenue for actual sales.3) You need to figure out what percent of 2010 sales occurred in November and December and use those sales as well.4) The market shares are calculated correctly, but the Incremental sales calculation doesn't work. You need to remove SportPet from the market and then figure out what percent of sales UCM would have made.You can think about the relative strength of Walmart v. Target, or the geographic sales split (you'll notice that the market shares for each geography don't add to 100%).One of the more difficult aspects of this case is that the evidence indicates that, if SportPet had not been in the market, UCM would not have made ALL of SportPet's sales.They were selling through different channels and had different geographic strengths.Thus, you're going to have to figure out some way to allocate SportPet's revenues as between UCM in the but-for world and other companies. UnderCover IncomeUnderCover MouseIncome Statement2009201020112012201320092010201120122013Units27,20097,017126,696115,74475,788Net Income$ (91,364)$ 220,604$ 216,280$ 113,584$ 39,318Wholesale2,74542,39451,01044,70528,220Discount Rate10.94117647060.88581314880.8337064930.7846649346Direct24,45554,62375,68771,03947,568Revenues$ 599,809$ 1,610,950$ 2,091,626$ 1,949,065$ 1,274,673Wholesale$ 32,133$ 356,025$ 426,243$ 412,685$ 254,311Direct$ 561,235$ 1,245,234$ 1,654,238$ 1,523,564$ 1,005,623Repair$ - 0$ - 0$ - 0$ - 0$ - 0Freight$ 6,441$ 9,691$ 11,145$ 12,816$ 14,739$ - 0COGS$ 294,192$ 811,790$ 1,073,656$ 996,310$ 655,945Manufacture$ 50,047$ 178,511$ 233,121$ 212,969$ 139,449Freight$ 112,187$ 278,870$ 380,377$ 354,546$ 236,067Labor$ 131,958$ 354,409$ 460,158$ 428,794$ 280,428Gross Margin$ 305,616$ 799,160$ 1,017,970$ 952,755$ 618,728SG&A$ 375,326$ 443,296$ 668,004$ 759,977$ 540,385Advertising$ 251,235$ 158,782$ 351,254$ 412,584$ 178,542Bank$ 192$ 212$ 212$ 212$ 212Dues$ 160$ 160$ 160$ 160$ 160Equipment Rental$ - 0$ 227$ 227$ 227$ 227Freight - ...

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Given a newly created Binary Search Tree with the following numerical key input sequence (from left to right): 9, 4, 12, 7, 5, 2, 20, 14, 11, 13, 19, 16, 17, 42, 24 a.) We are asked to add a function Position lowestKey(const Position& v) const to the class SearchTree. Function prototype: Position lowestKey(const Position& v); input argument: position of the starting node in the binary search tree to calculate from. For the whole tree this would be the position for the root of the tree. return value: position of the node having the lowest key value. i. Describe strategy with reasonings and examples to convince that it will work. ii. Implement the function using C++ and show the source code listing. b) We are also asked to add a function Position highestKey(const Position& v) const to the class SearchTree. Function prototype: Position highestKey(const Position& v) const; input argument: position of the node in the binary search tree to calculate from. For the whole tree this would be the position for the root of the tree. return value: position of the node having the highest key value. i. Describe strategy with reasonings and examples to convince that it will work. ii. Implement the function using C++ and show the source code listing. Solution #include #include /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } /* Give a binary search tree and a number, inserts a new node with the given number in the correct place in the tree. Returns the new root pointer which the caller should then use (the standard trick to avoid using reference parameters). */ struct node* insert(struct node* node, int data) { /* 1. If the tree is empty, return a new, single node */ if (node == NULL) return(newNode(data)); else { /* 2. Otherwise, recur down the tree */ if (data <= node->data) node->left = insert(node->left, data); else node->right = insert(node->right, data); /* return the (unchanged) node pointer */ return node; } } /* Given a non-empty binary search tree, return the minimum data value found in that tree. Note that the entire tree does not need to be searched. */ int minValue(struct node* node) { struct node* current = node; /* loop down to find the leftmost leaf */ while (current->left != NULL) { current = current->left; } return(current->data); } /* Driver program to test sameTree function*/ int main() { struct node* root = NULL; root = insert(root, 4); insert(root, 2); insert(root, 1); insert(root, 3); insert(root, 6); insert(root, 5); printf(\"\ Minimum value in BST is %d\", minValue(root)); getchar(); return 0; } #include #include /* A binary tree node has data, pointer to left child .

Given a newly created Binary Search Tree with the following numerica.pdf

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Gather information on the following two disasters: Space shuttle \"Challenger\", Jan 1986. Space shuttle \"Columbia\", Jan 2003. Describe (may still be speculative for Columbia) what (materials & design as well as political/management factors) caused the disasters. Solution Cause of Space shuttle Challenger disaster, Jan 1986: It was determined that the two rubber O-rings, which had been designed to separate the sections of the rocket booster, had failed due to cold temperatures on the morning of the launch. This seal, called an O-ring, on the shuttle\'s right solid rocket booster had failed at liftoff, allowing pressurized hot gas to escape from inside the booster. This vaporized material impinged on the strut connecting the solid rocket booster to the shuttle\'s huge orange external tank, causing both pieces of hardware to break down. About 72 seconds into Challenger\'s flight, there was a massive, almost explosive, burning of the hydrogen that was streaming from the failed tank bottom, combined with liquid oxygen leaking from a part of the fuel tank known as the intertank which eventually lead to the biggest disaster in the history. Cause of Space shuttle Columbia disaster, Jan 2003: This space shuttle Columbia broke apart returning to earth killing all the seven crew members on the board. An investigation determined that the disaster was caused by a problem that took place shortly while a large piece of foam fell from the shuttle’s propellant tank and fatally damaged the edge of the shuttle’s left wing.They lost temperature readings from sensors located on the left wing. Then, tire pressure readings from the left side also vanished. It was later found that a hole on the left wing allowed atmospheric gases to bleed into the shuttle as it went through its fiery re-entry, leading to the loss of the sensors and eventually, Columbia itself.This Columbia disaster was the second major tragedy in the history of the space shuttle program after the space shuttle Challenger broke apart shortly after launch on January 28, 1986..

Gather information on the following two disasters Space shuttle Ch.pdf

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Determine the Laplace transform, F(s), of the following time signal. Solution we\'ll be interested in signals de¯ned for t ¸ 0 the Laplace transform of a signal (function) f is the function F = L(f) de¯ned by F(s) = Z 1 0 f(t)e¡st dt for those s 2 C for which the integral makes sense ² F is a complex-valued function of complex numbers ² s is called the (complex) frequency variable, with units sec¡1; t is called the time variable (in sec); st is unitless ² for now, we assume f contains no impulses at t = 0 common notation convention: lower case letter denotes signal; capital letter denotes its Laplace transform, e.g., U denotes L(u), Vin denotes L(vin), etc..

Determine the Laplace transform, F(s), of the following time signal..pdf

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Describe the difference between a monitor and a semaphore Solution Monitor: Monitor is a highly structured programming language construct.It consists of Private variables and private procedures that can only be used within a monitor. Constructors that initialize the monitor.A number of public monitor procedures that can be invoked by users. These monitors have no public data.A monitor is a mini OS with monitor procedures as system calls. Semaphore: Semaphore is a data structure that is used to provide mutual exclusion to critical sections. Semaphores mainly support two operations called wait and Signal . The wait operation blocks a process until the semaphore is open and the signal operation allows another process to enter.MonitorSemaphoreCondition variables only uses these MonitorsThese can be used in anywhere but not in Monitorswait() always blocks it\'s callerwait() does not always blocks it\'s callersignal() either releases a process or the signal is lost as if it never occurssignal() either releases a process or increases Semaphore counterif signal() releases a process,either the caller or the released continues,but not bothif signal() releases a process,the caller and the released both continued.

Describe the difference between a monitor and a semaphoreSolutio.pdf

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Create a link list. Add some nodes to it, search and delete nodes from it. Solution #include #include using std::cout; using std::endl; /* A linked list is a list constructed using pointers. It is not fixed in size and could grow and shrink while the program is running. A typical defination of list nodes contains at least two parts, both the content or date of each element and the pointer to the next element, which is shown in the figure below. +---------+ | Data | -----> holds the data of this element in the list. +---------+ | pointer | -----> is a pointer to the next element in the list. +---------+ ***Attention***: The pointer holds the address of the next element, not the address of the data in the next element, even though they are the same in value sometimes. And It should be set to NULL while acting as the last node of the list. Implementation of the single linked list: +---------+ --->+---------+ --->+---------+ | Data | | | Data | | | Data | +---------+ | +---------+ | +---------+ | pointer |----- | pointer |----- | pointer | +---------+ +---------+ +---------+ */ /* definition of the list node class */ class Node { friend class LinkedList; private: int _value; /* data, can be any data type, but use integer for easiness */ Node *_pNext; /* pointer to the next node */ public: /* Constructors with No Arguments */ Node(void) : _pNext(NULL) { } /* Constructors with a given value */ Node(int val) : _value(val), _pNext(NULL) { } /* Constructors with a given value and a link of the next node */ Node(int val, Node* next) : _value(val), _pNext(next) {} /* Getters */ int getValue(void) { return _value; } Node* getNext(void) { return _pNext; } }; /* definition of the linked list class */ class LinkedList { private: /* pointer of head node */ Node *_pHead; /* pointer of tail node */ Node *_pTail; public: /* Constructors with No Arguments */ LinkedList(void); /* Constructors with a given value of a list node */ LinkedList(int val); /* Destructor */ ~LinkedList(void); /* Function to add a node at the head of a linked list */ void headInsert(int val); /* Function to add a node in the middle of a linked list */ void insertAfter(Node* afterMe, int val); /* Function to append a node to the end of a linked list */ void tailAppend(int val); /* Function to get the node in the specific position */ Node* getNode(int pos); /* Traversing the list and printing the value of each node */ void traverse_and_print(); }; LinkedList::LinkedList() { /* Initialize the head and tail node */ _pHead = _pTail = NULL; } LinkedList::LinkedList(int val) { /* Create a new node, acting as both the head and tail node */ _pHead = new Node(val); _pTail = _pHead; } LinkedList::~LinkedList() { /* * Leave it empty temporarily. * It will be described in detail in the example \"How to delete a linkedlist\". */ } void LinkedList::headInsert(int val) { /* The list is empty? */ if (_pHead == NULL) { /* the same to create a new list with a given value */ _pTail = _pHead = new Node(val); .

Create a link list. Add some nodes to it, search and delete nodes fro.pdf

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A speedometer is a measuring instrument. What are the independent and dependent variables when it is used? Give an example of an extraneous variable for this measurement. Solution Input time is the independent variable where as out put mileage is the dependent variable. Mileage is a function which takes something as input that cant be controlled by us. Example for this is robotic speedometer. It is clear that the dependent variables are derived from independent variables. Here time is independent variable, the dependent variables can be extended to derivatives such as velocity,acceleration,etc..

A speedometer is a measuring instrument. What are the independent and.pdf

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choice one correct answerRocks containing iron, calcium, magnesium.pdf

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1.write down the RTL for the SUBT instructions. 2.Draw a flowchart to indicate SUBT instructions is processed. Solution 1) RTL is a easy human understandable language to specify or to perform the operations SUBT is instruction which is used to perform or subtracts the Source operand from destination operand and the subtracted result is stored into destination operand Conditions Source operand should be a register or a memory location or may be a immediate address Destination operand should be a Register or memory location but not be a Immediate address At a single time its not possible if both source and destinations to be a memory locations General format of a SUB RTL(Register Transfer Language) format SUB DESTINATION OPERAND,SOURCE OPERAND; eg SUB AX,18H; SUB AX,BX; SUB AX,[2000H]; 2) FLOW CHART to indicate SUBT instruction process Instruction Cycle Implementation 1) Fetching of instructions will be performed at first 2) Updation of PC(Program counter will takes place) in second stage 3) Decoding of fetched instructions will be takes place in next stage 4) Execution of given eithr arthemetic or logic or conditional or unconditional or anyother operations will takes place in the later stage If Incase of SUB instruction Both Source and destinations datas are retrieved then it will check Sign of both datas,If.

1.write down the RTL for the SUBT instructions. 2.Draw a flowchart.pdf

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Write a Java application that asks for an integer and returns its factorization into prime factors including their powers. The prime factors must appear in increasing order. For example if 120 (= 2*2*2*3*5) in the input number, then your program should output 120 = 2^3 * 3^1 * 5^1 The factors should appear in increasing order and should be separated from each other by asterisks (*). Solution import java.util.Scanner; public class factorialFactorization { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.print(\"Enter a number:\"); int n = scan.nextInt(); int fact=1; for(int i=2;i<=n;i++) { fact = fact*i; } int count=0,hat=0; System.out.print(fact); System.out.print(\" = \"); for(int i=2;fact>1;i++) { count = 0; while(true) { if(fact%i==0) { count++; fact = fact/i; } else { break; } } if(count>0) { if(hat>0) { System.out.print(\" * \"); } System.out.print(i); System.out.print(\"^\"); System.out.print(count); hat ++; } } } }.

Write a Java application that asks for an integer and returns its fac.pdf

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Where should seaweeds and kelps be classified? Give characteristics of these organisms which you used to place them in a taxonomic unit and the source you used to find their characteristics. Solution Seaweeds are classified into three phylum\'s based on the colour of the thallus, biochemical compositions, cell wall structures, photosynthetic pigments and chloroplasts. The three kingdoms of seaweeds are, Kingdom bacteria, Kindom Plantae and Kingdom Chromista. The three phylum\'s are, Phylum Rhodophyta (red alage), Phylum Chlorophyta (green algae) and Phylum Ochrophyta (brown algae). Kelps are the brown algae that belong to the order Laminariales, genus Laminaria..

Where should seaweeds and kelps be classified Give characteristics .pdf

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What is the name of the connective tissue that divides the se.pdf

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What is glass-transition temperature? Give a material example. How many types of polymer structures do you know? Briefly explain each type of polymer structures. b) What is creep and what is stress relaxation? c) What are the two structure components in reinforced plastics and what are the functions of each of them? What are the advantages and disadvantages of reinforced plastics comparing with the regular plastics? Solution a) The glass–liquid transition or glass transition for short is the reversible transition in amorphous materials (or in amorphous regions within semicrystalline materials) from a hard and relatively brittle \"glassy\" state into a molten or rubber-like state, as the temperature is increased. An amorphous solid that exhibits a glass transition is called a glass. The glass-transition temperatureTg of a material characterizes the range of temperatures over which this glass transition occurs. It is always lower than the melting temperature, Tm, of the crystalline state of the material, if one exists. Examples Rubber elastomers like polyisoprene and polyisobutylene are used above their Tg\'s, that is, in the rubbery state, where they are soft and flexible. Typical polymer structures are b) creep is the tendency of a solid material to move slowly or deform permanently under the influence of mechanical stresses. It can occur as a result of long-term exposure to high levels of stress that are still below the yield strength of the material. Creep is more severe in materials that are subjected to heatfor long periods, and generally increases as they near their melting point. Stress relaxation is defined as a gradual decrease in stress with time under a constant deformation or strain. c) Fibre-reinforced plastic is a composite material made of a polymer matrix reinforced with fibres. The fibres are usually glass, carbon,aramid, or basalt. Rarely, other fibres such as paper or wood or asbestos have been used. The polymer is usually an epoxy, vinylester or polyester thermosetting plastic, andphenol formaldehyde resins are still in use. Advantages Disadvantages.

What is glass-transition temperature Give a material example. How ma.pdf

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What are some contributions to the development of industrialization made by the steam engine. In other words what role did the steam engine play in the development of the industrial revolution What are some contributions to the development of industrialization made by the steam engine. In other words what role did the steam engine play in the development of the industrial revolution What are some contributions to the development of industrialization made by the steam engine. In other words what role did the steam engine play in the development of the industrial revolution Solution Answer: The steam engine was one of the most important technologies of the Industrial Revolution, although steam did not replace water power in importance in Britain until after the Industrial Revolution. From Englishman Thomas Newcomen\'s atmospheric engine, of 1712, through major developments by Scottish inventor and mechanical engineer James Watt, the steam engine began to be used in many industrial settings, not just in mining, where the first engines had been used to pump water from deep workings. Early mills had run successfully with water power, but by using a steam engine a factory could be located anywhere, not just close to water. Water power varied with the seasons and was not available at times due to freezing, floods and dry spells The technological advances of the Industrial Revolution happened more quickly because firms often shared information, which they then could use to create new techniques or products. From mines to mills, steam engines found many uses in a variety of industries. The introduction of steam engines improved productivity and technology, and allowed the creation of smaller and better engines. The development of the stationary steam engine was an essential early element of the Industrial Revolution, however it should be remembered that for most of the period of the Industrial Revolution the majority of industries still relied on wind and water power as well as horse and man-power for driving small machines. Transportation was one of those important beneficiaries. By the early 1800s, high-pressure steam engines had become compact enough to move beyond the factory, prompting the first steam-powered locomotive to hit the rails in Britain in 1804. For the first time in history, goods were transported over land by something other than the muscle of man or animal. The railroad rapidly spread and sped transportation. This created new jobs, forcing people to move into cities. Cities like London, for example, the steam engine and locomotive propelled countless other industries such as driving bellows and rollers, spinning machines, and weaving machines. And, for the first time in history, people had the ability to quickly and cheaply move to big cities to find work, as opposed to working on a farm. Subsequently, it had a major impact on businesses, as it boosted production of natural resources, which were vital to society, and created numerou.

What are some contributions to the development of industrializat.pdf

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The “creative revolution” that handed more authority to agency art directors and copywriters in the 1960s led to key shifts in the appearance and message of mainstream advertising. Describe these changes and how they continue to influence advertising today. Solution The key shift in terms of appearance and message of mainstream advertising was that the commercials started using irony and irreverence to reach out to the young customers. The situation was that the young customers were indifferent to advertisements by companies at that time. Advertisements were designed, both in terms of its message and appearance, to connect with the youth. Secondly, the creative revolution also saw social messages and issues being used in commercials and advertisements. These changes continue to influence advertising today. Advertisements, be it print or any other media, increasingly are targeting the youths. Marketers are not afraid of using self deprecating humor or using social messages in today\'s advertisements and commercials..

The “creative revolution” that handed more authority to agency art d.pdf

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The three main coefficients (not properties or dimensionless numbers) interphase transport are For momentum transport For heat transport For mass transport Write two assumptions of Navier Stoke equation. What is the assumption of Euler\'s equation? Solution Assumption for navier stroke: 1. Continuous media 2. Viscous constant 3. Incompressible 4. Steady state 5. Fully developed Assumption of euler equation: The fluid is non-viscous (i,e., the frictional losses are zero). The fluid is homogeneous and incompressible (i.e., mass density of the fluid is constant). The flow is continuous, steady and along the streamline..

The three main coefficients (not properties or dimensionless numbers).pdf

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The project will study the coordination of multiple threads using semaphores. The design should consist of two things: (1) a list of every semaphore, its purpose, and its initial value, and (2) pseudocode for each function. Code Your code should be nicely formatted with plenty of comments. The code should be easy to read, properly indented, employ good naming standards, good structure, and should correctly implement the design. Your code should match your pseudocode. Project Language/Platform This project must target a Unix platform and execute properly on our cs1 or csgrads1 Linux server. The project must be written in Java. Elevator Simulation In this project threads are used to simulate people using an elevator to reach their floor. The threads to be used are as follows: Person: 1) 49 people are in line at the elevator at the beginning of the simulation (1 thread per person). 2) Each person begins at floor 1. 3) Each person randomly picks a floor from 2 to 10. 4) A person will wait for an elevator to arrive at floor 1. 5) A person will board the elevator only if there is room. 6) Once at the destination floor, the person exits the elevator. Elevator: 1) There is 1 elevator (1 thread for the elevator). 2) The elevator can only hold 7 people. 3) The elevator begins on floor 1. 4) The elevator leaves after the 7th person enters. Main 1) Creates all threads and joins all person threads. 2) When last person reaches their floor, the simulation ends. Other rules: 1) Each activity of each thread should be printed with identification (e.g., person 1). 2) A thread may not use sleeping as a means of coordinating with other threads. 3) Busy waiting (polling) is not allowed. 4) Mutual exclusion should be kept to a minimum to allow the most concurrency. 5) The semaphore value may not obtained and used as a basis for program logic. 6) All activities of a thread should only be output by that thread. Sample output: Your project’s output should match the wording of the sample output: Elevator door opens at floor 1 Person 0 enters elevator to go to floor 5 Person 1 enters elevator to go to floor 2 Person 2 enters elevator to go to floor 8 Person 3 enters elevator to go to floor 4 Person 4 enters elevator to go to floor 6 Person 5 enters elevator to go to floor 7 Person 6 enters elevator to go to floor 2 Elevator door closes Elevator door opens at floor 2 Person 1 leaves elevator Person 6 leaves elevator Elevator door closes Elevator door opens at floor 4 Person 3 leaves elevator Elevator door closes Elevator door opens at floor 5 Person 0 leaves elevator Elevator door closes Elevator door opens at floor 6 Person 4 leaves elevator Elevator door closes Elevator door opens at floor 7 Person 5 leaves elevator Elevator door closes Elevator door opens at floor 8 Person 2 leaves elevator Elevator door closes Elevator door opens at floor 1 … Simulation done Solution public class Person implements Runnable{ //possible actions public static final int WAITING = 1; publ.

The project will study the coordination of multiple threads using se.pdf

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The positive selection process for T lymphocytes that occurs In the thymus Involves selection of T lymphocytes that have a high affinity for MHC have a high affinity for MHC and self peptide have a low affinity for MHC and self peptide have a low affinity for MHC Solution Answer c: Positive selection select cells with a T cell receptor able to bind MHC class I and II molecules with at least a weak affinity.hence answer c is correct..

The positive selection process for T lymphocytes that occurs In the t.pdf

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The local public health agency has received reports of an outbreak of Salmonella gastroenteritis among attendees at a city-sponsored chili cook-off. What techniques from recombinant DNA technology would the agency likely use to investigate this outbreak? Describe the expected results from these techniques. Solution Please note that the investigation of the outbreak would require molecular biology techniques, not recombinant DNA technology! There would be two steps in the investigation: 1. Establishing that the gastroenteritis is due to Salmonella: Molecular identification techniques such as FISH (Fluorescence in situ hybridization) or Enterobacterial Repetitive Intergenic Consensus (ERIC)-PCR can be used. 2. Establishing that the Salmonella among the attendees is of the same strain: Molecular phylogenetic techniques such as 16S rRNA sequencing or MLST (Multilocus Sequence Typing) can be used..

The local public health agency has received reports of an outbreak o.pdf

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how are musical sounds produced? Explain your reasoning. Solution Almost all musical programs are produced basically in two stages: (1) Sound from each individual intrument is recorded in an acoustically inert studio on a single track of a multitrack tape recorder (2) Signals from each track are manipulated by the sound engineer by adding special audio effects and then combined in a mix-down system to generate the final stereo recording on a two- track tape recorder The special audio effects are either generated using time-domain operations or frequency-domain operations. Commonly used time-domain operations are echo generation,reverberation,flanging,chorus generation, and phasing. Commonly used frequency-domain operations are filters and equalizers.

how are musical sounds produced Explain your reasoning.Solution.pdf

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