Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf
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Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding. In Mendez’s first four years of operation, its board of directors paid the following cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000. Determine the dividends per share and total cash dividends paid to the preferred and common stockholders during each of the four years. Solution The dividends per year on preferred stock = 10,000 x $ 100 x 7 % = $ 10,000 x $ 100 x 0.07 = $ 70,000 Dividend distribution: Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000 Explanation: First dividend is paid to Preferred stock and then excess amount goes to common share holder. For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent years and is added to that year\'s dividend. Preferred Dividends unpaid in 2011 = $70,000 Preferred Dividends on 2012 = $70,000 + $ 70,000 = $ 140,000 Preferred Dividends paid in 2012 = $ 120,000 Preferred Dividends shortage on 2012 = $ 20,000 As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any dividend during 2011 & 2012. Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000 Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000 Preferred Dividends paid in 2014 = $ 70,000 Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000 Computation of Dividends per share: 2011 2012 2013 2014 Preferred Stock (Dividends per share) $0 $12.00 $9.00 $7.00 Common Stock (Dividends per share) $0 $0 $1.00 $1.40 Explanation: Dividends per share = Dividend paid/No. of shares Year 2012: Dividends per share for Preferred Stock = $120,000/10,000 = $ 12 Dividend per share for Common Stock = $ 0 Year 2013: Dividends per share for Preferred Stock = $90,000/10,000 = $ 9 Dividends per share for Common Stock = $ 50,000/50,000 = $ 1 Year 2014: Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7 Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40 Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000.
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Given below is the partial stockholders equity section of the Balance.docx
Given below is the partial stockholders equity section of the Balance Sheet (as of 12/31/2014) for Toys and Stuff, Inc.
Paid-in-Capital
4% Preferred shares ($100 par) $500,000
Common shares ($1 par) $500,000
Assume that the preferred shares are cumulative, and that no dividends had been paid for 2012 and 2013. If total dividends to be paid for 2014 is $200,000, how much dividends will be paid to the common stockholders?
Please show work.
Solution
Annual Dividend paid to Preferred Stockholders: 500,000 x 4% = $20,000 Per Year
Total Unpaid Preference Dividend for 2012 and 2013 = 20,000 x 2 = $40,000
+ Preference Dividend Payable for 2014 = $20,000
Total Preferred Dividend Payable for 2014 = $60,000
.
YearTotal DividendPreferred dividend ($)Per shareCommon divi.pdf
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Average Dividend per share for common stock six years is $0.4
Year
Total dividend ($)
Preferred dividend
Common dividend
Total ($)
Paid ($)
Per share ($)
Total ($)
Per share
2009
24000
50,000
-26,000
2
2010
10000
50,000
-40,000
2
2011
126000
50,000
116,000
2
10,000
0.1
2012
100000
50,000
50,000
2
50,000
0.5
2013
125000
50,000
50,000
2
75,000
0.75
2014
125000
50,000
50,000
2
75,000
0.75
Average dividend per share for common stock is $ 0.35 and for preferred stock is $2
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Solution
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Average Dividend per share for common stock six years is $0.4
Year
Total dividend ($)
Preferred dividend
Common dividend
Total ($)
Paid ($)
Per share ($)
Total ($)
Per share
2009
24000
50,000
-26,000
2
2010
10000
50,000
-40,000
2
2011
126000
50,000
116,000
2
10,000
0.1
2012
100000
50,000
50,000
2
50,000
0.5
2013
125000
50,000
50,000
2
75,000
0.75
2014
125000
50,000
50,000
2
75,000
0.75
Average dividend per share for common stock is $ 0.35 and for preferred stock is $2
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4.
At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx
At December 31, 2013, Hansen Corporation had 42,000 shares of common stock and 6,000 shares of 9%, $100 par cumulative preferred stock outstanding. No dividends were declared or paid in 2013. Net income was reported as $220,000. What is basic EPS? (Round your answer to 2 decimal places.)
Solution
Basic EPS is the amount of company\'s profit taht is available to its common stock shareholders. As we pay preferred shareholders before common stock holders , so we deduct the preferred dividend.
Basic EPS = (Net income - preferred dividend)/ weighted average number of shares
Net Income = $220000
Preferred dividend = 600000 x .09 = $54000
weighted number of shares ( assuming that all the shares were held by company for the whole year) = 42000
therefore,
Basic EPS = (220000 - 54000)/42000 = $3.95
.
Cleary- Wasser- and Nolan formed a partnership on January 1- 2010- wit.docx
Cleary, Wasser, and Nolan formed a partnership on January 1, 2010, with investments of $100,000, $150,000, and $200,000, respectively. For division of income, they agreed to (1) interest of 10% of the beginning capital balance each year, (2) annual compensation of $10,000 to Wasser, and (3) sharing the remainder of the income or loss in a ratio of 20% for Cleary, and 40% each for Wasser and Nolan. Net income was $150,000 in 2010 and $180,000 in 2011. Each partner withdrew $1,000 for personal use every month during 2010 and 2011.
1.What was Wasser\'s total share of net income for 2010? (answer should be $63,000)
2. What was Nolan\'s capital balance at the end of 2010?(answer should be $246,000)
3.What was wasser\'s total share of net income for 2011? (answer should be $75,000)
please show your work- thank you
Solution
Year 2011
.
Accounting Principles, 12th Edition Ch14
Accounting Principles, 12th Edition
International student version, Willy
WEYGANDT
KIMMEL
KIESO
Recommended
Given below is the partial stockholders equity section of the Balance.docx
Given below is the partial stockholders equity section of the Balance Sheet (as of 12/31/2014) for Toys and Stuff, Inc.
Paid-in-Capital
4% Preferred shares ($100 par) $500,000
Common shares ($1 par) $500,000
Assume that the preferred shares are cumulative, and that no dividends had been paid for 2012 and 2013. If total dividends to be paid for 2014 is $200,000, how much dividends will be paid to the common stockholders?
Please show work.
Solution
Annual Dividend paid to Preferred Stockholders: 500,000 x 4% = $20,000 Per Year
Total Unpaid Preference Dividend for 2012 and 2013 = 20,000 x 2 = $40,000
+ Preference Dividend Payable for 2014 = $20,000
Total Preferred Dividend Payable for 2014 = $60,000
.
YearTotal DividendPreferred dividend ($)Per shareCommon divi.pdf
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Average Dividend per share for common stock six years is $0.4
Year
Total dividend ($)
Preferred dividend
Common dividend
Total ($)
Paid ($)
Per share ($)
Total ($)
Per share
2009
24000
50,000
-26,000
2
2010
10000
50,000
-40,000
2
2011
126000
50,000
116,000
2
10,000
0.1
2012
100000
50,000
50,000
2
50,000
0.5
2013
125000
50,000
50,000
2
75,000
0.75
2014
125000
50,000
50,000
2
75,000
0.75
Average dividend per share for common stock is $ 0.35 and for preferred stock is $2
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Solution
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4
Average Dividend per share for common stock six years is $0.4
Year
Total dividend ($)
Preferred dividend
Common dividend
Total ($)
Paid ($)
Per share ($)
Total ($)
Per share
2009
24000
50,000
-26,000
2
2010
10000
50,000
-40,000
2
2011
126000
50,000
116,000
2
10,000
0.1
2012
100000
50,000
50,000
2
50,000
0.5
2013
125000
50,000
50,000
2
75,000
0.75
2014
125000
50,000
50,000
2
75,000
0.75
Average dividend per share for common stock is $ 0.35 and for preferred stock is $2
Year
Total Dividend
Preferred dividend ($)
Per share
Common dividend
Per share
2009
24000
45,000
-21000
1.8
2010
10000
45,000
-35000
1.8
2011
126000
45,000
101000
1.8
25,000
0.25
2012
100000
45,000
45,000
1.8
55,000
0.55
2013
125000
45,000
45,000
1.8
80,000
0.8
2014
125000
45,000
45,000
1.8
80,000
0.8
1.8
240,000
0.4.
At December 31- 2013- Hansen Corporation had 42-000 shares of common s.docx
At December 31, 2013, Hansen Corporation had 42,000 shares of common stock and 6,000 shares of 9%, $100 par cumulative preferred stock outstanding. No dividends were declared or paid in 2013. Net income was reported as $220,000. What is basic EPS? (Round your answer to 2 decimal places.)
Solution
Basic EPS is the amount of company\'s profit taht is available to its common stock shareholders. As we pay preferred shareholders before common stock holders , so we deduct the preferred dividend.
Basic EPS = (Net income - preferred dividend)/ weighted average number of shares
Net Income = $220000
Preferred dividend = 600000 x .09 = $54000
weighted number of shares ( assuming that all the shares were held by company for the whole year) = 42000
therefore,
Basic EPS = (220000 - 54000)/42000 = $3.95
.
Cleary- Wasser- and Nolan formed a partnership on January 1- 2010- wit.docx
Cleary, Wasser, and Nolan formed a partnership on January 1, 2010, with investments of $100,000, $150,000, and $200,000, respectively. For division of income, they agreed to (1) interest of 10% of the beginning capital balance each year, (2) annual compensation of $10,000 to Wasser, and (3) sharing the remainder of the income or loss in a ratio of 20% for Cleary, and 40% each for Wasser and Nolan. Net income was $150,000 in 2010 and $180,000 in 2011. Each partner withdrew $1,000 for personal use every month during 2010 and 2011.
1.What was Wasser\'s total share of net income for 2010? (answer should be $63,000)
2. What was Nolan\'s capital balance at the end of 2010?(answer should be $246,000)
3.What was wasser\'s total share of net income for 2011? (answer should be $75,000)
please show your work- thank you
Solution
Year 2011
.
Accounting Principles, 12th Edition Ch14
Accounting Principles, 12th Edition
International student version, Willy
WEYGANDT
KIMMEL
KIESO
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Accounting for Partnerships
Accounting for formation,operation, and changes in membership of partnerships
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a 1:4:5 ratio. After lengthy
disagreements among the partners and several unprofitable periods, the partners decided to
liquidate the partnership. Before the liquidation, the partnership balance sheet showed the
following:
Cash $10,000
Total “other assets,” $106,000
Total liabilities, $88,000
Sand, Capital, $1,200
Mell, Capital, $11,700
Rand, Capital, $15,100
The “other assets” were sold for $ 85,000. Proceeds from the sale of other assets were used to
payoff existing liabilities.
Determine the following:
1)The gain (or loss) realized on the sale of the assets and recording of liabilities’ payment.
2)The balances in the partners’ capital accounts after the distribution of this gain or loss to the
capital accounts.
3)Assume that if any capital deficits exist, they are not made up and the deficient partner pays
down his or her deficit to zero. How much cash will each of the partners receive in the final
liquidation?
Solution
Balance Sheet Assets Amount Liabilities Amount Cash $ 10,000.00 Total
liabilities $ 88,000.00 Other Assets $ 106,000.00 Partner\'s Capital A/c Sand $
1,200.00 Mell $ 11,700.00 RAND $ 15,100.00 Total $ 116,000.00 Total $
116,000.00 Journal Entries Particular Amount(DR) Amount(CR) 1) Cash
A/c $ 85,000.00 Loss on realization $ 21,000.00 To Other Assets $
106,000.00 (Being amount realized on sale of assets) Sand=($21000*1/10) $
2,100.00 Mell=($21000*4/10) $ 8,400.00 Rand=($21000*5/10) $ 10,500.00
To Loss on realization $ 21,000.00 (Being loss on realization distribute to
partner\'s in their profit sharing ratio) Total Liabilities $ 88,000.00 To Cash
$ 88,000.00 (Being amount paid to total liabilities) 2) Sand\'s Capital
A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $
2,100.00 By Balance B/d $ 1,200.00 By Cash A/c $ 900.00 Total $ 2,100.00
Total $ 2,100.00 Mell\'s Capital A/c Particular Amount(Dr) Particular
Amount(Cr) To Loss on realization $ 8,400.00 By Balance B/d $ 11,700.00 To
Cash $ 3,300.00 Total $ 11,700.00 Total $ 11,700.00 Rand\'s Capital A/c
Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 10,500.00 By
Balance B/d $ 15,100.00 To Cash $ 4,600.00 Total $ 15,100.00 Total $
15,100.00 3) Cash A/c Particular Amount(Dr) Particular Amount(Cr) To
Balance b/d $ 10,000.00 By Total Liabilities $ 88,000.00 To Other assets $ 85,000.00
By Mell\'s Capital A/c $ 3,300.00 To Sand\'s Capital A/c $ 900.00 By Rand\'s
Capital A/c $ 4,600.00 Total $ 95,900.00 Total $ 95,900.00.
What are this companys dividends per share for 2017Income State.pdf
What are this company\'s dividends per share for 2017?
Income Statement ($000s) Balance Sheet ($000s)
2017 2016 2017
2016
Revenues $ 200 $175 Cash $ 10 $150
Cost of Goods Sold 110 100 Accounts Receivable 30
40
Operating Expenses 30 28 Inventory 20 40
Interest Expense 20 12 Plant, Property, Equipment 180 200
Net Income 40 35 Total Assets $ 240
$430
Dividends 20 20 Accounts Payable $ 15 $ 25
Bank Debt Due in 6 Months 15 50
Long term Debt 80 245
Common Stock 60 60
Retained Earnings 70 50
Total Liabilities & Equity $ 240
$430
The company has 1,000,000 shares of common stock authorized, 500,000 shares issued, and
200,000 outstanding as of both 12/31/2017 and 12/31/2016.$1.00
Solution
The company\'s dividends per share for 2017 is $0.10 which is calculated as below:-
Dividend per share = (Common Dividends / Common Shares Outstanding)
= (20,000 / 200,000) = $0.10.
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We present a data analysis of the annual fund for a non-profit client and the resulting action plan that comes out of the study.
Herbie Corp-'s Balance Sheet showed the following at 12-31-13 Common S.docx
Herbie Corp\'s Balance Sheet showed the following at 12/31/13
Common Stock, $10 par $100,000
APIC $50,000
Retained Earnings $20,000
A cash dividend is declared on 12/31/13, and is payable on 1/20/14 to shareholders of record as of 1/10/14.
Show all appropriate journal entires assuming a cash dividend of $1 per share
Show all appropriate entries for a $5 per share dividend
Solution
1. Journal entry to record dividend of $1 per share. No. of shares is 10,000 ($100,000/$10)
31-12-2013 - Entry for dividend declaration will be recorded
Retained earnings (10,000X1) Debit 10,000
To Dividend payable Credit 10,000
20-1-2014, Entry to record dividend payment. There will be no entry on 10-1-2014 i.e. the record date. As record date merely determines the names of the stakeholders that will receive dividend.
Dividend payable Debit 10,000
To Cash Credit 10,000
2. Cash dividend of $5 per share. Retained earnings can be used for dividend payments. APIC cannot be used because the Company cannot declare dividends more than retained earnings. In this case, Retained earnings balance will be negative.
31-12-14
Retained earnings Debit 50,000
To Dividend payable Credit 50,000
20-1-14
Dividend payable Debit 50,000
To cash Credit 50,000
.
The following transactions relate to Penelope Inc. Prepare journal e.pdf
The following transactions relate to Penelope Inc. Prepare journal entries for each transaction.
Prepare the equity section of the balance sheet at each year-end, December 31. Assume 2013 was
Penelope’s first year of operations.
1.Penelope issued 1,000 shares of $1 par value common stock for $70 per share on January 1,
2013.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
2.Penelope issued 500 shares of no par value, $5, non-cumulative preferred stock for $50 per
share on January 1, 2013.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
1.Penelope reported net income of $2,000 during 2013 and paid no dividends. Prepare the
company’s equity section of the December 31, 2013, balance sheet.
Stockholders’ Equity:
Total Stockholders’
Equity
2.On January 1, 2014, Penelope purchased 200 shares of its common stock for $60 per share.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/14
3.On December 15, 2014, Penelope declared and paid dividends totaling $6,600 during 2014.
Prepare the company’s equity section from December 31, 2014.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
12/15/14
Class of Stock Total Dividend Dividend Per Share
Preferred
Common
4.Penelope reported net income of $20,000 during 2014. Prepare the company’s stockholders’
equity section of the December 31, 2014, balance sheet.
Stockholders’ Equity:
Total Stockholders’
Equity
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
Solution
1)a.
Cash Debit 70,000 [70 * 1000]
Common sto credit 1,000 [1* 1000 ]
Additional paid in capital credit 69,000 [69* 1000]
[being common stock issued at premium recorded]
Asset = liabilities + equity
+ 70,000 = 0 + +70,000
1b) cash Debit 25,000 [500*50]
Preferred stock credit 25,000
[Being preferred stock issued recorded]
1c)Stockholders equity
2) Treasury stock Debit 12,000 [60*200]
cash credit 12,000
[being stock repurchased recorded]
Asset = Liabilities + equity
- 12000 = 0 + (- )12000
3)Retained earnings Debit 6,600
Preferred stock dividend payable credit 2500 [5*500 ]
Common stock dividend payable credit 4100 [6600 -2500]
[being dividend declared recorded]
b)Preferred stock dividend payable Debit 2500 [5*500 ]
Common stock dividend payable Debit 4100
cash credit 6600
[being dividend paid recorded]
4)Preferred stock25,000Common stock1,000Additional paid in capital -common
stock69,000Retained earnings2,000stockholders equity97,000.
XYZ stock price and dividend history are as follows Year Beginning-.pdf
XYZ stock price and dividend history are as follows: Year Beginning-of-Year Price Dividend
Paid at Year-End 2010 $ 140 $ 4 2011 $ 159 $ 4 2012 $ 132 $ 4 2013 $ 137 $ 4 An investor buys
five shares of XYZ at the beginning of 2010, buys another three shares at the beginning of 2011,
sells one share at the beginning of 2012, and sells all seven remaining shares at the beginning of
2013. a. What are the arithmetic and geometric average time-weighted rates of return for the
investor? (Do not round intermediate calculations. Round your answers to 2 decimal places.)
Arithmetic mean 2.93 % Geometric mean 2.08 % b-1. Prepare a chart of cash flows for the four
dates corresponding to the turns of the year for January 1, 2010, to January 1, 2013. (Negative
amounts should be indicated by a minus sign.) Date Cash Flow 1/1/2010 $ -700 1/1/2011 -457
1/1/2012 164 1/1/2013 987 b-2. What is the dollar-weighted rate of return? (Hint: If your
calculator cannot calculate internal rate of return, you will have to use a spreadsheet or trial and
error.) (Negative value should be indicated by a minus sign. Round your answer to 4 decimal
places.) Rate of return %
Solution
a) Year Beginning-of-Year Price Dividend Paid at Year-End 2010 $140.00 $4.00
2011 $159.00 $4.00 2012 $132.00 $4.00 2013 $137.00 $4.00 Return = [(Capital
Gains + Dividend)/Price] Year 2010-2011 (($159 - $140)+$4)/$140 16.43% Year 2011-
2012 (($132 - $159)+$4)/$132 -14.47% Year 2012-2013 (($137 - $132)+$4)/$132 6.82%
Arithmetic mean: = (16.43%+(14.47%)+6.82%)/3 2.93% Geometric Mean Return
1+ return% 16.43% 1.164 -14.47% 0.855 6.82% 1.068 Geometric Mean =(
(1.164 x .855 x 1.068)^(1/3 ))-1 2.08% b-1. Prepare a chart of cash flows for the four
dates corresponding to the turns of the year for January 1, 2010, to January 1,2013.
Date 1/1/2010 1/1/2011 1/1/2012 1/1/2013 Cash Flow -$700.00 -$457.00 $164.00 $987.00
Date Cash Flow 1/1/2010 -$700.00 Purchase of 5 shares at $140 per share 1/1/2011 -
$457.00 Purchase of 3 shares at $159, plus dividend income on 5 shares held 1/1/2012
$164.00 Sale of 1 share at $132, plus dividend income on 8 shares held 1/1/2013 $987.00
Sale of 7 share at $137, plus dividend income on 7 shares held b-2 Date Cash Flow
1/1/2010 -$700.00 1/1/2011 -$457.00 1/1/2012 $164.00 1/1/2013 $987.00
IRR = Rate of Return 0.21%.
Which of the following are true when comparing TCPIP to the OSI Ref.pdf
Which of the following are true when comparing TCP/IP to the OSI Reference Model? (Choose
two.)
A. The TCP/IP model has seven layers, while the OSI model has only four layers.
B. The TCP/IP model has four layers, while the OSI model has seven layers.
C. The TCP/IP Application layer maps to the Application, Session, and Presentation layers of the
OSI Reference Model.
D. The TCP/IP Application layer is virtually identical to the OSI Application layer.
Solution
B and C are correct. The TCP/IP model has four layers. The Application layer maps to the top
three layers of the OSI Reference Model.
A and D are incorrect. A and D are incorrect because they are contrary to B and C..
What is the reason for having most of the blood volume distributed w.pdf
What is the reason for having most of the blood volume distributed within the venous vessels at
any given time?
Solution
The thin, non-elastic wall of veins (as compared to the thick, elastic, muscular walls of arteries)
allows greater volumes of blood to be accommodated within the lumen of the veins. This
accounts for most of the blood volume distributed within the venous vessels at any given time..
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Similar to Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf
Build Your Own Pension FPA Webinar
Slides from the 9/28/2011 FPA webinar "Build Your Own Pension." With the decline of pensions, clients will look to their own accounts to provide predictable retirement income. We show how advisors can create pension-like income using institutional liability driven investing (LDI) strategies.
Accounting for Partnerships
Accounting for formation,operation, and changes in membership of partnerships
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a 1:4:5 ratio. After lengthy
disagreements among the partners and several unprofitable periods, the partners decided to
liquidate the partnership. Before the liquidation, the partnership balance sheet showed the
following:
Cash $10,000
Total “other assets,” $106,000
Total liabilities, $88,000
Sand, Capital, $1,200
Mell, Capital, $11,700
Rand, Capital, $15,100
The “other assets” were sold for $ 85,000. Proceeds from the sale of other assets were used to
payoff existing liabilities.
Determine the following:
1)The gain (or loss) realized on the sale of the assets and recording of liabilities’ payment.
2)The balances in the partners’ capital accounts after the distribution of this gain or loss to the
capital accounts.
3)Assume that if any capital deficits exist, they are not made up and the deficient partner pays
down his or her deficit to zero. How much cash will each of the partners receive in the final
liquidation?
Solution
Balance Sheet Assets Amount Liabilities Amount Cash $ 10,000.00 Total
liabilities $ 88,000.00 Other Assets $ 106,000.00 Partner\'s Capital A/c Sand $
1,200.00 Mell $ 11,700.00 RAND $ 15,100.00 Total $ 116,000.00 Total $
116,000.00 Journal Entries Particular Amount(DR) Amount(CR) 1) Cash
A/c $ 85,000.00 Loss on realization $ 21,000.00 To Other Assets $
106,000.00 (Being amount realized on sale of assets) Sand=($21000*1/10) $
2,100.00 Mell=($21000*4/10) $ 8,400.00 Rand=($21000*5/10) $ 10,500.00
To Loss on realization $ 21,000.00 (Being loss on realization distribute to
partner\'s in their profit sharing ratio) Total Liabilities $ 88,000.00 To Cash
$ 88,000.00 (Being amount paid to total liabilities) 2) Sand\'s Capital
A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $
2,100.00 By Balance B/d $ 1,200.00 By Cash A/c $ 900.00 Total $ 2,100.00
Total $ 2,100.00 Mell\'s Capital A/c Particular Amount(Dr) Particular
Amount(Cr) To Loss on realization $ 8,400.00 By Balance B/d $ 11,700.00 To
Cash $ 3,300.00 Total $ 11,700.00 Total $ 11,700.00 Rand\'s Capital A/c
Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 10,500.00 By
Balance B/d $ 15,100.00 To Cash $ 4,600.00 Total $ 15,100.00 Total $
15,100.00 3) Cash A/c Particular Amount(Dr) Particular Amount(Cr) To
Balance b/d $ 10,000.00 By Total Liabilities $ 88,000.00 To Other assets $ 85,000.00
By Mell\'s Capital A/c $ 3,300.00 To Sand\'s Capital A/c $ 900.00 By Rand\'s
Capital A/c $ 4,600.00 Total $ 95,900.00 Total $ 95,900.00.
What are this companys dividends per share for 2017Income State.pdf
What are this company\'s dividends per share for 2017?
Income Statement ($000s) Balance Sheet ($000s)
2017 2016 2017
2016
Revenues $ 200 $175 Cash $ 10 $150
Cost of Goods Sold 110 100 Accounts Receivable 30
40
Operating Expenses 30 28 Inventory 20 40
Interest Expense 20 12 Plant, Property, Equipment 180 200
Net Income 40 35 Total Assets $ 240
$430
Dividends 20 20 Accounts Payable $ 15 $ 25
Bank Debt Due in 6 Months 15 50
Long term Debt 80 245
Common Stock 60 60
Retained Earnings 70 50
Total Liabilities & Equity $ 240
$430
The company has 1,000,000 shares of common stock authorized, 500,000 shares issued, and
200,000 outstanding as of both 12/31/2017 and 12/31/2016.$1.00
Solution
The company\'s dividends per share for 2017 is $0.10 which is calculated as below:-
Dividend per share = (Common Dividends / Common Shares Outstanding)
= (20,000 / 200,000) = $0.10.
Financial Tools for Start-ups
Stuart Reid, Executive Director, Food Co-op Initiative (FCI), presented to start-up staff, steering committee, and Board members at NFCA's Sixth Annual Meeting. This presentation reviews some of the most important financial tools that every start-up food co-op needs. This includes Sources & Uses budget, financial pro formas, and market studies, and focus on how to understand these tools and use the information to strengthen your co-op's plans.
Finding and acting on key donor renewal metrics
We present a data analysis of the annual fund for a non-profit client and the resulting action plan that comes out of the study.
Herbie Corp-'s Balance Sheet showed the following at 12-31-13 Common S.docx
Herbie Corp\'s Balance Sheet showed the following at 12/31/13
Common Stock, $10 par $100,000
APIC $50,000
Retained Earnings $20,000
A cash dividend is declared on 12/31/13, and is payable on 1/20/14 to shareholders of record as of 1/10/14.
Show all appropriate journal entires assuming a cash dividend of $1 per share
Show all appropriate entries for a $5 per share dividend
Solution
1. Journal entry to record dividend of $1 per share. No. of shares is 10,000 ($100,000/$10)
31-12-2013 - Entry for dividend declaration will be recorded
Retained earnings (10,000X1) Debit 10,000
To Dividend payable Credit 10,000
20-1-2014, Entry to record dividend payment. There will be no entry on 10-1-2014 i.e. the record date. As record date merely determines the names of the stakeholders that will receive dividend.
Dividend payable Debit 10,000
To Cash Credit 10,000
2. Cash dividend of $5 per share. Retained earnings can be used for dividend payments. APIC cannot be used because the Company cannot declare dividends more than retained earnings. In this case, Retained earnings balance will be negative.
31-12-14
Retained earnings Debit 50,000
To Dividend payable Credit 50,000
20-1-14
Dividend payable Debit 50,000
To cash Credit 50,000
.
The following transactions relate to Penelope Inc. Prepare journal e.pdf
The following transactions relate to Penelope Inc. Prepare journal entries for each transaction.
Prepare the equity section of the balance sheet at each year-end, December 31. Assume 2013 was
Penelope’s first year of operations.
1.Penelope issued 1,000 shares of $1 par value common stock for $70 per share on January 1,
2013.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
2.Penelope issued 500 shares of no par value, $5, non-cumulative preferred stock for $50 per
share on January 1, 2013.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
1.Penelope reported net income of $2,000 during 2013 and paid no dividends. Prepare the
company’s equity section of the December 31, 2013, balance sheet.
Stockholders’ Equity:
Total Stockholders’
Equity
2.On January 1, 2014, Penelope purchased 200 shares of its common stock for $60 per share.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/14
3.On December 15, 2014, Penelope declared and paid dividends totaling $6,600 during 2014.
Prepare the company’s equity section from December 31, 2014.
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
12/15/14
Class of Stock Total Dividend Dividend Per Share
Preferred
Common
4.Penelope reported net income of $20,000 during 2014. Prepare the company’s stockholders’
equity section of the December 31, 2014, balance sheet.
Stockholders’ Equity:
Total Stockholders’
Equity
DATE
ACCOUNT NAME
DEBIT
CREDIT
BALANCE SHEET
INCOME STMT
A
=
L
+
E
R
-
E
1/1/13
Solution
1)a.
Cash Debit 70,000 [70 * 1000]
Common sto credit 1,000 [1* 1000 ]
Additional paid in capital credit 69,000 [69* 1000]
[being common stock issued at premium recorded]
Asset = liabilities + equity
+ 70,000 = 0 + +70,000
1b) cash Debit 25,000 [500*50]
Preferred stock credit 25,000
[Being preferred stock issued recorded]
1c)Stockholders equity
2) Treasury stock Debit 12,000 [60*200]
cash credit 12,000
[being stock repurchased recorded]
Asset = Liabilities + equity
- 12000 = 0 + (- )12000
3)Retained earnings Debit 6,600
Preferred stock dividend payable credit 2500 [5*500 ]
Common stock dividend payable credit 4100 [6600 -2500]
[being dividend declared recorded]
b)Preferred stock dividend payable Debit 2500 [5*500 ]
Common stock dividend payable Debit 4100
cash credit 6600
[being dividend paid recorded]
4)Preferred stock25,000Common stock1,000Additional paid in capital -common
stock69,000Retained earnings2,000stockholders equity97,000.
XYZ stock price and dividend history are as follows Year Beginning-.pdf
XYZ stock price and dividend history are as follows: Year Beginning-of-Year Price Dividend
Paid at Year-End 2010 $ 140 $ 4 2011 $ 159 $ 4 2012 $ 132 $ 4 2013 $ 137 $ 4 An investor buys
five shares of XYZ at the beginning of 2010, buys another three shares at the beginning of 2011,
sells one share at the beginning of 2012, and sells all seven remaining shares at the beginning of
2013. a. What are the arithmetic and geometric average time-weighted rates of return for the
investor? (Do not round intermediate calculations. Round your answers to 2 decimal places.)
Arithmetic mean 2.93 % Geometric mean 2.08 % b-1. Prepare a chart of cash flows for the four
dates corresponding to the turns of the year for January 1, 2010, to January 1, 2013. (Negative
amounts should be indicated by a minus sign.) Date Cash Flow 1/1/2010 $ -700 1/1/2011 -457
1/1/2012 164 1/1/2013 987 b-2. What is the dollar-weighted rate of return? (Hint: If your
calculator cannot calculate internal rate of return, you will have to use a spreadsheet or trial and
error.) (Negative value should be indicated by a minus sign. Round your answer to 4 decimal
places.) Rate of return %
Solution
a) Year Beginning-of-Year Price Dividend Paid at Year-End 2010 $140.00 $4.00
2011 $159.00 $4.00 2012 $132.00 $4.00 2013 $137.00 $4.00 Return = [(Capital
Gains + Dividend)/Price] Year 2010-2011 (($159 - $140)+$4)/$140 16.43% Year 2011-
2012 (($132 - $159)+$4)/$132 -14.47% Year 2012-2013 (($137 - $132)+$4)/$132 6.82%
Arithmetic mean: = (16.43%+(14.47%)+6.82%)/3 2.93% Geometric Mean Return
1+ return% 16.43% 1.164 -14.47% 0.855 6.82% 1.068 Geometric Mean =(
(1.164 x .855 x 1.068)^(1/3 ))-1 2.08% b-1. Prepare a chart of cash flows for the four
dates corresponding to the turns of the year for January 1, 2010, to January 1,2013.
Date 1/1/2010 1/1/2011 1/1/2012 1/1/2013 Cash Flow -$700.00 -$457.00 $164.00 $987.00
Date Cash Flow 1/1/2010 -$700.00 Purchase of 5 shares at $140 per share 1/1/2011 -
$457.00 Purchase of 3 shares at $159, plus dividend income on 5 shares held 1/1/2012
$164.00 Sale of 1 share at $132, plus dividend income on 8 shares held 1/1/2013 $987.00
Sale of 7 share at $137, plus dividend income on 7 shares held b-2 Date Cash Flow
1/1/2010 -$700.00 1/1/2011 -$457.00 1/1/2012 $164.00 1/1/2013 $987.00
IRR = Rate of Return 0.21%.
Similar to Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf (20)
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
What are this companys dividends per share for 2017Income State.pdf
What are this companys dividends per share for 2017Income State.pdf
Herbie Corp-'s Balance Sheet showed the following at 12-31-13 Common S.docx
Herbie Corp-'s Balance Sheet showed the following at 12-31-13 Common S.docx
The following transactions relate to Penelope Inc. Prepare journal e.pdf
The following transactions relate to Penelope Inc. Prepare journal e.pdf
XYZ stock price and dividend history are as follows Year Beginning-.pdf
XYZ stock price and dividend history are as follows Year Beginning-.pdf
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Which of the following are true when comparing TCPIP to the OSI Ref.pdf
Which of the following are true when comparing TCP/IP to the OSI Reference Model? (Choose
two.)
A. The TCP/IP model has seven layers, while the OSI model has only four layers.
B. The TCP/IP model has four layers, while the OSI model has seven layers.
C. The TCP/IP Application layer maps to the Application, Session, and Presentation layers of the
OSI Reference Model.
D. The TCP/IP Application layer is virtually identical to the OSI Application layer.
Solution
B and C are correct. The TCP/IP model has four layers. The Application layer maps to the top
three layers of the OSI Reference Model.
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What is the reason for having most of the blood volume distributed w.pdf
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Solution
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What is eutectic reaction? What is eutectoid reaction? What is perifectic reaction? Are these
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Solution
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containing two elements soluble in one another cools down to a level where the solutbility is
limited, then the excess element is sandwiched between the zones containing the mixture. This
reaction is Eutectic reaction. In the Iron-Carbide diagram above, it is represented by the point D.
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What are the possible types of interfund transactions What are.pdf
What are the possible types of interfund transactions?
What are the possible types of interfund transactions?
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These are transactions that would be treated as revenues, expenditures or expenses if they
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What are Sociocultural issues when it come to body imageSolutio.pdf
What are Sociocultural issues when it come to body image?
Solution
Answer:
Sociocultural issues such as dissatisfaction in women and men each they are obese due to eating
disorders & media aspects and cultural aspects often have meticulous influence on individual
body image finally may lead to “development of depressive disorders”. Therefore, unrealistic
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Water can absorb and store a large amount of heat while increasing on.pdf
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temperature. The heat must first be used to break the covalent bonds rather than raise the
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Thanks for the question
5.A The hydrogen bonds are reponsible for the high boiling points of water.
6.C an acid is a substance which donates H.
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True or false1 Anabolism results in increased production of biomas.pdf
True or false
1 Anabolism results in increased production of biomass.
2 Oxygenic phototrophs can use hydrogen sulfide as an electron donor.
3 Oxidation is the process of accepting electrons.
4 Water appeared on the earth as a result of bombardment by asteroids.
5 The highest abundance of microbial life in soils appears in the O horizon.
6 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during
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7 Freshwater represents about 10% of the total water on Earth.
8 Oligotrophic organisms thrive in low nutrient conditions.
9 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level
phosphorylation.
10 Siderphores are used to increase availability of manganese in aerobic environments.
Solution
1. Anabolism: The process of formation of biomolecules during metabolism such as proteins,
polysaccharides, lipids, and nucleic acids etc is known as anabolism, this results in an increase in
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Therefore, option is: True
2.Oxygenic phototrophs use oxygen as electron donar, while as anoxygenic phototrophs uses
hydrogen sulfide as electron donor.
Therefore, option is: False
3. Oxidation is a process in which a substance looses one or more electrons, while as reduction is
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4. The earth got most of the water about 4.6 billion years ago from the asteroid impacts.
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The diploid number of the hypothetical animal Geneticus introductus is 2n= 44. Each diploid
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Part A : What amount of DNA is contained in each nucleus at the end of S phase? What amount
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a)There are 22 chromosomes containing 2.5 ng of DNA.
b)There are 44 chromosomes containing 5 ng of DNA.
c)There are 44 chromosomes containing 10 ng of DNA.
d)None of the above.
Part B: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
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Part C: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
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Part D: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase II of meiosis (after cytokinesis).
Solution
The cell cycle of actively dividing eukaryotic cell consists of gap phases (G 1 and G2), S phase
and M phase. In G 1 phase the cell prepares for cell division .In S phase DNA replication occurs
and the DNA content doubles. In G 2 phase, the cell gets ready for mitosis.
In mitosis, two daughter cells with same genetic content as parent are produced. Mitosis occurs
through five stages: prophase, prometaphase, metaphase, anaphase and telophase.
Prophase: This stage extends over half of mitosis. The nuclear membrane disintegrates. The
centrosome of chromosome duplicates to form two daughter centrosomes and migrates to
opposite poles of the cell. The centrosomes organise the production of microtubules, from which
the spindle fibres arise. The chromosomes condense and consist of two identical chromatids (or
sister chromatids) held together at the centromere.
Prometaphase: The chromosomes, move to the middle of the cell. The spindle fibres bind to
centromere of each chromosome at kinetochore.
Metaphase: The chromosomes align along the metaphase plate of the spindle apparatus.
Anaphase: This is the shortest stage of mitosis. The centromeres divide, and the sister chromatids
of each chromosome are separated and move to the opposite poles of the cell. The separated
sister chromatids ordaughter chromosomesreceives a copy of each parental chromosome.
Telophase: This is the final stage of mitosis. The nuclear membrane reforms, the chromosomes
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Cytokinesis: the membrane appears to divide the two daughter cells with same chromosome
number and the DNA content as the parent cell.
The cell then enters interphase - the interval between mitotic divisions.
Meiosis is the cell division occuring in sex cells or gametes (which contain a single copy of each
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The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For
temperatures near the normal boiling point:
a. Derive an expression that gives the behavior of Hvap as a function of temperature which is
thermodynamically consistent with the Antoine equation.
Solution
The Antoine equation is a mathematical expression of the relation between the vapour pressure
and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french
researcher, in 1888. The basic form of the equation is:
lnP = A - B / ( T + C )
and it can be transformed into this temperature-explicit form:
T = B ( A - lnP ) - C
where: P is the absolute vapor pressure of a substance
T is the temperature of the substance
A, B and C are substance-specific coefficients (i.e., constants or parameters
log is typically either log10 or loge
A simpler form of the equation with only two coefficients is sometimes used:
lnP = A - ( B \\ T )
which can be transformed to:
T = B ( A - lnP )
The Antoine equation is a semi-empirical equation which expresses vapour pressure as a
function of temperature. A new, rapid and highly accurate method for obtaining its three
constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic
substances where p is the vapour pressure, T is temperature and A, B and C are component-
specific constants.
Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure
curve from the triple point to the critical point, because it is not flexible enough. Therefore,
multiple parameter sets for a single component are commonly used. A low-pressure parameter
set is used to describe the vapour pressure curve up to the normal boiling point and the second
set of parameters is used for the range from the normal boiling point to the critical point.
The Antoine Equation is a vapor pressure equation and describes the relation between vapor
pressure and temperature for pure water between 0°C and 373.946°C.
For T in Kelvins and P in kPa:
273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724
373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665
Antoine(T) = exp(A B / (T + C))
For temperatures above the critical temperature (373.946°C / 647.096K), where water is a
superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e.
dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements
that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can
write
P(T Tc) = Antoine(T)
P(T > Tc) = 235.88 (T Tc) + Antoine(Tc)
Hence, the behaviour of delta hvap as a function of temperature which is termodynamically
consistent with the the antoine equation is derived ..
The converson of the English to Christianity began a rich period of .pdf
The converson of the English to Christianity began a rich period of artistic production in
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ecclesiastical treatises, d. gothic cathedrals
Solution
b. illuminated manuscripts
Elaborately decorated manuscripts were prepared after the conversion. Thus b is the correct
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Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa
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(flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with
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Solution
3. Bilateral symmetry
5.Value of body cavity:
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turgescent state of nematode body.
b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic
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Question 12 (5 points)The reason that mitosis can result in the pr.pdf
Question 12 (5 points)
The reason that mitosis can result in the production of haploid gametes in alternation of
generations is because
Question 12 options:
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Save
Question 13 (5 points)
In the fern life cycle, how does fertilization occur?
Question 13 options:
The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf
and find each other on the ground to fuse
The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming
through water
The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
There is no sexual reproduction in ferns
Save
Question 14 (5 points)
The earliest land plants were most likely to have lived in which of the following habitats?
Question 14 options:
arid desert environments
Salt water marshes
Tropical rainforests
fresh water wetlands
Save
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Solution
Q12 Eggs and sperms are produced by haploid gametophyte by mitosis.
The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells
undergo meiosis to create haploid cells. However, these cells do not immediately unite with
others to recreate the diploid state. Instead, they are shed as spores and germinate into small
haploid organisms. Because the diploid organism creates spores, it is called the sporophyte
generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg
and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called
the gametophyte generation of the life cycle. The male gametes (sperm) are then released and
swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing
the life cycle.
Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These
antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through
water; they do not travel great distances and are only released when free water is available. Only
a continuous film of water is needed. They antherozoids are chemically attracted to the
developing flask-like female organs or archegonia where fertilization of the single egg cell
embedded in each archegonium takes place.
The sporophyte has a double set of chromosomes (i.e. diploid) and i.
peripheral circuit functions built-in to the HC12 microcontroller, t.pdf
peripheral circuit functions built-in to the HC12 microcontroller, that are not part of the CPU
itself or the memories. Access to these is via external pins on the Microcontroller IC package.
What are some of these built-in circuit functions called, and what do they do?
Solution
The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS.
The bus is a set of shared communication lines that serves as the central nervous system of the
computer. Data, address, and control signals are shared by all system components via the bus.
Each component connected to the bus communicates information to and from the bus via its own
bus controller. The data lines,address lines, and control lines allow a specific component to
access data addressed to that component. The data lines are used to communicate words to and
from data.
registers in the various system components such as memory, CPU, and input/output
(I/O) peripherals. The address lines are used to select devices on the bus or specific
data locations within memory. Devices usually have a combinational logic address
decoder circuit that identifies the address code and activates the device..
If the dpy-11 and unc-31 genes are linked on the same chromosome, how.pdf
If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes
would the F_2 progeny show? If the two genes are on different chromosomes and therefore
independently assorting, how many phenotypes would you see in the F_2 worms? List the
phenotype(s). Strawberry plants have varieties that flower when the days are still short, and
other varieties that flower only when the days get longer. The long-day allele is dominant to the
short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion.
When you cross between heterozygous plants and count the offspring in the F_1 generation, this
is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short
days What would you expect the ratio of phenotypes be? What is your Chi squared value?
Show your work What does this mean?
Solution
2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs
independent assortment occurs. No new characters appear. Where as gegenes are on different
chromosomes, they are non linked. Crossing over or independent assortment takes place. New
characters appear along with parental characters. So dpy-11 & unc-31 present on same
chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the
said genes present on different chromosomes, they give 4 different phenotypes
3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between
these genes
Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short.
Chi square value can be calculated by using the formula.
(O-E)^2/E
O = observed phenotypes; E = Expected phenotypes.
How does a wave formation influence the resultant waveSolution.pdf
How does a wave formation influence the resultant wave?
Solution
Waves are usually formed by distant wind.Generally larger waves are more powerful than small
waves . Waves can pile up at shore lines and create a rip current.The resultant wave heights Hi
and the wave period Ti are related to the factors that influence the energy transfer between wind
and the ocean..
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How are health services paid for? Provide a definition for the term \"third-party payer\", discuss
the different payers currently operating in the market, and assess the importance of each. Please
use a table format to list the information.
Solution
There are two main payers types: single payer and third-party payer. Single payer is when the
person that obtains the health service is the same which pay for it, here relationship is direct taker
and provider of health services.
However, it is more common the presence of a third party payer, which is an external agent that
covers the other health expenses or part of them, they are insurance companies or public social
protection systems.
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage.
Given the information regarding mitosis and meiosis, clearly and con.pdf
Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
First, look to determine what is the earliest era that fossils have .pdf
First, look to determine what is the earliest era that fossils have been found here in Georgia.
Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of
the fossil, when it was found, and where the fossil was found. Give a brief description of the
organism that was fossilized.
Solution
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Translation is an operation which is uesd in geometry to move an object in a fixed direction and
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explain why right and why wrong Indentify configuration (below) whic.pdf
explain why right and why wrong Indentify configuration (below) which I s not a correct
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Solution
option C is not correct one.
Explanation:
If we stretch all right hand fingers in the direction of current direction given then the thumb gives
us into the board (into symbol) or out of the board (dot symbol).
In the option C the field should be out of the board, so it is not correct one..
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