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Free_Kalibri [48]
3 years ago
15

The balance sheet of Messi Services included the following shareholders' equity section at December 31, 2018: ($ in millions) Co

mmon stock ($1 par, authorized 200 million shares, issued and outstanding 180 million shares) $ 180 Paid-in capital—excess of par 1,080 Retained earnings 560 Total shareholders’ equity $ 1,820 On January 5, 2019, Holmes purchased 2 million treasury shares for $9 million. Immediately after the purchase of the shares, the balances in the paid-in capital- excess of par and retained earnings accounts are: Paid-in capital—excess of par Retained earnings a. $ 1,068 $ 556 b. $ 1,064 $ 560 c. $ 1,080 $ 560 d. $ 1,080 $ 544
Business
2 answers:
Yuri [45]3 years ago
6 0

Answer: The answer is c $1,080 $560

Explanation:

The journal entry will be

Dr: common stock $200 million

Dr: paid in capital $180 million

In the stockholders equity section , the treasury stock is seen as a separate line item in the stockholders equity. The treasury stock will be deducted from the total stockholders equity. The treasury stock is not a part of paid in capital nor part of the retained earning.

Therefore the balance in the paid in capital excess of par Retained Earnings is 1,080 $560

Lady_Fox [76]3 years ago
4 0

Answer:

             a.$1068 $556

Explanation:

The treasury shares are sold below par value therefore it will reduce equity.

The reduction will be recorded as=$1,080/180=$6 per share

When 2 million shares are sold=$6*2=$12

Paid up capital in excess of par will be =$1,080-$12=$1,068

Retained Earnings=$560-$4=$556

Please note that treasury shares are always part of common stock.

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julia-pushkina [17]

Answer:

a-The present value of revenue in the first year is $61,085.92.

b-The total time it would take to pay for its price is 2.44 years of 29.33 months.

Explanation:

a-

Let the function of the revenue earned is given as

S(t)=\left \{ {{66000t+38000} {\ \ 0The present value is given as [tex]PV=\int\limits^a_b {S(t)e^{-rt}} \, dt

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    b-

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    PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt+\int\limits^t_1 {S(t)e^{-0.05t}} \, dt\\PV=61085.9225+\int\limits^{t}_{1}{(71000)e^{-0.05t}} \, dt

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Answer:

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