Questions
Pin Corporation paid $4,500,000 for a 90 percent interest in San Corporation on January 1, 2016;...

Pin Corporation paid $4,500,000 for a 90 percent interest in San Corporation on January 1, 2016; San’s total book value was $4,500,000. The excess was allocated as follows: $150,000 to undervalued equipment with a three-year remaining useful life and $350,000 to goodwill. The income statements of Pin and San for 2016 are summarized as follows (in thousands):

                                                                                    Pin                               San

                        Sales                                                    $9,000                         $5,200

                        Income from San                                      450                         

                        Cost of sales                                         (5,000)                         (3,600)

                        Depreciation expense                         (1,000)                            (520)

                        Other expenses                                   (1,450)                            (580)

                        Net income                                         $2,000                             500

Required:

  1. Calculate the goodwill that should appear in the consolidated balance sheet of Pin and Subsidiary at December 31, 2016.
  2. Calculate the following 3 items for 2016:
  • Consolidated net income for 2016
  • Controlling share of consolidated net income for 2016
  • Noncontrolling share of consolidated net income for 2016

In: Accounting

The PFK1 enzyme is regulated by a number of factors in the cell, including the level...

The PFK1 enzyme is regulated by a number of factors in the cell, including the level of ATP. ATP is not only

involved in the reaction the enzyme catalyses but also binds to an allosteric site leading to a reduction in

activity.

The binding of ATP can be measured using a fluorescently labelled derivative of ATP – TNP-ATP. The data

presented below represents the determination of the binding of ATP to PFK. The change in fluorescence (?

fluor.) represents the concentration of PFK1 with TNP-ATP bound. This value was determined at 5

different concentrations of TNP-ATP, and the experiment was carried out in triplicate.

? fluor. (arbitrary units)

[TNP-ATP] (µM)   

Experiment 1

Experiment 2

Experiment 3

60

0.078

0.081

0.085

120

0.131

0.139

0.140

240

0.240

0.241

0.250

480

0.325

0.335

0.319

960

0.416

0.442

0.409

a) Create a double reciprocal plot of this data (1/[TNP-ATP] vs. 1/ ? fluor.). Plot the data from

each experiment as a separate dataset on the same graph. Only plot the data points – do

not join data points with lines. Ensure that each dataset is distinct from the others (use

alternate colours and/or shapes for the data points).

b) Fit an equation to each dataset separately. Do not show these on your plots. For each

experiment give the fitted equation and the value of Kd. Show these in a table. Ensure you

include the units of this value. Show your working for the Experiment 1 data.

c) Determine the mean and standard deviation of the data for each concentration of TNPATP.

Show these in a table. Transform this data as you have done in (a). Fit an equation to

the transformed data. Give the fitted equation and the calculated Kd for the averaged

data. Add a trendline defined by this equation to your graph.

d) With reference the data you plotted and Kd values you determined, explain the positive

and negative aspects of using a double reciprocal plot.

In: Biology

Calculate the 5-day and 50-day simple MA for the stock price from 28 Oct 2016 to...

Calculate the 5-day and 50-day simple MA for the stock price from 28 Oct 2016 to 14 June 2017. Plot the MA5 and MA50. How many golden crosses (if any) are there?

Date Time Closing Price 5-day MA 50-day MA
2016/10/28 1 204
2016/10/31 2 205.399994
2016/11/1 3 206
2016/11/2 4 203
2016/11/3 5 200.199997
2016/11/4 6 199.699997
2016/11/7 7 202.600006
2016/11/8 8 203.600006
2016/11/9 9 202.399994
2016/11/10 10 204.800003
2016/11/11 11 206.199997
2016/11/14 12 204.199997
2016/11/15 13 203.399994
2016/11/16 14 203.399994
2016/11/17 15 204
2016/11/18 16 203
2016/11/21 17 202.399994
2016/11/22 18 203.800003
2016/11/23 19 203.600006
2016/11/24 20 203.800003
2016/11/25 21 205
2016/11/28 22 204.800003
2016/11/29 23 203.600006
2016/11/30 24 204
2016/12/1 25 204.199997
2016/12/2 26 202.800003
2016/12/5 27 197.399994
2016/12/6 28 196.399994
2016/12/7 29 193.600006
2016/12/8 30 195
2016/12/9 31 192.699997
2016/12/12 32 189.600006
2016/12/13 33 189.600006
2016/12/14 34 190
2016/12/15 35 186.899994
2016/12/16 36 185.100006
2016/12/19 37 182.600006
2016/12/20 38 182.100006
2016/12/21 39 182.100006
2016/12/22 40 181
2016/12/23 41 179.899994
2016/12/28 42 180.699997
2016/12/29 43 180.899994
2016/12/30 44 183.199997
2017/1/3 45 184.300003
2017/1/4 46 182.600006
2017/1/5 47 185.300003
2017/1/6 48 185.399994
2017/1/9 49 185.600006
2017/1/10 50 187.5
2017/1/11 51 188.300003
2017/1/12 52 186.600006
2017/1/13 53 186.5
2017/1/16 54 185
2017/1/17 55 185.899994
2017/1/18 56 187.899994
2017/1/19 57 187.199997
2017/1/20 58 185.5
2017/1/23 59 185.199997
2017/1/24 60 185.199997
2017/1/25 61 185.800003
2017/1/26 62 187.800003
2017/1/27 63 188.699997
2017/2/1 64 188.5
2017/2/2 65 187.300003
2017/2/3 66 186.399994
2017/2/6 67 187.199997
2017/2/7 68 186.699997
2017/2/8 69 195.5
2017/2/9 70 195.5
2017/2/10 71 195.800003
2017/2/13 72 198.699997
2017/2/14 73 197.800003
2017/2/15 74 202
2017/2/16 75 202.399994
2017/2/17 76 200.199997
2017/2/20 77 200.800003
2017/2/21 78 199
2017/2/22 79 200
2017/2/23 80 200.399994
2017/2/24 81 198
2017/2/27 82 195.399994
2017/2/28 83 192.699997
2017/3/1 84 193.5
2017/3/2 85 192
2017/3/3 86 191.399994
2017/3/6 87 191.699997
2017/3/7 88 192.199997
2017/3/8 89 191.800003
2017/3/9 90 190.399994
2017/3/10 91 190.100006
2017/3/13 92 191.800003
2017/3/14 93 192
2017/3/15 94 194.699997
2017/3/16 95 197.399994
2017/3/17 96 196.5
2017/3/20 97 200.399994
2017/3/21 98 199.699997
2017/3/22 99 196.399994
2017/3/23 100 196.399994
2017/3/24 101 196.399994
2017/3/27 102 195
2017/3/28 103 195.399994
2017/3/29 104 194.600006
2017/3/30 105 194.399994
2017/3/31 106 195.600006
2017/4/3 107 196.100006
2017/4/5 108 197.399994
2017/4/6 109 196
2017/4/7 110 195.899994
2017/4/10 111 195.5
2017/4/11 112 193.800003
2017/4/12 113 195.199997
2017/4/13 114 194.199997
2017/4/18 115 191.699997
2017/4/19 116 190.800003
2017/4/20 117 191.5
2017/4/21 118 190.800003
2017/4/24 119 190.800003
2017/4/25 120 192
2017/4/26 121 193.699997
2017/4/27 122 194.100006
2017/4/28 123 191.600006
2017/5/2 124 191.600006
2017/5/4 125 190.600006
2017/5/5 126 189.300003
2017/5/8 127 189.600006
2017/5/9 128 194.399994
2017/5/10 129 194.199997
2017/5/11 130 195.699997
2017/5/12 131 195.5
2017/5/15 132 197.100006
2017/5/16 133 196.699997
2017/5/17 134 197.699997
2017/5/18 135 195.300003
2017/5/19 136 195.699997
2017/5/22 137 196.399994
2017/5/23 138 195.199997
2017/5/24 139 194.899994
2017/5/25 140 196.300003
2017/5/26 141 196
2017/5/29 142 197.199997
2017/5/31 143 196.699997
2017/6/1 144 196.800003
2017/6/2 145 205
2017/6/5 146 203.800003
2017/6/6 147 205
2017/6/7 148 204.199997
2017/6/8 149 205.600006
2017/6/9 150 205.399994
2017/6/12 151 199.699997
2017/6/13 152 201
2017/6/14 153 200.600006

In: Finance

Pension Expense and Liability On December 31, 2016, Robey Company accumulated the following information for 2016...

Pension Expense and Liability

On December 31, 2016, Robey Company accumulated the following information for 2016 in regard to its defined benefit pension plan:

Service cost $113,390
Interest cost on projected benefit obligation 11,810
Expected return on plan assets 11,250
Amortization of prior service cost 2,130

On its December 31, 2015, balance sheet, Robey had reported an accrued/prepaid pension cost liability of $14,100.

Required:

1. Compute the amount of Robey’s pension expense for 2016.
2. Prepare all the journal entries related to Robey’s pension plan for 2016 if it funds the pension plan in the amount of (a) $116,080, (b) $115,170, and (c) $119,920.
3. Next Level Assuming Robey’s beginning 2016 Accumulated Other Comprehensive Income: Prior Service Cost balance was $55,640 what would be its ending balance?
4. Next Level How much would Robey need to fund its pension plan for 2016 in order to report an accrued/ prepaid pension cost asset of $4,870 at the end of 2016?

General Journal

Assume Robey Company funds the pension plan in the amount of $116,080. Prepare the entries to record the pension expense for 2016 on December 31 and the amortized prior service cost for 2016 on December 31.

PAGE 1

GENERAL JOURNAL

DATE ACCOUNT TITLE POST. REF. DEBIT CREDIT

1

2

3

4

Assume Robey Company funds the pension plan in the amount of $115,170. Prepare the entries to record the pension expense for 2016 on December 31 and the amortized prior service cost for 2016 on December 31.

PAGE 1

GENERAL JOURNAL

DATE ACCOUNT TITLE POST. REF. DEBIT CREDIT

1

2

3

4

5

Assume Robey Company funds the pension plan in the amount of $119,920. Prepare the entries to record the pension expense for 2016 on December 31 and the amortized prior service cost for 2016 on December 31.

PAGE 1

GENERAL JOURNAL

DATE ACCOUNT TITLE POST. REF. DEBIT CREDIT

1

2

3

4

5

Calculations

Compute the amount of Robey’s pension expense for 2016.

Next Level: Assuming Robey’s beginning 2016 Accumulated Other Comprehensive Income: Prior Service Cost balance was $55,640 what would be its ending balance?

How much would Robey need to fund its pension plan for 2016 in order to report an accrued/ prepaid pension cost asset of $4,870 at the end of 2016?

In: Accounting

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10-2 cm

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10-2 cm, mounted coaxially within a cylindrical anode with a radius of 0.5580 cm. The potential difference between the anode and cathode is 275 V. An electron leaves the surface of the cathode with zero initial speed v(initial) = 0). Find its speed v(final) when it strikes the anode.

 

In: Physics

An electron leaves the surface of the cathode with zero initial speed v(initial) = 0). Find its speed v(final) when it strikes the anode.

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10-2 cm, mounted coaxially within a cylindrical anode with a radius of 0.5580 cm. The potential difference between the anode and cathode is 275 V. An electron leaves the surface of the cathode with zero initial speed v(initial) = 0). Find its speed v(final) when it strikes the anode.

 

In: Physics

A concentration cell based on the following half reaction at 309 K Ag+ + e- →...

A concentration cell based on the following half reaction at 309 K Ag+ + e- → Ag SRP = 0.80 V has initial concentrations of 1.37 M Ag+, 0.269 M Ag+, and a potential of 0.04334 V at these conditions. After 9.3 hours, the new potential of the cell is found to be 0.01406 V. What is the concentration of Ag+ at the cathode at this new potential?

In: Chemistry

Consider the galvanic cell described by (N and M are metals): N(s)|N2+(aq)||M+(aq)|M(s) If Eocathode = 0.182...

Consider the galvanic cell described by (N and M are metals):

N(s)|N2+(aq)||M+(aq)|M(s)

If Eocathode = 0.182 V and Eoanode = 1.411 V, and [N2+(aq)] = 0.931 M and [M+(aq)] = 0.655 M, what is Ecell, using the Nernst equation? ____ V

(the answer is supposed to be 1.233 but I don't understand how)

In: Chemistry

A concentration cell based on the following half reaction at 299 K Zn2+ + 2 e-...

A concentration cell based on the following half reaction at 299 K

Zn2+ + 2 e- → Zn       SRP = -0.760 V

has initial concentrations of 1.27 M Zn2+, 0.319 M Zn2+, and a potential of 0.01780 V at these conditions. After 5.6 hours, the new potential of the cell is found to be 0.002816 V. What is the concentration of Zn2+ at the cathode at this new potential?

In: Chemistry

A concentration cell based on the following half reaction at 318 K Cu2+ + 2 e-...

A concentration cell based on the following half reaction at 318 K

Cu2+ + 2 e- → Cu SRP = 0.340 V

has initial concentrations of 1.31 M Cu2+, 0.291 M Cu2+, and a potential of 0.02061 V at these conditions. After 8.1 hours, the new potential of the cell is found to be 0.006576 V. What is the concentration of Cu2+ at the cathode at this new potential?

In: Chemistry