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RUDIKE [14]
1 year ago
9

lark Bell started a personal financial planning business when he accepted $36,000 cash as advance payment for managing the finan

cial assets of a large estate. Bell agreed to manage the estate for a one-year period beginning June 1, Year 1. Required Show the effects of the advance payment and revenue recognition on the Year 1 financial statements using the following horizontal statements model. In the Cash Flows column, use OA to designate operating activity, IA for investing activity, FA for financing activity, and NC for net change in cash. If the account is not affected, leave the cell blank. How much revenue would Bell recognize on the Year 2 income statement
Business
1 answer:
just olya [345]1 year ago
5 0

The Effects of the Advance Payment (Receipt) on Lark Bell's Year 1 Financial Statements are:

                    Balance Sheet                                                                                                                      

              Assets =  Liabilities                                  + Equity  

Cash +$36,000 = Unearned revenue +$15,000 + Service Revenue +$21,000

                                 Income Statement                              Cash Flow

                     Revenue - Expense = Income                         Statement

Service Revenue +$21,000                                 Cash inflow +$36,000 OA

In Year 1, the Assets (Cash) will increase by $36,000.  There is a corresponding increase in Liabilities (Unearned Revenue) of $15,000 and an increase in Equity (Service Revenue) of $21,000.

Thus, the amount of revenue that Bell would recognize on the Year 2 income statement from this transaction in Year 1 is $15,000.  This covers 5 months from January to May.

Learn more about the effects of advance payment and revenue at brainly.com/question/24300418

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The compensation that can be provided to the employer over and above employee's income could be an Insurance payment or accumulated Tax refund.

<h3>What is Insurance?</h3>

Insurance is a financial plan that promises to mitigate any loss suffers.

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2 years ago
A process that produces computer chips has a mean of .04 defective chip and a standard deviation of .003 chip. The allowable var
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Answer:

a) 0.667

b) Yes

Explanation:

Data provided in the question:

Mean = 0.04

Standard Deviation = 0.003

Upper Specification Limit, USL = 0.046

Lower Specification Limit, LSL = 0.034

Now,

a) Capability Index is given as:

Cp = \frac{(USL-LSL)}{(6\sigma)}

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Cp = \frac{(0.046-0.034)}{(6\times0.003)}

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Cp = 0.667

Also,

Cpk = min(\frac{(USL-Mean)}{(3\sigma)},\frac{(Mean-LSL)}{(3\sigma)}

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Cpk = min(\frac{(0.046-0.04)}{(3\times0.003)},\frac{(0.04-0.034)}{(3\times0.003)}

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6 0
3 years ago
Suppose that a company needs new equipment, and that the machinery in question earns the company revenue at a continuous rate of
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

Here

  • a and b are the limits of integral which are 0 and 1 respectively
  • r is the rate of interest which is 5% or 0.05
  • S(t) is the function of value which is S(t)=\left \{ {{66000t+38000} {\ \ 0So the equation becomes[tex]PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t+38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t)e^{-0.05t}} \, dt+\int\limits^{0.5}_0 {(38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=8113.7805+18764.4669+34207.6751\\PV=61085.9225

    So the present value of revenue in the first year is $61,085.92.

    b-

    The time in which the machine pays for itself is given as

    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

    The present value is set equal to the value of machine which is given as

    $160,000 so the equation becomes:

    PV=61085.9225+\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt\\160000=61085.9225+\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt\\\int\limits^{t}_{0}{(71000)e^{-0.05t}} \, dt=160000-61085.9225\\\int\limits^{t}_{1}{(71000)e^{-0.05t}} \, dt=98914.07\\\\t=-\dfrac{\ln \left(0.93034\right)}{0.05}\\t=1.44496

    So the total time it would take to pay for its price is 2.44 years of 29.33 months.

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