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maw [93]
2 years ago
14

Shue, a partner in the Financial Brokers Partnership, has a 30 percent share in partnership profits and losses. Shue's capital a

ccount had a net decrease of $100,000 during 20X8. During 20X8, Shue withdrew $240,000 as withdrawals and contributed equipment valued at $50,000 to the partnership. What was the net income of the Financial Brokers Partnership for 20X8?
A. $633,334
B. $466,666
C. $300,000
D. $190,000
Business
1 answer:
mr Goodwill [35]2 years ago
4 0

Answer:

C. $300,000

Explanation:

Shue Capital Account:

contribution                          50,000

partnership income x 30%

withdrawals                       (240,000)

change in capital account (100,000)

50,000 + Shue profits - 240,000 = -100,000

Shue profit = 240,000 - 100,000 - 50,000

Shue profit =   90,000

Partnership profit:

90,000 / 0.30 = 300,000

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Each of two stocks, C and D, are expected to pay a dividend of $3 in the upcoming year. The expected growth rate of dividends is
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Answer:

Intrinsic value of Stock C is 300

Explanation:

given data

expected pay dividend = $3

growth rate of dividends = 9%

stock C require a rate of return = 10%

stock D require a rate of return = 13%

solution

we get here intrinsic value by the DDM method

intrinsic value = Upcoming Dividend ÷ ( Required rate of return - Growth rate of stock )  .................1

intrinsic value = \frac{3}{(0.10-0.09)}    

intrinsic value = \frac{3}{0.01}  

intrinsic value = 300

so intrinsic value of Stock C is 300

8 0
2 years ago
Scenario: An organization has recently suffered a series of security breaches that have significantly damaged its reputation. Se
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Answer:

Vulnerability analysis

Explanation:

Since the organization has hired a security consultant to help them reduce their risk from future attacks, What the consultant would use to identify potential attackers is vulnerability analysis.

Vulnerability Analysis is a vulnerability assessment which entails an in-depth analysis of the building functions, systems, and site characteristics to identify: 1. Weaknesses in the system and

2. Determine mitigation or corrective actions that can be designed and implemented to eradicate vulnerability or reduce the vulnerabilities.

3 0
2 years ago
Joy is a supervisor over Elias. She repeatedly solicits sexual behavior from Elias and does other inappropriate actions that a r
SVEN [57.7K]

Answer: C) guilty of hostile work environment sexual harassment.

Explanation: A hostile work environment is a work environment where one or more persons act in such a way to offend or provoke anger in other persons. The activities of Joy towards Elias is hostile as it is against the will of Elias.

Sexual harassment is a term used to describe any action taken to solicit for sex,sexual behaviour or touch the private parts of others in embarrassing manner. THE ACTIVITIES OF JOY MAKING HER TO SOLICIT SEXUAL BEHAVIOUR FROM ELIAS IS A SEXUAL HARASSMENT AND WHEN PROVEN IN THE COURTS SHE SHALL BE FOUND GUILTY.

3 0
1 year ago
Larry Ellison starts a company that manufactures high-end custom leather bags. He hires two employees. Each employee only begins
HACTEHA [7]

Answer:

12.55 days

Explanation:

<em><u>Provided information </u></em>

Number of employees 2

Average production time=1.8 days

Standard deviation=2.7 days

Inter-arrival time= 1 day

Coefficient of variation= 1 day

Standard deviation of inter-arrival time= 1 day

The coefficient of variations

<u>Inter-arrival coefficient of variation </u>

C_{vi}=\frac {\sigma}{T} where \sigma is standard deviation of inter-arrival time, T is inter-arrival time and C_v is coefficient of variation of inter-arrival time

C_{vi}=\frac {1 day}{1 day}=1

<u>Production time coefficient of variation </u>

C_{vp}=\frac {2.7}{1.8}=1.5

<u><em>Total utilization time </em></u>

U=\frac {T}{n*T_i} where T is the time of production, n is number of employees, U is utilization, T_i is inter-arrival time

U=\frac {1.8}{2*1}=0.9

Therefore, utilization time by 2 employees is 0.9

<u>Expected average waiting time </u>

T_e=(\frac {T}{n*T_i})*0.5(C_{vi}^{2}+C_{vp}^{2})*(\frac{U^{\sqrt{2(n+1)}-1}}{1-U})

Where T_e is expected average waiting time and the other symbols as already defined

Substituting 1.5 for C_{vp}, 1 for C_{vi}, 0.9 for U, 2 for n, 1 for T_iand 1.8 for T

T_e=(\frac {1.8}{2*1})*0.5(1^{2}+1.5^{2})*(\frac{0.9^{\sqrt{2(2+1)}-1}}{1-0.9})

T_e=0.9*1.625*8.583709=12.55367 days  and rounding off to 2 decimal places we obtain 12.55 days

Therefore, expected duration between order received and beginning of production is approximately 12.55 days

4 0
1 year ago
The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eatin
Sunny_sXe [5.5K]

Answer:

probability  = 0.3557

Explanation:

given data

young adults ages =  20 to 39

skip eating breakfast p = 0.238

random sample of size n = 500

to find out

we find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122

solution

we use here Normal Approximation to Binomial Distribution

so first consider random variable = x

so

x~ Bin (n,p)   .............1

and here Normal Approximation will be

x~ Normal Approx (np, npq)    .................2

so it will be

x~ (500, 0.238)  

as here we know q will be

q = 1 - p

q = 1 - 0.238

q = 0.762    .............3

so

here x~ Normal Approx (119, 90.678)

and now we get P(X > 122)

so

We will convert it to Z by as that

z = \frac{x-\mu}{\sigma}     ................4

and here

mean  \mu = np

and standard deviation \sigma =  \sqrt{npq}

so here for P(X > 122)

P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})     ............5

and it is  P(Z>0.37)

so

probability  = 1 - P(Z<0.37)

now we use here z table for value

probability  = 1-0.6443

probability  = 0.3557

7 0
1 year ago
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