Answer:
Determining when the cumulative total of net cash flows reaches zero.
Explanation:
Payback calculates the amount of time it takes to recover the amount invested in a project from it cumulative cash flows
Assume 20,000 was invested in a project, Cash flows in year 1 = 10,000 cash flow in year 2 = 20,000
Payback = 1.5 years
Amount invested = -20,000
Amount recovered in year 1 = -20,000 + 10,000= -10,000
Amount recovered in year 2 = -10,000 + 15,000 = 5000
Payback = 1 + 10,000 / 15,000 = 1.5
Architect, they have to have four more years of education.
Even though I'd think a therapist makes more.
Answer:
#1 Former name was the American Industrial Arts Association (AIASA)
#2 Founded in November 19, 2001 and it was formed in response to the September 11 attacks to prevent from future events of that type.
#3 "Learning to lead in a technical world."
#4 " TSA's mission is to protect the nation's transportation systems to ensure freedom of movement for people and commerce"
Answer:
a. Cash basis - Service revenue is $900
b. Accrual basis - Service revenue is $2,100 (which is $1200 + $900)
Explanation:
In accounting, there are 2 basis for recognizing transactions; these are cash basis and accrual basis.
In cash basis, sales and expenses are not recorded unless cash has been collected and paid respectively. In the accrual basis of accounting, expenses and sales are recorded when incurred and earned respectively.
Revenue earned under the accrual basis would therefore include the revenue for which cash has been collected and those for which cash is yet to be collected.
Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.
<u>Explanation:</u>
The problem is that of Multiple-server Queuing Model.
Number of servers, M = 2.
Arrival rate, 
= 6 boats per hour.
Service rate, 
= 10 boats per hour.
Probability of zero boats in the system,![\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\}](https://tex.z-dn.net/?f=%5Cmathrm%7BPO%7D%3D1%20%2F%5C%7B%5B%281%20%2F%200%20%21%29%20%5Ctimes%286%20%2F%2010%29%200%2B%281%20%2F%201%20%21%29%20%5Ctimes%286%20%2F%2010%29%201%5D%2B%5B%286%20%2F%2010%29%202%20%2F%282%20%21%20%5Ctimes%281-%286%20%2F%282%20%5Ctimes%2010%29%29%29%5D%5C%7D)
= 0.5385
<u>Average number of boats waiting in line for service:</u>
Lq =![[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0](https://tex.z-dn.net/?f=%5B%5Clambda.%5Cmu.%28%20%5Clambda%20%2F%20%5Cmu%20%29M%20%2F%20%7B%28M%20%E2%80%93%201%29%21%20%28M.%20%5Cmu%20%E2%80%93%20%5Clambda%20%292%7D%5D%20x%20P0)
= ![[\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385](https://tex.z-dn.net/?f=%5B%5C%7B6%20%5Ctimes%2010%20%5Ctimes%286%20%2F%2010%29%202%5C%7D%20%2F%5C%7B%282-1%29%20%21%20%5Ctimes%28%282%20%5Ctimes%2010%29-6%29%202%5C%7D%5D%20%5Ctimes%200.5385)
= 0.0593 boats.
The average time a boat will spend waiting for service, Wq = 0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.
The average time a boat will spend at the dock, W = 0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.
