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.
The answer is 3 most definitely three lol 1+1=3 yup you bet lol
The stock is now trading at $52.16 per share.
The current value of an annuity of n regular payments of P at r% with yearly payments is provided by:
PV = P × (1 -(
÷r))
Estes Park Corp. distributes a fixed rate of a dividend of P = $7.80 per share on its shares. The corporation will retain this dividend for the following n = 13 years before ceasing dividend payments permanently. If the necessary returns on this stock are not metis r = 11.2% = 0.112.
The actual share price is calculated as follows:
Current share price = $7.80 × (1 -(
÷0.112))
$7.80 × ((1 - 0.251) ÷ 0.112)
$52.16
Therefore, the current share price is $52.16
Read more about the stock price at
brainly.com/question/15327515?referrer=searchResults
#SPJ4
Answer:
Shoes from the Thai and korean firm is part of imports
Imports=40+1240=1280
Domestic consumption=220
Security check is part of government spending=1500
GDP=1500+220-1280=440
Answer:
The 1-year HPR for the second stock is <u>12.84</u>%. The stock that will provide the better annualized holding period return is <u>Stock 1</u>.
Explanation:
<u>For First stock </u>
Total dividend from first stock = Dividend per share * Number quarters = $0.32 * 2 = $0.64
HPR of first stock = (Total dividend from first stock + (Selling price after six months - Initial selling price per share)) / Initial selling price = ($0.64 + ($31.72 - $27.85)) / $27.85 = 0.1619, or 16.19%
Annualized holding period return of first stock = HPR of first stock * Number 6 months in a year = 16.19% * 2 = 32.38%
<u>For Second stock </u>
Total dividend from second stock = Dividend per share * Number quarters = $0.67 * 4 = $2.68
Since you expect to sell the stock in one year, we have:
Annualized holding period return of second stock = The 1-year HPR for the second stock = (Total dividend from second stock + (Selling price after six months - Initial selling price per share)) / Initial selling price = ($2.68+ ($36.79 - $34.98)) / $34.98 = 0.1284, or 12.84%
Since the Annualized holding period return of first stock of 32.38% is higher than the Annualized holding period return of second stock of 12.84%. the first stock will provide the better annualized holding period return.
The 1-year HPR for the second stock is <u>12.84</u>%. The stock that will provide the better annualized holding period return is <u>Stock 1</u>.
