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.
Answer: 15.35%
Explanation:
The total nominal return over the two years if inflation is 2.4% in the first year and 4.4% in the second year will be calculated thus:
= (1+Interest rate)² -1
= (1 + 7.4%) - 1
= (1 + 0.074)² - 1
= 1.074² - 1
= 1.153476 - 1
= 0.153476
= 15.35% over the two years
Answer:
Whoever reads the chapter of the book the fastest is crowned the chapter king. It can be very competitive, and the competitors have to be very good readers.
Answer:
b. $50.00
Explanation:
Intrinsic per share stock price immediately after the repurchase will be $50
First, you have to calculate the amount of tuition when the student reaches age 18. Do this by multiplying $11,000 by 1.07 each year from age 12 until it reaches age 18. Thus, 7 times.
At age 18: 16,508
At age 19: 17,664
At age 20: 18,900
At age 21: 20,223
Then, we use this formula:
A = F { i/{[(1+i)^n] - 1}}
where A is the monthly deposit each year, F is the half amount of the tuition each year illustrated in the first part of this solution, n is the number of years lapsed.
At age 18:
A = (16508/2) { 0.04/{[(1+0.04)^6] - 1}} = $1,244.389 deposit for the 1st year
Ate age 19
A = (17664/2) { 0.04/{[(1+0.04)^7] = $1,118 deposit for the 2nd year
At age 20:
A = (18900/2) { 0.04/{[(1+0.04)^8] = $1,025 deposit for the 3rd year
At age 21:
A = (18900/2) { 0.04/{[(1+0.04)^8] = $955 deposit for the 4th year
