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,
= 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)
=
= 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.
Itulah perbezaan antara tarif dan kuota.
Maaf lah bila tak saya tulis kat sini, sebab tak boleh hantar jawaban.
<em>Semoga </em><em>membantu </em><em>dan </em><em>bermanfaat </em><em>:</em><em>)</em>
Answer:
5,100 Consumers
Explanation:
The 17% of the total consumer recognize Flatfeet brand which means:
Consumers who recognize Flatfeet = Total Consumers * percentage of people that recognize the brand
Here
Total consumers are 30,000
And
Percentage of people that recognize the brand is 17%
By putting values, we have:
Consumers who recognize Flatfeet Brand = 30,000 * 17%
Consumers who recognize Flatfeet Brand = 5,100 Consumers
Answer:
Debit Cash $1,000 and credit Notes Receivable $1,000.
Explanation:
The adjusting entry is shown below:
Cash Dr $1,000
To Notes receivable $1,000
(Being the note receivable collected by the bank is recorded)
While recording the transaction, we debited the cash account as it increases the cash balance and credited the note receivable.
Hence, the second option is correct