<span>Assuming a normal distribution the noise level is below 44dB 90% of the time.
Looking up the standard deviation that goes along with .01 on a normal distribution chart, we find that 2.33 standard deviations is the answer. Given the standard deviation of six we turn this in to deviation in decibels from the mean:
6*2.33 = 13.98
Add this to the mean to get
30+13.98=43.98
which we round to 44dB</span>
Answer:
Must be journalized and posted.
Explanation:
Closing entries are journal entries that is made at the end of an accounting period. It involves the transfer of balances of a temporary account to a permanent account.
Organisations employ the use of closing entries to reset the balances of temporary accounts to zero.
Closing entries are carried out to bring back the revenue, expense, and drawing temporary account balances to zero in preparation for a fresh accounting period.
Answer:
The answer to this question is a= µ=60/12=5 students/min
Explanation:
Solution
Given that:
λ=4 students / min
The Waiting time in Queue= λ /µ(µ- λ )==4/(5*(5-4))=0.8 min
The Number of students in the line L(q)= λ *W(q)= 4*.8= 3.2 students
TheNumber of students in the system L(q)= λ /(µ- λ )=4/(5-40=4 students
Then,
The Probability of system to be empty= P0= 1-P= 1-0.8= 0.2
Now,
If the management decides to add one more cashier with the same efficiency then we have
µ= 6 sec/student= 10 students/min.
so,
P= λ /µ =4/10=0.4
Now,
The probability that cafeteria is empty= P0= 1-0.4= 0.6
If we look at the above system traits, it is clear that the line is not empty and the students have to standby for 0.8 in the queue waiting to place their order and have it, also on an average there are 3.2 students in the queue and in the entry cafeteria there are 4 students who are waiting to be served.
If the management decides to hire one more cashier with the same work rate or ability, then the probability of the cafeteria being free moves higher from 0.2 to 0.6 so it suggests that the management must hire one additional cashier.
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Answer:
We have:
Amount of principal = $268,000
Interest payment = $1,522.24
Explanation:
These can be calculated as follows:
Loan principal = Cost of the home * Percentage to borrow = $335,000 * 80% = $268,000
Interest payment = (Loan principal / $1,000) * $5.68 = ($268,000 / $1,000) * $5.68 = 268 * $5.68 = $1,522.24
Therefore, we have:
Amount of principal = $268,000
Interest payment = $1,522.24
