Answer:
A. 0.24
Explanation:
From the question ,
The probability that mutual funds A will rise is 40 % , i.e. , P ( A ) = 0.40
The second statement given is , the probability of rise in B with A , is 60% , i.e. , P ( B | A ) = 0.6
Therefore , to calculate the probability that both funds will increase is given by P( B n A ) .
Since ,
P ( B | A ) = P (B n A) / P(A)
Now, putting the respective values -
0.6 = P(B n A) / 0.4
rearranging ,
P (B n A) = 0.6 * 0.4
P(B n A) = 0.24
probability that both the fund A and fund B will rise in price = 0.24 .
Answer:
Purchasing Agent:
-Attend Conferences and trade shows to find new products
-Negotiate contracts with suppliers
Office Clerk:
-File paperwork according to a filing system in the office
-Type data into software to be used by the company
Freight Forwarder:
-Calculate the weight and volume of cargo
-Track shipments to make sure that they arrive on time
Explanation: I took the test.
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:
B. $129 million
Explanation:
bad debt expense for the year = balance in allowance at the end + write off - balance in allowance at the beggining
= $319 million + $137 million - $327 million
= $129 million
Therefore, Oracle Corporation report as bad debt expense for the year is $129 million.
A table is a systematic arrangement of data implementing columns and rows to display information which make it easier for better understanding.
Table number: A table should always be numbered for easy identification
Title of the table
Stubs: These refer to the headings of horizontal rows.
Captions: these refer to the headings of vertical columns
Clear, to the point and a suitable font/ size for the text. Information must be inserted in the same format into the appropriate cells
