The correct Option is D. OLAP is an example of a Predictive type of analytics.
Analytics is the systematic computational evaluation of statistics or information. it is used for the discovery, interpretation, and conversation of significant styles in statistics. It additionally involves making use of facts styles toward effective decision-making. it is able to be precious in areas rich with recorded statistics; analytics relies on the simultaneous utility of information, laptop programming, and operations research to quantify overall performance.
Records analytics is a multidisciplinary field. there is extensive use of pc abilities, mathematics, records, the use of descriptive techniques, and predictive fashions to gain treasured knowledge from records through analytics. humans Analytics won't certainly belong within Human resources in companies.
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Answer: The segment margin is obtained by deducting the common fixed costs that have been allocated to a segment from that segment's contribution margin
Explanation:
Segment margin is referred to the net profit or the net loss that a particular segment of a business makes. Segment margin is used to know segments that are performing well.
It is also used to know the long-run profitability of a particular segment as it shows the margin that is available after the cost has been covered by a segment.
Based on the above illustration, the statement that isn't true will be "the segment margin is obtained by deducting the common fixed costs that have been allocated to a segment from that segment's contribution margin".
This is false as segment margin is gotten after the traceable fixed costs of a segment has been subtracted from the contribution margin of that particular segment.
The new equilibrium price is higher and the new equilibrium quantity is higher
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.
