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NEW QUESTION: 1
Which definition best describes Account Planning?
A. a prescribed set of activities and tasks to help plan an account for a customer opportunity within the plan phase
B. a prescribed set of activities and tasks to help identify and qualify an account for a customer opportunity within the plan phase
C. a prescribed set of activities and tasks to help plan an account for a customer opportunity within the prepare phase
D. a prescribed set of activities and tasks to help identify and qualify an account for a customer opportunity within the prepare phase
E. a minimum set of activities and tasks required to successfully qualify and plan for a customer opportunity within the plan phase
Answer: C
NEW QUESTION: 2
ESXi 6.xホストのライセンスを取得するときに正しい2つのステートメントはどれですか？ （2つ選択してください。）
A. バルクライセンスでは、vSphere Auto Deployを有効にする必要があります。
B. ESXiホストクライアントを使用してESXiライセンスを割り当てることができます。
C. 一括ライセンスにはvSphere PowerCLIが必要です。
D. ESXiライセンスは、vCenter Serverに接続されているESXiホストにのみ適用できます。
Answer: B,C
NEW QUESTION: 3
Sixty percent of the customers of a fast food chain order the Whopper, fries and a drink. If a random sample of 15 cash register receipts is selected, what is the probability that 10 or more will show that the above three food items were ordered?
A. 0.403
B. 0.186
C. None of these answers
Answer: A
Explanation:
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by:
r (n-r)
n!(p )[q ]/r!(n-r)! Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
10 5
p(10 successes) = 15!(0.6 )(0.4 )/10!(15-10)! = 0.1859
11 4
p(11 successes) = 15!(0.6 )(0.4 )/11!(15-11)! = 0.1268
p(12 successes) = 0.0634
p(13 successes) = 0.0219
p(14 successes) = 0.0047
p(15 successes) = 0.00047
The sum of all the probabilities is 0.403.