<h3>Answer:</h3>
- A — 1
- B — ∑[k=0..2] 15Ck(.24^k)(.76^(15-k))
- D — 12 and 3.1
- A — 0.1091
- B — 0.83 ...
<h3>Explanation:</h3>
1. The probability is 0.52 that the officer will pull over a driver and then the expected number more. So, if x is the expected number of drivers pulled over until one is not texting, we have ...
... x = 0.52(1+x)
... 0.48x = 0.52
... x = 0.52/0.48 = 13/12 = 1 1/12 ≈ 1 . . . . matches selection A
(<em>Comment on this result</em>: I find it interesting that these are the odds in favor of finding a driver who texts. That is, if the probability of texting is 0.98, the odds are 49:1 that a driver will be texting, and the expected number of pull-overs is 49.)
2. The probability of at most 2 being cured is the probability of 0, 1, or 2 being cured. You need to add up those probabilities. The sum in answer selection B does that.
3. The mean of a binomial distribution is ...
... μx = np = 60·0.2 = 12
... σx = √(np(1-p)) = √(12·0.8) ≈ 3.0984
These match selection D.
4. 20C14(0.8^14)(0.2^6) = 38760·.043980·0.000064 ≈ 0.109100 . . . matches A
5. mean(x) = 0.94; mean(x^2) = 1.58, so ...
... σx = √(1.58 -0.94²) ≈ 0.83 . . . . matches selection B
Answer:
he used distribution method.
for step 1.as he distribute
2(x-1)=2x-2
Answer:
D. desire to surprise his friend and his friend’s failure to recognize him
Explanation:
Answer D
Correct. In these sentences, the author presents a humorous reversal that emerges from the ironic incongruity between the traveler’s plan to “overpower” his old friend with an excess of pleasure and the anticlimactic outcome of the surprise visit. As it turns out, the friend experiences no immediate pleasure from the visit because he fails to recognize the traveler and can only be made to remember him after the traveler gives a “gradual (in this context, methodical) explanation” of who he is.