<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
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Answer:
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Step-by-step explanation:
Answer:
Step-by-step explanation:
Final step: Check your answer.
Answer:
Step-by-step explanation:
Begin the solution by squaring both sides of the given equation. We get:
(3x - 4)^2 = 2x^2 - 2x + 2, or:
9x^2 - 24x + 16 = 2x ^2 - 2x + 2
Combining like terms results in:
7x^2 - 22x + 14 = 0
and the coefficients are a = 7, b = -22, c = 14, so that the discriminant of the quadratic formula, b^2 - 4ac becomes (-22)^2 - 4(7)(14) = 92
According to the quadratic formula, the solutions are
-b ± √discriminant -(-22) ± √92 22 ± √92
x = ------------------------------- = ----------------------- = ------------------------
2a 14 14
