<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.
</span>
Answer:
-12= -12
Step-by-step explanation:
You're correct dw!
Answer:
Slope (M): 3
Y-Int: 2
Standard Form: -3x+y=2
can go through points: (5,1), (-1, -1),...
I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs
are:
cos(x)tan(x) – ½ = 0
→ π/6 and 5π/6
cos(π/6)tan(π/6) – ½ = 0
cos(5π/6)tan(5π/6) – ½ = 0
sec(x)cot(x) + 2 =
0 → 7π/6 and 11π/6
sec(7π/6)cot(7π/6) + 2 = 0
sec(11π/6)cot(11π/6) + 2 = 0
sin(x)cot(x) +
1/sqrt2 = 0 → 3π/4 and 5π/4
sin(3π/4)cot(3π/4) + 1/sqrt2 = 0
sin(5π/4)cot(5π/4) + 1/sqrt2 = 0
csc(x)tan(x) – 2 = 0 → π/3 and 5π/3
csc(π/3)tan(π/3) – 2 = 0
csc(5π/3)tan(5π/3) – 2 = 0
Answer:
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