Answer:
52.56% probability that eight or more of the flights will arrive on time.
Step-by-step explanation:
For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
At a certain airport, 75% of the flights arrive on time.
This means that ![p = 0.75](https://tex.z-dn.net/?f=p%20%3D%200.75)
A sample of 10 flights is studied.
This means that ![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
Find the probability that eight or more of the flights will arrive on time.
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.75%29%5E%7B8%7D.%280.25%29%5E%7B2%7D%20%3D%200.2816)
![P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.75%29%5E%7B9%7D.%280.25%29%5E%7B1%7D%20%3D%200.1877)
![P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.75%29%5E%7B10%7D.%280.25%29%5E%7B0%7D%20%3D%200.0563)
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.2816%20%2B%200.1877%20%2B%200.0563%20%3D%200.5256)
52.56% probability that eight or more of the flights will arrive on time.
Answer:
The answer is X=9
Step-by-step explanation:
Remove the parentheses
Collect the like terms, subtract
Move the constant to the right
Subtract the leftover numbers
X=9
Answer:
percent of paint cost: 24.45%
percent of hardware cost: 8.14%
total percent of paint and hardware cost: 32.59%
Step-by-step explanation:
Add the cost of the lumber, the paint and the hardware:
![total\ cost=\$145.86+\$52.91+\$17.63\\total\ cost=\$216.4](https://tex.z-dn.net/?f=total%5C%20cost%3D%5C%24145.86%2B%5C%2452.91%2B%5C%2417.63%5C%5Ctotal%5C%20cost%3D%5C%24216.4)
If $216.4 is the total cost of the supplies bought by the theater group to construct the set of a play, then the percent of the cost of the paint is:
![\%paint\ cost=\frac{52.91*100}{216.4}=24.45\%](https://tex.z-dn.net/?f=%5C%25paint%5C%20cost%3D%5Cfrac%7B52.91%2A100%7D%7B216.4%7D%3D24.45%5C%25)
And the percent of the cost of the hardware is:
![\%hardware\ cost=\frac{17.63*100}{216.4}=8.14\%](https://tex.z-dn.net/?f=%5C%25hardware%5C%20cost%3D%5Cfrac%7B17.63%2A100%7D%7B216.4%7D%3D8.14%5C%25)
And the total percent of paint cost and hardware cost is:
![=\%24.45+\%8.14=\%32.59](https://tex.z-dn.net/?f=%3D%5C%2524.45%2B%5C%258.14%3D%5C%2532.59)
Answer:
15%
Step-by-step explanation:
![\frac{3}{20} = 0.15\\](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B20%7D%20%3D%200.15%5C%5C)
![0.15 = 15%](https://tex.z-dn.net/?f=0.15%20%3D%2015%25)