What is the equation of a line that is parallel to the line y=3/4x+2 and passes through the point (-2, 8)?
2 answers:
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Answer:
The answer to the question is;
The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.
Step-by-step explanation:
Let the sample size =n = 100
The success probability = 5 % = 0.05
Number of tickets sold = 105 tickets
In the case where there the airline has found that 5 % will not show up, then every passenger should have a seat, we have
A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails
However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus
n·p = 105×0.05 = 5.25 ≥ 5
and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5
As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution
= - 0.3358
We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307
Another way to solve the question is as follows
p = 0.95 q = 0.05
μ = np = 0.95*105 = 99.75, σ = = 2.233
P (x≤100) = P = P(z<0.34) = 0.63307.
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus![\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ab%3D%5Ctextit%7Brate%20of%20the%20boat%7D%5C%5C%0Ar%3D%5Ctextit%7Brate%20of%20the%20river%7D%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26distance%26rate%26time%28hrs%29%5C%5C%0A%26----%26----%26----%5C%5C%0Aupstream%2648%26b-r%264%5C%5C%0Adownstream%2648%26b%2B4%263%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0A48%3D%28b-r%29%284%29%5Cto%2048%3D4b-4r%5C%5C%5C%5C%0A%5Cfrac%7B48-4b%7D%7B-4%7D%3Dr%5C%5C%0A--------------%5C%5C%0A48%3D%28b%2Br%29%283%29%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Athus%5C%5C%5C%5C%0A48%3D%5Cleft%5B%20b%2B%5Cleft%28%5Cboxed%7B%5Cfrac%7B48-4b%7D%7B-4%7D%7D%5Cright%29%20%5Cright%5D%20%283%29%0A%5Cend%7Bcases%7D)
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"