STAN walk 10 km to the grocery store. He stops then walks 10 km east back to his house. What distance does he cover? what was hi
1 answer:
STAN walk 10 km to the grocery store.
He stops then walks 10 km east back to his house.
Let us draw this situation to better understand the problem.
As you can see, Stan first walked to the grocery store and covered a distance of 10 km then he came back to the house from the grocery store and covered a distance of 10 km.
So the total distance is 10 + 10 = 20 km
On the other hand, notice that Stan is exactly at the point where he first started that is his house, therefore, the displacement is 0 in this case.
Recall that the displacement is a vector quantity so the direction of travel also matters.
From house to grocery store = 10 km
From the grocery store to house = -10 km
Overall displacement = 10 km - 10 km = 0 km
Therefore,
Distance = 20 km
Displacement = 0 km
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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.