A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that
the angle of elevation to the top of the mountain is 31. From a point 1000 feet closer to the mountain along the plain, they find that the angle of elevation is 34. How high (in feet) is the mountain?
1 answer:
Answer:
5507.79 feet
Step-by-step explanation:
To find the height of the mountain, we can draw triangles as in the image attached.
Let's call the height of the mountain 'h', and the distance from the first point (31 degrees) to the mountain 'x'.
Then, we can use the tangent relation of the angles:
tan(34) = h/x
tan(31) = h/(x+1000)
tan(31) is equal to 0.6009, and tan(34) is equal to 0.6745, so:
h/x = 0.6745 -> x = h/0.6745
using this value of x in the second equation:
h/(x+1000) = 0.6009
h/(h/0.6745 + 1000) = 0.6009
h = 0.6009 * (h/0.6745 + 1000)
h = 0.8909*h + 600.9
0.1091h = 600.9
h = 600.9 / 0.1091 = 5507.79 feet
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Answer:
a) 85.31%
b) 56 minutes
c) 17 days
d) 0.1721
Step-by-step explanation:
In order to make the calculations easier, let's <em>standardize the curve</em> by doing the change
where is the average trip-time and
the standard deviation and
P(X≤ t) is the probability that the trip takes less than t minutes.
a)
Here we are looking for the value of P(X>20).
For X=20 we have Z = (20-24)/3.8 = -1.05
So, we want the area under the standard normal curve for Z > -1.05
that we can compute either using a table or a computer and we find this area equals 0.8531
So, he arrives late to work 85.31% of the times.
(See picture 1 attached)
b)
In this case we are looking for a value t of time such that
P(X≥ t) = 20% = 0.2
So, we are seeking a value Z such that the area under the normal curve to the left of Z equals 0.2
By using a table or a computer we find Z = 0.842
(See picture 2 attached)
By inserting this value in the equation on standardization
So 20% of the longest trips takes 55.996 ≅ 56 minutes
c)
Since the average of days he arrives late is 85.31%, it is expected that in 20 work days he arrives late 85.31% of 20, which equals 17 days.
d)
Since the probability that he does not arrive late is 1-0.8531 = 0.1469 and the probability of arriving early is independent of the previous trip,
the probability of arriving early n days in a row is
0.1496*0.1496*...*0.1496 n times and
the probability of being early most than 10 trips in 20 days is
