Given that,
The camera sights the stadium at a 7 degree angle of depression.
The altitude of the blimp i slide 300 m.
To find,
The line of sight distance from the camera to the stadium.
Solution,
If we consider a right angled triangle. Let x is its hypotenuse i.e. the line of sight distance from the camera to the stadium. Using trigonometry :
So, the line of sight is at a distance of 2461.65 m from the camera to the stadium.
The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].
According to the statement
we have to find that the standard deviation, mean and the intervals from the given data.
So, According to the given data from the method of tree ring dating
The value of mean is
x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD
And now we find standard deviation :
s = √∑(xi - x-bar) / (N - 1)
∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2
∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2
∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4
∑(xi - x-bar)^2 = 10,659
Now,
s^2 = 10659/8 = 1332
s = 37 years
So, standard deviation is 37 years.
We need the t-distribution table since the standard deviation is unknown. Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860. Set up the confidence interval for the mean:
[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]
[x-bar ± t*(s/√n)] = [1250,1296]
So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].
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Question:
The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.
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