A car can travel 400 miles on a full tank of petrol. A more efficient car can travel 30% further. How many miles further can it
1 answer:
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Answer:
18.14% probability that you would get at least 12 questions correct.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that you would get at least 12 questions correct?
This is 1 subtracted by the pvalue of Z when X = 12. So
has a pvalue of 0.8186.
So there is a 1-0.8186 = 0.1814 = 18.14% probability that you would get at least 12 questions correct.
The complete question in the attached figure
step 1
find the tree height for an angle of elevation of 55°<span>
tan 55</span>°=opposite side angle 55°/adjacent side angle 55°
opposite side angle 55°=the tree height h1
adjacent side angle 55°=10 ft
so
tan 55°=h/10-------> h=10*tan 55°-----> h=14.28 ft
step 2
find the tree height for an angle of elevation of 75°
tan 75°=opposite side angle 75°/adjacent side angle 75°
opposite side angle 75°=the tree height h2
adjacent side angle 75°=10 ft
so
tan 75°=h2/10-------> h2=10*tan 75°-----> h2=37.32 ft
therefore
<span>the height of the tree is within the range [14.26,37.32]
</span>
Part 1) 4.2 ft
<span>the value is not in the interval [14.26,37.32]
therefore
it is </span>Not Reasonable
Part 2) 14.7 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 3) 24.4 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 4) 33.9 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 5) 39.1ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
Part 6) 58.7 ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
step 1
find the tree height for an angle of elevation of 55°<span>
tan 55</span>°=opposite side angle 55°/adjacent side angle 55°
opposite side angle 55°=the tree height h1
adjacent side angle 55°=10 ft
so
tan 55°=h/10-------> h=10*tan 55°-----> h=14.28 ft
step 2
find the tree height for an angle of elevation of 75°
tan 75°=opposite side angle 75°/adjacent side angle 75°
opposite side angle 75°=the tree height h2
adjacent side angle 75°=10 ft
so
tan 75°=h2/10-------> h2=10*tan 75°-----> h2=37.32 ft
therefore
<span>the height of the tree is within the range [14.26,37.32]
</span>
Part 1) 4.2 ft
<span>the value is not in the interval [14.26,37.32]
therefore
it is </span>Not Reasonable
Part 2) 14.7 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 3) 24.4 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 4) 33.9 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 5) 39.1ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
Part 6) 58.7 ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable