A blue jay lives in a nest that is 12 meters high in a tree the blue jay flies 15 meters to get from its nest to a flower on the
ground how far is the flower from the base of the tree
1 answer:
Answer:
The distance between the base of the tree and the flower is 9m
Step-by-step explanation:
Here, we have to paint a picture.
The flower is on the ground, the height of the tree is 12m.
The distance from the nest to the flower on the floor is 15m
Indisputably, what we have is a right angled triangle, with the height being 12m, the length of the hypotenuse being 15 and we are asked to calculate the adjacent which represents the distance from the base of the tree to the flower
To get this distance, we simply apply the Pythagoras’ theorem which states that the square of the hypotenuse(longest side of the triangle) is equal to the sum of the squares of the other two sides.
Thus mathematically, we know that our hypotenuse is 15m and the height is 12m
The length we are to calculate is the adjacent and it is equal to;
15^2 - 12^2
= 225 - 144
= 81
The length is thus
√(81) = 9m
Please check attachment for a diagrammatic picture of the triangle
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Answer:
We conclude that there is no difference between the two classes.
Step-by-step explanation:
We are given that two statistics teachers both believe that each has a smarter class.
A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17
Let = mean age of student cars.
= mean age of faculty cars.
So, Null Hypothesis, :
{means that there is no difference in the two classes}
Alternate Hypothesis, :
{means that there is a difference in the two classes}
The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;
T.S. = ~
where, = sample mean age of student cars = 8 years
= sample mean age of faculty cars = 5.3 years
= sample standard deviation of student cars = 3.6 years
= sample standard deviation of student cars = 3.7 years
= sample of student cars = 110
= sample of faculty cars = 75
Also, =
= 17.491
So, <u><em>the test statistics</em></u> = ~
= 0.422
The value of t-test statistics is 0.422.
<u>Now, the P-value of the test statistics is given by;</u>
P-value = P( > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.
Since the P-value of our test statistics is way more than the level of significance of 0.04, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as our test statistics will not fall in the rejection region.
Therefore, we conclude that there is no difference between the two classes.