This equation would be written as 2(n-3)=8.
In order to satisfy the requirement, this are the following that is required;First, at least one observation must be above or below 90 seconds. Second is ether the population is normally distributted or greater than (>) 30, or maybe both.
The answer in this question is A and B.
Answer:
Let's call the side of square b x. Therefore, the side of square a is 0.5x, making its area 0.25x² and the area of square b is x². This means that the percentage is 25%.
sin2x =12/13
cos2x = 5/13
tan2x = 12/5
STEP - BY - STEP EXPLANATION
What to find?
• sin2x
,
• cos2x
,
• tan2x
Given:
tanx = 2/3 = opposite / adjacent
We need to first make a sketch of the given problem.
Let h be the hypotenuse.
We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.
Using the Pythagoras theorem;
hypotenuse² = opposite² + adjacent²
h² = 2² + 3²
h² = 4 + 9
h² =13
Take the square root of both-side of the equation.
h =√13
This implies that hypotenuse = √13
We can now proceed to find the values of ainx and cosx.
Using the trigonometric ratio;
![\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}](https://tex.z-dn.net/?f=%5Csin%20x%3D%5Cfrac%7Bopposite%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D)
![\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}](https://tex.z-dn.net/?f=%5Ccos%20x%3D%5Cfrac%7Badjacent%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D)
And we know that tanx =2/3
From the trigonometric identity;
sin 2x = 2sinxcosx
Substitute the value of sinx , cosx and then simplify.
![\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})](https://tex.z-dn.net/?f=%5Csin%202x%3D2%28%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%28%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29)

Hence, sin2x = 12/13
cos2x = cos²x - sin²x
Substitute the value of cosx, sinx and simplify.
![\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%202x%3D%28%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%5E2-%28%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%5E2%20%5C%5C%20%20%5C%5C%20%3D%5Cfrac%7B9%7D%7B13%7D-%5Cfrac%7B4%7D%7B13%7D%20%5C%5C%20%3D%5Cfrac%7B5%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Hence, cos2x = 5/13
tan2x = 2tanx / 1- tan²x






OR

Hence, tan2x = 12/5
Therefore,
sin2x =12/13
cos2x = 5/13
tan2x = 12/5
Pi r² is circle's area expression.
So the r is radius,.