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The first step to solving this expression is to factor out the perfect cube
![\sqrt[3]{m^{2} n^{3} X n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bm%5E%7B2%7D%20%20n%5E%7B3%7D%20X%20n%5E%7B2%7D%20%20%20%7D%20)
The root of a product is equal to the product of the roots of each factor. This will make the expression look like the following:
![\sqrt[3]{ n^{3} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20n%5E%7B3%7D%20%7D%20)
![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%20%7D%20)
Finally,, reduce the index of the radical and exponent with 3
n
This means that the correct answer to your question is n![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%7D%20)
.
Let me know if you have any further questions
:)
The root of a product is equal to the product of the roots of each factor. This will make the expression look like the following:
Finally,, reduce the index of the radical and exponent with 3
n
This means that the correct answer to your question is n
Let me know if you have any further questions
:)
Answer:
the plane's current height above the ocean is 3.42 miles
Step-by-step explanation:
Given that the diagonal distance between the plane and the aircraft carrier is 10 miles.
What we need to find is the vertical distance between the plane and the aircraft carrier which is the plane's current height above the ocean. This can be gotten using sine rule.
For sine rule, you need a side and its opposite angle. For this question the triangle formed is a right angle triangle with diagonal distance, vertical distance and horizontal distance.
let a = diagonal distance = 10 miles
let the opposite angle to the diagonal distance be A = 90° (right angle)
let the vertical distance = b
let the opposite angle to the vertical distance be B = 20°
Sine rule states that
∴
⇒ b = 3.42 miles
Therefore the plane's current height above the ocean is 3.42 miles