Answer:
5
Step-by-step explanation:
1) We calculate the volume of a metal bar (without the hole).
volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²
Volume=(0.935 m²)(2 m)=1.871 m³
2) we calculate the volume of the parallelepiped
Volume of a parallelepiped= area of the section x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³
3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar - volume of a parallelepiped
Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³
4) we calculate the mass of the metal bar
density=mass/ volume ⇒ mass=density *volume
Data:
density=8.10³ kg/m³
volume=1.551 m³
mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg
answer: The mas of the metal bar is 12.408x10³ kg or 12408 kg
The required value after simplification of the s = -16/3. None of these are correct.
Given that,
To simplify
and to find the value of s in
.
<h3>What is simplification?</h3>
The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.
Simplification,
![=[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2\\= \frac{x^{4/3}x^{-1}}{x^2x^3*{x}^{2/3}}\\= \frac{x^{1/3}}{x^{17/3}}\\=x^{-16/3}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7Bx%5E%7B2%2F3%7Dx%5E%7B-1%2F2%7D%7D%7Bx%5Csqrt%7Bx%5E3%7D%5Csqrt%5B3%5D%7Bx%7D%7D%5D%5E2%5C%5C%3D%20%5Cfrac%7Bx%5E%7B4%2F3%7Dx%5E%7B-1%7D%7D%7Bx%5E2x%5E3%2A%7Bx%7D%5E%7B2%2F3%7D%7D%5C%5C%3D%20%5Cfrac%7Bx%5E%7B1%2F3%7D%7D%7Bx%5E%7B17%2F3%7D%7D%5C%5C%3Dx%5E%7B-16%2F3%7D)
Comparing with 
s = -16/3
Thus, the required value of the s = -16/3. None of these are correct.
Learn more about simplification here: brainly.com/question/12501526
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Statistic significance requires that your sample be representative of the population, but I'm struggling with this because if the class is an elective then the answer would be: <span>No, it is not a valid inference because she asked all 22 students in her science class instead of taking a sample of the students in her school.
BUT if the class is required, as most science classes are, then it WOULD be a random sample of the school. So the last option would be correct.
My guess though is that the teacher is looking for answer B.
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