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Luden [163]
3 years ago
8

Which unit of measure would be appropriate for the volume of a cube with sides of 2 meters?

Mathematics
2 answers:
algol133 years ago
7 0

Answer:

m³

Step-by-step explanation:

Side of the cube is 2 meters

Volume of the cube is:

  • (2 m)³ = 8 m³

Unit of measure is cubic meter or  m³

Ilya [14]3 years ago
3 0

Answer: 8m³

Step-by-step explanation: to workout the volume, you have to use the formula for a cube which is a³ where a is the one side. 2³ = 8.

The unit of measurement would be m³.

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Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
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Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

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Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

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Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

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Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

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