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soldier1979 [14.2K]
3 years ago
13

You have 1,000,000 number cubes, each measuring one inch on a side. if you layered the cubes to make one big cube what would be

the dimensions of the big cube?
Mathematics
1 answer:
azamat3 years ago
3 0

Answer:

100 x 100 x 100 number cubes

Step-by-step explanation:

It's basically a volume problem.  The question is asking if you made a big cube filled with smaller cubes , specifically 1,000,000 of them, what are the dimensions of this big cube.  Or in other words the volume is 1,000,000.  Now how do we find the volume of a cube?  length times width times height, and witht he cubes each of those are the same so we can call them all x.

Now we just set up the equation where we have the equation whatever one side is times itself  three times, (or x cubed) it will equal 1,000,000 or x^3=1,000,000,000.  Now you just take the cube root of 1,000,000 which gets us 100.  so the length, width and height are 100 small cubes.

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Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

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Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

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