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n200080 [17]
2 years ago
6

The line segments of a cube include edges, the diagonals of the faces, and the diagonals through the interior of the cube. Which

is the longest? Which is the shortest? Explain.
Mathematics
2 answers:
Ksenya-84 [330]2 years ago
0 0

They diagonal through the interior of the cube is the longest. The hypotenuse is the longest side of a right triangle. The diagonal of the cube is the hypotenuse of a right angle with legs that are diagonal of a face and an edge. The edges are the shortest. They are legs of a right triangle with the diagonal of a face as the hypotenuse.


nata0808 [166]2 years ago
0 0

Answer:

The diagonal through the interior of the cube is the longest. The hypotenuse is the longest side of a right triangle. The diagonal of the cube is the hypotenuse of a right triangle with legs that are the diagonal of a face and an edge. The edges are the shortest. They are legs of a right triangle with the diagonal of a face as the hypotenuse.

Step-by-step explanation:its the sample response

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Assume the blood pressures of 10 people were measured before and after sleeping for the night. What is the corresponding critica
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For the critical value we need to calculate the degrees of freedom given by:

df = n-1= 10-1=9

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

|t_{crit}| =1.833

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value with right arm , y = test value with left arm  

The system of hypothesis for this case are:  

Null hypothesis: \mu_y- \mu_x = 0  

Alternative hypothesis: \mu_y -\mu_x \neq 0  

The first step is calculate the difference d_i=y_i-x_i

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}  

The third step would be calculate the standard deviation for the differences, and we got:  

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}  

The 4 step is calculate the statistic given by :  

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}  

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df = n-1= 10-1=9

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

|t_{crit}| =1.833

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