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Usimov [2.4K]
2 years ago
10

Translate "The product of the cube root of 5 and X^ prime prime into an algebraic expression

Mathematics
1 answer:
svp [43]2 years ago
5 0

Answer:

Le produit de la racine cubique de 5 et de X ^ prime prime en une expression algébrique: french, youre welcome.

Step-by-step explanation:

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Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b
kramer

Answer:

(a)

\bar x = 0.0075

s = 0.0049

(b)

B. The sample is too small to make judgments about skewness or symmetry.

Step-by-step explanation:

Given:

n = 8

\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 \ \end{array}

Solving (a):

First, calculate the difference between the recorded TBBMC for both operators:

\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 &{|1 - 2|} &0.003 & 0.015 & 0.006 & 0.004 & 0.002 & 0.010 & 0.005 & 0.015 \ \end{array}

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.

The mean is then calculated as:

\bar x = \frac{\sum x}{n}

\bar x = \frac{0.003 + 0.015 + 0.006 + 0.004 + 0.002 + 0.010 + 0.005 + 0.015}{8}

\bar x = \frac{0.06}{8}

\bar x = 0.0075

Next, calculate the standard deviation (s).

This is calculated using:

s = \sqrt{\frac{\sum (x - \bar x)^2}{n}}

So, we have

s = \sqrt\frac{(0.003 - 0.0075)^2 + (0.015 - 0.0075)^2+ (0.006- 0.0075)^2+ ....... + (0.005- 0.0075)^2+ (0.015- 0.0075)^2 }{8}s = \sqrt\frac{0.00019}{8}

s = \sqrt{0.00002375

s = 0.0049

Solving (b):

Of the given options (A - E), option B is correct because the sample is actually too small

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