Alan

To find the sample mean add them all up and divide by how many there are:

$${\mathtt{mean}} = {\frac{\left({\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{17}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}}\right)}{{\mathtt{11}}}} \Rightarrow {\mathtt{mean}} = {\mathtt{14.818\: \!181\: \!818\: \!181\: \!818\: \!2}}$$

To find the sample standard deviation use the following formula:

$$sd = \sqrt{\frac{\sum_{i=1}^n(x_i-mean)^2}{n-1}}$$

where n is the number of items (11 here), xi are the individual data values (15, 18 ... etc.), and $$\sum_{i=1}^n$$ means sum all of the squared bracketed terms that follow.  

I'll leave you to plug in the numbers.

You give no information about what comparison is to be done, so there is no way I can say anything about what the null hypothesis and its alternate might be! 

Anonymous's answer assumed the radius is the apothem, which is the radius of the maximum inscribed circle.

If the radius is that of the minimum circumscribed circle (imagine a circle completely surrounding the polygon and shrinking it until it just touches all the vertices, but doesn't cross to the inside of the polygon anywhere) then, assuming we have a regular polygon with n sides, chop the polygon into n equal triangles.  Find the area of each triangle from (1/2)base*height, where base = 2*radius*sin(180°/n) and height = radius*cos(180°/n).

So total area of polygon = n*radius²*sin(180°/n)*cos(180°/n) = (n/2)*radius²*sin(360°/n)

Frictional force is given by $$F=\mu W$$ where μ is the coefficient of friction and W is the weight (13552.2N here).

The centripetal force required to prevent slipping is $$F=m\frac{v^2}{r}$$ where m is the mass (=W/g), v is speed and r is radius.

Equating these two forces we have: $$\mu W = \frac{W}{g}\frac{v^2}{r}$$  so  $$\mu = \frac{v^2}{gr}$$  

v = 52.2*10³/3600 m/s, r = 1.68*102m and g = 9.81m/s². 

$${\frac{{\left({\frac{{\mathtt{52\,200}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{9.81}}{\mathtt{\,\times\,}}{\mathtt{1.68}}{\mathtt{\,\times\,}}{\mathtt{102}}\right)}} = {\mathtt{0.125\: \!071\: \!265\: \!339\: \!299\: \!2}}$$

So  $$\mu = 0.125$$  approx.

See the answers to the same question here: 

and here:    

To be "technically" correct, I suppose it should be included!!

Not if we are talking prime factors only. The number 1 is no longer considered to be a prime number (hasn't been since the end of the 19th century) - that's why the factorize keyword doesn't list it!

If we are talking about all factors of 2310, then 1 is included, as is 2310 itself, and all the other numbers (combinations of the prime factors) that divide 2310 exactly, such as 6, 10, 14, 22, 15, ...etc.

If you have found the mean value, μ, and standard deviation, s, of a sample data set, and the data is distributed like a normal (Gaussian) distribution, then the confidence interval can be expressed as the interval between μ-z*s/√n and  μ+z*s/√n, where n is the number of data points in your sample and z is a number that depends on the degree of confidence you require.  For example, if you want a 95% confidence interval then z is 1.96.

Divide both sides by 16;

$$x^2=\frac{1000}{16}$$

Take the square root of both sides:

$$\\
x=\pm\sqrt{\frac{1000}{16}}\\
x=\pm\frac{\sqrt{1000}}{\sqrt16}\\
x=\pm\frac{\sqrt{1000}}{4}\\
x=\pm\frac{10\sqrt{10}}{4}\\
x=\pm\frac{5\sqrt{10}}{2}$$

Use the keyword factorize here:

$${factorize}{\left({\mathtt{2\,310}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{2}}\\
{\mathtt{3}}\\
{\mathtt{5}}\\
{\mathtt{7}}\\
{\mathtt{11}}\\
\end{array} \right\}$$

This gives you the prime factors.

Mea culpa.  Hmm! Seems we can't delete posts!

The image below should help.  B is a reflection of A in the y-axis.