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Luke R.
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Linear Algebra
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Question:

Find the determinant of the 2x2 matrix A. where $A= \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} Luke R. Answer: The determinant of a 2x2 matrix, labelled$detA is found using the formula, detA=$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}=ad-bc$. So here detA=$\det\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ $=(1)x(3)-(2)x(2)=3-4$ =-1

Calculus
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Question:

Differentiate the following equation with respect to $x$. $y=x^2(x-2)$

Luke R.

To differentiate this equation we need to use the product rule. The formula for using product rule is given by, $\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$ So here lets let $f(x)=x^2$ and $g(x)=(x-2)$ and differentiate then separately. $\frac{d}{{dx}}f\left( x \right)=2x$ and $\frac{d}{{dx}}g\left( x \right)=1$. So putting what we know into the formula for product rule, $f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$ $=(x^2)(1)+(2x)(x-2)$ $=x^2+2x(x-2)$

Statistics
TutorMe
Question:

Calulate the mean and variance from the following list of 10 numbers. 45.2, 38.0, 62.7, 48.7, 29.9, 66.5, 52.0, 61.3, 57.4, 41.8

Luke R.

The mean (also known as the expected value), E[x], is the average value of a group of $n$ numbers. Put simply, we calculate the mean by adding up all the numbers and dividing by $n$. The sum of a group of numbers is symbolized by $\sum_x$. The formula for finding the mean is E[x]=$\frac{\sum{x}}{n}}$. So here, we have 10 numbers, so $n=10$. Adding up the ten numbers we get 503.5. Putting these values into our formula, E[x]=$\frac{503.5}{10}=50.35$. Next we need to find the variance, labelled Var[x], of the ten numbers. The variance is one measure we can use to analyse the spread of the data. The formula for variance is Var[x]=E[$x^2$]-$(E[x])^2$. So first we need to add up the squares of the individual 10 numbers, and divide it by $n=10$. E[$x^2$]=26609.97 $\frac{E[x^2]}{n}=2660.997$ Then we are left take away the mean (that we have already found) squared. $(E[x])^2$=2535.1225 Thus the variance is given by, Var[x]=E[$x^2$]-$(E[x])^2$=2660.997-2535.1225=125.8745.

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