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Luke R.

Mathematics Undergraduate

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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.

Answer:

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

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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.

Answer:

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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