Alan

a)  The range is simply the difference between the largest and the smallest.

b)  I'm sure you can add them up to find Σx and add up the squares to find Σx² by yourself!

c)  The sample variance is \(s^2=\frac{\sum{(x-\mu)^2}}{n-1}\quad\text{ where }\mu\text{ is the mean and n is the number of terms}\\s \text{ is simply the square root of }s^2\)

d) The population variance is similar to the sample variance except that the divisor is just n instead of n-1.

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e^z = 1 + z + z²/2! + z³ /3! + z⁴/4! + ...

e^-z = 1 - z + z²/2 - z³/3! + z⁴/4! - ...

So (e^z + e^-z)/2 = 1 + z²/2! + z⁴/4! + ...

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4*5^2 + 1*5^1 + 0*5^0 = 105, so 410₅ = 105₁₀ 

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Your last step has the closing bracket in the wrong place Melody! It should be:

\(\frac{u}{s^2}(\ln u-1)+c\)

(The constant shouldn't be multiplied by u, a function of t).

This also affects Max's extension below (now corrected by Max).

As follows:

\(2x^2+5x+12=19-7x\rightarrow 2x^2+12x-7=0\\ \text{ use the quadratic formula to find x } x= \frac{-12\pm \sqrt{12^2+4\times 2\times 7}}{2\times2}\\ x = \frac{-12\pm\sqrt{200}}{4}\\ \text{ positive difference: }\sqrt{200}/2\rightarrow 5\sqrt2\)

Let u = e^-st  then du/dt = -se^-st or e^-stdt = -(1/s)du so we can write the integral as:

\(\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})\)

I assume you want to find x

\(\log_4(x+2)+\log_4(x-2)= 2\\ \log_4((x+2)(x-2))=2 \\ (x+2)(x-2)=4^2\\x^2-4=16 \\ x^2= 20 \\ x=\pm2\sqrt5\)

However, the negative value is not valid in the original equation, so \(x = 2\sqrt5\)

The value of cos must lie between 1 and -1

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Prob first child is a girl = 8/12

Prob second child is a girl = 7/11

So prob of two girls = (8/12)*(7/11) → 14/33 or 0.424242...

The n'th term of an arithmetic progression is given by a + (n-1)d, where a is the first term and d is the common difference. So here we have:

a + (3-1)*d = 29   and

a + (29-1)*d = 289

Subtract the first from the second:  26*d = 260   so d = 10

Put this in the first equation to get a + 2*10 = 29   so a = 9

Hence the fifth term is 9 + (5-1)*10   or 49

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