# Tutor profile: Mark W.

## Questions

### Subject: Calculus

A ball is thrown. The height of the ball in meters at any given time can be found by the function $$ h(t) = -5t^2 + 4t + 10$$ where t is in seconds. a) At what moment has the ball reached its highest point? b) What is the acceleration of the ball at this point?

a) At the highest point, the rate of change of the equation of height, the derivative, will be 0. The derivative is $$\frac{dh}{dt} = -10t +4$$ This is 0 when $$t = 0.4$$ b) The acceleration of the ball is the second derivative of $$h(t)$$ This is: $$\frac{d^2h}{dt^2} = -10$$ Hence the acceleration of the ball is a constant, of -10. This is what we would expect, as objects falling under the force of gravity have constant acceleration towards the ground.

### Subject: Statistics

In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that a) exactly 5 have access to the internet. b) at least 6 students have access to the internet.

First, we need to assume that the school is large enough, that removing any particular student does not significantly change the 60% probability that the remaining students have internet at home. So each of the 8 students chosen will have the same 60% chance. a) Let p = the probability any randomly chosen student has access to the internet = 0.6 q = the probability any randomly chosen student does not have access to the internet = 0.4 The probability that exactly 5 have access to the internet is the probability that 5 do, and 3 do not, which is $$ p^5 * q^3 * \left(\begin{array} {c}8\\5 \end{array}\right)$$ since there are $$\left(\begin{array} {c}8\\5 \end{array}\right)$$ ways of ordering the five who do have internet. So, this is $$0.6^5 * 0.4^3 * \frac{8!}{5! 3!} = 0.279$$ b) The probability that at least 6 students have access to the internet is P (exactly 6 have access) + P (exactly 7 have access) + P (exactly 8 have access) As before, we can work these out: $$ p^6 * q^2 * \left(\begin{array} {c}8\\6 \end{array}\right)+ p^7 * q^1 * \left(\begin{array} {c}8\\7 \end{array}\right) + p^8 * q^0 * \left(\begin{array} {c}8\\8 \end{array}\right)$$ $$ = 0.6^6 * 0.4^2 * \frac{8!}{6! 2!} + 0.6^7 * 0.4^1 * \frac{8!}{7! 1!} + 0.6^8 * 0.4^0 * \frac{8!}{8! 0!}$$ $$ = 0.6^6 * 0.4^2 * 28 + 0.6^7 * 0.4^1 * 8 + 0.6^8 * 0.4^0 * 1$$ $$ = 0.2090 + 0.0896 + 0.0168 = 0.3154$$

### Subject: Algebra

For a general quadratic equation $$ax^2 + bx + c = 0$$ show that the general solution is: $$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$

$$ax^2 + bx +c = 0$$ divide by a: $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$ Completing the square for the first two terms, we notice that, when expanded: $$(x + \frac{b}{2a})^2 = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}$$ So, adjusting for the final constant: $$x^2 + \frac{b}{a}x + \frac{c}{a} = (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$$ $$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$$ $$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$

## Contact tutor

needs and Mark will reply soon.