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Mark W.

Math and Statistics Tutor

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Calculus

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Question:

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?

Mark W.

Answer:

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.

Statistics

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Question:

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.

Mark W.

Answer:

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: $$ p^6 * q^2 * \left(\begin{array} {c}8\\6 \end{array}\right)+ p^7 * q^1 * \left(\begin{array} {c}8\\7 \end{array}\right) + p^8 * q^0 * \left(\begin{array} {c}8\\8 \end{array}\right)$$ $$ = 0.6^6 * 0.4^2 * \frac{8!}{6! 2!} + 0.6^7 * 0.4^1 * \frac{8!}{7! 1!} + 0.6^8 * 0.4^0 * \frac{8!}{8! 0!}$$ $$ = 0.6^6 * 0.4^2 * 28 + 0.6^7 * 0.4^1 * 8 + 0.6^8 * 0.4^0 * 1$$ $$ = 0.2090 + 0.0896 + 0.0168 = 0.3154$$

Algebra

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Question:

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

Mark W.

Answer:

$$ax^2 + bx +c = 0$$ 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: $$x^2 + \frac{b}{a}x + \frac{c}{a} = (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$$ $$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$$ $$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$

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