Mathematics Questions and Answers

The first five prime numbers are 2, 3, 5, 7 and 11

M.D = \(\frac {\sum (x - x̄)}{n}\)

n = 5

\(x̄ = \frac {\sum x}{n} = \frac {2 + 3 + 5 + 7 + 11}{5} = \frac {28}{5} = 5.6\)

M.D = \(\frac {(2 - 5.6) + (3 - 5.6) + (5 - 5.6) + (7 - 5.6) + (11 - 5.6)}{5}\)

= \(\frac {(-3.6) + (-2.6) + (-0.6) + (1.4) + (5.4)}{5} = \frac {13.6}{5}\)

Take the absolute value of all numbers in bracket

= \(\frac {3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac {13.6}{5}\)

\(\therefore M.D = 2.72\)

y = \(\sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)\)

= \(2x^{5/3} - x^{8/3}\)

Now, we can differentiate the function

\(\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}\)

\(3x + 2y + 4 = 0\)

Rearrange:

\(2y = -3x - 4\)

Divide both sides by 2

y = \(\frac {-3 \times - 4}{2}\)

y = \(\frac {-3}{2} \times - 2\)

∴ the gradient of the line 3x + 2y + 4 = 0 is \(\frac {-3}{2}\)

If two lines are perpendicular to each other ∴ \(m_1 x m_2\) = -1

Let \(m_1 = \frac {-3}{2} \therefore m_2 = \frac {-1}{m_1} = \frac {-1}{-3/2} = \frac {2}{3}\)

From the equation of a line which is given as m = \(\frac {y - y_1}{x - x_1} where (x_1, y_1) = (2,3)\)

\(\therefore \frac {2}{3} = \frac {y - 3}{x - 2}\)

=3(y - 3) = 2(x - 2)

=3y - 9 = 2 x -4

=3y = 2 x -4 + 9

∴ 3y = 2x + 5

Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

Total number of possible outcomes = 2 x 2 x 2 = 8

favourable number of outcomes = 7

∴Pr(at least one head) = \(\frac {7}{8}\)

tan θ = \(\frac {opp}{adj} = \frac {RQ}{QP} = \frac {6}{8}\)

tan θ = 0.75

θ = tan\(^{-1} (0.75) = 36.87^o\)

∴ The bearing of R from P = 360\(^o\) - 36.87\(^o\) = 323\(^o\) (to the nearest degree)