Correct Answer: A. \(\frac{2}{11}\)

Explanation

Let the number of people that speak both English and French = x

Then (7 - x) + x + (6 - x) = 11

13 - x = 11 \(\implies\) x = 2.

\(\therefore\) P(picking a person that speaks both languages) = 2/11

Explanation

(a) \(n(U) = 40\)

\(n(M) = 18\) \(n(E) = 16\)

\(n(A) = 19\) \(n(M \cap E) = 5\)

\(n(A \cap E) = 2\) \(n(M) only = 6\)

\(n(A) only = 9\)

Accounts:

x + 5 + 2 + 9 = 19

x + 16 = 19

x = 19 - 16 = 3.

Mathematics :

y + 6 + x + 5 = 18

y + 6 + 3 + 5 = 18

y + 14 = 18

y = 18 - 14 = 4.

Economics :

z + y + x + 2 = 16

z + 4 + 3 + 2 = 16

z + 9 = 16

z = 16 - 9 = 7.

Let the number of students that failed be r.

40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7

40 - r = 36

r = 40 - 36 = 4.

(b) Percentage that failed at least one of Economics and Mathematics

= 100% - (% of people that passed at least one of Economics and Mathemics)

Number that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M)

= 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics

% passed = \(\frac{17}{40} \times 100% = 42.5%\)

Percentage that failed at least one of Economics and Mathematics = 100% - 42.5% = 57.5%

(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.

P(passed Accounts) = \(\frac{n(A)}{n(U)}\)

= \(\frac{19}{40}\)

P(failed accounts) = \(1 - \frac{19}{40}\)

= \(\frac{21}{40} = 0.525\)

Correct Answer: D. 9\(\sqrt{3m}\)

Explanation

tan 30\(^o\) = \(\frac{h}{27}\)

h = 27 tan 30\(^o\)

= 27 x \(\frac{1}{\sqrt{3}}\)

= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{27 \sqrt{3}}{3}\)

= 9\(\sqrt{3m}\)

Correct Answer: D. 1.20

Explanation

\(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\)

\(\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}\)

Equating bases, we have

\(\frac{5}{6} - \frac{1}{n} = 0\)

\(\frac{5n - 6}{6n} = 0\)

\(5n - 6 = 0 \implies 5n = 6\)

\(n = \frac{6}{5} = 1.20\)

Correct Answer: C. \(\sqrt{\frac{3}{3}}\)

Correct Answer: A. 0.58

Explanation

When x = 2

y = 2(2)\(^3\) - 2(2)\(^2\)

-5(2) + 5

y = 16 - 8 - 10 + 5

y = 3

when x = 2.05

y = 2(2.105)\(^3\) - 2(2.05)\(^2\)

- 5(2.05) + 5

y = 17.23 - 8.405 - 10.25 + 5

y = 3.575 - 3

Difference = 3.575 - 3

= 0.58

Correct Answer: A. {a, b, d, e}

Correct Answer: (b) 12

Correct Answer: D. Rhombus

Explanation

(a) \(-3 \ast 2 = \frac{(-3)^{2} - (2)^{2}}{2(-3)(2)}\)

= \(\frac{9 - 4}{-12}\)

= \(-\frac{5}{12}\).

(b) The operation \(\ast\) is associative if given a, b and c which are real numbers, then

\((a \ast b) \ast c = a \ast (b \ast c)\)

Let a = -3 ; b = 2 ; c = 1.

\((-3 \ast 2) \ast 1 = (-\frac{5}{12}) \ast 1\)

= \(\frac{(-\frac{5}{12})^{2} - 1^{2}}{2(-\frac{5}{12})(1)}\)

= \(\frac{\frac{25}{144} - 1}{-\frac{5}{6}}\)

= \(\frac{-\frac{119}{144}}{\frac{-5}{6}}\)

= \(\frac{119}{120}\)

\(-3 \ast (2 \ast 1) \)

\(2 \ast 1 = \frac{2^{2} - 1^{2}}{2(2)(1)}\)

= \(\frac{3}{4}\)

\(-3 \ast \frac{3}{4} = \frac{(-3)^{2} - (\frac{3}{4})^{2}}{2(-3)(\frac{3}{4})}\)

= \(\frac{9 - \frac{9}{16}}{-\frac{9}{2}}\)

= \(-\frac{15}{8}\)

\((-3 \ast 2) \ast 1 \neq -3 \ast (2 \ast 1)\)

The operation \(\ast\) is not associative.