\(\frac{1}{10}\)
\(\frac{3}{10}\)
\(\frac{9}{20}\)
\(\frac{11}{20}\)
Correct answer is A
Prob.(John pass)\(\frac{3}{4}\) prob.(John fail) \(=1-\frac{3}{4}=\frac{1}{4}\)
Prob.(James pass) \(\frac{3}{5}\) Prob.(James fail) \(=1-\frac{3}{5}=\frac{2}{5}\)
Prob(both boys fail)\(\frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\)
2
3
4
5
Correct answer is B
\(x^2+(x-1)^2 = (\sqrt{13})^2 \Rightarrow x^2 + x^2 - 2x + 1 = 13\\
2x^2 - 2x - 12 = 0\) dividing through by 2
\(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 \)
Make f the subject of the relation \(v = u + ft\)
\(\frac{v-u}{t}\)
\(\frac{u-v}{t}\)
\(t(v+u\)
\(\frac{v}{u}-t\)
Correct answer is A
\(v=u+ft \Rightarrow v-u=ft\Rightarrow f=\frac{v-u}{t}\)
\(2x^2 + 13xy - 15y^2\)
\(2x^2 - 13xy - 15y^2\)
\(2x^2 + 13xy + 15y^2\)
\(2x^2 - 13xy + 15y^2\)
Correct answer is D
\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\\
=2x^2 - 13xy + 15y^2\)
Solve the equation \(2^7 = 8^{5-x}\)
\(\frac{5}{8}\)
\(\frac{8}{3}\)
\(\frac{3}{2}\)
\(\frac{15}{4}\)
Correct answer is B
\(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x\\
\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\)