Further Mathematics Questions and Answers

\(P(A)=\frac{1}{6},P(T)=\frac{1}{9}\)

Probability that only one of them will hit the target = \(P(A)\times P( \bar T ) + P( \bar A )\times P(T)\)

Where \(P( \bar T )\) is the probability that Tunde will not hit the target and \(P( \bar A )\) is the probability that Atta will not hit the target

\(P( \bar T )=1-\frac{1}{9}=\frac{8}{9}\)

\(P( \bar A )=1-\frac{1}{6}=\frac{5}{6}\)

Pr(only one) =\((\frac{1}{6}\times\frac{8}{9}) + (\frac{5}{6} \times \frac{1}{9}) =\frac{4}{27} + \frac{5}{54}\)

\(\therefore\) pr (only one) = \(\frac{13}{54}\)

\(f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?\)

Let \(f :x=y\)

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

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

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

\(=x(1-y)=-3y-2\)

\(=x=\frac{-3y - 2}{1 - y}=\frac{-(3y + 2)}{- (y - 1)}\)

\(=x=\frac{3y + 2}{y - 1}\)

\(∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1\)

\(P(X⋃Y)=\frac{5}{8}\)

\(P(X⋂Y)=P(X)\times P(Y)\)

Since X and Y are independent events, the probability of their union (X ⋃ Y) can be calculated as:

\(P(X⋃Y)=P(X)+P(Y)-P(X⋂Y)\)

\(=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{1}{2}=P(Y)(1-\frac{1}{8})\)

\(=\frac{1}{2}=P(Y)(\frac{7}{8})\)

\(=P(Y)=\frac{1}{2}÷\frac{7}{8}\)

\(∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}\)

\(y^2 + xy = 5\)

By implicit differentiation

\(=2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0\)

\(=2y\frac{dy}{dx}+x\frac{dy}{dx}=-y\)

Factor out \(\frac{dy}{dx}\)

\(=\frac{dy}{dx}(2y+x)=-y\)

\(∴\frac{dy}{dx}=\frac{-y}{2y + x}\)

\(P : (x, y) → (2x + y, -2y)\)

\(p\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 2x & y\\0 &-2y\end{bmatrix}\)

\(\therefore p = \begin{bmatrix} 2 & 1\\0 &-2\end{bmatrix}\)

\(\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) \(\begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) = \(\begin{bmatrix} 4&0\\0&4\end{bmatrix}\)