\(\frac{1}{54}\)
\(\frac{41}{54}\)
\(\frac{20}{27}\)
\(\frac{13}{54}\)
Correct answer is D
\(P(A)=\frac{1}{6},P(T)=\frac{1}{9}\)
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) = \(\frac{13}{54}\)
A function \(f\) is defined by \(f :x→\frac{x + 2}{x - 3},x ≠ 3\).Find the inverse of \(f\) .
\(\frac{x + 3}{x - 2},x ≠ 2\)
\(\frac{x - 3}{x + 2},x ≠ -2\)
\(\frac{3x - 2}{x+1},x ≠ -1\)
\(\frac{3x + 2}{x - 1},x ≠ 1\)
Correct answer is D
\(f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?\)
Let \(f :x=y\)
\(y=\frac{x + 2}{x - 3}\)
\(=x+2=y(x-3)\)
\(=x-xy=-3y-2\)
\(=x(1-y)=-3y-2\)
\(=x=\frac{-3y - 2}{1 - y}=\frac{-(3y + 2)}{- (y - 1)}\)
\(=x=\frac{3y + 2}{y - 1}\)
\(∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1\)
\(\frac{1}{6}\)
\(\frac{4}{7}\)
\(\frac{4}{21}\)
\(\frac{3}{7}\)
Correct answer is B
\(P(X⋃Y)=\frac{5}{8}\)
\(P(X⋂Y)=P(X)\times P(Y)\)
Since X and Y are independent events, the probability of their union (X ⋃ Y) can be calculated as:
\(P(X⋃Y)=P(X)+P(Y)-P(X⋂Y)\)
\(=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)\)
\(=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)\)
\(=\frac{1}{2}=P(Y)(1-\frac{1}{8})\)
\(=\frac{1}{2}=P(Y)(\frac{7}{8})\)
\(=P(Y)=\frac{1}{2}÷\frac{7}{8}\)
\(∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}\)
Given that \(y^2 + xy = 5,find \frac{dy}{dx}\)
\(\frac{y}{2y + x}\)
\(\frac{-y}{2y + x}\)
\(\frac{-y}{2y - x}\)
\(\frac{y}{2y + x}\)
Correct answer is B
\(y^2 + xy = 5\)
By implicit differentiation
\(=2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0\)
\(=2y\frac{dy}{dx}+x\frac{dy}{dx}=-y\)
Factor out \(\frac{dy}{dx}\)
\(=\frac{dy}{dx}(2y+x)=-y\)
\(∴\frac{dy}{dx}=\frac{-y}{2y + x}\)
A linear transformation on the oxy plane is defined by \(P : (x, y) → (2x + y, -2y)\). Find \(P^2\)
\(\begin{bmatrix} 4&0\\1&4\end{bmatrix}\)
\(\begin{bmatrix} 4&4\\0&0\end{bmatrix}\)
\(\begin{bmatrix} 4&0\\0&4\end{bmatrix}\)
\(\begin{bmatrix} 4&1\\0&4\end{bmatrix}\)
Correct answer is C
\(P : (x, y) → (2x + y, -2y)\)
\(p\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 2x & y\\0 &-2y\end{bmatrix}\)
\(\therefore p = \begin{bmatrix} 2 & 1\\0 &-2\end{bmatrix}\)
\(\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) \(\begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) = \(\begin{bmatrix} 4&0\\0&4\end{bmatrix}\)