The probability distribution of a random variable \( X \) is given as follows. Then, \( P(X = 50) - \frac{P(X \leq 30)}{P(X \geq 20)} \) = 
We are given the probability distribution for a random variable \( X \) with values at \( X = 10, 20, 30, 40, 50 \) and corresponding probabilities.
The total sum of probabilities must be 1, so we start by solving for \( k \) using the equation: \[ k + 2k + 3k + 4k + 5k = 1 ⇒ 15k = 1 ⇒ k = \frac{1}{15}. \] Thus, the individual probabilities are: \[ P(X = 10) = \frac{1}{15}, P(X = 20) = \frac{2}{15}, P(X = 30) = \frac{3}{15}, P(X = 40) = \frac{4}{15}, P(X = 50) = \frac{5}{15}. \] Now, to find \( P(X = 50) - \frac{P(X \leq 30)}{P(X \geq 20)} \), we first calculate the cumulative probabilities: \[ P(X \leq 30) = P(X = 10) + P(X = 20) + P(X = 30) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15} = \frac{6}{15} = \frac{2}{5}, \] \[ P(X \geq 20) = P(X = 20) + P(X = 30) + P(X = 40) + P(X = 50) = \frac{2}{15} + \frac{3}{15} + \frac{4}{15} + \frac{5}{15} = \frac{14}{15}. \] Substitute these values into the expression: \[ P(X = 50) - \frac{P(X \leq 30)}{P(X \geq 20)} = \frac{5}{15} - \frac{\frac{2}{5}}{\frac{14}{15}} = \frac{5}{15} - \frac{2}{5} \times \frac{15}{14} = \frac{5}{15} - \frac{6}{14} = \frac{1}{12}. \]
The rank of matrix \(\begin{bmatrix} k & -1 & 0 \\[0.3em] 0 & k & -1 \\[0.3em] -1 & 0 & k \end{bmatrix}\) is 2, for \( k = \)
If \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\), then \(A^{-1} =\)
The supply voltage magnitude \( |V| \) of the circuit shown below is ____ .
A two-port network is defined by the relation
\(\text{I}_1 = 5V_1 + 3V_2 \)
\(\text{I}_2 = 2V_1 - 7V_2 \)
The value of \( Z_{12} \) is: