Step 1: Write the equation.
The given equation is:
\[
\frac{1}{9!} + \frac{1}{10!} = \frac{x}{11!}
\]
Step 2: Simplify the terms.
We can factor \( \frac{1}{9!} \) out of the left-hand side of the equation:
\[
\frac{1}{9!} \left( 1 + \frac{1}{10} \right) = \frac{x}{11!}
\]
\[
\frac{1}{9!} \left( \frac{10 + 1}{10} \right) = \frac{x}{11!}
\]
\[
\frac{1}{9!} \times \frac{11}{10} = \frac{x}{11!}
\]
Step 3: Further simplify.
Now, multiply both sides by \( 11! \) to isolate \( x \):
\[
\frac{11!}{9! \times 10} = x
\]
\[
x = \frac{11 \times 10!}{10} = 120
\]
Thus, the value of \( x \) is 120. Therefore, the correct answer is 2. 120.
In C language, mat[i][j] is equivalent to: (where mat[i][j] is a two-dimensional array)
Suppose a minimum spanning tree is to be generated for a graph whose edge weights are given below. Identify the graph which represents a valid minimum spanning tree?
\[\begin{array}{|c|c|}\hline \text{Edges through Vertex points} & \text{Weight of the corresponding Edge} \\ \hline (1,2) & 11 \\ \hline (3,6) & 14 \\ \hline (4,6) & 21 \\ \hline (2,6) & 24 \\ \hline (1,4) & 31 \\ \hline (3,5) & 36 \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match LIST-I with LIST-II
Choose the correct answer from the options given below:
Consider the following set of processes, assumed to have arrived at time 0 in the order P1, P2, P3, P4, and P5, with the given length of the CPU burst (in milliseconds) and their priority:
\[\begin{array}{|c|c|c|}\hline \text{Process} & \text{Burst Time (ms)} & \text{Priority} \\ \hline \text{P1} & 10 & 3 \\ \hline \text{P2} & 1 & 1 \\ \hline \text{P3} & 4 & 4 \\ \hline \text{P4} & 1 & 2 \\ \hline \text{P5} & 5 & 5 \\ \hline \end{array}\]
Using priority scheduling (where priority 1 denotes the highest priority and priority 5 denotes the lowest priority), find the average waiting time.