Step 1: The probability that the student studied for at most 2 hours is given by: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). \] Step 2: Substitute the given values: \[ P(X \leq 2) = 0.1 + k(1) + k(2). \] Step 3: Substitute \( k = 0.15 \): \[ P(X \leq 2) = 0.1 + 0.15(1) + 0.15(2). \] Step 4: Compute the value: \[ P(X \leq 2) = 0.1 + 0.15 + 0.3 = 0.55. \] Thus, the probability that the student studied for at most 2 hours is 0.55.
Find the solution to the following linear programming problem (if it exists) graphically:
Maximize \( Z = x + y \)
Subject to the constraints \[ x - y \leq -1, \quad -x + y \leq 0, \quad x, y \geq 0. \]
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]