Step 1: Understanding the problem.
The number of shifts corresponding to different numbers of patients per shift is given in the bar graph. The amount Dr. Gita earns is \(1000(X - 0.2)\), where \(X\) is the number of patients per shift. The data from the graph is as follows:
For \(X = 5\), the number of shifts is 20.
For \(X = 6\), the number of shifts is 40.
For \(X = 7\), the number of shifts is 30.
For \(X = 8\), the number of shifts is 10.
Step 2: Calculating the total earnings. For \(X = 5\): \[ \text{Earnings} = 1000 \times (5 - 0.2) \times 20 = 1000 \times 4.8 \times 20 = 96,000. \] For \(X = 6\): \[ \text{Earnings} = 1000 \times (6 - 0.2) \times 40 = 1000 \times 5.8 \times 40 = 232,000. \] For \(X = 7\): \[ \text{Earnings} = 1000 \times (7 - 0.2) \times 30 = 1000 \times 6.8 \times 30 = 204,000. \] For \(X = 8\): \[ \text{Earnings} = 1000 \times (8 - 0.2) \times 10 = 1000 \times 7.8 \times 10 = 78,000. \] Step 3: Calculating the total earnings and average earnings.
Total earnings for all 100 shifts:
\[ \text{Total Earnings} = 96,000 + 232,000 + 204,000 + 78,000 = 610,000. \] The average earnings per shift: \[ \text{Average Earnings} = \frac{610,000}{100} = 6,100. \] Thus, the average earnings per shift are ₹6,100, which corresponds to Option (A).
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?
A quadratic polynomial \( (x - \alpha)(x - \beta) \) over complex numbers is said to be square invariant if \[ (x - \alpha)(x - \beta) = (x - \alpha^2)(x - \beta^2). \] Suppose from the set of all square invariant quadratic polynomials we choose one at random. The probability that the roots of the chosen polynomial are equal is ___________. (rounded off to one decimal place)
Consider a directed graph \( G = (V,E) \), where \( V = \{0,1,2,\dots,100\} \) and
\[ E = \{(i,j) : 0 < j - i \leq 2, \text{ for all } i,j \in V \}. \] Suppose the adjacency list of each vertex is in decreasing order of vertex number, and depth-first search (DFS) is performed at vertex 0. The number of vertices that will be discovered after vertex 50 is:
Create empty stack S Set x = 0, flag = 0, sum = 0 Push x onto S while (S is not empty){ if (flag equals 0){ Set x = x + 1 Push x onto S } if (x equals 8): Set flag = 1 if (flag equals 1){ x = Pop(S) if (x is odd): Pop(S) Set sum = sum + x } } Output sumThe value of \( sum \) output by a program executing the above pseudocode is:
def f(a, b): if (a == 0): return b if (a % 2 == 1): return 2 * f((a - 1) / 2, b) return b + f(a - 1, b) print(f(15, 10))The value printed by the code snippet is 160 (Answer in integer).
Consider the following tables, Loan and Borrower, of a bank.
Query: \[ \pi_{\text{branch\_name}, \text{customer\_name}} (\text{Loan} \bowtie \text{Borrower}) \div \pi_{\text{branch\_name}}(\text{Loan}) \] where \( \bowtie \) denotes natural join. The number of tuples returned by the above relational algebra query is 1 (Answer in integer).