.02(5000)+.1(1500)+0.3(0)+$50=$300
On Jan 12, 4:47 am, snehal jain <learner....@gmail.com> wrote:
> An insurance company issues a policy on a small boat under the following
> conditions:
> The replacement cost of $5000 will be paid for a total loss. If it is not a
> total loss,
> but the damage is more than $2000, then $1500 will be paid. Nothing will be
> paid for damage costing $2000 or less and of course nothing is paid out if
> there
> is no damage. The company estimates the probability of the first three
> events
> as .02, .10, and .30 respectively. The amount the company should charge for
> the
> policy if it wishes to make a profit of $50 per policy on average is:
> a) $250
> b) $201
> c) $300
> d) $1200
>
>
>
>
>
>
>
> On Wed, Jan 12, 2011 at 3:15 PM, snehal jain <learner....@gmail.com> wrote:
> > A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 ns
> > (nano seconds)
> > respectively. Registers that are used between the stages have a delay
> > of 5 ns each. Assuming
> > constant clocking rate, the total time taken to process 1000 data
> > items on this pipeline will
> > approximately be
> > a. 120 us (micro seconds)
> > b. 165 us
> > c. 180 us
> > d. 175 us
>
> > Which of the following statements are true?
> > I Shortest remaining time first scheduling may cause starvation
> > II Preemptive scheduling may cause starvation
> > III Round robin is better than first come first serve in terms of
> > response time
> > a) I only
> > b) I and III only
> > c) II and III only
> > d) I, II and III
>
> > Increasing the RAM of a computer typically improves performance
> > because:
> > a. Virtual memory increases
> > b. Larger RAMs are faster
> > c. Fewer segmentation faults occur
> > d. Fewer page faults occur
>
> > Suppose we want to synchronize two concurrent processes P and Q using
> > binary
> > semaphores S and T. The code for the processes P and Q is shown below.
> > Process P:
> > while (1) {
> > W:
> > print '0’;
> > print '0';
> > X:
> > }
> > Process Q:
> > while (1) {
> > Y:
> > print '1'
> > print '1'
> > Z:
> > }
> > Synchronization statements can be inserted only at points W, X, Y and
> > Z.
> > V() increments the semaphore, whereas P() decrements it
> > Which of the following will always lead to an output staring with
> > '001100110011' ?
> > a) P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
> > b) P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T
> > initially 0
> > c) P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1
> > d) P(S) at W, V(S) at X, p(T) at Y, V(T) at Z, S initially 1, and T
> > initially 0
>
> > can any1 suggest some source for such kind of questions????????
>
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