A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the same d
irection and make in barely in time. He can also run backwards towards the train and also barely make it. How fast is the train going relative to the man.
1 answer:
Answer:
The train is twice fast going relative to the man.
Step-by-step explanation:
Consider the provided information.
Let the distance between the train and beginning of bridge is x and length of bridge is y.
A man is on a 1/4 on a bridge. Thus, the 1/4 of y is y/4.
The train is going x distance in time man runs y/4 distance.
Also if the train is going x + y in time man runs the distance 3y/4.
For better understanding refer the figure 1:
So, if train goes x+y-x distance in time man covers the distance 3y/4 - y/4
Now solve 3y/4 - y/4 = 2y/4
The train covers y distance in the time man runs 2y/4 = y/2
That means train covers 2 times of the distance cover by the man or the train goes twice as fast as man.
Hence, train is twice fast going relative to the man.
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The question is incomplete. Here is the complete question:
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit 10 such targets in a row.
Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?
Answer:
40.13%
Step-by-step explanation:
Let 'A' be the event of not missing a target in 10 attempts.
Therefore, the complement of event 'A' is
Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.
Now,
We know that the sum of probability of an event and its complement is 1.
So,
Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.
Answer:
a. 12
b. 7.200 and 2.683
Step-by-step explanation:
The computation is shown below:
Given that
P = 0.40 and n = 30
a)
The expected value of received e-mails is
= n × p
= 30 × 0.4
= 12
b)
The variance of emails received is
= n × p × (1 - p)
= 30 × 0.4 × 0.6
= 7.200
Now
The standard deviation of emails received is
= sqrt(variance)
= 2.683
We simply applied the above formula
