Simplify the following expression:
1 answer:
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Part A:
Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85
Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:
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</span>Part B:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15
Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:
Part C:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The mean is given by:
Part D:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:
</span>![P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]](https://tex.z-dn.net/?f=P%28X%5C%20%5Ctextgreater%20%5C%20%5Cmu%29%3DP%28X%5C%20%5Ctextgreater%20%5C%201.9125%29%20%5C%5C%20%20%5C%5C%201-%5BP%280%29%2BP%281%29%5D)
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Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85
Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:
</span>
<span>
</span>Part B:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15
Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:
Part C:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The mean is given by:
Part D:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:
</span>
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We know that speed (rate) is equal to distance over time.
The current unit of distance is miles, and we want to convert it to feet. The unit of time is hours, and we want seconds. In order to do this, we need to create a conversion factor.
Since
and 
, we will use these are our conversion factors.
The goal in a dimensional analysis problem like this is for your units to work out. Remember, we want feet per second.
We write
.
You may wonder why in the conversion factor, we chose to put feet over miles and hours over seconds. This is again because we want our units to cancel. If you have one unit the numerator and the same one in the denominator, the units cancel out.
In sum, the answer is 22 ft/s.
The current unit of distance is miles, and we want to convert it to feet. The unit of time is hours, and we want seconds. In order to do this, we need to create a conversion factor.
Since
The goal in a dimensional analysis problem like this is for your units to work out. Remember, we want feet per second.
We write
You may wonder why in the conversion factor, we chose to put feet over miles and hours over seconds. This is again because we want our units to cancel. If you have one unit the numerator and the same one in the denominator, the units cancel out.
In sum, the answer is 22 ft/s.