19.
2568.75 - 2500 = 68.75;
( ? / 100 ) * 2500 = 68.75 ;
? * 25 = 68.75;
? = 68.75 ÷ 25
? = 2.75;
r = 2.75 %
Answer:
15+6
Step-by-step explanation:
You would need to multiply 3*5 then 3*2 and then put 15+6
Answer:
x=14
PQ=46
Step-by-step explanation:
PT and TQ are equal, so we can set up the equation like this: PT=TQ
23=2x-5
calculate to find x
x=14
PQ is just PT+TQ so 23+23=46
That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:
-- 0:
A number greater than 10 with a 0 in the units place is a multiple of
either 5 or 10, so it's not a prime number.
-- 1:
A number greater than 10 with a 1 in the units place could be
a prime (11, 31 etc.) but it doesn't have to be (21, 51).
-- 2:
A number greater than 10 with a 2 in the units place has 2 as a factor
(it's an even number), so it's not a prime number.
-- 3:
A number greater than 10 with a 3 in the units place could be
a prime (13, 23 etc.) but it doesn't have to be (33, 63) .
-- 4:
A number greater than 10 with a 4 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 5:
A number greater than 10 with a 5 in the units place is a multiple
of either 5 or 10, so it's not a prime number.
-- 6:
A number greater than 10 with a 6 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 7:
A number greater than 10 with a 7 in the units place could be
a prime (17, 37 etc.) but it doesn't have to be (27, 57) .
-- 8:
A number greater than 10 with a 8 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 9:
A number greater than 10 with a 9 in the units place could be
a prime (19, 29 etc.) but it doesn't have to be (39, 69) .
So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.
It CAN't have a 0, 2, 4, 5, 6, or 8 .
The only choice that includes all of the possibilities is 'A' .
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:
</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>
![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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</span>

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