Answer:
![[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]](https://tex.z-dn.net/?f=%5Bp-%7Cp%7C%2A10%5E%7B-3%7D%20%5C%2C%20%2C%20%5C%2C%20p%2B%7Cp%7C%2A%2010%5E-3%5D)
Step-by-step explanation
The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is
r = |p*-p|/|p|
we want r to be at most 10⁻³, thus
|p*-p|/|p| ≤ 10⁻³
|p*-p| ≤ |p|* 10⁻³
therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]
Answer:
c= -20h + 90
Step-by-step explanation:
618 hope this helps you have an amazing week
Answer:a is the answer I think..........................
<h2>
Answer:</h2>
The probability is:

<h2>
Step-by-step explanation:</h2>
It is given that:
An urn contains balls numbered 1 through 20.
A ball is chosen, returned to the urn, and a second ball is chosen.
This means that this is a case of a replacement.
Hence, one of the event is independent of the other.
Now we know that the probability to get a particular number of ball is:

( Since, there are total 20 balls and a ball of one particular number is just unique )
i.e. 
Hence, the probability that the first and second balls will be a 8 is:
