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 ]
I think the answer to you question is 35 i might be wrong tho but it may work.
Answer:
Let x = number of regular tickets sold.
Let y represent the number of student tickets sold.
12x+8y≤100012x+8y≤1000
x+y≥200x+y≥200
12x+8y≤20012x+8y≤200
x+y≤200x+y≤200
12x+8y≥1000
hope this helps
Step-by-step explanation:
Answer:
11
Step-by-step explanation:
The distance that a is to b is b-a=17-2=15.
The line segment from a=2 to b=17 has length 15.
We need to know what is 3/5 of 15.
3/5 of 15 means what is 3/5 times 15?
(3/5)(15)=3(3)=9
So this means we are looking to make a line segment that is 9 units from 2 which is 2 to 11.
So 11 is 3/5 the way from a=2 to b=17.
Let's check from 2 to 11 that is a length of 9 and from 2 to 17 that is a length of 15.
Is 9/15 equal to 3/5?
Yes, 9/15 can be reduced to 3/5.
The distance between (5,5) and (1,5) is 4 blocks
