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There are 180 integers between 100 and 999 with AT LEAST 1 zero. So, the probability is:

180 / 900 = 1 / 5

If they are independent events, multiply the probabilities.

The reason that adding is not appropriate:

-- consider this case:

    -- assume that the probability of A not appearing is 0.90 

    -- assume that the probability of B not appearing is 0.80

-- if you added them together, you would get a probability of 1.70, an impossibilit.

 Solve for x:

x^2+1/x^2==171

Bring x^2+1/x^2 together using the common denominator x^2:

x^4+1/x^2==171

Multiply both sides by x^2:

x^4+1==171 x^2

Subtract 171 x^2 from both sides:

x^4-171 x^2+1==0

Substitute y==x^2:

y^2-171 y+1==0

Subtract 1 from both sides:

y^2-171 y==-1

Add 29241/4 to both sides:

y^2-171 y+29241/4==29237/4

Write the left hand side as a square:

(y-171/2)^2==29237/4

Take the square root of both sides:

y-171/2==(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

Add 171/2 to both sides:

y==171/2+(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

Substitute back for y==x^2:

x^2==171/2+(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

Take the square root of both sides:

x==Sqrt[171/2+(13 Sqrt[173])/2] or x==-Sqrt[171/2+(13 Sqrt[173])/2] or y-171/2==-((13 Sqrt[173])/2)

171/2+(13 Sqrt[173])/2 == 169/4+(13 Sqrt[173])/2+173/4 == 169+26 Sqrt[173]+173/4 == 169+26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]+13)^2/4:

x==1/2 (Sqrt[173]+13) or x==-Sqrt[171/2+(13 Sqrt[173])/2] or y-171/2==-((13 Sqrt[173])/2)

171/2+(13 Sqrt[173])/2 == 169/4+(13 Sqrt[173])/2+173/4 == 169+26 Sqrt[173]+173/4 == 169+26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]+13)^2/4:

x==1/2 (13+Sqrt[173]) or x==-1/2 (Sqrt[173]+13) or y-171/2==-((13 Sqrt[173])/2)

Add 171/2 to both sides:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or y==171/2-(13 Sqrt[173])/2

Substitute back for y==x^2:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x^2==171/2-(13 Sqrt[173])/2

Take the square root of both sides:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x==Sqrt[171/2-(13 Sqrt[173])/2] or x==-Sqrt[171/2-(13 Sqrt[173])/2]

171/2-(13 Sqrt[173])/2 == 169/4-(13 Sqrt[173])/2+173/4 == 169-26 Sqrt[173]+173/4 == 169-26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]-13)^2/4:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x==1/2 (Sqrt[173]-13) or x==-Sqrt[171/2-(13 Sqrt[173])/2]

171/2-(13 Sqrt[173])/2 == 169/4-(13 Sqrt[173])/2+173/4 == 169-26 Sqrt[173]+173/4 == 169-26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]-13)^2/4:

x==1/2 (13+Sqrt[173])     or     x==1/2 (-13-Sqrt[173])     or     x==1/2 (Sqrt[173]-13)     or     x==-1/2 (Sqrt[173]-13)

[1/2 (13+Sqrt[173])]  -  1 / [1/2 (13+Sqrt[173])] = 13   or    [1/2 (Sqrt[173]-13)]  -  1 / [1/2 (Sqrt[173]-13)] = -13

there is way more

303,323,343,353,363,373,383,393

so the answer is 18+3=21

Another way:

5!  -  4!  = 96 arrangements.

thank you so much Melody!!! I really appreciate it! :)

You can select them in : 20C2 ways =190 ways.

There are actually 22 of them:

133 , 233 , 303 , 313 , 323 , 330 , 331 , 332 , 333 , 334 , 335 , 336 , 337 , 338 , 339 , 343 , 353 , 363 , 373 , 383 , 393 , 433 , Total =  22

13 = 133, 233, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 433.

13 = 133, 233, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 433.

Oh. And for those of you who are confused about writing an expression for the first part, remember PA is the distance from P to A. How can you find their distance? If you are still confused, try to draw an example which has easy numbers and coordinates to be working with (ex. (1,1), length of 2, etc...)

Hope this helps! :)

For those of you (if anyone) who need help on this, here are some hints: 

Hint 1: let P be (x,y) and find an expression for PA^2 + PB^2 + PC^2

WARNING: remember what you got is an EXPRESSION and not an equation!!! So don't start to simplify it like it's an equation! (I almost fell for this part)

Hint 2: Simplify the expression and try to resemble whatever you got into the form 3PQ^2 + k, and remember k is a constant!

I hope this helps. :)

Lol I almost forgot to reply. I don't know if you did this on purpose or you just made a mistake, but your answer is completely wrong. I don't know how you got it and I SPECIFICALLY asked to explain. Glad I only used your answer as verification and didn't use it immediately. I found a way to do this after I solved it.

The original roll has a diameter of 5 unts   --->   radius = 2.5 units.

It's total area is  pi·2.5²  =  6.25pi units².

However, not all of that is paper; there is a hollow core of diameter 1 unit   --->   radius = 0.5 units.

The area of the core is  pi·0.5²  =  0.25pi units².

Subtracting the area of the core from the total area gives  6.25pi units² - 0.25pi units²  =  6.00pi units².

To get twice the amount of paper:  2 x 6.00pi units²  =  12.00pi units².

However, we can't forget the core, so we need  12.00pi units² + 0.25pi units²  =  12.25pi units.

Finding the square root of 12.25, we get 3.5, so the roll needs a radius of 3.5 units,

which is a diameter of 7 units.

so the prime numbers list under 100 is

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 and we have to have prime numbers that are 0,2, and 4 more than itself are only (3,5,7) so the answer (3,5,7)

As far as I know there is only one such example. And that is:

3, 5, 7.  Other than this exception, every 3rd number in the form: n, (n + 2), (n + 4) is always divisible by 3. Here are a few examples of "twin primes":  (11, 13) , (17, 19) , (29, 31) , (41, 43) , (59, 61) , (71, 73) , (101, 103) , (107, 109) . In each of these examples, adding 2 to the 2nd prime or adding 4 to the first prime, it becomes divisible by 3. Always!.

Since the chairs and the table have no arms, the 10 arms belong to 5 people (assuming that each person has 2 arms).

Therefore, the 5 persons also have 10 legs (again assuming that each person has 2 legs).

Since there is a total of 18 legs, 8 legs belong to the table and the chairs.

Since 4 of the legs belong to the one table, the remaining 4 legs belong to 1 chair.

We have 5 persons and 1 chair; it's time to get 4 more chairs.

Q:What were the religious practices of the following groups?

A:They had a wide pantheon of gods and they have lots of gods

If n = 3, then n + 2 = 5, and n + 4 = 7; all primes.

69 nice. 

The answer is 69, because 69 nice

Never, mind, I solved it.