Suppose a number is chosen at random from the set {0,1,2,3,...,1721}. What is the probability that the number is a perfect cube?
Round your answer to 6 decimal places as needed.

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Explanation:

One way to go about this is to list out all the perfect cubes. A perfect cube is the result of taking any whole number and multiplying it by itself 3 times.

1 cubed = 1^3 = 1*1*1 = 1

2 cubed = 2^3 = 2*2*2 = 8

3 cubed = 3^3 = 3*3*3 = 27

4 cubed = 4^3 = 4*4*4 = 64

and so on. We stop once we reach 1721, or if we go over. Ignore any values larger than 1721. You'll find that 11^3 = 1331 and 12^3 = 1728. So we stop here and exclude 1728 as that is larger than 1721.

A quick way to see where we should stop is to apply the cube root to 1721 and we get

The approximate result of 11.98377 tells us that 1721 is between the perfect cubes of 11^3 = 1331 and 12^3 = 1728

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So effectively, we have 11 perfect cubes in the set {0, 1, 2, 3, ..., 1721} and this is out of 1722 numbers in that same set. Note how I added 1 onto 1721 to get 1722. I'm adding an extra number because of the 0. If 0 wasn't part of the set, then we would have 1721 values total inside.

In summary: There are 11 values we want (11 perfect cubes) out of 1722 values total.

Divide 11 over 1722 to get

11/1722 = 0.00638792102207

which rounds to 0.006388

y + 11 = (1/4)(x + 16)

step-by-step explanation: