If $$r = 98.3$$, which of the following best approximates the value of $$\frac{r}{2r+2}$$?

Correct.
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If you use $$r=100$$, the expression equals
>$$\frac{100}{2(100)+2} = \frac{100}{202}$$,
which is **slightly less than** $$\frac{1}{2}$$. That means you can eliminate D and E. The first answer less than $$\frac{1}{2}$$ is $$\frac{1}{3}$$, but that is much too small. $$\frac{100}{202}$$ is much closer to $$\frac{100}{200} = \frac{1}{2}$$ than $$\frac{100}{300} = \frac{1}{3}$$.

Incorrect.
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If you use $$r=100$$, the expression equals
>$$\frac{100}{2(100)+2} = \frac{100}{202}$$,
which is slightly _less_ than $$\frac{1}{2}$$. This answer choice is slightly *larger* than $$\frac{1}{2}$$, so it is incorrect.

Incorrect.
[[Snippet]]
If you use $$r=100$$, the expression equals
>$$\frac{100}{2(100)+2} = \frac{100}{202}$$,
which is slightly _less_ than $$\frac{1}{2}$$. This answer choice is slightly *larger* than $$\frac{1}{2}$$, so it is incorrect.

Incorrect.
[[Snippet]]
If you use $$r=100$$, the expression equals
>$$\frac{100}{2(100)+2} = \frac{100}{202}$$,
which is not really close to $$\frac{1}{3}$$ (which would be $$\frac{100}{300}$$ instead).

Incorrect.
[[Snippet]]
If you use $$r=100$$, the expression equals
>$$\frac{100}{2(100)+2} = \frac{100}{202}$$,
which is not really close to $$\frac{1}{4}$$ (which would be $$\frac{100}{400}$$ instead).

$$\frac{1}{4}$$

$$\frac{1}{3}$$

$$\frac{1}{2}$$

$$\frac{5}{9}$$

$$\frac{3}{5}$$