Question

Which set of numbers is correctly ordered from greatest to least?

Answer 1

Answer:

Step-by-step explanation:

Given numbers: $7\tfrac{1}{3}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,16\sqrt{2e}$

We need to arrange these numbers from greatest to least .

On writing these numbers in simplified form, we get

$7\tfrac{1}{3}=\frac{22}{3}=7.33\\\frac{221}{30}=7.367\\7.\overline{36}=7.3636...\\16\sqrt{2e}=16\times \sqrt{5.437}=16\times 2.332=37.312$

Here,$7.\overline{36}$ has a non-terminating repeating decimal expansion .

Vale of e = 2.718$16\sqrt{2e}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,7\tfrac{1}{3}$

The greatest number is $16\sqrt{2e}$

Then number less than $16\sqrt{2e}$ is $\frac{221}{30}$

Then number less than $\frac{221}{30}$ is $7.\overline{36}$

Then number less than $7.\overline{36}$ is $7\tfrac{1}{3}$

i.e $16\sqrt{2e}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,7\tfrac{1}{3}$

So, answer is option 4.

Answer 2

The set with the numbers ordered from greatest to least is given by:

$3.16\sqrt{2e}, \frac{221}{30}, 7.363636,7\frac{1}{3}$, given by fourth option.

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• The first number is: $7\frac{1}{3} = 7 + \frac{1}{3} = 7 + 0.3333 = 7.3333$

• The second number is 7 with repeating 36 decimal, that is: $7.363636$
• The third number is: $\frac{221}{30} = 7.36363637$
• The fourth number is: $3.16\sqrt{2e} = 7.368$

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• Looking at the decimal numbers, ordered from greatest to least, they are:

$3.16\sqrt{2e}, \frac{221}{30}, 7.363636,7\frac{1}{3}$, given by fourth option.

A similar problem is given at https://brainly.com/question/17407153