Which set of numbers is correctly ordered from greatest to least?
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Answer:
Step-by-step explanation:
Given numbers: [tex]7\tfrac{1}{3}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,16\sqrt{2e}[/tex]
We need to arrange these numbers from greatest to least .
On writing these numbers in simplified form, we get
[tex]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[/tex]
Here,[tex]7.\overline{36}[/tex] has a non-terminating repeating decimal expansion .
Vale of e = 2.718[tex]16\sqrt{2e}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,7\tfrac{1}{3}[/tex]
The greatest number is [tex]16\sqrt{2e}[/tex]
Then number less than [tex]16\sqrt{2e}[/tex] is [tex]\frac{221}{30}[/tex]
Then number less than [tex]\frac{221}{30}[/tex] is [tex]7.\overline{36}[/tex]
Then number less than [tex]7.\overline{36}[/tex] is [tex]7\tfrac{1}{3}[/tex]
i.e [tex]16\sqrt{2e}\,,\,\frac{221}{30}\,,\,7.\overline{36}\,,\,7\tfrac{1}{3}[/tex]
So, answer is option 4.
The set with the numbers ordered from greatest to least is given by:
[tex]3.16\sqrt{2e}, \frac{221}{30}, 7.363636,7\frac{1}{3}[/tex], given by fourth option.
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[tex]3.16\sqrt{2e}, \frac{221}{30}, 7.363636,7\frac{1}{3}[/tex], given by fourth option.
A similar problem is given at https://brainly.com/question/17407153