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50%

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= 5730/11460 * 100%

= 0.5 * 100%

= 50%

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Q: What is 5730 out of 11460 as a percent?
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What is 57.3 as a percent?

5730%


The half-life is C-14 is 5730 years if a skeleton is found within 25 percent of the level it occurs at today how old is the skeleton?

5730/25% = 5730/0.25 = 22,920 years


The half-life of carbon-14 is 5730 years. How long will it take for the number of carbon-14 nuclei in a sample to drop to a quarter of the original number?

It will take two half-lives for the number of carbon-14 nuclei to drop to a quarter of the original number. Since the half-life of carbon-14 is 5730 years, it will take 2 * 5730 years = 11460 years for this to occur.


The half-life of carbon-14 is 5730 years After 11460 years how much of the original carbon-14 remains?

Since the radioactive decay of 14C is first order, one can use the equation offraction remaining = 0.5^n where n is the number of half lives. 28650yr/5730 yr = 5 half lives. Thus..fraction remaining = 0.5^5 = 0.03125 or 3.125% would be remaining.


Carbon 14 has a half life of 5730 In a plant fossil you find that this has decayed to a quarter of the original amount How long ago was this plant alive?

2 Half-Lifes= 1/4 Carbon 14 remaining. So 2 half lives have past. You add the 5730 years together to get 11460 years or 1.15*10^4 years.


What are the release dates for Days of Our Lives - 1965 1-11460?

Days of Our Lives - 1965 1-11460 was released on: USA: 15 November 2010


What is one fourth of 5730?

1/4 of 5730 = 1432.5


How do you write 5730 in exponential notation?

5730 in exponential notation is 5.73 x 10^3


What is 5730 divided by 100?

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What is the half life of an Egyptian papyrus with 63 percent of its original carbon-14 atoms?

The question is improperly asked. Papyrus does not have a half-life. Carbon-14 does. Half-life is constant, not variable, in this case, about 5730 years.So, the question would be more properly asked as "what is the age of an Egyptian papyrus with 63 percent of its original carbon-14 atoms?"The equation for radioactive decay is...AT = A0 2(-T/H)... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.Substituting what we know, we get...0.63 = (1) 2(-T/5730)Solve for T...log2(0.63) = -T/5730T = -log2(0.63)(5730)T = 3819 (conservatively rounded, T = 3800)


What is the refresh code of Nokia 5730?

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5730+53 its divide?

ok