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2 years ago

16/21 as a decimal rounded to the nearest hundredth

Mathematics

2 answers:

Vikki [24]2 years ago

7 0

16 / 21 as Decimal, Rounded to the Nearest Hundredth, is 0.76

16 / 21 ==> Decimal: 0.7619047619047619

0.7619047619047619, Round to the Nearest Hundredths: ==> 0.76

0.7619047619047619 ===> Answer: 0.76

Therefore, 16 /21 ==>, Decimal: ==> 0.7619047619047619, Rounded to the nearest hundredths = 0.76

Zolol [24]2 years ago

5 0

The answer to this question is .76.

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Brenna found 10% of 60. Her work is shown below. 10% of 60 = (60 ÷ 2) = 30 Is Brenna’s answer correct? No. Brenna should have di

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Minute Maid states that a bottle of juice contains 473 mL. Consumer groups are interested in determining if the bottles contain

olga2289 [7]

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Option C) The average contents of all bottles of juice in the population, which is 473 mL

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 473 mL

Sample mean, $\bar{x}$ = 472 mL

Sample size, n = 30

Sample standard deviation, σ = 0.2

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mLinches}$

Representation of $\mu$

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2 years ago

solve "in muddy lake, 600 fish were tagged. later, 12 out 480 fish were found tagged. about how many fish are in the lake?"

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We estimate that 12/480=1/40 fish are tagged, so 600 is 1/40 of the total number of fish. This means that there are about 600*40=24000 fish total in the pond.

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2 years ago

An anthropologist finds that a prehistoric bone contains less than 8.1% of the amount of Carbon-14 the bones would have containe

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Answer:

20,944 years

Step-by-step explanation:

The formula you use for this type of decay problem is the one that uses the decay constant as opposed to the half life in years. We are given the k value of .00012. If we don't know how much carbon was in the bones when the person was alive, it would be safer to say that when he was alive he had 100% of his carbon. What's left then is 8.1%. Because the 8.1% is left over from 100% after t years, we don't need to worry about converting that percent into a decimal. We can use the 8.1. Here's the formula:

$N(t)=N_{0} e^{-kt}$

where N(t) is the amount left over after the decay occurs, $N_{0}$ is the initial amount, -k is the constant of decay (it's negative cuz decay is a taking away from as opposed to a giving to) and t is the time in years. Filling in accordingly,

$8.1=100e^{-.00012t}$

Begin by dividing the 100 on both sides to get

$.081=e^{-.00012t}$

Now take the natural log of both sides. Since the base of a natual log is e, natural logs and e "undo" each other, much like taking the square root of a squared number.

ln(.081)= -.00012t

Take the natual log of .081 on your calculator to get

-2.513306124 = -.00012t

Now divide both sides by -.00012 to get t = 20,944 years

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2 years ago