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uysha [10]
3 years ago
7

Due to employee safety negligence at a nuclear waste facility, 2000 tons of a radioactive element is spilled into the nearby pon

d. The half-life of the radioactive element is 36 days. In order to be declared safe for swimming, based on its size and the amount of water, there must be less than 100 tons of the material found in the pond. How long, to the nearest day, until it is safe to swim again?
Mathematics
1 answer:
HACTEHA [7]3 years ago
6 0

How long, to the nearest day, until it is safe to swim again will be 156 days

Let x represent number of day until it is safe to swim again

First step

=2000 *1/2^x

100=2000 *0.5^x

0.05=0.5^x

Second step

Log 0.05=xLog 0.5

Log 0.05/L0g 0.5=x

x=36 days* Log 0.05/L0g 0.5

x=36 days*43.22

x=156 days

Inconclusion How long, to the nearest day, until it is safe to swim again will be 156 days

Learn more about  radioactive element here:

brainly.com/question/13812761

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Answer:

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Step-by-step explanation:

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You go to Ihop and order two servings of pancakes and a
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Answer: $4.76

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Anyone know this geometry question? Will give brainiest
Dafna11 [192]
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3 years ago
The mean consumption of bottled water by a person in the United States is 28.5 gallons per year. You believe that a person consu
Over [174]

Answer:

t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707    

p_v =P(t_{99}

If we compare the p value with a significance level for example \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

Step-by-step explanation:

Data given and notation    

\bar X=27.8 represent the mean for the account balances of a credit company

s=4.1 represent the population standard deviation for the sample    

n=1000 sample size    

\mu_o =28.5 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:    

Null hypothesis:\mu \geq 28.5    

Alternative hypothesis:\mu < 28.5    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=100-1=99

Since is a one-side lower test the p value would be:    

p_v =P(t_{99}

In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"  

Conclusion    

If we compare the p value with a significance level for example \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

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