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Tju [1.3M]
3 years ago
5

Situation:

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0
N=NOe^-kt
N=mass at time t
NO = initial mass
k= 0.1476
t= time, in days

We are asked to find the half-life, which means you want to find how long it will take for half of the substance to decay/disappear (depending on situation). 

If we are looking for half-life, we can simply set N to half of NO (which we are given a value of 40grams for)
Therefore:
N = 20
NO = 40
plugging these values and the value given for k back into the equation you get:

20 = 40e^-0.1476(t)
We are looking for t, so we have to manipulate the formula to get t by itself on one side of the equation.
We can start by dividing 40 from both sides, and you get:
0.5 = e^-0.1476(t)

We have the exponential function "e".
To get rid of e, we can use natural log (ln)
if e^y=x then ln (x) = y
look back at our equation we can set
0.5 = x
-0.1476(t) = y

Rewriting it in natural log form:
ln (0.5) = -0.1476(t)
Plug in ln (0.5) on a calculator to find its value and we get:
-0.693147 = -0.1476(t)
*Note: normally, getting a negative value would suggest that we did something wrong, because you cannot have a negative value as your t (you cannot have negative days), but because there is a negative on both sides of the equation, they will cancel out in this case. 

The last step is to simply divide both sides by -0.1476
therefore:
T = 4.696119
But it asks you for the answer to the nearest tenth (one place after decimal pt) so 
T (half life) = 4.7 days

Hope that helps :)
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Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

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No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

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No relief                    11                     11                       11               33

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Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

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And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

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