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fiasKO [112]
3 years ago
6

A radio active isotope decays according to the exponential decay equation where t is in days. Round to the thousandths place. Fo

r the half life: The half life is the solution (t) of the equation : a2=ae−7.571t a 2 = a e − 7.571 t
Mathematics
1 answer:
vovikov84 [41]3 years ago
6 0
The decay function is
a(t)=a_{0} e^{-7.571t}
where
a₀ = initial mass
t = time, days

At half life, a(t) = a₀/2, therefore the time required to achieve half life is given by the equation
\frac{a_{0}}{2} =a_{0} e^{-7.751t}\\ \frac{1}{2} =e^{-7.571t}

Take natural log of each side.
ln(1/2) = -7.571 t

Divide each side by -7.571 to obtain
t= \frac{ln(1/2)}{-7.571} =0.0916

Answer: 0.0192 days (nearest thousandth)
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4 years ago
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to mo
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This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

       Observed    Expected  

Mon   Freq (Oi)   Freq Ei       (Oi-Ei)^2 /Ei

Jan      11    16.0833        1.606649

Feb      13    16.0833        0.591105

Mar      22    16.0833        2.176598

Apr       11    16.0833        1.606649

May     15    16.0833        0.072971

Jun     29    16.0833        10.373489

Jul     15    16.0833        0.072971

Aug       9    16.0833        3.119603

Sep      13    16.0833        0.591105

Oct      15    16.0833        0.072971

Nov      22    16.0833        2.176598

Dec      18    16.0833       0.228411

Total:   193     93                22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

7 0
3 years ago
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