The age of the radioactive artifact is approximately 19029.6 years old
<h3>Radioactivity</h3>
Radioactivity is the phenomenon of the spontaneous disintegration of unstable atomic nuclei to atomic nuclei to form more energetically stable atomic nuclei. Radioactive decay is a highly exoergic, statistically random, first-order process that occurs with a small amount of mass being converted to energy.
To solve this problem, we have to find the disintegration constant and we can use the formula of half-life of the sample to do that.
The half-life of a sample is given as;
![T_\frac{1}{2}=\frac{ln}{\lambda}\\T_\frac{1}{2} =\frac{ln}{5715} \\ T_\frac{1}{2} = 0.000121\\](https://tex.z-dn.net/?f=T_%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7Bln%7D%7B%5Clambda%7D%5C%5CT_%5Cfrac%7B1%7D%7B2%7D%20%3D%5Cfrac%7Bln%7D%7B5715%7D%20%5C%5C%20T_%5Cfrac%7B1%7D%7B2%7D%20%3D%200.000121%5C%5C)
The formula of radioactivity is given as
![N = N_oe^(^-^\lambda ^t)](https://tex.z-dn.net/?f=N%20%3D%20N_oe%5E%28%5E-%5E%5Clambda%20%5Et%29)
Substituting the values into the equation
![N = N_oe^(^-^\lambda ^t)\\\frac{N}{N_o} = e^-^(^\lambda ^t)\\0.1 = e^-^(^0^.^0^0^0^1^2^1^*^t^)\\t = 19029.6 years](https://tex.z-dn.net/?f=N%20%3D%20N_oe%5E%28%5E-%5E%5Clambda%20%5Et%29%5C%5C%5Cfrac%7BN%7D%7BN_o%7D%20%3D%20e%5E-%5E%28%5E%5Clambda%20%5Et%29%5C%5C0.1%20%3D%20e%5E-%5E%28%5E0%5E.%5E0%5E0%5E0%5E1%5E2%5E1%5E%2A%5Et%5E%29%5C%5Ct%20%3D%2019029.6%20years)
The artifact is approximately 19029.6 years.
Learn more on radioactivity here;
brainly.com/question/26626062
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Answer: ![(15.263,\ 17.537)](https://tex.z-dn.net/?f=%2815.263%2C%5C%2017.537%29)
Step-by-step explanation:
According to the given information, we have
Sample size : n= 50
![\overline{x}=16.40](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D16.40)
![s=4.00](https://tex.z-dn.net/?f=s%3D4.00)
Since population standard deviation is unknown, so we use t-test.
Critical value for 95 percent confidence interval :
![t_{n-1,\alpha/2}=t_{49, 0.025}= 2.009575\approx2.010](https://tex.z-dn.net/?f=t_%7Bn-1%2C%5Calpha%2F2%7D%3Dt_%7B49%2C%200.025%7D%3D%202.009575%5Capprox2.010)
Confidence interval : ![\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20t_%7Bn-1%2C%20%5Calpha%2F2%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
![16.40\pm (2.010)\dfrac{4}{\sqrt{50}}\\\\=16.40\pm1.13702770415\\\\=16.40\pm1.1370\\\\=(16.40-1.1370,\ 16.40+1.1370)\\\\=(15.263,\ 17.537)](https://tex.z-dn.net/?f=16.40%5Cpm%20%282.010%29%5Cdfrac%7B4%7D%7B%5Csqrt%7B50%7D%7D%5C%5C%5C%5C%3D16.40%5Cpm1.13702770415%5C%5C%5C%5C%3D16.40%5Cpm1.1370%5C%5C%5C%5C%3D%2816.40-1.1370%2C%5C%2016.40%2B1.1370%29%5C%5C%5C%5C%3D%2815.263%2C%5C%2017.537%29)
Required 95% confidence interval : ![(15.263,\ 17.537)](https://tex.z-dn.net/?f=%2815.263%2C%5C%2017.537%29)
Answer:
2x
Step-by-step explanation:
x ≥ 0
| x | = x .... absolute value will think all numbers are positive (or 0)
x + | x | = 2x
The answer is 4g^2-12g+9
(2g+-3)(2g+3)
(2g)(2g)+(2g)(-3)+(-3)(2g)+(-3)(-3)
Answer:
y = 52500(.91)^x
Step-by-step explanation:
Exponential functions are written y = ab^x where a is the starting point and b is the common ratio. We know the initial worth is 52500 and each year the worth decreases 9% of the original year, making it work 91% of the previous value.