Yes, both are repeating fractions
hope this helps
Answer:
.0222
The probability of selecting a nickle then a dime without replacement is 2.22%.
(1/10)*(2/9)= .0222
Answer: (17.42, 20.78)
Step-by-step explanation:
As per given , we have
Sample size : n= 9
years
Population standard deviation is not given , so it follows t-distribution.
Sample standard deviation : s= 1.5 years
Confidence level : 99% or 0.99
Significance level : ![\alpha= 1-0.99=0.01](https://tex.z-dn.net/?f=%5Calpha%3D%201-0.99%3D0.01)
Degree of freedom : df= 8 (∵df =n-1)
Critical value : ![t_c=t_{(\alpha/2,\ df)}=t_{0.005,\ 8}= 3.355](https://tex.z-dn.net/?f=t_c%3Dt_%7B%28%5Calpha%2F2%2C%5C%20df%29%7D%3Dt_%7B0.005%2C%5C%208%7D%3D%203.355)
The 99% confidence interval for the population mean would be :-
![\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20t_c%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
![19.1\pm (3.355)\dfrac{1.5}{\sqrt{ 9}}\\\\=19.1\pm1.6775\\\\=(19.1-1.6775,\ 19.1+1.6775)\\\\=(17.4225,\ 20.7775)\approx(17.42,\ 20.78)](https://tex.z-dn.net/?f=19.1%5Cpm%20%283.355%29%5Cdfrac%7B1.5%7D%7B%5Csqrt%7B%209%7D%7D%5C%5C%5C%5C%3D19.1%5Cpm1.6775%5C%5C%5C%5C%3D%2819.1-1.6775%2C%5C%2019.1%2B1.6775%29%5C%5C%5C%5C%3D%2817.4225%2C%5C%2020.7775%29%5Capprox%2817.42%2C%5C%2020.78%29)
Hence, the 99% confidence interval for the population mean is (17.42, 20.78) .
Yes the answer is correct
The degree of this polynomial is 4!!!