The correlation coefficient is calculated by the given formula
![r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{{(n\sum x^{2}-(\sum x)^{2})(n\sum y^{2}-(\sum y)^{2})}}}](https://tex.z-dn.net/?f=%20r%3D%5Cfrac%7Bn%28%5Csum%20xy%29-%28%5Csum%20x%29%28%5Csum%20y%29%7D%7B%5Csqrt%7B%7B%28n%5Csum%20x%5E%7B2%7D-%28%5Csum%20x%29%5E%7B2%7D%29%28n%5Csum%20y%5E%7B2%7D-%28%5Csum%20y%29%5E%7B2%7D%29%7D%7D%7D%20)
Here we have to find correlation coefficient for three ordered pairs, so n=3
![\sum x = 2+3+5=10](https://tex.z-dn.net/?f=%20%5Csum%20x%20%3D%202%2B3%2B5%3D10%20)
![\sum y = 26+2+16=44](https://tex.z-dn.net/?f=%20%5Csum%20y%20%3D%2026%2B2%2B16%3D44%20)
![\sum xy = 52+6+80 =138](https://tex.z-dn.net/?f=%20%5Csum%20xy%20%3D%2052%2B6%2B80%20%3D138%20)
![\sum x^{2} = 4+9+25 =38](https://tex.z-dn.net/?f=%20%5Csum%20x%5E%7B2%7D%20%3D%204%2B9%2B25%20%3D38%20)
![\sum y^{2} =676+4+256 =936](https://tex.z-dn.net/?f=%20%5Csum%20y%5E%7B2%7D%20%3D676%2B4%2B256%20%3D936%20)
Substituting all the values in the above formula,
![r=\frac{3(138)-440}{\sqrt{(38(3)-100)(3(936)-1936)}}](https://tex.z-dn.net/?f=%20r%3D%5Cfrac%7B3%28138%29-440%7D%7B%5Csqrt%7B%2838%283%29-100%29%283%28936%29-1936%29%7D%7D%20)
![r= \frac{-26}{\sqrt{14 \times 872}}](https://tex.z-dn.net/?f=%20r%3D%20%5Cfrac%7B-26%7D%7B%5Csqrt%7B14%20%5Ctimes%20872%7D%7D%20)
r= -0.24
Answer:
hi
Step-by-step explanation:
Here, exterior angles are 1, 2, 7 & 8 as they are outside the parallel lines, & among them, alternates are: 1 with 7 and 2 & 8. 1 & 7 is not in option but 2 and 8 is there in C
In short, Your Answer would be Option C
Hope this helps!
Answer: 39.308 pounds
Step-by-step explanation:
We assume that the given population is normally distributed.
Given : Significance level : ![\alpha: 1-0.98=0.02](https://tex.z-dn.net/?f=%5Calpha%3A%201-0.98%3D0.02)
Sample size : n= 12, which is small sample (n<30), so we use t-test.
Critical value (by using the t-value table)=![t_{n-1, \alpha/2}=t_{11,0.01}=2.718](https://tex.z-dn.net/?f=t_%7Bn-1%2C%20%5Calpha%2F2%7D%3Dt_%7B11%2C0.01%7D%3D2.718)
Sample mean :
Standard deviation : ![\sigma= 20](https://tex.z-dn.net/?f=%5Csigma%3D%2020)
The lower bound of confidence interval is given by :-
![\overline{x}-t_{(n-1,\alpha/2)}\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D-t_%7B%28n-1%2C%5Calpha%2F2%29%7D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
i.e. ![55-(2.718)\dfrac{20}{\sqrt{12}}](https://tex.z-dn.net/?f=55-%282.718%29%5Cdfrac%7B20%7D%7B%5Csqrt%7B12%7D%7D)
![=55-15.6923803166\approx55-15.692=39.308](https://tex.z-dn.net/?f=%3D55-15.6923803166%5Capprox55-15.692%3D39.308)
Hence, the lower bound for the 98% confidence interval for the mean yearly sugar consumption= 39.308 pounds
Subtract the smaller amount ($23.04) from the larger ($29.40), and then divide the difference by 12 pounds:
$6.36
-------- = $0.53 / lb
12 lb