Answer:
Yes,there is a significant association shell weight and the widths of the opercula
Step-by-step explanation:
Using a correlation Coefficient calculator :
Given the data above :
The Coefficient of correlation(r) obtained is :
0.7632
Obtaining the test statistic :
T = r² / √(1 - r²) / (n - 2)
T = 0.7632² / √(1 - 0.7632²) / (10 - 2)
T = 0.58247424 / 0.2284528
Test statistic = 2.550
The Pvalue from r score , N = 10
Pvalue(0.7632, 10) = 0.01022
α = 0.05
If Pvalue < α ; reject H0
Pvalue < α ; We conclude that there is a significant association shell weights and the widths of the opercula
15-7=8. 8 would be your answer
C is the best answer because it makes sence
I think it is D for the answer
My first answer. Lots of parts for five minutes. Took me that long to decode the question. It looks like your multiplying where you're taking roots.
![(\sqrt[4]{7})^5 = (7^{\frac 1 4})^5 = 7^{\frac 5 4} ](https://tex.z-dn.net/?f=%20%28%5Csqrt%5B4%5D%7B7%7D%29%5E5%20%3D%20%287%5E%7B%5Cfrac%201%204%7D%29%5E5%20%3D%207%5E%7B%5Cfrac%205%204%7D%0A)
<span>Choice A. Next
</span>
![(3^{2/3})^{1/6} = 3^{ \frac 2 3 \times \frac 1 6 } = 3^{\frac 1 9} = \sqrt[9]{ 3} ](https://tex.z-dn.net/?f=%283%5E%7B2%2F3%7D%29%5E%7B1%2F6%7D%20%3D%203%5E%7B%20%5Cfrac%202%203%20%5Ctimes%20%5Cfrac%201%206%20%7D%20%3D%203%5E%7B%5Cfrac%201%209%7D%20%3D%20%5Csqrt%5B9%5D%7B%203%7D%20%0A%0A%0A)
Choice B.
![\dfrac {2^{3/4} } {2 ^{1/2} } = 2^{3/4 - 1/2} = 2^{1/4} = \sqrt[4] {2} ](https://tex.z-dn.net/?f=%5Cdfrac%20%7B2%5E%7B3%2F4%7D%20%7D%20%7B2%20%5E%7B1%2F2%7D%20%7D%20%20%3D%202%5E%7B3%2F4%20-%201%2F2%7D%20%3D%202%5E%7B1%2F4%7D%20%3D%20%5Csqrt%5B4%5D%20%7B2%7D%0A)
Choice C