Answer:
hehejejejejejejejejeheheheheheStep-by-step explanation:
Answer: ( A ) cleavage so the first one
Explanation: I hope this helps you
if the diameter of a circle is 15, its radius is half that or 7.5.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=7.5 \end{cases} A=\pi (7.5)^2\implies A=56.25\pi \implies \stackrel{\pi =3.14}{A=176.625}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D7.5%20%5Cend%7Bcases%7D%20A%3D%5Cpi%20%287.5%29%5E2%5Cimplies%20A%3D56.25%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7BA%3D176.625%7D%20)
Amalia. Phones rule is only true with numbers between 1 and 1,000.
Answer:
the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
Step-by-step explanation:
Given the data in the question;
we know that;
the coefficient of determination is r²
while the correlation coefficient is defined as r = √(r²)
The coefficient of determination tells us the percentage of the variation in y by the corresponding variation in x.
Now, given that class attendance explained 16% of the variation in grade index among the students.
so
coefficient of determination is r² = 16%
The correlation coefficient between percent of classes attended and grade index will be;
r = √(r²)
r = √( 16% )
r = √( 0.16 )
r = 0.4
Therefore, the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
