So hmm check the picture below
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\qquad \begin{cases} h=height\\ r=radius\\ ------\\ h=2(2r) \end{cases}\implies V=\cfrac{\pi r^2[2(2r)]}{3} \\\\\\ V=\cfrac{4\pi r^3}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ah%3Dheight%5C%5C%0Ar%3Dradius%5C%5C%0A------%5C%5C%0Ah%3D2%282r%29%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%5B2%282r%29%5D%7D%7B3%7D%0A%5C%5C%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D)
Answer:
Step-by-step explanation:
Given that for a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears.
r = linear correlation coeff = 0.952
H_0: r =0\\
H_1 :r\neq 0
(Two tailed test)
r difference = 0.952
n=8
Std error = \sqrt{\frac{1-r^2}{n-2} } =0.12496
Test statistic t = 0.952/0.12496 = 7.618
Alpha = 0.05
df = 6
p value = 0.000267
This implies H0 is rejected.
There exists a linear relation between the variables and r cannot be 0
0.952^2 = 0.906=90.6% of variation in weight can be explained by the linear relationship between weight and chest size
I’m very confused what are you trying to do here ?
The average number of people who entered the stadium per minute was 4,160
we know that
To find out the average number of people who entered the stadium per minute, divide the total number of people by the time required to fill the stadium
Let
x ----> the total number of people
y ----> the time required to fill the stadium
we have
1.04*10 5/25 = 4,160 people/mintute
so
therefore
The average number of people who entered the stadium per minute was 4,160
I hope this helped! Plz, comment in the paragraphs soon!
