Answer:
The hypothesis test is right-tailed
Step-by-step explanation:
To identify a one tailed test, the claim in the case study tests for the either of the two options of greater or less than the mean value in the null hypothesis.
While for a two tailed test, the claim always test for both options: greater and less than the mean value.
Thus given this: H0:X=10.2, Ha:X>10.2, there is only the option of > in the alternative claim thus it is a one tailed hypothesis test and right tailed.
A test with the greater than option is right tailed while that with the less than option is left tailed.
Answer:supplementary
Step-by-step explanation:
Opposite angles in any quadrilateral inscribed in a circle are supplements of each other.
Answer:
28.27 cm/s
Step-by-step explanation:
Though Process:
- The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
- the radius
- Punch is being poured into the bowl
- The height at which the punch is increasing in the bowl is
![\frac{dh}{dt} = 1.5](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%201.5)
- the exposed area is a circle, (since the bowl is a hemisphere)
- the radius of this circle can be written as
!['a'](https://tex.z-dn.net/?f=%27a%27)
- what is being asked is the rate of change of the exposed area when the height
- the rate of change of exposed area can be written as
. - since the exposed area is changing with respect to the height of punch. We can use the chain rule:
![\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%20%5Cfrac%7BdA%7D%7Bdh%7D%20.%20%5Cfrac%7Bdh%7D%7Bdt%7D)
- and since
the chain rule above can simplified to
-- we can call this Eq(1)
Solution:
the area of the exposed circle is
![A =\pi a^2](https://tex.z-dn.net/?f=A%20%3D%5Cpi%20a%5E2%20)
the rate of change of this area can be, (using chain rule)
we can call this Eq(2)
what we are really concerned about is how
changes as the punch is being poured into the bowl i.e ![\frac{da}{dh}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdh%7D)
So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:
![r = \frac{a^2 + h^2}{2h}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Ba%5E2%20%2B%20h%5E2%7D%7B2h%7D)
and rearrage the formula so that a is the subject:
![a^2 = 2rh - h^2](https://tex.z-dn.net/?f=a%5E2%20%3D%202rh%20-%20h%5E2)
now we can derivate a with respect to h to get ![\frac{da}{dh}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdh%7D)
![2a \frac{da}{dh} = 2r - 2h](https://tex.z-dn.net/?f=2a%20%5Cfrac%7Bda%7D%7Bdh%7D%20%3D%202r%20-%202h)
simplify
![\frac{da}{dh} = \frac{r-h}{a}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdh%7D%20%3D%20%5Cfrac%7Br-h%7D%7Ba%7D)
we can put this in Eq(1) in place of ![\frac{da}{dh}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdh%7D)
and since we know ![\frac{dh}{dt} = 1.5](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%201.5)
![\frac{da}{dt} = \frac{(r-h)(1.5)}{a}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%28r-h%29%281.5%29%7D%7Ba%7D%20)
and now we use substitute this
. in Eq(2)
![\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%202%20%5Cpi%20a%20%5Cfrac%7B%28r-h%29%281.5%29%7D%7Ba%7D)
simplify,
![\frac{dA}{dt} = 3 \pi (r-h)](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%203%20%5Cpi%20%28r-h%29)
This is the rate of change of area, this is being asked in the quesiton!
Finally, we can put our known values:
![r = 5cm](https://tex.z-dn.net/?f=r%20%3D%205cm)
from the question
![\frac{dA}{dt} = 3 \pi (5-2)](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%203%20%5Cpi%20%285-2%29)
![\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%209%20%5Cpi%20cm%2Fs%2F%2F%20or%2F%2F%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%2028.27%20cm%2Fs)