Answer:
(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Step-by-step explanation:
Let the random variable <em>X</em> follow a Normal distribution with parameters <em>μ</em> = 155.4 and <em>σ</em> = 49.5.
(a)
Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:
*Use a standard normal table.
Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b)
A sample of <em>n</em> = 246 is selected.
Compute the probability that a sample mean is between 158.6 and 159.2 as follows:
*Use a standard normal table.
Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Answer:
Step-by-step explanation:
In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).
The formula we will use for this problem is the following:
where:
a=0
so the volume becomes:
This can be simplified to:
and the integral can be rewritten like this:
which is a standard integral so we solve it to:
![V=9\pi[tan y]\limits^\frac{\pi}{3}_0](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20y%5D%5Climits%5E%5Cfrac%7B%5Cpi%7D%7B3%7D_0)
so we get:
![V=9\pi[tan \frac{\pi}{3} - tan 0]](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20%5Cfrac%7B%5Cpi%7D%7B3%7D%20-%20tan%200%5D)
which yields:
]
Answer:the slope would be 7/3
Step-by-step explanation:
This is because you write the slope in rise over run form and use the two points provided to get 14/6 and reduce hat number down to 7/3.
I think your answer is D) i am not 100% sure tho.
Answer:
0.28cm/min
Step-by-step explanation:
Given the horizontal trough whose ends are isosceles trapezoid
Volume of the Trough =Base Area X Height
=Area of the Trapezoid X Height of the Trough (H)
The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)
The Volume of water in the trough at any time
=8h(8+2x)
V=64h+16hx
We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles
x/h=1/4
4x=h
x=h/4
Substituting x=h/4 into the Volume, V
h=3m,
dV/dt=25cm/min=0.25 m/min
=0.002841m/min =0.28cm/min
The rate is the water being drawn from the trough is 0.28cm/min.
