Answer:
60 minutes for the larger hose to fill the swimming pool by itself
Step-by-step explanation:
It is given that,
Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool.
takes 30 minutes for the larger hose to fill the swimming pool by itself
Let x be the efficiency to fill the swimming pool by larger hose
and y be the efficiency to fill the swimming pool by larger hose
<u>To find LCM of 20 and 30</u>
LCM (20, 30) = 60
<u>To find the efficiency </u>
Let x be the efficiency to fill the swimming pool by larger hose
and y be the efficiency to fill the swimming pool by larger hose
x = 60/30 =2
x + y = 60 /20 = 3
Therefore efficiency of y = (x + y) - x =3 - 2 = 1
so, time taken to fill the swimming pool by small hose = 60/1 = 60 minutes
In this question, the Poisson distribution is used.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Parameter of 5.2 per square yard:
This means that
, in which r is the radius.
How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?
We want:

Thus:

We have that:


Then





Thus, the radius should be of at least 0.89.
Another example of a Poisson distribution is found at brainly.com/question/24098004
<span>Yes it is true that a continuous function that is never zero on an interval never changes sign on that interval. This is because of ever important Intermediate Value Theorem.</span>
I would say no thats my anwser
Step-by-step explanation:
You have found a function r(V(t)). We can see that this function is a one variable function. The variable is time.
So in this specific function we can call r(v(t)), r(t).
So:
![r(t) = \sqrt[3]{ \frac{3 \times (10 + 20t)}{4\pi} }](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020t%29%7D%7B4%5Cpi%7D%20%7D%20)
If α is the moment that the radius is 10 inches and since the function above gives radius in inches we have to solve the equation:

Which is the same as:
![\sqrt[3]{ \frac{3 \times (10 + 20 \alpha )}{4\pi} } = 10 \\ \frac{3 \times (10 + 20 \alpha )}{4\pi} = 1000 \\ (10 + 20 \alpha ) = \frac{4000\pi}{3} \\ 20 \alpha = \frac{(4000\pi - 30)}{3} \\ \alpha = \frac{(4000\pi - 30)}{60}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020%20%5Calpha%20%29%7D%7B4%5Cpi%7D%20%7D%20%20%3D%2010%20%5C%5C%20%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020%20%5Calpha%20%29%7D%7B4%5Cpi%7D%20%20%3D%201000%20%5C%5C%20%2810%20%2B%2020%20%5Calpha%20%29%20%3D%20%20%5Cfrac%7B4000%5Cpi%7D%7B3%7D%20%20%5C%5C%2020%20%5Calpha%20%20%3D%20%20%5Cfrac%7B%284000%5Cpi%20-%2030%29%7D%7B3%7D%20%5C%5C%20%20%5Calpha%20%20%3D%20%20%5Cfrac%7B%284000%5Cpi%20-%2030%29%7D%7B60%7D%20)