By the seventh day, there are zero rabbits remaining.
To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour
Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.
Please ask me any questions you may have!
There were 30 adults and 10 children, totaling 40 attendees. Just sum them together.
Answer:
Step-by-step explanation:
The probability distribution for the random variable X is provided.
X 4 5 6 7 Total
P 0.2 0.4 0.3 0.1 1
x*p 0.8 2 1.8 0.7 5.3
x^2*p 3.2 10 10.8 4.9 28.9
a) The expected value E(X), representing the mean of X, equates to 5.3.
Standard deviation is the square root of the variance, which amounts to 0.9.
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b) For the mean of the sample, we find:
Mean = 5.3
Variance = var(x)/n = 
c) 
