Answer:
(P(t)) = P₀/(1 - P₀(kt)) was proved below.
Step-by-step explanation:
From the question, since β and δ are both proportional to P, we can deduce the following equation ;
dP/dt = k(M-P)P
dP/dt = (P^(2))(A-B)
If k = (A-B);
dP/dt = (P^(2))k
Thus, we obtain;
dP/(P^(2)) = k dt
((P(t), P₀)∫)dS/(S^(2)) = k∫dt
Thus; [(-1)/P(t)] + (1/P₀) = kt
Simplifying,
1/(P(t)) = (1/P₀) - kt
Multiply each term by (P(t)) to get ;
1 = (P(t))/P₀) - (P(t))(kt)
Multiply each term by (P₀) to give ;
P₀ = (P(t))[1 - P₀(kt)]
Divide both sides by (1-kt),
Thus; (P(t)) = P₀/(1 - P₀(kt))
Answer:
See I don't know the answer but I did another one like this if it doesn't help you can report it no problem
Step-by-step explanation:
There are 2 different ways you can solve this problem and you'll get the same answer.
To help you visualize what you have to do, it might be easiest to translate your problem to a unit that easier to use.
Since you only have a ribbon 5 feet long, but you must cut it into 6 pieces, you know it won't even be 1 foot long.
One way to solve the problem would be to translate your 5-foot ribbon into a ribbon divided into inches.
1 foot = 12 inches
To find out how many inches long your ribbon is, you'll need to multiply the number of feet by 12.
Once you know that, then you can more easily divide it by 6. This will tell you how many inches each ribbon is.
Careful! the answer is to be put into feet so once you know how many inches long the ribbon is, you must then divide by 12 to get the correct answer in feet. Don't forget to round to the nearest tenth.
Another way to solve the problem is to simply divide the 5 feet by the number 6 and then round your number to the nearest tenth. It may be harder to visualize the answer this way, but you will get the same answer as if you translated your ribbon into inches and then divided by 12.
Good luck! I hope this helps!
Answer:
Slope = rise / run
= -12 / 1500 (It's -12 and not 12 because a drop means it's decreasing)
= -0.008
I say it is C. Basically I just eliminate the potential answered down. A could not be the one since not all student in 2nd period got 100% and average students got below 95%. B cannot be it since both box plot have the same median. I do not think D is it because of how the answer is told. "The 4th period class should get the reward. Their lowest score is an outlier, and should be thrown out," it sound childish and makeing a joke to put "and should be thrown out." I may be wrong but that is my opinion. The relatively the best and reasonable answer is C.
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