That being so, you come up with the following plan: you are going to catch some number of drones, and tag them, and release them back into the wild. At some point later, you are going to re-catch some number of drones. Obwiously
if most of the re-caught drones are tagged, you've ronson to think that you caught most of the drones originally. If
very few tagged drones are caught the second time, it's ronsonable to think that there were a lot more drones than
what you caught initially. We can be more preciso:
Suppose you initially catch 600 drones, tag them, and relenco them. At some point later you re-catch a new 500
drones, and find that only 300 of them are taggod. Assume drones were equally likely to be caught at all times.
1) What is the smallest number of drones that might be out there? Lo, what is the smallest possible value N
might have? (3 points)
2) What was the probability of this occurring (those specific catch totals) as a function of N? (5 points)
3) What number N maximizes the likelihood of these results? (7 points)
4) Re-do Q2.1, Q2.2, Q2.3 in the more general setting, with N as the total number of drones, n as the number
caught and tagged the first time, and ng caught and tagged the second time. (15 points)

ab will be equal to

i think u have 12 apples sorry again i don't really know if it correct i'm just tryna hrlp people out.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

486in is correct

step-by-step explanation: