Answer:
Population of mosquitoes in the area at any time t is:

Step-by-step explanation:
assume population at any time t = P(t)
population increases at a rate proportional to the current population:
⇒dP/dt ∝ P
----(1)
where k is constant rate at which population is doubled
solving (1)

---- (2)
initial population = 400,000
population is doubled every week
⇒P(1)=2P(0)
Using (2)


In presence of predators amount is decreased by 50,000 per day
Then amount decreased per week = 350,000
In this case (1) becomes
---(3)
solving (3) by calculating integrating factor

Multiplying I.F with all terms of (3)

Integrating w.r.to t




at t=0



So, population of mosquitoes in the area at any time t is

Answer:
Option C
Step-by-step explanation:
A and B are said to be independent of
P(A intersection B) = P(A)*P(B).
Or if drawing is done with replacement, etc.
A) P(drawing a king) = 4/52 = 1/13
IF we replace again drawing a kind is the same 1/13
Hence independent
B) Coin flipping each toss is obviously independent of the other as prob of getting tail in a fair coin is 1/2 irrespective of the previous outcomes
C) Without replacement is not independent
Since first prob = 4/52 and second would be 3/51 not the same as before
D) Rolling a die is independent as getting a 2 in any throw is always the same.
C is answer
Answer is D alternative form is 3787.99512
When you divide by a multiple of 10 or when you multiply by a multiple of 10.