Capture-recapture studies are common in ecology. One form of the study is conducted as follows. Suppose we have a population of N deer in a study area. Initially n deer from this population are captured, marked so that they can be identified as having been captured, and returned to the population. After the deer are allowed to mix together, m deer are captured from the population and the number k of these deer having marks from the first capture is observed. Required:
Assuming that the first and second captures can be considered random selections from the population and that no deer have either entered or left the study area during the sampling period, what is the probability of observing k marked deer in the second sample of m deer?

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation: