The pheasant population on an island is declining at a rate of 1.7%per year. the population was 4600 in the year 2009.what is the best prediction of the population in the year 2015

4150 pheasants in the year 2105

Step-by-step explanation:

First thing we are going to do is define the years.  We will call year 2009 our year t = 0.  Therefore, year 2015 is t = 6.

The decay equation that we need has a standard form of

where P(t) is the ending population after a certain amount of time goes by, a is the initial population at t = 0, r is the rate in decimal form, and t is time in years.  Fitting in our info gives us a standard form equation that looks like this:

This means that we are looking for the ending population at a rate of decay of 1.7% after 6 years have gone by.

Doing the subtraction in the parenthesis first simplifies it down a bit to

Raise .983 to the 6th power to get

P(t) = 4600(.9022379843)

and then multiply to get

P(t) = 4150

This means that in the year 2015 the population of pheasants is 4150, dropping from 4600 six years earlier.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

range =5

step-by-step explanation:

to find the range, we take the largest value and subtract the smallest value.

the largest value is 5

the smallest value is 0

the range is 5-0   =5