Me with the following questions. in advance!

1- 3407

use p(1-r/100) ^6

Okay. So you calculate by the year for each problem until you get to the desired amount. I suggest a better strategy. I will make an equation for each problem so that the calculations are much faster.

1. a(n) = 3800(0.982)^n-1

a(7) = 3800(0.982)^6 = 3408 lemurs approx.

2. a(n) = 35000(1+0.03/4)^4n

a(6) = 35000(1+0.03/4)^24 = $41874.47 approx.

Explanation for question 1: In both situations there is a starting amount which is applied a factor either increasing it or decreasing it. In the first equation, we have 3800 lemurs with 98.2% of the population surviving every year (because 1.8% dies every year). The "n-1" makes year 1 display 3800 lemurs. Any year after will apply the decline. So if we want the year 2015 which would correspond with year 7.

Explanations for question 2. The equation shows the starting amount (35000) which receives interest (3% compounded monthly). That is why 0.03 is divided by 4. But because interest is gained every quarter of a year... the interest rate is raised to the 4n instead of just n. "n" of course represents the number of years that interest has been gained. So, to get the balance after 6 years, we evaluate a(6).

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.