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).

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

125 yards

step-by-step explanation:

i take 75 plus 50.