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Marianna [84]
4 years ago
10

Please help me with the following questions. Thanks in advance!

Mathematics
2 answers:
Anastaziya [24]4 years ago
7 0

Answer:

1- 3407

use p(1-r/100) ^6

N76 [4]4 years ago
6 0

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).

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