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- Jul 17th 2006, 12:14 PMBrookePre-test 6b Help Hurry
questions are on attachment!!!

- Jul 17th 2006, 03:04 PMgalactus
Hey Brookester:

Let's do #4. The bacteria.

What you need to do is use the given data and solve for k.

Once you know k, you can use that to solve the problem.

Given:

$\displaystyle 125=50e^{3k}$

Solve for k:

$\displaystyle \frac{5}{2}=e^{3k}$

$\displaystyle ln(\frac{5}{2})=3k$

$\displaystyle k=\frac{ln(\frac{5}{2})}{3}$

Now use that in the problem to solve. The initial population is 50. Triple that is 150.

$\displaystyle 150=50e^{\frac{ln(\frac{5}{2})}{3}t}$

$\displaystyle 3=e^{\frac{ln(\frac{5}{2})}{3}t}$

$\displaystyle ln(3)=\frac{ln(\frac{5}{2})}{3}t$

$\displaystyle t=\frac{3ln(3)}{ln(\frac{5}{2})}\approx{3.60}$ - Jul 17th 2006, 04:08 PMSoroban
Hello, Brooke!

I believe we're suppose to use: $\displaystyle A \:= \:P\left(1 + \frac{i}{n}\right)^{nt}\quad\Rightarrow\quad P \;=\;\frac{A}{\left(1 + \frac{i}{n}\right)^{nt}}$

where $\displaystyle P$ is the principal investment,

$\displaystyle i$ is the annual interest rate,

$\displaystyle n$ is the number of compounding periods per year,

$\displaystyle t$ is the number of years.

Quote:

2. An investment with an interest rate of 7% per annum yields $4000 in 3.5 years.

What is the present value for each compounding period?

. . A: annually . . B: semi-annually . . C: quarterly . . D: monthly

(A) annually: $\displaystyle n = 1$

$\displaystyle P \;= \;\frac{4000}{(1 + 0.07)^{3.5}} \;=\;3156.579774 \:\approx\; \$3,156.58$

(B) semi-annually: $\displaystyle n = 2$

$\displaystyle P \;= \;\frac{4000}{\left(1 + \frac{0.07}{2}\right)^{(2)(3.5)}} \;= \; 3143.963843 \:\approx\:\$3143.96$

You can do the other two . . .