# Pre-test 6b Help Hurry

• July 17th 2006, 12:14 PM
Brooke
Pre-test 6b Help Hurry
questions are on attachment!!!
• July 17th 2006, 03:04 PM
galactus
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:

$125=50e^{3k}$

Solve for k:

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

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

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

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

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

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

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

$t=\frac{3ln(3)}{ln(\frac{5}{2})}\approx{3.60}$
• July 17th 2006, 04:08 PM
Soroban
Hello, Brooke!

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

where $P$ is the principal investment,
$i$ is the annual interest rate,
$n$ is the number of compounding periods per year,
$t$ 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: $n = 1$

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

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

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

You can do the other two . . .