# Help Hurry Pre-test 6a

• Jul 13th 2006, 07:48 PM
Lane
Help Hurry Pre-test 6a
questions on attachment!!!
• Jul 13th 2006, 08:54 PM
earboth
1,2,3 solved
Quote:

Originally Posted by Lane
questions on attachment!!!

Hello, Lane,

to 1) rearrange the expression:
$g=10*\left(\log_{10}P_o-\log_{10}P_i\right)=10*\log_{10}\left(\frac{P_0}{P _i}\right)$

Now plug in the given values:

$g=10*\log_{10}\left(\frac{30 W}{.06 W}\right)=10*\log_{10}\left(500\right)\approx 26.9897...$

to 2) Use the substitution $y=2^{x}$. You'll get a quadratic equation:
$y^2-13y+42=0\quad\Longrightarrow\quad y=6\vee y=7$

Re-substitute: $2^x=6\ \Leftrightarrow\ x=\log_{2}(6)=\frac{\log_{10}(6)}{\log_{10}(2)} \approx 2.5849...$ or $2^x=7\ \Leftrightarrow\ x=\log_{2}7=\frac{\log_{10}7}{\log_{10}2}\approx 2.8073...$

to 3) Multiply the equation by the denominator. You'll get:
$900=25+25e^{-x} \Leftrightarrow\ 35=e^{-x}$ $\Leftrightarrow\ e^x=\frac{1}{35}$

Therefore: $x=\ln\left(\frac{1}{35}\right)\ \Longrightarrow\ x\approx -3.555...$

Greetings

EB
• Jul 14th 2006, 06:37 AM
Soroban
Hello, Lane!

Quote:

4. The amount of power generated by a satellite's power supply is: $P\:=\:50e^{-\frac{t}{300}}$
where $P$ is the power in watts and $t$ is the time in days.
For how many days will 22 watts of power be available? .Round to the nearest whole day.

We are given: $P = 22$

The equation becomes: . $22\;=\;50e^{-\frac{t}{300}}$

Divide by 50: . $0.44\;=\;e^{-\frac{t}{300}}$

Multiply by $e^{\frac{t}{300}}:\;\;0.44e^{\frac{t}{300}}\;=\;1 \quad\Rightarrow \quad e^{\frac{t}{300}}\:=\:\frac{1}{0.44}$

Rewrite in log form: . $\frac{t}{300} \:=\:\ln\left(\frac{1}{0.44}\right)\:=\:-\ln(0.44)$

Therefore: . $t\:=\:-300\cdot\ln(0.44) \:=\:246.2941656\:\approx\:246\text{ days}$

Quote:

5. The mathematical model for learning an aseembly line procedure needed for assembling
one component of a manufactured item is: . $P\:=\:\frac{0.93}{1 + e^{-2n}}$
where $P$ is the proportion of correctly assembled compnents after $n$ practice sessions.

How many practice sessions are required to have at last 75% of the components correctly
assembled within the given time period? .Round any non-integer to the next higher whole number.

Is there a typo? . . . I'm getting a strange answer.

We want $P = 0.75$
The equation becomes: . $\frac{0.93}{1 + e^{-2n}}\:=\:0.75\quad\Rightarrow\quad0.93\:=\:0.75(1 + e^{-2n})$

. . . . $0.93\:=\:0.75 + 0.75e^{-2n}\quad\Rightarrow\quad 0.18\:=\:0.75e^{-2n}$

Multiply by $e^{2n}:\;\;0.18e^{2n}\:=\:0.75\quad\Rightarrow \quad e^{2n}\;=\;\frac{0.75}{0.18}\:=\:\frac{25}{6}$

Log form: . $2n \:=\:\ln\left(\frac{25}{6}\right)\quad\Rightarrow \quad n \:=\:\frac{1}{2}\ln\left(\frac{25}{6}\right)$

Then we have: . $n \:= \:0.713558178 \:=\:1$

An assembler achieves a 75% proficiency level during the first session?