
Substituting this into the other ODE gives

Since
, it follows that
. The ODE in
has characteristic equation

with roots
, admitting the characteristic solution

From the initial conditions we get



So we have

Take the derivative and multiply it by -1/4 to get the solution for
:

The correct answer is 64 bc 6 times 6 is 64
<u>Answer:</u>
<u>1. A. You allow the passenger to board his flight when the passenger has a weapon.</u>
<u>2. B. You select the passenger for further inspection when the passenger has no weapon.</u>
<u>Explanation:</u>
1. Remember, a Type I error in simple words means that the assumption "the passenger has a weapon" (null hypothesis) is <em>actually true,</em> but the airport security screener <em>incorrectly concluded it is false. </em>In other words, he assumed the passenger had no weapon and allowed the passenger to board his flight <u>when he actually did have one.</u>
<em>2. While, </em><em>a </em><em>Type II error </em><em>means that </em>the assumption "the passenger has a weapon" (null hypothesis) is <em>actually false, </em>but the airport security screener <em>incorrectly concluded it is true. </em>In other words, he assumed the passenger had a weapon and selected the passenger for further inspection <u>when he actually didn't have one.</u>
Answer:
the answer is c
Step-by-step explanation: