<span>From the message you sent me:
when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths
If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

Why does this work? Initially, you start with 500 mL of air that you breathe in, so

. After the second breath, you have 12% of the original air left in your lungs, or

. After the third breath, you have

, and so on.
You can find the amount of original air left in your lungs after

breaths by solving for

explicitly. This isn't too hard:

and so on. The pattern is such that you arrive at

and so the amount of air remaining after

breaths is

which is a very small number close to zero.</span>
Answer:
Labrador retrievers
Step-by-step explanation:
We know that the mean
is:

and the standard deviation
is:

The probability that a randomly selected Labrador retriever weighs less than 65 pounds is:

We calculate the Z-score for X =65

So

Looking in the table for the standard normal distribution we have to:
.
Finally the amount N of Labrador retrievers that weigh less than 65 pounds is:


Labrador retrievers
Answer:
$-6
Step-by-step explanation:
$(8-14)
= $-6
Hope this helped!
Draw a line through R parallel to the segment PQ
First choice.