Answer: 2500 years
Step-by-step explanation:
I'm not quite sure if I'm doing this right myself but I'll give it a shot.
We use this formula to find half-life but we can just plug in the numbers we know to find <em>t</em>.

We know half-life is 5730 years and that the parchment has retained 74% of its Carbon-14. For
let's just assume that there are 100 original atoms of Carbon-14 and for A(t) let's assume there are 74 Carbon-14 atoms AFTER the amount of time has passed. That way, 74% of the C-14 still remains as 74/100 is 74%. Not quite sure how to explain it but I hope you get it. <em>h</em> is the last variable we need to know and it's just the half-life, which has been given to us already, 5730 years, so now we have this.

Now, solve. First, divide by 100.

Take the log of everything

Divide the entire equation by log (0.5) and multiply the entire equation by 5730 to isolate the <em>t</em> and get

Use your calculator to solve that giant mess for <em>t </em>and you'll get that <em>t</em> is roughly 2489.128182 years. Round that to the nearest hundred years, and you'll find the hopefully correct answer is 2500 years.
Really hope that all the equations that I wrote came out good and that that's right, this is definitely the longest answer I've ever written.
Answer:
0.9953, 3.3629<x<4.0371
Step-by-step explanation:
Given that slader the internal revenue service claims it takes an average of 3.7 hours to complete a 1040 tax form, assuming th4e time to complete the form is normally distributed witha standard devait of the 30 minutes:
If X represents the time to complete then
X is N(3.7, 0.5) (we convert into uniform units in hours)
a) percent of people would you expect to complete the form in less than 5 hours
=
b) P(b<x<c) = 0.50
we find that here
c = 4.0371 and
b = 3.3629
Interval would be


Employ a standard trick used in proving the chain rule:

The limit of a product is the product of limits, i.e. we can write

The rightmost limit is an exercise in differentiating

using the definition, which you probably already know is

.
For the leftmost limit, we make a substitution

. Now, if we make a slight change to

by adding a small number

, this propagates a similar small change in

that we'll call

, so that we can set

. Then as

, we see that it's also the case that

(since we fix

). So we can write the remaining limit as

which in turn is the derivative of

, another limit you probably already know how to compute. We'd end up with

, or

.
So we find that