Question

Air pressure decreases exponentially with increases in elevation. The air pressure, y, (in atmosphere units) at a given elevation, x, (in meters) can be modeled using equation y equals e Superscript negative k x Baseline, where k is the decay constant.



At an elevation of 5486 meters where the air pressure is 0.5 atmospheres, what is the value of k?

Answer options with 4 options
A.
k equals negative StartFraction Natural Log left-parenthesis 5486 right-parenthesis EndLog over 0.5 EndFraction

B.
k equals negative StartFraction Natural Log left-parenthesis 0.5 right-parenthesis EndLog over 5486 EndFraction

C.
k equals negative Natural Log left-parenthesis StartFraction 5486 over 0.5 EndFraction right-parenthesis EndLog

D.
k equals negative Natural Log left-parenthesis StartFraction 0.5 over 5486 EndFraction right-parenthesis EndLog

Answer 1

u sure

Answer 2

The correct answer is C.

To find the value of k, we can rearrange the equation:
y = e^(-kx)
0.5 = e^(-k*5486)

Next, we take the natural logarithm of both sides to solve for k:
ln(0.5) = ln(e^(-k*5486))
ln(0.5) = -k*5486

Now we solve for k by dividing both sides by -5486:
k = -ln(0.5) / 5486

So the answer is k = -ln(0.5) / 5486, which is equivalent to option C.

Answer 3

Apologies for the previous error. The correct answer is B.

To find the value of k, we can rearrange the equation:
y = e^(-kx)
0.5 = e^(-k*5486)

Next, we take the natural logarithm of both sides to solve for k:
ln(0.5) = ln(e^(-k*5486))
ln(0.5) = -k*5486

Now we solve for k by dividing both sides by -5486:
k = -ln(0.5) / 5486

So the answer is k = -ln(0.5) / 5486, which is equivalent to option B.