Answer:
12 hours are required to be less than 10 milligrams of caffeine in Faye's body
Step-by-step explanation:
Geometric sequences
We can recognize a geometric progression, when we can get each member n as the previous member n-1 multiplied or divided by a constant value, called the common ratio.
Faye holds 120 milligrams of caffeine. We know each hour the amount of caffeine in her body will be 80% of the amount from the previous hour. For example, the first hour she will hold 80%(120)=96 milligrams of caffeine. Next hour it will be 80%(96)=76.8 and so on
We can see the sequence of contents of caffeine follows the rule of a geometric sequence and the common ratio is 80%, i.e. 0.8. The first term is 120, so the general term of the sequence is
[tex]a_t=120(0.8)^t\ ,\ t>=0[/tex]
We are required to find how much time is needed to be less than 10 milligrams of caffeine remaining in Faye's body. It can be stated that
[tex]120(0.8)^t<10[/tex]
Let's solve for t. Operating
[tex](0.8)^t<\frac{10}{120}[/tex]
Taking logarithms in both sides
[tex]ln(0.8)^t<ln\frac{10}{120}[/tex]
Applying the power property of logarithms
[tex]t\ ln(0.8)<ln\frac{10}{120}[/tex]
Solving for t. Since ln0.8 is negative, the direction of the inequality changes
[tex]t>\frac{ln\frac{10}{120}}{ln(0.8)}[/tex]
[tex]t>11.14[/tex]
Rounding to the next integer
12 hours are required to be less than 10 milligrams of caffeine in Faye's body
Answer: The other answer is right but for a quick answer it is in the image attached
Step-by-step explanation:
