Question

Bruce added some chlorine to the water in a pool. The chlorine evaporated at a fixed rate every week. The table below shows the amount of chlorine f(n), in ounces, that was left in the pool after n weeks:

Which function best shows the relationship between n and f(n)?
f(n) = 108(0.5)^n-1
f(n) = 54(0.5)^n
f(n) = 108(0.5)^n+1
f(n) = 54(0.5)^n-1

Answer 1

Is this f(n) = 108(0.5)^n-1 ---> f(n) = 108(0.5)^(n-1)

and NOT f(n) = 108(0.5)^n - 1

Answer 2

the n-1 isn't in parenthesis, if that's what you mean ?

Answer 3

I'm thinking that .5 is raised to the (n - 1) power.

Answer 4

In your book, the n-1 is probably a superscript.

Answer 5

Let's look at this:

f(n) = 54(0.5)^n-1

f(1) = 54(0.5)^(1-1)
= 54 * (.5)^0 = 54 * 1 = 54 so that checks.

Answer 6

oh, okay.

Answer 7

For n = 2,

f(2) = 54(0.5)^(2-1) --->You crank this one out, okay?

Answer 8

And,I'll do the next after that. We'll alternate and share the load.

Answer 9

uh, wait, how do i do it ?

Answer 10

i get that n = 2 so it's f(2) = 54(0.5)^(2-1) but i'm not sure how to solve it .... ?

Answer 11

f(2) = 54(0.5)^(2-1) = 54* (.5)^!

You are multiplying 54 times whatever .5 to the first power is.

Answer 12

What is 54 times .5 or 54 time one-half?

.5 to the first power is just .5

Answer 13

that's 27, right ?

Answer 14

Correct. And, that is what is in the table for n = 2.

Answer 15

n=3

f(3) = 54 * (0.5)^ (3-1)

= 54 * (.5) ^ 2

= 54 * .25

= 13.5

Answer 16

f(4) = 54 * (0.5) ^ (4-1)
= 54 * (0.5) ^ 3
= 54 * 0.125
= 6.75

??

Answer 17

Correct. And, that is the last option, I think.

Answer 18

thank you for your help !! c:

Answer 19

You are welcome. I enjoyed working with you.