Answer:
Lori will have [tex]1[/tex] cavity if she brushed for [tex]120\ sec[/tex] each night.
Step-by-step explanation:
As the question suggests cavities developed varies inversely with the number of seconds spent on brushing.
Lets take the terms as [tex]c=\frac{k}{t}[/tex],where [tex]k[/tex] is the constant of proportionality.
And
c = Number of cavities developed each year
t = Number of minutes spent on brushing per night
Now we will plugging the values of the given terms and find the value of [tex]k[/tex],for this we can re-arrange it as...[tex]k=c\times( t)[/tex]
[tex]t=30\ sec=\frac{30}{60} =0.5\ min[/tex] and [tex]c=4[/tex]
[tex]k=0.5\times 4=2[/tex]
Now as Lori will increase her brushing time so as the cavity will decrease.
So for this we have been given,[tex]t=120\ sec=\frac{120}{60} = 2\ min[/tex] and [tex]k=2[/tex].
As [tex]c=\frac{k}{t} =\frac{2}{2} =1[/tex]
So as per Paul's observation if Lori brushed for [tex]2\ mins[/tex] at night she will have [tex]1[/tex] cavity on that year.
Answer:
c=120/t Loris expected cavities: 1 cavity
Step-by-step explanation:
We are given that c, the number of cavities Lori will have, varies inversely with amount of time, t, she spent brushing, so c=k/t for some constant, k. To determine the value of k, substitute the known values, c=4 when t=30, to find that
4=k/30
Multiplying by 30 gives k=120, so an equation that relates c and t is
c=120/t
Substituting t=120 gives
120/120=1
So Paul would expect Lori to have 1 cavity if she had brushed her teeth for 120 seconds each night.
