Question

At a chess tournament, the number of competitors in each round is 50% of the number of competitors in the previous round. What type of relationship most appropriately models this situation?

Answer 1

Exponential Decay

Step-by-step explanation:

Answer 2

solution: It is a geometric progression. and most appropriate relationship is x×
`((1)/(2) )^(n-1)`.

Step-by-step explanation:

let the number of competitors in first round be x.

number of competitors in 2nd round will be
`(x)/(2)`.

number of competitors in 3rd round will be
`(((x)/(2) ))/(2)` =
`(x)/(4)`

similarly, number of competitors in next round will be
`(((x)/(4) ))/(2)` =
`(x)/(8)`

and so on.

so number of competitors in various rounds are forming a sequence

i.e.,
`x, (x)/(2) ,(x)/(4) ,(x)/(8) ,(x)/(16) ,...`

now ratio of second and first term =
`(((x)/(2) ))/(x)` =
`(1)/(2)`

similarly, ratio of third and fourth term =
`(((x)/(4) ))/((x)/(2))` =
`(1)/(2)`

and so on.

so, it is forming a geometric progression .

where the first term i.e.,a = x

and common ratio i.e., r =
`(1)/(2)`.

so here, most appropriate relationship is
`ar^(n-1)`

i.e., x×
`((1)/(2) )^(n-1)`.