In a national chess tournament, 40% of the players are eliminated each day. If the tournament starts with 2,000 players, to the nearest integer, how many players will be present at the start of the 5th day of play?
the common ratio here would be 1 / 1.4 I think
1 divided by 1.4 ?
what kind of equation do i use for this?
I like your username, irrod I mean times (5/7) same thing...
If 40% of the players are eliminate each day, it means 60% of the players are retained after each day. At the end of 1st day how many players will remain?
I like chess better than math (I used to make money in tournaments) no kidding.
thanks so is this like an An=A1+9n-1)d kind of equation or what?
This is a geometric progression question and not arithmetic progression. You have to find the 5th term in the GP.
okay , whats the equation for that?
A=ar^(n-1) ?
40% eliminated implies 60% (or 0.6) retained. Start of 1st day -- 2000 players Start of 2nd day --- 2000 x 0.6 Start of 3rd day --- 2000 x 0.6 x 0.6 ....
so when i list my variables does r=.6 or r=.4 ?
r = 0.6 a1 = 2000 a5 = ?
so when set up would it look like =2000(.6)^(5-1) ?
Yes.
and my answer is 259?
Yes.
I calculated 259 also
okay thanks guys
you are welcome.
@ranga you think you could help me w another ?
part 1: What are the possible number of positive, negative, and complex zeros of f(x) = –2x^3 – 5x^2 + 6x + 4 ? Part 2: Use complete sentences to explain the method used to solve this equation
Familiar with Descartes' Rule of Signs?
positives i got 1, negatives i got 2 or 0, i don't know how to find the complex ze
Correct. This has 1 positive zero and 2 or 0 negative zeros. Since this is a third degree polynomial it must have a total of 3 roots. Put all the possibilities in a table form:|dw:1396832674948:dw|
Join our real-time social learning platform and learn together with your friends!