Seven members of the track team run the mile. Their mile times are all faster than 7 minutes but not faster than 6 minutes. Can you conclude that there are two runners whose times are less than nine seconds apart

There are at least two runners whose times are less than 9 seconds apart.

Step-by-step explanation:

Let's assume that Tₙ is the time of the n-th runner, we know that:

6 min < Tₙ < 7 min

knowing that:

1 min = 60 s

We can rewrite this as:

6*60 s < Tₙ < 7*60 s

360 s < Tₙ < 420 s

We know that there are 7 runners, and we want to see if we can conclude that there are two runners whose times are less than nine seconds.

So, the smallest time allowed in seconds is 361 seconds (the first value larger than 360 seg) while the largest time allowed is 419 seconds (the largest time allowed smallest than 420 seconds).

Now, let's assume that the first runner has the smallest time:

then:

T₁ = 361 s

Now let's add 9 seconds to the time of each runner (here we want to check that we can have all the runners with exactly 9 seconds apart in their times, so we will prove that the statement is false), then:

T₂ = 361s + 9s = 370s

T₃ = 370s + 9s = 379s

T₄ = 379s + 9s = 388s

T₄ = 388s + 9s = 397s

T₅ = 397s + 9s = 406s

T₆ = 406s + 9s = 415s

T₇ =  415s + 9s = 424s

But 424s > 420s

So this is not allowed (as the maximum time allowed was 419 s), so at least two of the runners must have times that are less than 9 seconds apart.

Then;  Can you conclude that there are two runners whose times are less than nine seconds apart? Yes.

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

⊂⊃⊂⊃ω

step-by-step explanation:

i hope this : )