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Dovator [93]
2 years ago
10

How do i do this?????????

Mathematics
1 answer:
Gnom [1K]2 years ago
8 0

Answer:

x≥5

Step-by-step explanation:

-1 ≥ x/-5

x/-5 ≤ -1

x(-1)/5 ≥ (-1) (-1)

x/5 ≥ 1

5x/5 ≥ 1 ×5

x ≥ 5

I'm not sure but I tried.

For the graph: you can draw a line about 15cm, put 0 at the start then after 5cm put 5 then 10 at 10cm and 15 at the end. At 5 put • that's the starting and go all the way to 15 and add > to show that the line continues. I'm not good at explaining but I hope it helps.

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There were 23 students running in a race. How many different arrangements of first, second, and third place are possible?
Alex787 [66]

Answer:

10626 different arrangements

Step-by-step explanation:

As we have a total of 23 students and we want to form groups of 3, where the order of the 3 students matters, we can solve this problem using a permutation of 23 choose 3.

The formula for a permutation of n choose p is:

P(n,p) = n! / (n-p)!

So, using n = 23 and p = 3, we have:

P(23,3) = 23! / (23-3)! = 23! / 20! = 23 * 22 * 21 = 10626

So we have a total of 10626 different arrangements of first, second and third place.

7 0
3 years ago
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Answer:

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