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elena55 [62]
2 years ago
9

Solve 4 + 3n ≥ 1 for n.

Mathematics
1 answer:
ludmilkaskok [199]2 years ago
4 0

Hi,

4 + 3n ≥ 1

3n ≥ - 3

n ≥ -3/3

n ≥ - 1

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The maximum value of that range would be simply  μ + 2s, where μ is the mean and s the standard deviation. In the same way, the minimum value would be μ - 2s:

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