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3 years ago

(x+3)(x−2)/(x+7)^2 ≥0

Mathematics

1 answer:

LiRa [457]3 years ago

6 0

Answer:

(x+3)(x−2)/(x+7)^2 ≥0 when x ≤ -3 or x ≥ 2.

Step-by-step explanation:

(x+3)(x−2)/(x+7)^2 ≥0

The critical values are (x + 3)(x - 2) = 0.

That is when x = -3 and x = 2.

Set up a table:

x < - 3 x = -3 -3 < x < 2 x = 2 x > 2

x + 3 -ve 0 +ve 0 +ve

x - 2 -Ve -ve -ve 0 +ve

(x+7)^2 +ve +ve +ve +ve +ve

Whole

function + ve 0 - ve 0 +ve

So the function is ≥ 0 when x ≤ -3 and x ≥ 2.

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b) There is a 41.07% probability that a person who walks into the store will buy something.

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Step-by-step explanation:

This a probability problem.

The probability formula is given by:

$P = \frac{D}{T}$

In which P is the probability, D is the number of desired outcomes and T is the number of total outcomes.

The problem states that:

123 people walked by the store.

56 people came into the store.

23 bought something in the store.

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$P = \frac{D}{T} = \frac{56}{123} = 0.4553$

There is a 45.53% probability that a person who walks by the store will enter the store.

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$P = \frac{D}{T} = \frac{23}{56} = 0.4107$

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There is a 18.70% probability that a person who walks by the store will come in and buy something.

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$P = \frac{D}{T} = \frac{33}{56} = 0.5893$

There is a 58.93% probability that a person who comes into the store will buy nothing.

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