For what real numbers x is the value of x2 – 6x + 9 negative

2 answers:

7 0

This is the concept of quadratic equations. The equation has the original form of y=ax^2+bx+c. Our equation is given by y=x^2-6x+9. Since our leading coefficient is positive, then it means that our function will not be negative at any point. In other words, the range is [0,∞). Hence we conclude that the function will be no where negative.

Nutka1998 [239]3 years ago

3 0

A real numbers are numbers that can be found on a number line, they include both rational and irrational numbers.
We can first solve the equation;
x² - 6x + 9 =0, using completing square method;
x²- 6x + (-3)² = -9 + (-3)²
(x-3)² = 0
x = 3
therefore, both values of x are 3, thus the expression has no real number x is negative.

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As a practical equation, this one doesn't make much sense -- why would the profit per person have a term proportional to the number of people? Let's just go with it.

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Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

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This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .