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Nostrana [21]
2 years ago
9

What are the zeros of the quadratic function f(x)=8^2-16-15

Mathematics
1 answer:
Free_Kalibri [48]2 years ago
6 0

Answer:

C)

Step-by-step explanation:

f(x)=8x^{2} -16x-15
8x^{2} -16x-15=0
x=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}  ,  a=8, b= - 16, c= - 15

x=\frac{-(-16)+-\sqrt{(-16)^2-4*8*(-15)} }{2*8} = \frac{16+-\sqrt{736} }{16} =\frac{16+-4\sqrt{46} }{16} = 1 +- \frac{\sqrt{46} }{4} = 1+- \frac{\sqrt{23} \sqrt{2}}{2\sqrt{2} \sqrt{2} } =1+-\frac{\sqrt{23} }{2\sqrt{2} } = 1+-\frac{\sqrt{23} }{\sqrt{8} }

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g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib
algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125

This means that n = 125, s = \frac{7320}{\sqrt{125}}

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{39000 - 39725}{\frac{7320}{\sqrt{125}}}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

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