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Gennadij [26K]
3 years ago
9

Determine the zero of F(x)=x^2+3x-5

Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

The answer to your question is:

Step-by-step explanation:

                                             x² + 3x - 5

                                    x² + 3x + (3/2)² = 5 + (3/2)²

                                         (x + 3/2)² = 5 + 9/4

                                         (x + 3/2)² = (20 + 9) /4

                                          (x + 3/2)² = 29/4

                                          x + 3/2 = ±√29 / 2

                                     

x 1 = -3/2 + √29/2                               x2 = -3/2 - √29/2

x1 = 1.19                                                x2 = -4.19

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The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 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}

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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
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Alexxandr [17]

|a| -b\\\\|a|>b\iff a>b\ or\ a

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4 0
4 years ago
An object is dropped from 43 feet below the tip of the pinnacle atop a 527​-ft tall building. The height h of the object after t
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Answer:

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Step-by-step explanation:

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marta [7]

Answer:

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nikdorinn [45]
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