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2 years ago

Find the solution to Y=-x^2+3 for X=-3,0, and 3

Mathematics

2 answers:

Alecsey [184]2 years ago

8 0

For this case we must evaluate the following quadratic equation, $y = -x ^ 2 + 3,$ for $x = -3$ , $x = 0$ and $x = 3$

For $x = -3$ :

$y = - (- 3) ^ 2 + 3\\y = -9 + 3 =\\y = -6$

For $x = 0$ :

$y = - (0) ^ 2 + 3\\y = 0 + 3\\y = 3$

For $x = 3$ :

$y = - (3) ^ 2 + 3\\y = -9 + 3\\y = -6$

Answer:

$(-3, -6)\\(0,3)\\(3, -6)$

BabaBlast [244]2 years ago

7 0

Answer:

3-,3,0,3

Step-by-step explanation:

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Answer:

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 are 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 $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63$

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 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

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

$Z = \frac{120.1 - 119}{1.63}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 117.9

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

$Z = \frac{117.9 - 119}{1.63}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

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

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Answer:

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