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3 years ago

****-2x^2(-5x^2+4x^3) in standard form?

2 answers:

nataly862011 [7]3 years ago

8 0

$-2 x^{2} (-5 x^{2} +4 x^{3} )$ Use distributive property

$-2 x^{2} (-5 x^{2}) +-2 x^{2} (4 x^{3} )$

$10 x^{2+2} +-8 x^{2+3}$

$10 x^{4}+ -8 x^{5}$

$-8 x^{5}+ 10 x^{4}$ <-answer

Lyrx [107]3 years ago

3 0

-2x^2(-5x^2+4x^3)=10x^4-8x^5

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3 years ago

Quadrilateral ABCD quadrilateral WXYZ. If AD = 12, DC = 6, and ZY = 36, find WZ.

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2 years ago

Read 2 more answers

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points.

Artyom0805 [142]

Answer:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)$

And we can find this probability on this way:

$P(Z>-0.75)=1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problm

Let X the random variable that represent the scores on an exam of a population, and for this case we know the distribution for X is given by:

$X \sim N(75,8)$

Where $\mu=75$ and $\sigma=8$

We are interested on this probability

$P(X>69)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)$

And we can find this probability on this way:

$P(Z>-0.75)=1-P(Z$

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2 years ago

Evaluate the expression when x = 24 and p = -4.

Stels [109]

Answer:

The correct answer is 3. -88

Step-by-step explanation:

24 / 3 + 24 x -4

24 x -4 = -96

24 / 3 = 8

8 + -96 = -88

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2 years ago

In statistics, what term means "a numerical range around a sample mean accompanied with a statement of how confident one is that

Alex Ar [27]

Answer:

Confidence Interval for the mean

Step-by-step explanation:

Confidence interval is made using the observations of a sample of data obtained from a population, so it is constructed in such a way, that, with a certain level of confidence (this is the statement mentioned in the question), that is, one could have a percentage of probability that the interval, or range around the value obtained, frequently 95%, contains the true value of a population parameter (in this case, the population mean).

It is one way to extract information from a population using a sample of it. This kind of information is what inference statistic is always looking for.

An approximation about how to construct this interval or range:

Select a random sample.
For the specific case of a mean, you need to calculate the mean of the sample (sample mean), and, if standard deviation is unknown or not mentioned, also calculate the sample standard deviation.
With this information, and acknowledged that these values follows a standard normal distribution (a normal distribution with mean 0 and a standard deviation of 1), represented by random variable Z, one can use all this information to calculate a confidence interval for the mean, with a certain confidence previously choosen (for example, 95%), that the population mean must be in this interval or range around this sample mean.

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2 years ago