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mash [69]
4 years ago
11

A non-terminating non-repeating decimal is always rational. true or false

Mathematics
2 answers:
soldier1979 [14.2K]4 years ago
5 0

Answer:

True

Step-by-step explanation:

nevsk [136]4 years ago
4 0

Answer:

false

Step-by-step explanation:

these can not b represented as fractions therefore are not rationale

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Cube A has an edge length of 3 inches. Cube B has an edge length double that of Cube A.
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Todd Holland from NY grows pumpkins. He studied his past records carefully and concluded that his biggest pumpkins are distribut
ehidna [41]

Answer:

P(X>1000)

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>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)

And we can find this probability using the complement rule and the normal standard table and we got:

P(z>1)=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 problem

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

X \sim N(950,50)  

Where \mu=950 and \sigma=50

We are interested on this probability

P(X>1000)

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>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)

And we can find this probability using the complement rule and the normal standard table and we got:

P(z>1)=1-P(z

5 0
3 years ago
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