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3 years ago

What is 492.3069 rounded to nearest thousandth

Mathematics

2 answers:

WARRIOR [948]3 years ago

7 0

492.3069 rounded to the nearest thousandth is 492.307 :-)

Leokris [45]3 years ago

6 0

Answer:

492.307

Step-by-step explanation:

492.3069 rounded to nearest thousandth

To round to nearest thousandth , we look at the number in ten thousandth place.

3 is in tenths place

0 is in hundredth place

6 is in thousandth place

9 is in ten thousandth place

9 in ten thousandth place is greater than or equal to 5. So, we add 1 in thousandth place.

6 is in thousandth place. 6+1 is 7

So the given number is rounded to 492.307

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Answer:

$x = \frac{4}{9}$

Step-by-step explanation:

Given

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Required

Represent as a fraction

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To do this, we simply subtract 1 from the denominator

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3 years ago

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Answer:

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Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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:

And we can find this probability using the complement rule:

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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 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:

$X \sim N(114.8,13.1)$

Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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: $P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140- 1114.8}{2.6})=P(z>1.924)$ And we can find this probability using the complement rule:

$P(z>1.924)=1-P(z$

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4 years ago