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3 years ago

1.0•10^-10 standard form

Engineering

1 answer:

Drupady [299]3 years ago

3 0

Answer:

$1.0 * 10^{-10} = 0.0000000001$

Explanation:

Given

$1.0 * 10^{-10}$

Required

Convert to standard form

$1.0 * 10^{-10}$

From laws of indices

$a^{-x} = \frac{1}{a^x}$

So, $1.0 * 10^{-10}$ is equivalent to

$1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}$

$1.0 * 10^{-10} = 1.0 * 0.0000000001$

$1.0 * 10^{-10} = 0.0000000001$

Hence, the standard form of $1.0 * 10^{-10}$ is $0.0000000001$

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Rama09 [41]

Answer:

See explaination

Explanation:

#include<stdio.h>

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3 years ago

In manufacturing a particular set of motor shafts, only shaft diameters of between 38.00 and 37.50 mm are usable. If the process

Masteriza [31]

Answer:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

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 diameters of a population, and for this case we know the distribution for X is given by:

$X \sim N(37.8,0.12)$

Where $\mu=37.8$ and $\sigma=0.12$

We are interested on this probability

$P(37.5$

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:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

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2 years ago