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son4ous [18]
3 years ago
5

Simplify the expression.

Mathematics
2 answers:
astra-53 [7]3 years ago
8 0

a^{\log_a b}=b\\\\12^{\log_{12}24}=24

Pachacha [2.7K]3 years ago
8 0

<u>Answer:</u>

The correct answer option is A) 24.

<u>Step-by-step explanation:</u>

We are given the following log expression and we are to simplify it:

1 2 ^ { log _ { 1 2 } } ^ { 2 4 }

Here, we are going to apply the rule for solving a log problem:

a^{log_a^{(b)} = b

So if 1 2 ^ { log _ { 1 2 } } ^ { 2 4 }, then it would be equal to 24.

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Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure
Marina CMI [18]

Answer:

a) P(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

b) P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< 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".  

Part a

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

X \sim N(0.12,0.009)  

Where \mu=0.12 and \sigma=0.009

We are interested on this probability

P(0.11

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(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want this probability:

P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< Z

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Remove parentheses.

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Make the denominators the same as the LCD.

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Answer:

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Step-by-step explanation:

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