All categories

1 year ago

Solve. 92.5÷105 brainliest

Mathematics

2 answers:

Karo-lina-s [1.5K]1 year ago

6 0

Answer:

$\frac{37}{42}$

Step-by-step explanation:

Given :

$\frac{92.5}{105}$

Multiply numerator and denominator by 2 :

$\frac{92.5}{105} \times\frac{2}{2}$

$\frac{185}{210}$ [Divide both sides by 5 as it is a common factor]

$\frac{185}{210} \div \frac{5}{5}$

$\frac{37}{42}$

Naddik [55]1 year ago

5 0

$\\ \rm\Rrightarrow 92.5\div 105$

$\\ \rm\Rrightarrow \dfrac{925}{10}\times \dfrac{1}{105}$

$\\ \rm\Rrightarrow \dfrac{925}{1050}$

$\\ \rm\Rrightarrow 0.88$

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And we can ue the z score formula given by:

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

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

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So we want to find this probability:

$P(-1.515$

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 annual income of a population, and for this case we know the following info:

$\mu=37$ and $\sigma=28$ and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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And we want to find this probability:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

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2 years ago