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2 years ago

I need helppp help meeeeeeee

Mathematics

1 answer:

Rama09 [41]2 years ago

8 0

1) $\frac{11-13}{4-7}= \frac{2}{3}$

2) $\frac{-2-7}{-10-(-7)}=\frac{-9}{-3} =3$

3) $\frac{9-3}{-11-(-14)} =\frac{6}{3}=2$

4) $\frac{8-4}{11-(-9)} =\frac{4}{20}= \frac{1}{5}$

ok done. Thank to me :>

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Naddik [55]

Answer:

$X \sim N(114.8,13.1)$

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:

And we can find this probability using the complement rule:

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

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 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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3 years ago

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TiliK225 [7]

Answer:

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Fraction = 4.1666666

Percentage = 416%

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2 years ago

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