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3 years ago

Please see the question and solve plz

Mathematics

1 answer:

Korolek [52]3 years ago

7 0

Answer:

Step-by-step explanation:

sin²β + sin²β×tan²β = tan²β

sin²β( 1 + tan²β ) = tan²β

~~~~~~~~~~~~~~~~

sin²β + cos²β = 1

$\frac{sin^2\beta }{cos^2\beta }$ + $\frac{cos^2\beta }{cos^2\beta }$ = $\frac{1}{cos^2\beta }$ ⇒ tan²β + 1 = sec²β ⇔ 1 + tan²β = sec²β

~~~~~~~~~~~~~~

1 + tan²β = $\frac{1}{cos^2 \beta }$

L.H. = sin²β ( $\frac{1}{cos^2 \beta }$ ) = tan²β

R.H. = tan²β

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$X \sim N(114.8,13.1)$

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:

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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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noname [10]

Answer:

Step-by-step explanation:

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Please see the question and solve plz​

Please see the question and solve plz