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3 years ago

Someone please help me. How do you find y

Mathematics

1 answer:

Schach [20]3 years ago

6 0

X and Y are vertical angles, which mean they are the same.

The 3 inside angles need to equal 180 degrees, so solve for X:

X = 180 - 35 - 65 = 80 degrees.

Y = 80 degrees.

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The sum of 2 complementary angles equals 90 degrees.

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3 years ago

What is the solution to -2(8x-4) 2x+5?

-Dominant- [34]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A woman is randomly selected from the 18–24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

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: