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vodka [1.7K]
3 years ago
5

Find the missing side. round to the nearest tenth

Mathematics
2 answers:
barxatty [35]3 years ago
3 0
So you would use the law of sines so set up the equation sin18/16 = sin72/x (sin 72 being the top angle) and then you cross multiply and solve from there:)
Hope this helps!
Firdavs [7]3 years ago
3 0

Answer:

6.0

Step-by-step explanation:

This is trigonometry so you use SOH CAH TOA, in this case you need TOA.

tan (18) x 16 = 5.19871514 which simplifies to 6.0

Hope this is correct and helpful

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vladimir1956 [14]
You would do 12.3/1.5= 8.2 years
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Consider the expression (x+3)(y+1)(x+2) applying the distributive property.
Naddik [55]
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If mWP= 56° and mKA= 162°, find m E.
inn [45]

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7 0
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F(n)=2n+5<br> g(n)=n²-1<br> Find f(-3) ⋅ g(-3)
kirill115 [55]

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Step-by-step explanation:

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3 years ago
For a normal distribution, is it likely that a data value selected at random is more than 2 standard deviations above the mean?
makkiz [27]

Answer:

P(X>\mu + 2\sigma)

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}

Using the z score formula we want this probability:

P(Z>2)

And using the complement rule we got:

P(Z>2) = 1-P(Z

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

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 interst of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

Where \mu the mean and \sigma  the deviation

We are interested on this probability

P(X>\mu + 2\sigma)

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}

Using the z score formula we want this probability:

P(Z>2)

And using the complement rule we got:

P(Z>2) = 1-P(Z

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

8 0
4 years ago
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