All equilateral triangles are isosceles? A. True B. False

Mathematics

2 answers:

mojhsa [17]3 years ago

8 0

Answer:

Yes, all equilateral triangles are also isosceles.

Step-by-step explanation:

An isosceles triangle has at least two sides of the triangle of equal measure and the two base angles are equal in degree measure. By definition all sides of an equilateral triangle are equal in length and all angles are equal in degree measure.

Using this info, I will say yes, all equilateral triangles are also isosceles. The correct answer would be A.) True.

Maru [420]3 years ago

7 0

Answer:

A. true

Step-by-step explanation:

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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: