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vitfil [10]
3 years ago
13

Determine if the following lengths could represent a triangle. If they can, determine if the triangle, is acute, obtuse, or righ

t.
A. 5, 8, 13
B. 7, 12, 26
C. 11, 15, 20
Mathematics
1 answer:
stiks02 [169]3 years ago
3 0

9514 1404 393

Answer:

  A. no

  B. no

  C. obtuse

Step-by-step explanation:

For side lengths to form a triangle, the sum of the shorter two must exceed the longest.

A. 5 + 8 = 13 . . . . a line segment, not a triangle

B. 7 + 12 < 26 . . . . no closure, not a triangle

C. 11 + 15 > 20 . . . . a triangle. A picture shows it to be obtuse

  You can also compare 11² +15² vs 20² ⇒ 346 vs 400. The long side is too long for a right triangle, so the triangle must be obtuse. (The Pythagorean theorem tells you a right triangle with those legs would have a long side of √346 = 18.6.)

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Answer:

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X =69 represent the sample mean for the sample  

\mu population mean

s=4.0 represent the sample standard deviation

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Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The sample mean \bar X have the following distribution:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

The margin of error for the sample mean is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Since we have the sample standard deviation we can estimate the margin of error like this:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025

We can find the degrees of freedom like this:

df=n-1=64-1=63

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

t_{\alpha/2}=\pm 1.998

And the margin of error would be :

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

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