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3 years ago

The vertices of a triangle are located at P(0, 0), Q(8, 6), and R(-3, 4). What is the perimeter of this triangle?

1 answer:

4 0

I graphed the vertices and got a triangle but its very hard to solve its perimeter. Therefore, I identified right triangles to help me solve for the perimeter making each side the hypotenuse of each right triangle.
Pythagorean Theorem: a² + b² = c²
a = short leg ; b = long leg ; c = hypotenuse

P(0,0) ; R(-3,4)
a = 3 ; b = 4 ;
3² + 4² = c²
9 + 16 = c²
25 = c²
√25 = √c²
5 = c

P(0,0) ; Q(8,6)
a = 6 ; b = 8
6² + 8² = c²
36 + 64 = c²
100 = c²
√100 = √c²
10 = c

Q(8,6) ; R(-3,4)
a = 2 ; b = 11
2² + 11² = c²
4 + 121 = c²
125 = c²
√125 = √c²
11.18 = c

Perimeter = 5 + 10 + 11.18 = 26.18

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Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

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