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miv72 [106K]
2 years ago
14

The points P(2, 1), Q(4, 5), and R(7, 4) are the midpoints of the sides of a triangle. Graph the three midsegments to form the m

idsegment triangle. Then use your graph and the properties of midsegments to draw the original triangle.
Mathematics
1 answer:
MariettaO [177]2 years ago
7 0

Answer:

  see attached

Step-by-step explanation:

We followed instructions to create the attached graph of the two triangles.

Each mid-segment is parallel to the side of the triangle whose midpoint is the vertex opposite that segment. So, we used a graphing tool to draw, for example, a line parallel to PQ through point R. The vertices of the original triangle are at the intersection points of these lines (or 1 mid-segment length from the midpoint).

__

Analytically, we can make use of the fact that each mid-segment is effectively a diagonal of a parallelogram. For example, PQ is a diagonal of APRQ. The other diagonal is AR. The two diagonals cross at the midpoint of each. This means we can find point A using the midpoint relation:

  midpoint AR = midpoint PQ

  (A +R)/2 = (P +Q)/2 . . . use the midpoint formula

  A +R = P +Q . . . . . . . . multiply by 2

  A = P +Q -R . . . . . . . . subtract R to get an expression for A

  A = (2, 1) +(4, 5) -(7, 4) = (2+4-7, 1+5-4) = (-1, 2)

The other vertices of ΔABC can be found in similar fashion:

  B = R +P -Q = (7, 4) +(2, 1) -(4, 5) = (7+2-4, 4+1-5) = (5, 0)

  C = Q +R -P = (4, 5) +(7, 4) -(2, 1) = (4+7-2, 5+4-1) = (9, 8)

So, the original triangle's vertices are ...

  A(-1, 2), B(5, 0), C(9, 8)

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Answer:

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Step-by-step explanation:

2:1 is the same as 2/1 so all you need to do is multiply 2 to the oxygen

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Answer:

frequency I

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According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard devi
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Answer:

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is mean amount of inches of rain and \sigma is the standard deviation.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Mean \mu, standard deviation \sigma

n days:

This means that s = \frac{\sigma}{\sqrt{n}}

Applying the Central Limit Theorem to the z-score formula.

Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?

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5 0
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Raise  

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