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valina [46]
1 year ago
15

Triangle BAC was rotated 90° clockwise and dilated at a scale factor of 2 from the origin to create triangle XYZ. Based on these

transformations, which statement is true?
coordinate plane with triangles ACB and XYZ with A at negative 4 comma 4, C at negative 1 comma 3, B at negative 4 comma 0, X at 0 comma 8, Y at 8, comma 8, and Z at 6 comma 2


∠C ≅ ∠X

∠C ≅ ∠Y

∠A ≅ ∠Y

∠A ≅ ∠X

Mathematics
2 answers:
ANEK [815]1 year ago
8 0

Answer:

The correct option is;

∠A ≅ ∠X

Step-by-step explanation:

The given coordinates of the points of triangle ACB are;

A(-4, 4), C(-1, 3), B(-4, 0)

The given coordinates of the points of triangle XYZ are;

X(0, 8), Y(8, 8), Z(6, 2), therefore, we have

The length. l. of segment is given by the following formula;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

For the length of the segment AC; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-1, 3), l = √(10)

For the length of the segment AB; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-4, 0), l = 4

For the length of the segment BC; (x₁, y₁) = (-4, 0), (x₂, y₂) = (-1, 3), l = 3·√2

For the length of the segment XY; (x₁, y₁) = (0, 8), (x₂, y₂) = (8, 8), l = 8

For the length of the segment XZ; (x₁, y₁) = (0, 8), (x₂, y₂) = (6, 2), l = 6·√2

For the length of the segment ZY; (x₁, y₁) = (6, 2), (x₂, y₂) = (8, 8), l = 2·√(10

Therefore;

XY ~ AB, XZ ~ BC, ZY ~ AC

Which gives;

∠A ≅ ∠X, ∠B ≅ ∠Y, ∠C ≅ ∠Z

kompoz [17]1 year ago
7 0

Answer: D) ∠A ≅ ∠X is the true statement.

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For each resident, there is only two possible outcomes. Either they can name the news anchor on the evening news at their station, or they cannot.

This means that the binomial probability distribution will be used in our solution.

However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

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