△ABC is the image of △XYZ after a sequence of rigid transformations.

Mathematics

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

Answer:D.

Step-by-step explanation:I took the test

Korvikt [17]3 years ago

3 0

Answer:

D. ∠B≅∠Y, ∠A≅∠X, ∠C≅∠Z

Step-by-step explanation:

Corresponding angles are in the same order in both triangle names. That is, the corresponding angles are ...

(A, X), (B, Y), (C, Z)

Rigid transformations do not change any angles, so the angles of the image are the same as the corresponding angles of the original.

∠A≅∠X, ∠B≅∠Y, ∠C≅∠Z . . . . . matches selection D

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In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1

Serga [27]

Answer:

The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

1999:

20 out of 100 in the bottom third, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

2001:

10 out of 100 in the bottom third, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Test if proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

At the null hypothesis, we test if the proportion is still the same, that is, the subtraction of the proportions in 1999 and 2001 is 0, so:

$H_0: p_1 - p_2 = 0$

At the alternative hypothesis, we test if the proportion has been reduced, that is, the subtraction of the proportion in 1999 by the proportion in 2001 is positive. So:

$H_1: p_1 - p_2 > 0$