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3 years ago

Please Help! LOOK BELOW! The first hexagon is dilated to form the second hexagon.

Mathematics

2 answers:

mote1985 [20]3 years ago

8 0

Answer:

Part 1) Option B. $4$

Part 2) Option A. An enlargement

Step-by-step explanation:

we know that

The scale factor is equal to the length side of the dilated figure divided to the length side of the original figure

$scale\ factor=\frac{0.8}{0.2} =4$

The dilation represent an enlargement because the scale factor is greater than $1$

$4> 1$ -------> represent an enlargement

Zolol [24]3 years ago

6 0

The scale factor is B. 4. To get the second shape, you multiply the sides of the first shape by 4.
This is an A. enlargement. To get the second shape, the sides of the first shape must be increased/enlarged.

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2 years ago

To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data

nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

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

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

This Zscore is how many standard deviations the value of the measure X 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.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

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

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$