Use the grouping method to factor this polynomial.

Point C is twice as close to point B as it is to point A. If point A is at -30 and point B is at 30, what are two possible value

arsen [322]

different between A and B

=30-(-30)

=60

position of C from A or B

=(60÷3)×2

=40

possible value of c

=-30+40

=10

possible value of c

=30-40

=-10

8 0

3 years ago

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Which phrase best describes the graph of a proportional relationship

Setler [38]

In proportional, it is y/x=k or y=kx
basically it is like y=mx+b, it is a line

4 0

3 years ago

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Kona and doug found 9 rock crystals. kona kept 5 fewer than doug. How many crystals does kona have?

Zarrin [17]

Simple, just do 9 - 5, Kona has 4 rock crystals. are you new to math?

8 0

4 years ago

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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 $Z = -3$ .

So

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

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

$3\sigma = 0.5$

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

$\sigma = 0.167$

(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 $\sigma = 2.8$