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Zielflug [23.3K]
3 years ago
7

What is the length of segment CD question mark

Mathematics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

CD = two square root of 10 end square root

Step-by-step explanation:

To find the length of a segment, use the distance formula. Substitute the order pairs for the endpoints of the segment. CD has the end points (-7, -4) and (-1, -2).

d = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} \\\\d = \sqrt{(-4--2)^2 + (-7--1)^2} \\\\d = \sqrt{-2^2 + -6^2}\\\\d = \sqrt{4 + 36}\\\\d=\sqrt {40} = 2\sqrt{10}

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Step-by-step explanation:

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A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
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Answer:

a) \hat p=\frac{471}{1024}=0.460

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And the margin of error is given by:

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b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

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Data given and notation  

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X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

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Answer:

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Step-by-step explanation:

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Multiplying this out, we get ...

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