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jasenka [17]
3 years ago
5

Write an equation parallel to x - 3y = 9 that passes through the point ( 3, -1 )

Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

Step-by-step explanation:

Use the point-slope formula.

y - y_1 = m(x - x_1)   ....   x_1 = 3 and y_1 = - 1

m is the slope to  x - 3y = 9 because (lines are  parallel )

calculate : m   by equation   x - 3y = 9 :

x = 3y +9

divid by : 3      1/3 x = y + 3

y =  (1/3) x - 3      so : m = 1/3

an equation parallel to x - 3y = 9 that passes through the point ( 3, -1 ) is :

y +1 = (1/3)(x - 3)

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Which expression can be used to approximate the expression below, for all positive numbers a, b, and x, where a Not-equals 1 and
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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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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".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Data given and notation  

n=1024 represent the random sample taken    

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

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

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