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alexandr402 [8]
2 years ago
14

P(x)=3x^2+12x+7 find coordinates of vertex and intercepts

Mathematics
1 answer:
kotegsom [21]2 years ago
8 0
Te x intercepts are where the graph crosses the x axis or wher y=0
the y intercept is where the graph crosses the y axis or where x=0

to find vertex, here is a hack
the x coordinate of the vertex for an equation in form ax^b+bx+c=y is -b/2a

so
y=3x^2+12x+7
-b/2a=-12/2(3)=-12/6=-2
sub that back
y=3(-2)^2+12(-2)+7
y=3(3)-24+7
y=9-17
y=-8

vertex is (-2,-8)

intercepts
x intercept is where y=0

0=3x^2+12x+7
using quadratic formula
x=(-6-√15)/3 and (-6+√15)/3
xints at ((-6-√15)/3,0) and ((-6+√15)/3,0)

yint is where x=0
set x=0
y=7
yint is at y=7 or (0,7)


vertex at (-2,-8)
xints at x=\frac{-6+ \sqrt{15} }{3} and \frac{-6- \sqrt{15} }{3} or at the points ( \frac{-6+ \sqrt{15} }{3} , 0) and (\frac{-6- \sqrt{15} }{3},0)
yint at y=7 or at (7,0)
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The cost of the sheets is less for the second example, the second example is the better deal.

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Here, A brand of sheets is on sale for “Buy one get one at 50% off”. If each sheet costs $18.00.

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And since the other sheet is at 50% off .

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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

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

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".  

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

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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)

8 0
2 years ago
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