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3 years ago

Find the factors from the roots, then multiply the factors together. y=x3−3.5x2+3.5x−1

Mathematics

1 answer:

neonofarm [45]3 years ago

8 0

Answer:

hello :

y=x3−3.5x²+3.5x−1

y = (x3 - 1 ) - 3.5 ( x² - 1)

y= (x -1 ) ( x² + x + 1 ) + 3.5 ( x - 1 ) (x + 1)

y = ( x - 1 ) ( x² + x + 1 + 3.5x -3.5)

y = (x - 1 ) ( x² + 4.5x - 2.5)

the dicriminent of x² + 4.5x - 2.5 is : Δ = (4.5)²- 4(1)(- 2.5) = 30.25 = (5.5)²

x1 = ( - 4.5 - 5.5)/2 x2 = ( - 4.5 + 5.5)/2

x1 = - 5 x2 = 1/2 = 0.5

y = (x - 1 )( x +5) (x - 0.5)

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3 years ago

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3 years ago

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

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.

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The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

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 margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

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Hope this helps!

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