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2 years ago

What are the solutions of 2x2 + 16x + 34 = 0?

2 answers:

8 0

2x^2 + 16x + 34 = 0
x^2 + 8x + 17 = 0
x^2 + 8x + 16 = -17 + 16 = -1
(x + 4)^2 = -1
x + 4 = sqrt(-1) = ±i
x = -4 <span>±</span> i

Scorpion4ik [409]2 years ago

3 0

Answer:

$\text{The solution is }x=-4\pm i$

Step-by-step explanation:

Given the equation

$2x^2+16x+34=0$

we have to find the solutions of above equation

$2x^2+16x+34=0$

Comparing it with standard quadratic equation $ax^2+bx+c=0$

a=2, b=16, c=34

The solution is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-16\pm \sqrt{16^2-4(2)(34)}}{2(2)}$

$x=\frac{-16\pm \sqrt{-16}}{4}$

$x=-4\pm i$

$\text{The solution is }x=-4\pm i$

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

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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.

$z_{\alpha/2}=1.96$

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).

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).

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2 years ago

100 points if you show how you got the answer!

ludmilkaskok [199]

First, simplify each one.

9.98 x 10^6 = 9980000

7.3 x 10^7 = 73000000

Next, subtract the freight from the aircraft

73000000 - 9980000 = 63020000

Round the decimal point to the first significant digit, and place the amount of place values the decimal point moved to the left as a power sign, over 10.

63020000 = 6.302 x 10^7

6.302 x 10^7 is your answer

hope this helps

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