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3 years ago

PLEASE PLEASE HELP!!

Mathematics

2 answers:

Sloan [31]3 years ago

3 0

Answer:

Step-by-step explanation:

11)

A= (B x H)/2

A= (2.9x3.2)/2= 4.64cm²

12)

A= (B+b)H /2

A= (12+10)5.6 /2 = 61.6in

MatroZZZ [7]3 years ago

3 0

Answer:

I think that the answer might be 9.30 for the first one and the second one I think is 670.

Step-by-step explanation:

To find the area, you have to multiply the height and width for the two figures.

3.2 × 2.9 = 9.28

12 × 10 × 5.6 = 672

I think you round those to the nearest tenth. 9.28 would be 9.30 and 672 would be 670. I hope this answers your question.

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Solve for x -12+4=100

strojnjashka [21]

Answer by JKismyhusbandbae:

$x-12+4=100\\\mathrm{Add/Subtract\:the\:numbers:}\:-12+4=-8\\x-8=100\\\mathrm{Add\:}8\mathrm{\:to\:both\:sides}\\x-8+8=100+8\\\mathrm{Simplify}\\x=108$

3 0

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.

$z_{\alpha/2}=1.96$

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).

5 0

2 years ago

-10k+1=40-7k<br><br><br> Please show work