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3 years ago

Help please, it’s for average rate of change.

Mathematics

1 answer:

Nana76 [90]3 years ago

8 0

Answer in fraction form = -3/2

Answer in decimal form = -1.5

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Work Shown:

$m = \frac{g(b)-g(a)}{b-a}\\\\m = \frac{g(1)-g(-3)}{1-(-3)}\\\\m = \frac{0-6}{1-(-3)}\\\\m = \frac{0-6}{1+3}\\\\m = \frac{-6}{4}\\\\m = -\frac{3}{2}\\\\m = -1.5\\\\$

Effectively, we're finding the slope through the points (-3, 6) and (1, 0)

The g(b)-g(a) represents how y changes when going from x = a to x = b.

Read more about probability at:

brainly.com/question/11234923

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7 0

2 years ago

Efficiency is a long term measure of how well organization uses it's resources true or false?

Digiron [165]

Answer: That is true , it could mean alot of things good and bad , but mostly good. How efficient is the company on disposing waste.How efficient is the company good on firing someone after they made a mistake. Sad but true

Step-by-step explanation:

3 0

3 years ago

50 points!!! help fast! 50 points!!!

Gekata [30.6K]

Answer:

What's d question It all letters

6 0

2 years ago

Read 2 more answers

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

3 years ago

Write the point-slope form of the equation of the line through the given point with the given slope. (-5,0) slope= 2/3

amid [387]

Answer:

Step-by-step explanation:

y - 0 = 2/3(x + 5)

y = 2/3x + 10/3

8 0

3 years ago