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1 year ago

Mrs. Sparks is a 9th-grade Algebra 1 teacher. She teaches three

Mathematics

1 answer:

soldi70 [24.7K]1 year ago

7 0

Answer:

Imagine that scientists have just discovered a non-bird dinosaur skeleton. They want to know whether the dinosaur was closely related to birds. What features in ...

Step-by-step explanation:

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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

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

Knowing that P = X x i / 100, C = main, e = Rate. Calculate the rate for the formula: A principal of R $ 105,000.00 obtained a p

Brut [27]

Answer:

Interest expressed as a percent of the principal will be referred to as an ... earned on an investment of P at the annual interest rate r for a period of t ... (c) monthly, (d) weekly, (e) daily, and (f) continuously at a nominal annual ... and substitute x = 3−2y in for x in second equation and find 4y−3(3−2y) = 1.

Step-by-step explanation: sorry if that doesn't help :(

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2 years ago

Write (-14/25)(45/49) in simplest form

allochka39001 [22]

-18/35

is the answer hope I am helpful

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2 years ago

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Bad White [126]

Answer:

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2 years ago

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What is the solution of the following? *<br> -4 = 16<br> Infinite solutions<br> no solution

rjkz [21]

Answer: no solution

Hope this helps and good luck :)

Step-by-step explanation:

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2 years ago