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2 years ago

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Sydney and Tom each count the number of steps it takes for them to walk to school. They each count a 4 digit number of steps. To

tal number of steps is also 4 digit. What is the greatest possible digit in the thousands place for Sydney's or Tom's steps?

1 answer:

lesantik [10]2 years ago

7 0

Answer:

8

Step-by-step explanation:

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I need 1 answer for 99 points

Nataly [62]

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$7.45

Step-by-step explanation:

If it's constant , divide the amount he earns by the amount of hours. It should be the same every time.

44.70 divided by 6=7.45

and so on

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2 years ago

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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).

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2 years ago

A sailboat travels a distance of 2 1/2 miles in 1/6 of an hour. Which complex fraction represents the unit rate in miles per hou

Nikitich [7]

At first glance, prolly c

3 0

2 years ago

5.69÷7 you can round to nearest hundredth

pogonyaev

The answer would be .81 becuase you can't round the 1 up to 2 becuase the number in the thousandths place is not above 5.
Hope this helps!!

5 0

2 years ago