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Rus_ich [418]
3 years ago
12

Raquel is going on a vacation. She is trying to figure out how much money to bring along. She plans to spend $120 on hotels, $45

on gas for her car, and $65 on food. She is also going to go to an amusement park twice. Tickets to the amusement part cost $30 per day.
If she wants to have $30 left in her wallet after the trip is over, How much money should she bring?
Mathematics
1 answer:
algol [13]3 years ago
3 0

Answer:

$320

Step-by-step explanation:

she needs to take 320 dollars with her for the trip because she needs about 290 dollars for her expenses then she wants 30 dollars left over so she needs to take about 3320 dollars and none of that includes any tax.

hope this helps :)

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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.

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The margin of error for this case is given by:

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And replacing we got:

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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
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Look at your hundredth spot which is where the 7 is. So…

248, 739

Since 3 is not 5 or higher you cant round 7 up to 8. So your correct answer is 248,700

Hope this helps!

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