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Kipish [7]
3 years ago
10

On Monday, Elsbeth’s bank account balance was -$16.75. On Tuesday, she deposited a check for $5.72. On Thursday, she deposited $

16.75. What is her new balance?
-16.75 + (5.72 + 16.75)

Use the commutative property to get:

-16.75 + (16.75 + 5.72)

Use the associative property to get:

(-16.75 + 16.75) + 5.72

On Friday, Elsbeth deposited her pay check for $872.56. What was the account’s final total?
Mathematics
2 answers:
ZanzabumX [31]3 years ago
7 0
The answer is 22.47
erastova [34]3 years ago
3 0

Answer:

$5.72

Step-by-step explanation:

The easiest way to do this is take the two deposits and add them together then combine with the original balance.

Add 5.72 and 16.75

5.72+16.75=22.47

Then subtract -16.75 from 22.47

22.47-16.75=5.72

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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
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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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Step-by-step explanation:

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What makes a square a square <br> why isn't a rectangle?
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