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gizmo_the_mogwai [7]
3 years ago
12

At the beginning of March, a store bought a fancy watch at a cost of $250 and marked it up 20%. At the end of the month, the fan

cy watch had not sold, so the store marked it down 10%. What was the discounted price?
Mathematics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

$270

Step-by-step explanation:

Price after markup was 1.20($250) = $300

Price after discounting:  (1.00 - 0.10)($300) = $270

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<span>Let's solve your equation step-by-step.
-x^2 = -36

</span><span>Step 1: Divide both sides by -1.
</span><span>-x^2/-1 = -36/-1
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</span><span>Step 2: Take square root.
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Figure A maps to figure B with a scale factor of 2/3 -> Every single side of figure B equal to 2/3 of its corresponding side on triangle A.

So, we can find x by taking 10.5 x 2/3 = 7

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The health of the bear population in a park is monitored by periodic measurements taken from anesthetized bears. A sample of the
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Answer:

182.167-2.03\frac{114.05}{\sqrt{36}}=143.580  

182.167+2.03\frac{114.05}{\sqrt{36}}=220.754  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

Step-by-step explanation:

Assuming the following dataset:

77, 349,417,349, 167 , 225, 265, 360,205

145,335,40,139, 177,108, 163, 202, 22

123,439, 125,135, 86,43, 217,49, 156

119,178, 151, 61, 350, 312, 91, 89,89

We can calculate the sample mean with the followinf formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}= 182.167

And the sample deviation with:

s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}=114.05

The sample size on this case is n =36.

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=182.167 represent the sample mean  

\mu population mean (variable of interest)  

s=114.05 represent the sample standard deviation  

n=36 represent the sample size    

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

The point estimate of the population mean is \hat \mu = \bar X =182.167

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=36-1=35  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,35)".And we see that t_{\alpha/2}=2.03  

Now we have everything in order to replace into formula (1):  

182.167-2.03\frac{114.05}{\sqrt{36}}=143.580  

182.167+2.03\frac{114.05}{\sqrt{36}}=220.754  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

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