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yaroslaw [1]
3 years ago
14

Brianna went to the store to buy some clothes. She bought 9 pairs of pants at $20 per pair of pants, and 11 shirts that cost $15

each. Which of the following is a good estimation of how much money Brianna spent at the store?
Mathematics
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

$345 : whichever is closest

Step-by-step explanation:

Pants were $20 each and she bought 9 of them so multiply:

9x20=180

She paid $180 for 9 pants.

Shirts cost $15 each and she bought 11 of them so Mulitply:

15x11=165

She paid $165 for 11 shirts

Add both totals from pants and shirts

165+180=$345

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Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr
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Answer:

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation: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".  [tex]p_1 represent the real population proportion for 1

\hat p_1 =0.768 represent the estimated proportion for 1

n_1=92 is the sample size required for 1

p_2 represent the real population proportion for 2

\hat p_2 =0.646 represent the estimated proportion for 2

n_2=95 is the sample size required for 2

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}  

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

ME= \frac{0.2753-(-0.0313)}{2}=0.1533

And we know that the margin of erro is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}

We have all the values except the value for z_{\alpha/2}

So we can find it like this:

0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}

And solving for z_{\alpha/2} we got:

z_{\alpha/2}=2.326

And we can find the value for \alpha/2 with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98 or 98%

7 0
3 years ago
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snow_lady [41]
To do this, you first do 3^3. this is 27.
now because of the minus part, you have to do this under 1 (a reciprocal).
your answer is 1/27
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