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jok3333 [9.3K]
3 years ago
12

I need to know how to solve this question

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
4 0
9 + v = 0
2v + 5 = 0

v = -9
v = -5/2

The final solutions are
v1 = -9 and v2 = -5/2
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PLZ HELP!<br><br> Complete this proof. <br> Given: a | | b <br> Prove: ΔMOP ~ ΔRON
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According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S. cities, Washington, D.C., and New
tamaranim1 [39]

Answer:

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

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The (1 - <em>α</em>) % confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}

Here the population standard deviation (σ) is not provided. So the confidence interval would be computed using the <em>t</em>-distribution.

The (1 - <em>α</em>) % confidence interval for population mean (<em>μ</em>) using the <em>t</em>-distribution is:

CI=\bar x\pm t_{\alpha /2,(n-1)}\frac{s}{\sqrt{n}}

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\bar x=27.97\\s=10.04\\n=25\\t_{\alpha /2, (n-1)}=t_{0.01/2, (25-1)}=t_{0.005, 24}=2.797

*Use the <em>t</em>-table for the critical value.

Compute the 99% confidence interval as follows:

CI=27.97\pm 2.797\times\frac{10.04}{\sqrt{25}}\\=27.97\pm5.616\\=(22.354, 33.586)\\\approx(22.35, 33.59)

Thus, the 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

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