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wlad13 [49]
3 years ago
14

- A square rug has an area of 49x^2-56x+16.

Mathematics
1 answer:
CaHeK987 [17]3 years ago
8 0

(49x^2 - 56x + 16) - (16x^2 + 24x + 9)

49x^2 - 56x + 16 - 16x^2 - 24x - 9

33x^2 - 80x + 7

You could simply subtract the numbers as is, which I've shown above. Or, you may be able to factor first, then subtract.

Hope this helps!! :)

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An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : S
ziro4ka [17]

Answer:

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

Sample of 421 new car buyers, 75 preferred foreign cars. So n = 421, \pi = \frac{75}{421} = 0.178

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.151

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.205

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

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6383 to the nearest hundred<br>​
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Answer:

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Step-by-step explanation:

its 6400 to the nearest hundred

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