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Tema [17]
3 years ago
14

5-6(a+2)=7+a What is a

Mathematics
2 answers:
iogann1982 [59]3 years ago
5 0
Remmeber you can do anything to an equaiton as long as you do it to btoh sides
5-6(a+2)=7+a
distribute
5-6a-12=7+a
add like terms
-6a-7=7+a
add 6a both sides
-7=7+7a
minus 7 both sides
-14=7a
divide by 7
-2=a
a=-2
morpeh [17]3 years ago
4 0
The first step here would be to distribute the parenthesis to the -6.

5 - 6a - 12 = 7 + a

Next, combine all like terms. This means that the constants will combine, thus making the next step easier.

- 6a - 7 = 7 + a

Now, subtract the negative seven from the other side of the parenthesis. When I say add the negative, I mean that the sign (-) will flip, turning into a positive.

- 6a = 14 + a

Now do the same with the a, except this time the a will be subtracted from the other side due to it's positive sign being flipped into a negative.

- 7a = 14

Now you divide both sides by - 7. Almost done!!!

- 7a / - 7 = 14 / -7

Finally the equation simplifies to:

a= -2
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Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

<h3>z-distribution interval:</h3>

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

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

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For this problem:

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  • 33% was mislabeled or misidentified, hence p = 0.33.
  • 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

<h3>The lower limit of this interval is:</h3>

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036

<h3>The upper limit of this interval is:</h3>

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047

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