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GarryVolchara [31]
3 years ago
10

8 + x/2 = 12 need quick ☁

Mathematics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

8

Step-by-step explanation:

12-8=4

4 times 2 =8

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If restaurant served 517 meals at lunch. It then served 257 meals at dinner. How many meals did it serve in all
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X^2 - 9x -10 =0 has a discriminant of: a. 121 b. 72 c. -359 d. 41 e. 136
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In the standard form of quadratic
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3 years ago
The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses.
Luden [163]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 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 the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

n = 450, \pi = 0.47.

The lower and the upper bound of the interval are given, respectively, by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

5 0
1 year ago
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