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poizon [28]
2 years ago
15

Solve the equation: -6p + 14 and 3p - 2

Mathematics
1 answer:
sleet_krkn [62]2 years ago
5 0

Answer:

p = 16/9

Step-by-step explanation:

-6p+14=3p-2\qquad\text{subtract 14 from both sides}\\\\-6p=3p-16\qquad\text{subtract 3p from both sides}\\\\-9p=-16\qquad\text{divide both sides by (-9)}\\\\\boxed{p=\dfrac{16}{9}\to p=1\dfrac{7}{9}}

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3 years ago
25⋅5⋅5⋅4<br><br> A.500<br><br> B.2000<br><br> C.2500<br><br> D.5000
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c

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How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide
kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

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

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

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

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

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MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

       n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}

          =[\frac{1.96\times 60}{2}]^{2}

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Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

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2 years ago
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