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disa [49]
3 years ago
13

Write the expression as a single logarithm

Mathematics
1 answer:
Yakvenalex [24]3 years ago
4 0

\log _a(6w+1)^{7} + \log _a(w+7)^{\frac{1}{5}} \\ \log _a(6w+1)^{7} + \log _a\sqrt[5]{(w+7)}\\\log _a((6w+1)^{7} (\sqrt[5]{(w+7)}))

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Y=4x^2-36 solve the following quadratic function by utilizing the square root method<br><br> x=?
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This ans is easily solved by using algebraic identities a²- b² = (a+b) (a-b)

Step-by-step explanation:

1. convert into Identity

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In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct a 95% conden
S_A_V [24]

Answer:

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

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}.

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead.

This means that n = 42, \pi = \frac{26}{42} = 0.619

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 - 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.472

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 + 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.766

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

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try Jiskha Homework Help

Step-by-step explanation:

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3 years ago
Number 11 and 14, I need help on please.
Ulleksa [173]
Consider this option:
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