2 years ago

Expand: In (4a)^3 help please

1 answer:

4 0

Answer: 3ln4 + 3lna

Step-by-step explanation:

$\ln (4a)^{3}\\=3 \ln 4a\\=3(\ln 4+\ln a)\\=\boxed{3\ln 4+3\ln a}$

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lions [1.4K]

Answer:

Confidence interval: (0.04649,0.04913)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100,000

Number of people who donated, x = 4781

$\hat{p} = \dfrac{x}{n} = \dfrac{4781}{100000} = 0.04781
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95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get:

$0.04781
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(1-0.04781
)}{100000}})\\\\=0.04781
\pm 0.00132\\\\=(0.04649,0.04913)$

is the required 95% confidence interval for the true proportion of their entire mailing list who may donate.

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