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3 years ago

Express as the sum or difference of logarithms. log311y

Mathematics

1 answer:

vovikov84 [41]3 years ago

8 0

Answer:

$log311y=log311+logy$

Step-by-step explanation:

We can use the property of logarithms of sum of logarithms:

$log_{b} (M*N)=log_{b} M+log_{b}N$

So, if we consider $M=311$ and $N=y$

We replace

$log311y=log311+logy$

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3 years ago

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz. An ornithologi

il63 [147K]

Answer:

The formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.7}{2} = 0.15$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.15 = 0.85$ , so Z = 1.037.

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$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Assume that body masses of Goldfinch birds follow a normal distribution with standard deviation equal to 0.04 oz.

This means that $\sigma = 0.04$ .

Sample of 10 birds:

This means that $n = 10$ .

The margin of error is of:

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.037\frac{0.04}{\sqrt{10}}$

$M = 0.013$

The lower end of the interval is the sample mean of $\overline{x}$ subtracted by M.

The upper end of the interval is the sample mean of $\overline{x}$ added to M.

Then, the formula to generate 70% confidence interval is: $[\overline{x} - 0.013, \overline{x} + 0.013]$ , in which $\overline{x}$ is the sample mean.

8 0

3 years ago