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Sergeeva-Olga [200]
3 years ago
10

How to do the work for 0.0338÷1.3

Mathematics
1 answer:
Butoxors [25]3 years ago
8 0
Start dividing this by having 0.0339 over 1.3 then move each decimal over one space. Divide as normal.
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What is 5x + 3y = 12 written in slope intercept form
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y=4-5/3x

Step-by-step explanation:

Rearrange and simplify.

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-8=-(x+4). Please help I don't know the answer or steps to solve this equation.
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x = 4

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First u need to distribute the -1 to (x + 4), creating the equation, -8 = -x - 4. Then add the -4 to both sides, and you get, -4 = -x. Then divide by -1 to both sides, and x = 4.

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Answer the following questions about complementary angles. Complementary angles are 2 angles that add up to 90 degrees.
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2 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.

Now, find the margin of error M as such

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.

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