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DIA [1.3K]
1 year ago
5

Lines,rays,and segments

Mathematics
1 answer:
tino4ka555 [31]1 year ago
8 0

Answer: line: To represent straight objects in geometry, mathematicians introduced the concept of line or straight line

rays: vector, a ray is a vector from a point to a point. In geometry, a ray is usually taken as a half-infinite line (also known as a half-line) with one of the two points and taken to be at infinity

segments: The term line segment refers to a section of a line bounded by two distinct points. It is comprised of all points on the line between its endpoints. A closed line segment includes both endpoints, whereas an open line segment excludes both endpoints.

Step-by-step explanation:

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Answer:

x > -15

Step-by-step explanation:

-x / 3 < 5

(-x / 3) * 3 < 5 * 3

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x > -15 (Remember to flip the sign when multiplying or dividing an inequality by a negative number.)

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The line through (-2, 3) with slope 2/3 in point slope form​
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Answer:

y - 3 = 2/3(x + 2)

Step-by-step explanation:

slope = 2/3

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point slope form of the line is y - 3 = 2/3(x + 2)

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g A researcher wants to construct a 95% confidence interval for the proportion of children aged 8-10 living in Atlanta who own a
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Answer:

The sample size needed is 2401.

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}.

The margin of error of the interval is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

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.

Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

We have to find n, for which M = 0.02.

There is no estimate of p available, so we use \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}

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The sample size needed is 2401.

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Answer:

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Step-by-step explanation:

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